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# Generating Randomly distributed points inside a given triangle

Given the cartesian coordinates of three vertices of a triangle $P_1$, $P_2$, $P_3$ I know (have simulated) that I get randomly distributed points by using this protocol:

$s=\text{rand}(0,1)\quad t=\sqrt{\text{rand}(0,1)}$

$P = (1-t)\cdot P_1 + t\cdot\left( (1-s)\cdot P_2 + s\cdot P_3\right )$

where $\text{rand}()$ is the uniform random number generator.

Can someone explain why this works?

Intuitively, this seems like applying the lever rule on the edge between one pair of points, say, $P_2$ & $P_3$ to get a intermediate point $P'$ & then a second lever rule between $P'$ and $P_1$.

What I don't get is why the second lever is drawn from a square root of the first uniform distribution.

$$Q_s=(1-s) P_2 + s P_3$$ is a convex combination of $P_2$ and $P_3$ and $$P_{s,t} = (1-t) P_1 + t Q_s$$ is a convex combination of $P_1$ and $Q_s$, hence a point of the convex envelope of $P_1,P_2,P_3$, i.e. the original triangle. If we draw a random point $P$ in the triangle from a uniform distribution, $P=P_{s,t}$ for some $(s,t)\in[0,1]^2$, and the marginal distributions of $s$ and $t$ can be recovered in a straightforward way. For any $\lambda\in[0,1]$, both $$\mathbb{P}[s\leq \lambda]\qquad \mathbb{P}[t\leq \lambda]$$ can be computed in terms of triangle areas, but while $\mathbb{P}[s\leq \lambda]=\lambda$, $\mathbb{P}[t\leq \lambda]=\lambda^\color{red}{2}$.
So $s$ has a uniform distribution, but the distribution of $t$ is the square root of a uniform distribution.
• @curious_cat: Draw the entire triangle and the one for $t\le\lambda$ -- you'll see that the latter is the former scaled by $\lambda$ in both dimensions. – joriki May 16 '16 at 10:23
• @curious_cat I meant to say: sure you do not see it there.To keep uniformity, you have to combine the three components linearly (Tri-linearly). Your combination involves instead the product $t s$ of the parameters, since you made a double combination. Then Jack already answered about. – G Cab May 14 '16 at 16:51


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Question

# One byte is a unit of memory size used with computers. One megabyte contains approximately $10^6$ bytes and Paul’s computer hard-drive accesses the information at a rate of 150 megabytes per second. If Paul wants to work with 5 files on his computer that are $2,340 \times 10^6$ bytes each, approximately how long will it take his computer to access them?A) 15.6 secondsB) 9 secondsC) 1.3 minutesD) 78 minutes

Verified
129k+ views
Hint:The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value.
While using the unitary method, we have to find the amount needed for one unit of the substance and then we multiply by the quantity provided.
For example, if 3 boxes can be filled by 12 chocolates, 1 box will be filled by 4 chocolates and hence 5 jugs will be filled by 20 chocolates. It is simply done by calculating multiplication $5 \times 4$.

Complete step by step solution:
Step 1:
1 megabytes $= \mathop {{\text{10}}}\nolimits^{\text{6}} {\text{ }}$bytes
Step 2:
Also, $\mathop {{\text{10}}}\nolimits^{\text{6}} {\text{ }}$bytes = 1 megabytes
Hence, ${\text{1 byte=}}\dfrac{{\text{1}}}{{\mathop {{\text{10}}}\nolimits^{\text{6}} }}{\text{ megabytes}}$
$2340 \times 10^6 \text{bytes} = 2340 \times 10^6 \times \dfrac{1}{10^6} \text{megabytes}$
= 2,340 megabytes …… (1)
Step 3:
The rate at which the information is accessed on the computer = 150 megabytes per second
($\because$given)
$\Rightarrow$In one second, 150 megabytes of information is accessed on the computer.
Or, 150 megabytes of information is accessed in = 1 sec
1 megabyte of information is accessed in $= \dfrac{1}{{150}}$sec
2,340 megabytes of information is accessed in $= 2,340 \times \dfrac{1}{{150}}{\text{ sec}}$
${\text{ =2,34{0}}} \times \dfrac{1}{{15{0}}}{\text{ sec}} \\ {\text{ =}}\dfrac{{234}}{{15}}{\text{ sec}} \\$
${\text{ =15}}{\text{.6 sec}}$…… (2)
Step 4:
Size of one file$= 2,340 \times \mathop {10}\nolimits^6$bytes($\because$given)
= 2,340 megabytes(from (1))
Step 5:
Time to access one file on computer = 15.6 sec(from (2))
Time to access five files on computer $= 5 \times 15.6{\text{ sec}}$
= 78 sec
Step 6
1 minutes = 60 seconds
Or, 60 seconds = 1 minutes
${\text{1 seconds=}}\dfrac{{\text{1}}}{{{\text{60}}}}{\text{ minutes}}$
$\therefore {\text{78 seconds=}}\dfrac{{{\text{78}}}}{{{\text{60}}}}{\text{ minutes}}$
= 1.3 minutes
Step 7
Time to access five files on computer = 78 sec = 1.3 minutes.

$\therefore$ Time taken to access five files on the computer is 1.3 minutes. Hence, the correct answer is option (C).

$7$ minutes = $7 \times 60$ seconds

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In a process occurring in a closed system F, the heat transferred from F to the surroundings E is 600 J. If the temperature of E is 300 K and that of F is in the range 380 – 400 K, the entropy changes of the surroundings (${\Delta}S{_E}$) and systems (${\Delta}S{_F}$), in J/K, are given by

1. ${\Delta}S{_E}$ = 2, ${\Delta}S{_F}$ = -2
2. ${\Delta}S{_E}$ = -2, ${\Delta}S{_F}$ = 2
3. ${\Delta}S{_E}$ = 2, ${\Delta}S{_F}$ < -2
4. ${\Delta}S{_E}$ = 2, ${\Delta}S{_F}$ > -2


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## On the class of sketchability of modelizable (accessible) categories with products of two. (Sur le genre d’esquissabilité des catégories modelables (accessibles) possédant les produits de deux.)(French)Zbl 0899.18003

The modelizable categories used in the paper are also called sketchable categories or accessible categories. The modelizable categories which have binary products of objects are characterized up to an equivalence of categories. They are proved to be the categories of models in $${\mathcal S}et$$ of some small sketch whose distinguished co-cones have an index category $${\mathcal I}$$ satisfying the following property: for any pair of objects $$I_1, I_2$$ of $${\mathcal I}$$,
1) there exists some pair of morphisms of $${\mathcal I}$$ with common domain $$s_1: S\to I_1$$, and $$s_2: S\to I_2$$, and
2) two such pairs of morphisms are connected by a zig-zag of pairs of such morphisms.

### MSC:

 18B99 Special categories 18A30 Limits and colimits (products, sums, directed limits, pushouts, fiber products, equalizers, kernels, ends and coends, etc.) 18A10 Graphs, diagram schemes, precategories 18A25 Functor categories, comma categories
Full Text:

### References:

 [1] C. Lair. Catégories modelables et catégories esquissables. Diagrammes 6, Paris. 1981. Zbl0522.18008 MR684535 · Zbl 0522.18008 [2] R. Guitart et C. Lair. Calcul des modèles syntaxiques et calcul des formules internes. Diagrammes 4, Paris, 1980. Zbl0508.03030 · Zbl 0508.03030 [3] C. Lair. Catégories qualifiables et catégories esquissables. Diagrammes 17. Paris. 1987. Zbl0624.18003 MR918001 · Zbl 0624.18003 [4] C. Lair. Diagrammes localement libres, extensions de corps et théorie de Galois. Diagrammes 10. Paris, 1983. Zbl0569.18001 MR780448 · Zbl 0569.18001 [5] C. Ehresmann. Esquisses et types des structures algébriques. Bul. Instit. Polit. Iasi. XIV. 1968. Zbl0196.03102 · Zbl 0196.03102
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Solving projectile motion equations

Gold Member
Last edited by a moderator:

Staff Emeritus

Er... "s" is not an unknown. It is a unit for "seconds".

Zz.

Gold Member

Okay, but why is it in second degree? Why did they need to add ^2 into a unit?

Staff Emeritus

Okay, but why is it in second degree? Why did they need to add ^2 into a unit?

Er.. it seems that you appear to not know that units can be squared, or undergo the same operations as a variable.

An acceleration, for example, is $L/T^2$, where L is length, T is time. So since the time dimension is squared, the units is also squared. In SI units, it is $m/s^2$. If you don't like the square symbol, then write it as $m/ss$. There, no more squared.

Zz.

Mentor

If you don't like the square symbol, then write it as $m/ss$.

Or as (m/s)/s: (meters per second) per second.

Gold Member

Ah...a little deeper ponderage after your replies got me to see the logic :) mercy.


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## Introduction

With the success of transcatheter aortic valve (AV) implantation for patients at high surgical risk, attention has now been cast on transcatheter therapies for the mitral valve (MV)1. These techniques vary from balloon and self-expanding AV used in the mitral position, to newer devices specific to the MV, to Valve-in-Valve interventions for failed bioprosthetic devices2,3,4. Irrespective of the approach, it is clear that transcatheter mitral valve replacement (TMVR) poses a distinct set of challenges compared to AV implantation. The geometry of the mitral apparatus and its relationship to atrial and ventricular structures introduce significant concerns pertaining to device stability. Specifically, TMVR devices need to be oversized to ensure adequate anchoring while accommodating deformational changes of the valve annulus during the cardiac cycle. This, in conjunction with the displacement of the anterior MV leaflet towards the AV, introduces a risk of left ventricular outflow tract (LVOT) obstruction as it involves a taller frame protruding inside the left ventricle (LV)5,6.

The interplay between the protruding device, the septal wall, and the anterior leaflet results in the formation of a neo-LVOT with reduced cross-sectional area compared to the original outflow tract7. Outflow obstruction can in turn increase left ventricular afterload and introduce adverse hypertrophic LV remodeling, which correlates with high risks of ventricular failure especially in patients with underlying dysfunction8. Evidence also exists that TMVR is associated with risks of blood stagnation inside the ventricle, with elevated pressure gradients and less effective ejection after MV implantation identified as predisposing factors for early postoperative thrombi9,10,11.

Although pre-procedural planning aims to predict the likelihood of LVOT obstruction, existing techniques are largely based on standard geometric evaluation of pre-procedural imaging datasets and there is consensus on neither the effect of the cardiac phase at which measurements are taken, nor the procedural canting of the device on the prediction. It also remains unknown as to what threshold of LVOT obstruction is clinically significant in these patients. To enable a successful introduction of TMVR devices, it is clear that careful patient selection and advanced quantitative evaluation of both anatomy and hemodynamics will be vital. Advanced imaging evaluation can provide valuable insight into the patient pathophysiology12,13,14,15. In the current study we sought to evaluate the utility of combining this technique with computer models to simulate blood flow dynamics in TMVR patients and to predict the ventricular response to LVOT obstruction.

## Methods

### Patient data

Seven patients were included in the study (Tables 1 and 2): six were under consideration for TMVR with bioprosthetic AV devices, e.g. Sapien (Edwards Lifesciences, Irvine, CA, USA) and Lotus (Boston Scientific, Marlborough, MA, USA), and one was assessed for treatment under compassionate use with an Intrepid valve (Medtronic, Minneapolis, MN, USA). Patients proceeded to transapical implantation with the chosen device, except Cases 2 and 7 – unimplanted due to heart failure and predicted excessive LVOT obstruction, respectively. This study complies with the principles of the Declaration of Helsinki and the research protocol was approved by local ethics committees (Guy’s and St Thomas’ NHS Foundation Trust); all patients provided informed consent. Datasets were collected pre-operatively only, except Case 6 where the complete acquisition protocol was also applied post-TMVR for model validation.

### Imaging protocol

Multiphase contrast-enhanced computed tomography angiograms (CTA) were acquired in all patients using a Philips Brilliance iCT 256-slice MDCT scanner (Philips Healthcare, Best, The Netherlands). Intravenous metoprolol was used to achieve a heart rate of <65 beats/min (or <100 beats/min if in atrial fibrillation). Intravenous contrast (Omnipaque, GE Healthcare, Princeton, NJ, USA) was subsequently injected into the antecubital vein (5 ml/s for a total of 100 ml). Ascending aorta contrast triggered, retrospectively ECG-gated scanning with no dose modulation was performed in a single breath hold after 10–12 seconds with a heart rate-dependent pitch of 0.2–0.45, a gantry rotation time of 270 ms, a tube voltage of 100 or 120 kVp (depending on the patient’s body mass index) and a tube current of 125–300 mA (depending upon the thoracic circumference). Transthoracic and transesophageal echocardiography data were also acquired. The direction and maximum magnitude of blood velocity across the valves was evaluated from Color Doppler data, which was also used to estimate mean and peak LVOT and transaortic pressure gradients. For all Doppler measurements, the average of three signals was taken. The mitral valve annulus was defined as the hinge points between the left atrium and mitral valve leaflets. The area of the annulus was measured as an average of three estimates on a 3D volume acquisition of the left atrium and left ventricle in mid systole by multiplanar reformatted imaging.

### Model personalisation

The valve dimensions, the angle between the MV and AV planes (α), the distance between the MV coaptation point and the interventricular septum (C-sept), and LV aspect ratio (AR = LLV/DLV) were measured as shown in Fig. 1A–C.

Although the valve dimensions in the models were informed from specific devices, our results can be applied to any stent-based valve design within these size ranges. This study aims at identifying the degree of obstruction that causes significant changes in hemodynamic function, and not at investigating the valve design itself: therefore we chose a simplistic representation of the bioprostheses, with a rigid cylinder of diameter D and height H, respectively equal to the inner diameter and frame height of the commercial valve under consideration for implantation (Fig. 1C). A tetrahedral volume mesh of the LV was created based on the blood pool obtained from the manual segmentation of the LV cavity from CTA data (Fig. 1D). The portion of the valve frame lying inside the LV (HLV) is given by:

$${H}_{LV}=H-{H}_{LA}$$
(1)

where HLA is the portion of the valve frame in the left atrium relative to the mitral sewing cuff (Fig. 1E). HLA was chosen to vary between 3 and 5 mm, resulting in a range of HLV tested for each patient, in agreement with surgical recommendations to prevent late atrial migration16,17. The outflow obstruction fraction (OF) was then calculated as the normal projection of the valve frame on the AV plane, divided by the total outflow area (blue and red area respectively in Fig. 1F). In each model, boundary conditions were personalised by extracting information on wall and blood velocity from imaging data (Fig. 1G). Specifically, the wall velocity is derived from the wall motion tracked using an algorithm based on the temporal-sparse free form deformation18 and the transvalvular blood velocities were obtained from Doppler ultrasound data at the valve planes19,20. The finite-element software CHeart, which has been extensively validated for cardiovascular simulations21, was used in all flow simulations to compute the blood pressure and velocity inside the LV (Fig. 1H–J).

### Personalised CFD simulations

Different simulations were performed in each patient: first, the cardiac cycle was simulated in a model without implanted valve to provide baseline conditions, then additional simulations were performed using predictive models where a simplified valve of the same size of the commercial device under consideration was embedded in the LV mesh with varying levels of HLA (Table 3). In each scenario, three consecutive cardiac cycles were simulated to minimise any artifacts from initial conditions. The study comprised a total of 32 personalised simulations for 7 patients. The heterogeneity of this cohort allowed us to test the models under different conditions. In Cases 1 to 5, the anatomical characteristics of the LV allowed HLV to be increased above the frame height H of the commercial device without interfering with the endocardium. These “virtual” scenarios, effectively corresponding to a negative HLA in Eq. 1, were tested to assess the impact of a longer channel on the blood flow dynamics and are marked with an asterisk in Table 3. In Case 7, which was not implanted due to interference of the bioprosthesis with the septal wall, valve heights smaller than the commercially available sizes were tested.

### Anatomical and hemodynamics measurements

The pressure gradients in the LV were calculated at peak systole in each model. In addition to pressure calculations, a functional flow analysis was performed by splitting the blood flow into different components depending on the time spent inside the LV (Fig. 2B), with each component expressed as percentage of the end-diastolic volume22,23,24. The following hemodynamic parameters were tested:

1. 1.

PG-AV (mmHg): gradient between the mean pressures within a sphere of radius 4 mm in the apical region and on the AV plane (Fig. 1I).

2. 2.

PG-LVOT (mmHg): gradient between the mean pressures within a sphere of radius 1 mm placed 6 mm below the AV plane and on the AV plane (Fig. 1I).

3. 3.

Direct flow (DF, %): percentage of blood entering and exiting the LV within the same cardiac cycle.

4. 4.

Retained inflow (RI, %): percentage of blood that enters the LV during diastole and is not ejected in the following systole.

5. 5.

Residual volume (RV, %): percentage of blood that spends more than two cycles in the LV before being ejected.

6. 6.

RVU, RVL (%): residual volume of blood in the upper and lower LV respectively (Fig. 2A).

7. 7.

R: ratio of DF over RV at peak systole (measure of systolic flow efficiency).

In each patient, all hemodynamic parameters from the predictive models were normalised based on the corresponding value from the baseline model.

### Statistical analysis

Descriptive statistics are presented as mean ± SD. A simple regression model was used to predict the variations of each normalised hemodynamic parameter with OF. Pair-wise comparisons between predictive and baseline models were performed using a Wilcoxon test and differences were considered significant for p < 0.05. A piecewise linear regression of the hemodynamic parameter on OF was also applied to the data to detect the change point (OFC, %) at which a change in the slope of the regression line occurred.

### Model validation

The validation dataset (Case 6) included comprehensive imaging data acquired immediately before and after implantation, allowing us to create two independent models pre- and post-TMVR. When in its predictive capacity, the pre-TMVR model included a rigid cylinder with the same internal diameter, frame height and atrial positioning of the implanted valve, a 29-mm Medtronic Intrepid device. The mean and peak PG and obstruction fraction from this predictive model were validated against the values derived from both the post-TMVR imaging data and model (Table 4). A difference of 10%, 14%, and 8% between the imaging data and the predictive model are observed in the end-systolic values of the α angle, C-sept distance, and LV AR, respectively. The same parameters measured in the predictive and post-TMVR validation model showed a difference of 5%, 3%, and 6%, respectively. A comparison of the Doppler-derived data and the predictive model showed differences of 26% and 15% in the mean and maximum PG-LVOT, respectively. Similarly, when the predictive and validation model were compared, the mean and maximum PG-LVOT differed by 3% and 13%, respectively. The maximum flow velocity at the AV plane derived from the Doppler data was 12% lower than the corresponding value in the predictive model and 4% lower than the that from the validation model. Both models showed a maximum OF between 6% and 7%, which was in agreement with the difference in the effective orifice area measured from CTA data pre- and post-TMVR (approximately 9%).

## Results

### Impact of TMVR on intraventricular pressure gradients

Simple linear regression was performed to predict the effect of LVOT obstruction on PG-AV and PG-LVOT (Fig. 3A,B).

The normalised peak systolic PG-AV increased monotonically with OF in Fig. 3A (r2 = 0.74, p < 0.05) compared to the baseline values (normalised mean 1.18 ± 0.19, p = 0.002). Piecewise regression analysis identified a critical value OFC = 35%, which corresponded to an increase in PG-AV of 21%. Specifically, the mean increase in PG-AV from the baseline was 5.7% for OF <35% and 30.1% for OF >35% (normalised mean 1.06 ± 0.07, p = 0.020 vs. 1.30 ± 0.19, p = 0.028). A polynomial fitting analysis showed that the value OFC = 35% identified a switch from a linear to an exponential growth of PG-AV with OF (r2 = 0.91).

Contrary to the trend identified for PG-AV, in Fig. 3B the peak systolic PG-LVOT post-TMVR decreased from the baseline value (r2 = 0.53, p < 0.05) over the range of OF tested (normalised mean 0.85 ± 0.15, p = 0.000). A mean decrease of 6.6% for OF <35% and 23.5% for OF >35% was observed (normalised mean 0.93 ± 0.06, p = 0.046 vs. 0.76 ± 0.17, p = 0.036).

Pressure isocontours from the baseline and predictive models for Case 1 typify the disrupting effect of the obstruction on the PG-LVOT (Fig. 3C,D). The corresponding streamlines of blood flow show the velocity increase induced by the artificial narrowing of the LVOT due to the prosthetic valve frame (Fig. 3E,F).

In Case 2 and 6, where no obstruction was observed due to anatomical features, both peak systolic PG-AV and PG-LVOT after implantation were close to the baseline values.

### Impact of TMVR on blood flow components

Figure 4A shows that the DF component post-TMVR decreased on average by 13.9% from the baseline value in the range of OF tested (normalised mean 0.86 ± 0.13, p = 0.000). The RV and RI components post-TMVR increased on average by 35.6% and 10.3% respectively from the corresponding baseline values (normalised means 1.36 ± 0.50 and 1.10 ± 0.14, p < 0.005) (Fig. 4B,C).

For all flow components a change point was identified for OF values of approximately 29–30%. Above this OFC the slope of the regression line increased significantly by 46.8% and 37.5% for RV and RI when respectively compared to the corresponding slope of the overall linear regression, while a moderate increase of 3.5% was reported for DF. For OF <30% the mean increase from the baseline in the RV component was 4.4%, but this value raised significantly to 57.5% for OF >30% (normalised mean 1.04 ± 0.14 and 1.57 ± 0.56, p < 0.01). Mean increases from the baseline of 8.9% for OF <30% and of 14.1% for OF >30% were recorded for the RI component (normalised mean 1.09 ± 0.11, p = 0.0124 vs. 1.14 ± 0.15, p = 0.0009).

Overall the R ratio in the post-TMVR models decreased from the baseline by 36.7% with increasing OF (normalised mean 0.63 ± 0.22, p < 0.0001) (Fig. 4D). Piecewise regression identified an OFC = 34%, which corresponded to a decrease of 31%. The mean decrease was 17.4% for OF <34% and 46.2% for OF >34% (normalised mean 0.83 ± 0.15 vs. 0.54 ± 0.18, p < 0.0001).

The post-TMVR RVU component increased on averaged by 27.1% from the baseline in the range of obstructions tested (normalised mean 1.27 ± 0.43, p = 0.0014), while RVL component did not show any statistically significant variation (Fig. 5A). A comparison of the RV particle distribution in the baseline and predictive model for Case 1 showed a marked accumulation of RV particles between the septal wall and the valve frame and inside the valve frame in the predictive model, while a similar distribution was observed in the lower LV of both models (Fig. 5B,C).

## Discussion

Our results show that an LVOT obstruction greater than 35% is associated with deleterious effects on ventricular hemodynamics following TMVR, including a significant increase in ventricular afterload and in the blood volume residing inside the ventricle for more than two cycles.

Previous studies on LVOT obstruction found that a 24% reduction in the AV indexed effective orifice area (EOA) resulted in a 33% increase in transaortic pressure gradients, which increased exponentially with EOA decreasing below a threshold value25,26. An acute afterload increase of 30% was also associated with sudden death and profound functional and anatomical changes in 7 closed-chest porcine models27. Our results consistently show an exponential increase in afterload for LVOT area reductions greater than 35%, with a mean increment in the peak systolic pressure gradient of 30.1% in the upper range of obstruction tested (Fig. 3A). Since our predictive and baseline models are both based on the same wall motion (i.e. volume change) and systolic duration, in the presence of LVOT obstruction the systolic pressure gradient must increase to achieve the same cardiac output in the same length of time. If persistent, this compensatory mechanism can introduce adverse ventricular remodelling leading to progressive heart failure. In particular, increases in systolic pressure gradients and afterload can trigger adverse hypertrophic remodelling. In turn, a hypertrophic septum bulging into the ventricular cavity can exacerbate LVOT obstruction and further increase the afterload, causing a negative feedback mechanism. However follow-up studies are needed to identify the level of LVOT obstruction that promotes transition from an adaptive to a maladaptive ventricular remodeling.

Outflow obstruction also resulted in a localised decrease in the systolic pressure gradient in the LVOT that was not present in the baseline models, where the pressure changes occurred gradually in space (Fig. 3D). The predictive models showed that the narrowing due to the valve frame caused a local increase in the blood flow velocity, thereby generating a sharp pressure decrease in the bottleneck region as postulated by the Bernoulli principle (Fig. 3C,E). The magnitude of this pressure recovery depends on the amount of energy dissipated in the blood as it flows through the bottleneck, which in turn is affected by the neo-LVOT shape and the blood velocity. This spatial variation of the pressure gradient can represent a confounding factor in the LVOT assessment using Doppler measurements28.

The afterload increase necessary to maintain the cardiac output also affects the blood residence time in the LV: the degree of LVOT obstruction that triggered hemodynamic deterioration was consistent across blood residence times and pressure gradients estimates at 30–34% and 35% respectively. The blood residence time increased for all models following implant due to a shift from cardiac output being predominantly driven by the direct flow component to increasingly being a result of retained inflow and residual volume from earlier cycles being ejected. A significant increase of 57.5% in the RV component was observed in the higher OF range, indicating a deterioration of systolic flow efficiency that is proportional to the degree of LVOT obstruction. This finding is in agreement with results from previous in-vitro models, which showed blood accumulation sites near the valve structure and between the valve frame and the myocardium, creating potential sites for thrombus formation29. Such a conclusion is also supported by our finding of an increase in the number of RVU particles post-TMVR, which was not accompanied by a consistent decrease in the number of RVL particles. In the predictive models, these extra RVU particles accumulate in the sites between the myocardial wall and the valve or inside the valve frame itself (Fig. 5C), while no significant difference is observed between the baseline and TMVR groups in the accumulation site of the lower LV, i.e. the apical region.

The goal of this study was to quantify the level of LVOT obstruction that is likely to impair ventricular hemodynamics. In this context, the heterogeneity of our cohort in terms of baseline characteristics and type of implanted device does not subtract from our main finding. Simulation results are only dependent on the aspect ratio of the device (H/D) and do not account for specific manufacturing characteristics. This approximation allows the models to be consistent across implants and thus does not limit the generalisability of the results. The simplified valve represents the bulk of frame and leaflets without distinguishing between each structure: while such representation is an approximation of the real system, this is what eventually causes obstruction in the LVOT, whose prediction is the main goal of the study. However, further modelling work on a larger number of patients will be necessary to investigate how anatomical parameters such as the size of the LV cavity may affect the threshold OF for haemodynamics deterioration, since LVOT obstruction can impact the behaviour of the LV differently in small ventricular cavities. Similarly, future modelling studies with more realistic shapes of implanted device should also be performed to analyse the effect of the bioprosthesis geometry on ventricular haemodynamics. This will be highly relevant to the topic of LVOT obstruction, as new dedicated D-shaped MV bioprostheses that are able to mimic the original shape of the mitral annulus, and thus avoid high transvalvular gradients by maximising the inflow area, are also likely to protrude into the LVOT causing obstruction due to their large size. While smaller devices like the AV bioprostheses deployed in the mitral position may circumvent this risk, they are also prone to stability issues and patient-prosthesis mismatch, underscoring the need for robust means to match device model and size to the individual patient’s pathophysiology.

Finally, the ventricular geometry in our models did not include trabeculations, chordae and papillary muscles and thus the respective effects on the blood flow could not be investigated. However, the tracked wall motion from the CTA data incorporates the effects of these structures on the global and segmental wall motion, as well as those due to wall thickness, potential prior myocardial infarction or any other contraction defect, which would also result in an alteration of the motion pattern derived from the imaging data. Similarly, as direct in-vivo measurement of pressure were not available, we used experimentally measured velocity profiles at the valve planes that account for both preload, afterload, and potential altered blood flow dynamics such as aortic stenosis or concomitant mixed valve disease. While the present study only provides a proof of concept of the potential of this modelling approach, the impact of contraction patterns, ventricular loading, and geometries of the subvalvular apparatus on LV function post-TMVR can be investigated in further work using this methodology by modelling a cohort of patients of sufficient sample size to represent significant variations in all these factors.

## Conclusion

By combining imaging and computer modeling we have provided a tool to predict LVOT obstruction post-TMVR and to quantify the impact of varying degrees of obstruction on ventricular afterload and blood residence times, providing a link between anatomical and hemodynamics parameters. This approach overcomes some of the limitations of current methods for LVOT obstruction predictions, which largely rely on anecdotal experience and standard anatomical evaluation. The goal of our modelling technique is to create impact on patient care and procedural skills by informing the device choice and placement to lower the possibility of reinterventions. This is particularly relevant in TMVR, where reintervention is associated with high morbidity and mortality rates. Despite the small size and heterogeneity of the cohort analysed, these results show a clear correlation between the degree of LVOT obstruction and changes in ventricular afterload and blood residence times following TMVR. A LVOT obstruction between 30% and 35% has been identified as a threshold for hemodynamic deterioration, providing a quantitative criterion for intervention that has been lacking so far. The next step for an effective clinical translation of this technology will be to apply it to a large number of patients to generate detailed response maps, which could be used in the future to design and test new devices.


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# Equivalent condition for base of a root system

In Humphrey's Lie algebra, following is the definition of base of a root system.

Let $\Phi$ be a root system in standard Euclidean space $\mathbb{R}^n$. A subset $\Delta$ of $\Phi$ is called as a base of $\Phi$ if

(1) $\Delta$ is basis of the vector space $\mathbb{R}^n$.

(2) Every $\alpha$ in $\Phi$ is a $\mathbb{Z}$-combination of elements of $\Delta$ with coefficients either all non-negative or all non-positive.

The elements of a base have property that their inner product is non-positive: $(\alpha,\beta)\leq 0$ for all $\alpha,\beta\in\Delta$ with $\alpha\neq\beta$.

Question: Suppose $\Delta'$ is a subset of $\Phi$ with condition that

(1) $\Delta'$ is a basis of the vector space $\mathbb{R}^n$.

(2) For any $\alpha\neq \beta$ in $\Delta'$, $(\alpha,\beta)\leq 0$. Then is $\Delta'$ necessarily a base of $\Phi$?

• Try to think about some examples first. What about $B_2$ and $G_2$? – Claudius Jan 25 '17 at 13:44
• OK, got it (ans: not true)! Actually I thought only of $A_2$ and $A_1\times A_1$; thanks for suggestion. – Beginner Jan 25 '17 at 14:57


text:
# Limit

We won't study limits in this course (indeed, they are very non-discrete!), but this example demonstrates that the basic logical tools we study are used to define continuous concepts like limits.

We can use deeply quantified statements to defined complicated concepts like limits. The idea behind the limit of a function is that one can force the function's output to be equal to the limit at (within any given tolerance) by making the input sufficiently close to . This can be formalized as follows:

If is a function from to , we say that the limit of at is (written ) if for all tolerances , there exists a distance such that for any value that is within the range , the distance between and is smaller than the tolerance.

More concisely, means .

Since we have a formal definition using standard logical constructs, we can use our usual proof techniques to prove things about limits:

where .
Proof:
We want to show that .

Choose an arbitrary . Let ( since ). Choose an arbitrary , and assume that . We want to show that .

Well, by definition, , and by assumption, we have . But by definition; therefore as required.

In this example, you might conclude that since the limit is , it cannot be 1. But this assumes that the limit is unique. We can directly prove that the limit is not 1 by proving the logical negation of the fact that the limit is 1:

where .
Proof:
The logical negation of

is

Let . Choose an arbitrary .

Let be the minimum of and . Note that , and (you can prove this using case analysis).

Then we have . Moreover, so , and therefore , which is what we were trying to prove.


text:
×

# Results: Can You Outsmart Everyone Else?

Congrats to Mathh Mathh and Ahaan Rungta for being the closest!
They will be receiving a Brilliant T-shirt for outsmarting everyone else.

Number of participants: 25
Average: 18.944
W = 0.9 * ave : 17.0499

For the rules + actual game breakdown, see Can You Outsmart Everyone Else?

Raw results: Values have been rounded down to 3 decimal places

 Name Entry Absolute Difference from W Math Man 8.539 8.510 Sharky Kesa 22.722 5.672 Daniel Liu 9.869 7.180 Mietantei Conan 10 7.049 Yannick Yao 31.006 13.956 Anthony Susevski 13.413 3.636 Tan Li Xuan 21.415 4.365 Ahaan Rungta 16 1.049 Aneesh Kundu 19.911 2.861 Victor Song 38.660 21.610 Andy Hayes 14.771 2.278 Mathh Mathh 16.581 0.468 Raj Magesh 14.106 2.943 Victor Martin 29.1 12.050 Enrique Naranjo Bejarano 42 24.950 Samuraiwarm Tsunayoshi 42.377 25.327 Zhijie Goh 28 10.950 Bogdan Simeonov 2.685 14.364 Tan Wee Kean 20.678 3.628 Daniel Ploch 0 17.049 Pranshu Gaba 15.154 1.895 Justin Wong 1 16.049 John Muradeli 0.616 16.433 Chung Kevin 10 7.049 Ajala Singh 45 27.950

Entries which did not follow the participation rules have been ignored.

Here is a follow-up discussion:

1) No one voted above 45. Why?
2) Few people voted above 40. Does this make sense? Why, or why not?
3) Why didn't the rational strategy of "vote 0" win? It turned out to be the 5th worst performer.
4) If this game occurred again, how would your strategy change?
5) Is there an "ideal number" to submit?
6) What would be your best strategy in approaching this game?

Note by Calvin Lin
3 years, 1 month ago

Sort by:

I am happy I wasn't horribly off! Can we do another one of these? It was awesome! :D · 3 years, 1 month ago

After playing this game, I Googled it and stumbled upon this interesting Wikipedia article on the Keynesian Beauty Contest, in which, similar to this game, A attempts to reason what B would believe C would think, ad infinitum (ouch!). Here's the link.

Why didn't "Vote 0!" win? Possibly because you phrased the question as "Can you outsmart everyone else?", so the contest had two aims: to get the number closest to 0.9 times the average, and to make sure no one else got any closer than you, making 0 an irrational choice. If the aim had been simply to get the number closest to 0.9 times the average, sans any competition, the rational choice would have been zero, in which case everyone would have won and you would have run out of T-shirts!

Also, since we were able to see the other participants' entries, we could guesstimate the value we would need to enter to win. I wonder if the results would have been more different if no one knew anyone else's entry (i.e. assuming everyone else was perfectly rational). Then this becomes analogous to the Keynesian beauty contest.

What if, on the other hand, you gave us, say, 50 random nonzero values and told us that these would be included in calculating the average, in addition to our entries? What would the optimal strategy be then?

After a lot of random Wikipedia surfing, I chanced upon the Monte Carlo method, and began wondering if it would make sense to apply it here. After all, if we repeatedly use large, random sets of data, eventually a small range of numbers would emerge as the clear winners, making it optimal to select these. But then again, if everyone starts doing this, the pattern is upset once more... We can't attempt to hack the system without messing up the system. · 3 years, 1 month ago

The "official" name of this game is P-beauty, where "P" stands for the proportion of the average. In this case, we played a 0.9-Beauty game.

Typically, the game involved blind bidding by everyone, so I had to tweak it slightly to suit our current system. I think that the random time cutoff added an interesting element to this game, where withholding your entry gave you more information, but you could potentially lose on the ability to submit an entry.

In almost all simulations run by game theorists, they were unable to find a scenario where "Many people bid 0-1 (close to 0)", and there were few cases where a 0 to 1 bid actually won. There are various explanations for this, and it's worthwhile to find a "rational" explanation why a bid of 10 could make sense.

Possible explanation: You may only assume that you are rational. You do not know to what degree someone else is rational.

Possible explanation: Some people are just out to screw everyone else. Staff · 3 years, 1 month ago

1) In my opinion, when the first few numbers were put in, most people saw the scale of these numbers and went for smaller numbers.

2) Same as first.

3) I think that everyone wanted their own answer, unique and different. We had plenty of quirky as well as standard numbers. So really, we all overcame conformity.

4) I would have picked a slightly smaller number but who knows how if this game would be played if played again. A strategy everyone could have used, however, is to keep al the numbers roughly equivalent.

5) In my opinion, there really is no 'ideal number'. People picked completely random numbers which all happened to be mainly small.

6) Equivalence means that the average is closer to your number. However, Randomness means that you will get an average closer to you anyway. So really, there is no best strategy. They all work fine if you work in the same scale.

Your opinions? :D · 3 years, 1 month ago

That's much more a game of luck than of math! There could be some game changer person who doesn't want a Tshirt (already has one) could post an extreme value! Then that could change the winner from one person to the other! You need luck if you want a Tshirt, but moreover, you need luck to see that something like this has started ! For example me, I saw this discussion of results before I saw that original post i.e., after it has been closed ! $$:($$ · 3 years, 1 month ago

It is partially luck, but remember, this is a game after all and it's for fun. You haven't lost anything so there's no need to sweat about it. I bet it would be less enjoyable if it was less luck-based. Also, such a game poses some nice questions, which you can't get if you make it extremely simple and 0% luck. · 3 years, 1 month ago

Truly said, I agree ! You're right ! · 3 years ago

Well, you need luck to win the lottery, but if you do win, it's not like you have to divide it evenly between the people who were a number off. · 3 years, 1 month ago

$$\Huge{\color{Red}{\textbf{LOL}}}$$ · 3 years ago

There were no stakes involved initially, other than "bragging rights" and the fun of participation. See my reply to Justin's comment.

Check back often on Brilliant! You get exciting problems posted by your friends, and who knows what else you may see :) Staff · 3 years, 1 month ago

ooh darn I thought it was every person posts 0.9 of what you post or something WHOA I would NOT have posted $$\frac{\pi^2}{16}$$! I would've posted 25. Oh well. · 3 years, 1 month ago

I'm horrible at math, but this generally is my opinion.

1) And 2) The first few posts were around pi and e, thus nobody wanted to post anything much higher than that (since the aim was to get 0.9x the mean)

3) People are selfish. They don't want everyone to win, they want the whole thing to themselves

4) And 6) Hmm. I don't know, check the trend, if it happens to be around 10, play about 10 too, as not many people would want tout something much higher than the average

5) Hmm. Not really. At least I don't think so. · 3 years, 1 month ago

Wait what... Oh it autocorrected the headings Should read: First line for 1 and 2 Second line for 3 Third line for 4 and 6 Fourth line for 5 · 3 years, 1 month ago

When people type 1., 2., 3., we automatically assume that they want it to be a sequential list and rename it as such. I've edited your response to 1), 2), 3), to fool our system. Staff · 3 years, 1 month ago

@mathh mathh , @Ahaan Rungta Can you email me your mailing address + T-shirt sizes? Thanks! Staff · 3 years, 1 month ago

Wow. Just wow. · 3 years, 1 month ago

I didn't participated in the contest but why "For those of you who were waiting to snipe at the last possible moment, tough luck."? · 1 year, 5 months ago

Assuming that everyone is rational, it would make sense that they would take the average of everyone else's answers multiplied by 0.9 as their own answer, to have a higher chance of winning. However, when I did this simulation on excel ,(assuming that there were 25 people, and the first answer was 45),I found that the average was 33.85101, and when this was multiplied by 0.9, the result was 30.46591. This is way off from the results of the experiment , which shows that not everyone is rational. · 2 years, 8 months ago

Hmm... I saw the first few posts and thought people would post things related to pi and e. So I thought the average would be around 16. Then, I remember seeing someone posting a number ~40 and I thought it would increase the average a lot.

I guess the best strategy would be to get as much information as possible? Perhaps if we knew when @Calvin Lin is usually online, we could then guess when the discussion would be locked XD. I went to this website before entering my results. The average then was around ~14. · 3 years ago

1) 45 was the first submission. In an attempt to be lower than the average at first sight (which started at 45) each consecutive person voted lower than the highest vote, creating a cascade of numbers securing the average around 15-20ish. I guess people weren't brave enough to vote higher because they feared there wouldn't be enough support in the higher tier - maybe that strategy could be easily foiled. Maybe early voters didn't predict future averages and followed the popular strategy of immediate averaging (at least I think that's what happened). · 3 years ago

:( Aww @Calvin Lin you overlooked my entry... (I thought it followed the rules?) Anyway it was terribly, terribly off by about $$8.574$$, so never mind :D · 3 years, 1 month ago

Sorry, I decided to ignore your entry as it violated the rule of "you may not edit your entry". Staff · 3 years, 1 month ago

Are these the only way to get Brilliant t-shirts? · 3 years, 1 month ago

As for now, winning this competition is the only existing way to win Brilliant t-shirts; and now that it's done, there are no existing ways to win Brilliant t-shirts. I guess you'll have to wait for the next competition.

@mathh mathh @Ahaan Rungta Congrats! But Ahaan, don't you already own a Brilliant.org t-shirt? I would expect you have already. · 3 years, 1 month ago

Oh, I just forgot I had claimed a brilliant t-shirt around 8-9 months ago, but I haven't gotten it yet, ohh well. · 3 years, 1 month ago


text:
# What does it mean to say that the electron is a near-perfect sphere?

It's announced that researchers at Imperial College London has found that the electron is almost a perfect sphere. The popular articles all have a nice photo of a billiard ball, etc. It is reported that they found this by measuring the "wobble" as the electron spins (and finding none).

What does it exactly mean that the electron is a sphere? Is it the wave function that is spherical? Measuring the spin wobble brings to mind a solid object, which I think the electron surely isn't. So, what would have been wobbling if it didn't have perfect spherical symmetry?

-
It means people still think of the electron in terms of classical charge distribution. It means they learnt nothing from QM. In QM the charge distribution is state-dependent. –  Vladimir Kalitvianski May 26 '11 at 16:45
Chad Orzel has a delightful writeup of the experiment here: scienceblogs.com/principles/2011/05/… –  Anonymous Coward May 27 '11 at 18:15

There's a Nature article that describes the experiment and the results, http://www.nature.com/nature/journal/v473/n7348/full/nature10104.html, but that's behind a paywall. The experiment is described in some detail in "Prospects for the measurement of the electron electric dipole moment using YbF", http://arxiv.org/abs/1103.1566 (I've only scanned the latter, but it looks to be quite informative).

From the Nature article, "In an atom or molecule with an unpaired valence electron, the interaction of the electron EDM [Electric Dipole Moment] with an applied electric field results in an energy difference between two states that differ only in their spin orientation. This energy difference is proportional to $d_e$ and changes sign when the direction of the field is reversed. A sensitive method of measuring this energy difference is to align the spin perpendicular to the field and measure its precession rate, which is proportional to the energy difference. An alternative description of the method is in terms of an interferometer. There is quantum interference between the two spin states, and the EDM appears as an interferometer phase shift that changes sign when the electric field is reversed."

An electron is not either a sphere or not-a-sphere, but we can introduce more or fewer internal degrees of freedom into the quantum fields that are used to describe experiments results that we attribute to the electron field. Introducing different degrees of freedom has consequences for the geometrical configurations of recorded experimental results. The Nature article is explicit in saying that this is intended to distinguish between different speculative quantum field theories, "many extensions to the standard model naturally predict much larger values of $d_e$ that should be detectable". This is an experimentalists' article, however, so they link to a theory paper on the subject (which I cannot access directly). If these fields give better descriptions than the standard model of particle physics, we expect to see different, less geometrically symmetrical statistics of events.

Many of the problems of reference here can be avoided if we talk about electron fields instead of about electrons. An "electron field" is less likely to be misrepresented as spherical or not spherical, but it can be associated with (representation spaces of) space-time symmetry groups (which describe in a systematic way how something deviates from being symmetrical). Care is needed because a quantum field is a more elaborate mathematical object than a classical field, but we can loosely think of a quantum field as a way to generate probabilities that the configuration of a classical field is one thing or another at any single time, while the details of quantum measurement are such that we can't talk about such probabilities at multiple times.

-
I think the distinction that you point out between electrons and (quantum) electron fields is crucial for a clear understanding of the result. Yet, this raises an interesting question: if these two are different things, than what does the word "electron" refer to? –  recipriversexclusion May 26 '11 at 18:45
We could say that "electron" is an adjective. An electron field as opposed to a "green" field. We see an "electron event" in a detector, not an electron, caused by the preparation of a specific quantum field state, and the detector being put where it is, not by an electron. Would you like a yellow? A yellow what? There are other ways to go with the interpretation of QM/QFT, of course. –  Peter Morgan May 26 '11 at 18:56

Well, one strong piece of evidence for this is the fact that the electric dipole moment

$${\vec D} \equiv \int d^{3}x' \rho({\vec r}-{\vec r_{0}})$$

of the electron is nearly zero. The electric dipole moment would measure a reflection asymmetry in a potential charge distribution of the electron. The fact that it takes on a zero means that the electron charge distribution is symmetric about all planes. Now, it is still possible that the electron may have nonzero values for it's higher multipole moments, while having a zero electric dipole moment which would indicate non-sphericity as well, but as far as I know, there is no evidence for this either, and finding these moments would be an even harder experiment.

-
Jerry Schirmer is correct, but just to rephrase his answer so it answers the question as asked: "What does it mean that the electron is [almost a perfect] sphere?", it means that the electric dipole moment has been measured to be very small. –  Anonymous Coward May 26 '11 at 17:34
Yes, they should have state this clearly, without mentioning the electron shape. –  Vladimir Kalitvianski May 26 '11 at 17:51
But they have to dumb it down so people that don't understand Quantum Mech or spherical harmonics can follow it. –  Omega Centauri May 26 '11 at 17:58
Maybe you should say what $\vec r_0$ is, @Jerry, because it does affect the result - because the total charge is nonzero. Is that the center of mass? Also, it's not clear why you integrated over $x'$ even though no $x'$ appears later. One more thing, the dipole moment has a direction that must be proportional to the spin because it's the only preferred direction of the electron. Finally, a dipole moment is prohibited by CP-symmetry, so a (large) dipole moment requires a (new) source of CP-violation. –  Luboš Motl May 26 '11 at 18:28
Vladimir: if the dipole moment wasn't parallel to the spin, the electron would have an additional degree of freedom. This would change the entire periodic table, since we could fit 4 electrons in the 1s orbital, etc. etc. –  Anonymous Coward May 26 '11 at 23:45

As the comment to your post says: This spherical charge distribution picture hasn't really much to do with modern QM physics.

In fact we are talking about different quantities. The charge distribution is given mathematically by $\bar{\psi}\gamma^o\psi$ while the relevant quantities are the components of the polarization/magnetization tensor $\bar{\psi}\sigma^{\mu\nu}\psi$

The electric polarization components are zero in the current (Dirac) theory (in the local restframe) while the magnetization components are non zero. The latter represent the magnetic moment of the electron. It is the Dirac equation which mathematically describes the precession of the electron searched for in the experiment.

A non-zero electric dipole moment would give the electron the very special property of either going one of two different ways in a non-homogeneous electric field. Just like the famous Stern Gerlach experiment shows in the case of the magnetic moment.

Such a behavior can of course never be explained by a classical non spherical charge distribution...

-

It is probably better to talk about deviation from spherical symmetry. If you want to think of it as a sphere then the classical radius computed from equating the mass-energy of the electron to its field $$mc^2 = \frac{1}{4\pi\epsilon_0}\frac{e^2}{r}$$ and with variations on this way of computing this. If there were some deviation from spherical symmetry it turns out to be on part $10^{16}$ of this classical electron radius.

-


text:
# systune

Tune fixed-structure control systems

## Syntax

• ```[CL,fSoft] = systune(CL0,SoftReqs)``` example
• ```[CL,fSoft,gHard] = systune(CL0,SoftReqs,HardReqs)``` example
• ```[CL,fSoft,gHard] = systune(CL0,SoftReqs,HardReqs,options)```
• ```[CL,fSoft,gHard,info] = systune(___)```

## Description

example

``````[CL,fSoft] = systune(CL0,SoftReqs)``` tunes the free parameters of the control system model, `CL0`, to best meet the soft tuning requirements. The best achieved soft constraint values are returned as `fSoft`. Note:   For tuning Simulink® models with `systune`, use `slTuner` to create an interface to your Simulink model. You can then tune the control system with `systune` for `slTuner` (requires Simulink Control Design™).```

example

``````[CL,fSoft,gHard] = systune(CL0,SoftReqs,HardReqs)``` tunes the control system to best meet the soft tuning requirements subject to satisfying the hard tuning requirements (constraints). It returns the best achieved values for the soft and hard constraints. ```
``````[CL,fSoft,gHard] = systune(CL0,SoftReqs,HardReqs,options)``` specifies options for the optimization.```
``````[CL,fSoft,gHard,info] = systune(___)``` also returns detailed information about each optimization run. All input arguments described for the previous syntaxes also apply here.```

## Examples

collapse all

### Tune Control System to Soft Requirements

Tune a cascaded control system to meet requirements of reference tracking and disturbance rejection.

The cascaded control system of the following illustration includes two tunable controllers, the PI controller for the inner loop, , and the PID controller for the outer loop, .

The blocks and mark analysis-point locations. These are locations at which loops can be opened or signals injected for the purpose of specifying requirements for tuning the system.

Tune the free parameters of this control system to meet the following requirements:

• The output signal, , tracks the reference signal, , with a response time of 10 seconds and a steady-state error of 1%.

• A disturbance injected at is suppressed at by a factor of 10.

Create tunable Control Design Blocks to represent the controllers, and numeric LTI models to represent the plants. Also, create `AnalysisPoint` blocks to mark the points of interest in each feedback loop.

```G2 = zpk([],-2,3); G1 = zpk([],[-1 -1 -1],10); C20 = ltiblock.pid('C2','pi'); C10 = ltiblock.pid('C1','pid'); X1 = AnalysisPoint('X1'); X2 = AnalysisPoint('X2'); ```

Connect these components to build a model of the entire closed-loop control system.

```InnerLoop = feedback(X2*G2*C20,1); CL0 = feedback(G1*InnerLoop*C10,X1); CL0.InputName = 'r'; CL0.OutputName = 'y'; ```

`CL0` is a tunable `genss` model. Specifying names for the input and output channels allows you to identify them when you specify tuning requirements for the system.

Specify tuning requirements for reference tracking and disturbance rejection.

```Rtrack = TuningGoal.Tracking('r','y',10,0.01); Rreject = TuningGoal.Gain('X2','y',0.1); ```

The `TuningGoal.Tracking` requirement specifies that the signal at `'y'` track the signal at `'r'` with a response time of 10 seconds and a tracking error of 1%.

The `TuningGoal.Gain` requirement limits the gain from the implicit input associated with the `AnalysisPoint` block, `X2`, to `'y'`. (See `AnalysisPoint`.) Limiting this gain to a value less than 1 ensures that a disturbance injected at `X2` is suppressed at the output.

Tune the control system.

```[CL,fSoft] = systune(CL0,[Rtrack,Rreject]); ```
```Final: Soft = 1.24, Hard = -Inf, Iterations = 105 ```

`systune` converts each tuning requirement into a normalized scalar value, f. The command adjusts the tunable parameters of `CL0` to minimize the f values. For each requirement, the requirement is satisfied if f < 1 and violated if f >1. `fSoft` is the vector of minimized f values. The largest of the minimized f values is displayed as `Soft`.

The output model `CL` is the tuned version of `CL0`. `CL` contains the same Control Design Blocks as `CL0`, with current values equal to the tuned parameter values.

Validate that the tuned control system meets the tracking requirement by examining the step response from `'r'` to `'y'`.

```stepplot(CL) ```

The step plot shows that in the tuned control system, `CL`, the output tracks the input with approximately the desired response time.

Validate the tuned system against the disturbance rejection requirement by examining the closed-loop response to a signal injected at `X2`.

```CLdist = getIOTransfer(CL,'X2','y'); stepplot(CLdist); ```

`getIOTransfer` extracts the closed-loop response from the specified inputs to outputs. In general, `getIOTransfer` and `getLoopTransfer` are useful for validating a control system tuned with `systune`.

You can also use `viewSpec` to compare the responses of the tuned control system directly against the tuning requirements, `Rtrack` and `Rreject`.

```viewSpec([Rtrack,Rreject],CL) ```

### Tune Control System to Both Hard and Soft Requirements

Tune a cascaded control system to meet requirements of reference tracking and disturbance rejection. These requirements are subject to a hard constraint on the stability margins of the inner and outer loops.

The cascaded control system of the following illustration includes two tunable controllers, the PI controller for the inner loop, , and the PID controller for the outer loop, .

The blocks and mark analysis-point locations. These are locations at which you can open loops or inject signals for the purpose of specifying requirements for tuning the system.

Tune the free parameters of this control system to meet the following requirements:

• The output signal, , tracks the reference signal at with a response time of 5 seconds and a steady-state error of 1%.

• A disturbance injected at is suppressed at the output, , by a factor of 10.

Impose these tuning requirements subject to hard constraints on the stability margins of both loops.

Create tunable Control Design Blocks to represent the controllers and numeric LTI models to represent the plants. Also, create `AnalysisPoint` blocks to mark the points of interest in each feedback loop.

```G2 = zpk([],-2,3); G1 = zpk([],[-1 -1 -1],10); C20 = ltiblock.pid('C2','pi'); C10 = ltiblock.pid('C1','pid'); X1 = AnalysisPoint('X1'); X2 = AnalysisPoint('X2'); ```

Connect these components to build a model of the entire closed-loop control system.

```InnerLoop = feedback(X2*G2*C20,1); CL0 = feedback(G1*InnerLoop*C10,X1); CL0.InputName = 'r'; CL0.OutputName = 'y'; ```

`CL0` is a tunable `genss` model. Specifying names for the input and output channels allows you to identify them when you specify tuning requirements for the system.

Specify tuning requirements for reference tracking and disturbance rejection.

```Rtrack = TuningGoal.Tracking('r','y',5,0.01); Rreject = TuningGoal.Gain('X2','y',0.1); ```

The `TuningGoal.Tracking` requirement specifies that the signal at `'y'` tracks the signal at `'r'` with a response time of 5 seconds and a tracking error of 1%.

The `TuningGoal.Gain` requirement limits the gain from the implicit input associated with the `AnalysisPoint` block `X2` to the output, `'y'`. (See `AnalysisPoint`.) Limiting this gain to a value less than 1 ensures that a disturbance injected at `X2` is suppressed at the output.

Specify tuning requirements for the gain and phase margins.

```RmargOut = TuningGoal.Margins('X1',18,60); RmargIn = TuningGoal.Margins('X2',18,60); RmargIn.Openings = 'X1'; ```

`RmargOut` imposes a minimum gain margin of 18 dB and a minimum phase margin of 60 degrees. Specifying `X1` imposes that requirement on the outer loop. Similarly, `RmargIn` imposes the same requirements on the inner loop, identified by `X2`. To ensure that the inner-loop margins are evaluated with the outer loop open, include the outer-loop analysis-point location, `X1`, in `RmargIn.Openings`.

Tune the control system to meet the soft requirements of tracking and disturbance rejection, subject to the hard constraints of the stability margins.

```SoftReqs = [Rtrack,Rreject]; HardReqs = [RmargIn,RmargOut]; [CL,fSoft,gHard] = systune(CL0,SoftReqs,HardReqs); ```
```Final: Soft = 1.13, Hard = 0.97731, Iterations = 182 Some closed-loop poles are marginally stable (decay rate near 1e-07) ```

`systune` converts each tuning requirement into a normalized scalar value, f for the soft constraints and g for the hard constraints. The command adjusts the tunable parameters of `CL0` to minimize the f values, subject to the constraint that each g < 1.

The displayed value `Hard` is the largest of the minimized g values in `gHard`. This value is less than 1, indicating that both the hard constraints are satisfied.

Validate the tuned control system against the stability margin requirements.

```figure; viewSpec(HardReqs,CL) ```

The `viewSpec` plot confirms that the stability margin requirements for both loops are satisfied by the tuned control system at all frequencies. The red liness represent the actual stability margins of the tuned system. The blue lines represent the margin used in the optimization calculation, which is an upper bound on the actual margin.

Examine whether the tuned control system meets the tracking requirement by examining the step response from `'r'` to `'y'`.

```figure; stepplot(CL,20) ```

The step plot shows that in the tuned control system, `CL`, the output tracks the input but the response is somewhat slower than desired and the tracking error may be larger than desired. For further information, examine the tracking requirement directly with `viewSpec`.

```figure; viewSpec(Rtrack,CL) ```

The actual tracking error crosses into the shaded area between 1 and 10 rad/s, indicating that the requirement is not met in this regime. Thus, the tuned control system cannot meet the soft tracking requirement, time subject to the hard constraints of the stability margins. To achieve the desired performance, you may need to relax one of your requirements or convert one or more hard constraints to soft constraints.

## Input Arguments

collapse all

### `CL0` — Control system to tunegeneralized state-space model | model array

Control system to tune, specified as a generalized state-space (`genss`) model or array of models with tunable parameters. To construct `CL0`:

1. Parameterize the tunable elements of your control system. You can use predefined structures, such as `ltiblock.pid`, `ltiblock.gain`, and `ltiblock.tf`. Alternatively, you can create your own structure from elementary tunable parameters (`realp`).

2. Build a closed-loop model of the overall control system as an interconnection of fixed and tunable components. To do so, use model interconnection commands such as `feedback` and `connect`. Use `AnalysisPoint` blocks to mark additional signals of interest for specifying and assessing tuning requirements.

Specify an array of tunable `genss` models that have the same tunable parameters for robust tuning of a controller against a set of plant models.

### `SoftReqs` — Soft tuning requirements (objectives)vector of `TuningGoal` requirement objects

Soft tuning requirements (objectives) for tuning the control system, specified as a vector of `TuningGoal` requirement objects, such as `TuningGoal.Tracking`, `TuningGoal.StepTracking`, or `TuningGoal.Margins`

`systune` tunes the tunable parameters of the control system to minimize the soft tuning requirements. This tuning is subject to satisfying the hard tuning requirements (if any).

### `HardReqs` — Hard tuning requirements (constraints)`[]` (default) | vector of `TuningGoal` requirement objects

Hard tuning requirements (constraints) for tuning the control system, specified as a vector of `TuningGoal` requirement objects, such as `TuningGoal.Tracking`, `TuningGoal.StepTracking`, or `TuningGoal.Margins`.

`systune` converts each hard tuning requirement to a normalized scalar value. `systune` then optimizes the free parameters minimize those normalized values. A hard requirement is satisfied if the normalized value is less than 1.

### `options` — Options for tuning algorithm`systuneOptions` object

Options for the tuning algorithm, specified as an options set you create with `systuneOptions`. Available options include:

• Number of additional optimizations to run. Each optimization starts from random initial values of the free parameters.

• Tolerance for terminating the optimization.

• Flag for using parallel processing.

## Output Arguments

collapse all

### `CL` — Tuned control systemgeneralized state-space model

Tuned control system, returned as a generalized state-space (`genss`) model. This model has the same number and type of tunable elements (Control Design Blocks) as `CL0`. The current values of these elements are the tuned parameters. Use `getBlockValue` or `showTunable` to access values of the tuned elements.

If you provide an array of control system models to tune as the input argument, `CL0`, `systune` tunes the parameters of all the models simultaneously. In this case, `CL` is an array of tuned `genss` models. For more information, see Robust Tuning Using Multiple Plant Models at Command Line.

### `fSoft` — Best achieved soft constraint valuesvector

Best achieved soft constraint values, returned as a vector. `systune` converts the soft requirements to a function of the free parameters of the control system. The command then tunes the parameters to minimize that function subject to the hard constraints. (See Algorithms.) `fSoft` contains the best achieved value for each of the soft constraints. These values appear in `fSoft` in the same order that the constraints are specified in `SoftReqs`. `fSoft` values are meaningful only when the hard constraints are satisfied.

### `gHard` — Best achieved hard constraint valuesvector

Best achieved hard constraint values, returned as a vector. `systune` converts the hard requirements to a function of the free parameters of the control system. The command then tunes the parameters to drive those values below 1. (See Algorithms.) `gHard` contains the best achieved value for each of the hard constraints. These values appear in `gHard` in the same order that the constraints are specified in `HardReqs`. If all values are less than 1, then the hard constraints are satisfied.

### `info` — Detailed information about optimization runsstructure

Detailed information about each optimization run, returned as a data structure. In addition to examining detailed results of the optimization, you can use `info` as an input to `viewSpec` when validating a tuned MIMO system. `info` contains scaling data that `viewSpec` needs for correct evaluation of MIMO open-loop requirements such as loop shapes and stability margins.

The fields of `info` are:

### `Run` — Run numberscalar

Run number, returned as a scalar. If you use the `RandomStart` option of `systuneOptions` to perform multiple optimization runs, `info` is a struct array, and `info.Run` is the index.

### `Iterations` — Total number of iterationsscalar

Total number of iterations performed during run, returned as a scalar. This value is the number of iterations performed in each run before the optimization terminates.

### `fBest` — Best overall soft constraint valuescalar

Best overall soft constraint value, returned as a scalar. `systune` converts the soft requirements to a function of the free parameters of the control system. The command then tunes the parameters to minimize that function subject to the hard constraints. (See Algorithms.) `info.fBest` is the maximum soft constraint value at the final iteration. This value is meaningful only when the hard constraints are satisfied.

### `gBest` — Best overall hard constraint valuescalar

Best overall hard constraint value, returned as a scalar. `systune` converts the hard requirements to a function of the free parameters of the control system. The command then tunes the parameters to drive those values below 1. (See Algorithms.) `info.gBest` is the maximum hard constraint value at the final iteration. This value must be less than 1 for the hard constraints to be satisfied.

### `fSoft` — Individual soft constraint valuesvector

Individual soft constraint values, returned as a vector. `systune` converts each soft requirement to a normalized value that is a function of the free parameters of the control system. The command then tunes the parameters to minimize that value subject to the hard constraints. (See Algorithms.) `info.fSoft` contains the individual values of the soft constraints at the end of each run. These values appear in `fSoft` in the same order that the constraints are specified in `SoftReqs`.

### `gHard` — Individual hard constraint valuesvector

Individual hard constraint values, returned as a vector. `systune` converts each hard requirement to a normalized value that is a function of the free parameters of the control system. The command then tunes the parameters to minimize those values. A hard requirement is satisfied if its value is less than 1. (See Algorithms.) `info.gHard` contains the individual values of the hard constraints at the end of each run. These values appear in `gHard` in the same order that the constraints are specified in `HardReqs`.

### `MinDecay` — Minimum decay rate of closed-loop polesvector

Minimum decay rate of closed-loop poles, returned as a vector.

By default, closed-loop pole locations of the tuned system are constrained to satisfy Re(p) < –10–7. Use the `MinDecay` option of `systuneOptions` to change this constraint.

### `Blocks` — Tuned values of tunable blocks and parametersstructure

Tuned values of tunable blocks and parameters in the tuned control system, `CL`, returned as a structure. You can also use `getBlockValue` or `showBlockValue` to access the tuned parameter values.

### `LoopScaling` — Optimal diagonal scaling for MIMO tuning requirementsstate-space model

Optimal diagonal scaling for evaluating MIMO tuning requirements, returned as a state-space model.

When applied to multiloop control systems, `TuningGoal.LoopShape` and `TuningGoal.Margins` can be sensitive to the scaling of the loop transfer functions to which they apply. This sensitivity can lead to poor optimization results. `systune` automatically corrects scaling issues and returns the optimal diagonal scaling matrix `d` as a state-space model in `info.LoopScaling`.

The loop channels associated with each diagonal entry of `D` are listed in `info.LoopScaling.InputName`. The scaled loop transfer is `D\L*D`, where `L` is the open-loop transfer measured at the locations `info.LoopScaling.InputName`.

collapse all

### Algorithms

x is the vector of tunable parameters in the control system to tune. `systune` converts each soft and hard tuning requirement `SoftReqs(i)` and `HardReqs(j)` into normalized values fi(x) and gj(x), respectively. `systune` then solves the minimization problem:

Minimize $\underset{i}{\mathrm{max}}{f}_{i}\left(x\right)$ subject to $\underset{j}{\mathrm{max}}{g}_{j}\left(x\right)<1$, for ${x}_{\mathrm{min}}.

xmin and xmax are the minimum and maximum values of the free parameters of the control system.

`systune` returns the control system with parameters tuned to the values that best solve the minimization problem. `systune` also returns the best achieved values of fi(x) and gj(x), as `fSoft` and `gHard` respectively.

For information about the functions fi(x) and gj(x) for each type of constraint, see the reference pages for each `TuningGoal` requirement object.

`systune` uses the nonsmooth optimization algorithms described in [1].

`systune` computes the H norm using the algorithm of [2] and structure-preserving eigensolvers from the SLICOT library. For more information about the SLICOT library, see http://slicot.org.

## References

[1] Apkarian, P. and D. Noll, "Nonsmooth H-infinity Synthesis." IEEE Transactions on Automatic Control, Vol. 51, No. 1, (2006), pp. 71–86.

[2] Bruisma, N.A. and M. Steinbuch, "A Fast Algorithm to Compute the H-Norm of a Transfer Function Matrix," System Control Letters, Vol. 14, No, 4 (1990), pp. 287–293.


text:
0
101 Feb 27, 2012 at 23:49

In a book I’m reading I learned that we use constant buffers in directX to minise the number of API calls needed to do state changes, which leads to better performance.

However, then I see the book using it like this:

The C++ application code typically needs to communicate with the effect; in particular, the C++ application usually needs to update variables in the constant buffers. For example, suppose that we had the following constant buffer defined in an effect file:

cbuffer cbPerObject
{
float4x4 gWVP;
float4 gColor;
float gSize;
int gIndex;
bool gOptionOn;
};

Through the ID3D10Effect interface, we can obtain pointers to the variables in the constant buffer:

ID3D10EffectMatrixVariable* fxWVPVar;
ID3D10EffectVectorVariable* fxColorVar;
ID3D10EffectScalarVariable* fxSizeVar;
ID3D10EffectScalarVariable* fxIndexVar;
ID3D10EffectScalarVariable* fxOptionOnVar;
fxWVPVar      = mFX->GetVariableByName("gWVP")->AsMatrix();
fxColorVar  = mFX->GetVariableByName("gColor")->AsVector();
fxSizeVar    = mFX->GetVariableByName("gSize")->AsScalar();
fxIndexVar  = mFX->GetVariableByName("gIndex")->AsScalar();
fxOptionOnVar = mFX->GetVariableByName("gOptionOn")->AsScalar();

The ID3D10Effect::GetVariableByName method returns a pointer of type ID3D10EffectVariable. This is a generic effect variable type; to obtain a pointer to the specialized type (e.g., matrix, vector, scalar), you must use the appropriate As***** method (e.g., AsMatrix, AsVector, AsScalar).
Once we have pointers to the variables, we can update them through the C++ interface. Here are some examples:

fxWVPVar->SetMatrix( (float*)&M ); // assume M is of type D3DXMATRIX
fxColorVar->SetFloatVector( (float*)&v ); // assume v is of type
// D3DXVECTOR4
fxSizeVar->>SetFloat( 5.0f );
fxIndexVar->SetInt( 77 );
fxOptionOnVar->SetBool( true );

But now they make separate calls, is this not beating the purpose of constant buffers?

If it isn’t, then why not?
If it is, then how do I set it correctly? (I assume this has something to do with “ID3D10EffectConstantBuffer::SetConstantBuffer()” but it seems to be poorly documented)

#### 2 Replies

0
167 Feb 28, 2012 at 05:54

I think it’s less to do with the number of C++ function calls and more to do with the amount of stuff in the GPU command buffer. On older architectures, each time you set a vertex or fragment parameter resulted in a separate command in the buffer and (I guess) that could be costly in some cases. With hardware constant buffers, ultimately you’re just assembling your data in a struct, then using one command to pass a pointer to that struct to the GPU.

You can do all this yourself: you’d make a struct in C++ that exactly matches the layout of the constant buffer as declared in the shader code, and populate it with your data. Create an ID3D10Buffer object for each constant buffer, map it using MAP_WRITE_DISCARD and copy your struct into it each frame (or whenever it needs to be updated). Then use SetConstantBuffer() to point the GPU at that buffer object before a draw call.

Most of these mechanics are done for you by the effect system, though. When you use the ID3D10EffectBlahVariable objects, It’s internally keeping track of the constant buffers, and when you do a SetFloat(), SetInt(), etc. probably all it’s doing is storing the value into a struct member, so those would be pretty lightweight calls. At some point, presumably when you do a draw call, it caches off a copy of the current constant buffer and points the GPU to that. At least that’s my guess for what it’s doing under the hood.

Doing it yourself might be a bit faster, but setting constants isn’t likely to be a bottleneck for a typical app, at least I wouldn’t think so. So it’s probably not worth worrying about.

0
102 Mar 05, 2012 at 00:26

In order to efficiently use constant buffers, you need to organize your data according to the data update frequency and cache the cbuffers. Essentially you want to minimize the number of cbuffer updates in order to minimize amount of data that needs to be transferred from CPU to GPU. In your example you have MVP matrix in per-object constant buffer, but this isn’t a good idea because you have to update the cbuffer per frame because of the camera changes. It’s better to hold only local->world matrix in per-object cbuffer, so that the cbuffer needs to be updated only when the object moves. Then have world->projection matrix in separate cbuffer and perform the matrix multiplication in shader instead (or rather transfer vector first to world with per-object matrix and then world->projection with per-view matrix).

In practice I use 3 different types of cbuffer in my DX11 code: per-view, per-object and per-material cbuffers. This minimizes the need to update cbuffers since per-view needs update only once per view. per-object cbuffer requires updates only if object moves (only small percentage of objects is expected to move per frame), and per-material cbuffer requires changes only upon material property updates (almost never).

Cheers, Jarkko


text:
My Math Forum Correlation

 May 11th, 2017, 05:53 AM #1 Newbie   Joined: May 2017 From: Belgium Posts: 2 Thanks: 0 Correlation I have to say if my correlation is significant or not so I did a t-test but I don't know what to do with the outcome. I calculated my correlation using excel = 0,754651674 R= 0,754651674 R square= 0,569499149 variance 57% -> strong connection Test statistic t = r/√1-r²/n-2 0,75/√1-(0.75)²/40-2 = 6.90 I don't know what to do with the 6,90 outcome?
 May 23rd, 2017, 02:18 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics $\displaystyle t_{0.05, df=38} = 1.685954$; your correlation is significant at the 0.05 level.

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# Near-quantum-limited amplification from inelastic Cooper-pair tunnelling

## Abstract

The readout of microwave quantum systems, such as spin or superconducting qubits, requires low-noise amplifiers with added noise as close as possible to the quantum limit. This limit has so far been approached only by parametric amplifiers that exploit nonlinearities in superconducting circuits and are driven by a strong microwave pump tone. However, this microwave drive makes the amplifiers much more difficult to implement and operate than conventional d.c.-powered amplifiers, which currently suffer from much higher noise. Here, we show that a simple d.c.-powered set-up can provide amplification close to the quantum limit. Our amplification scheme is based on the stimulated microwave photon emission accompanying inelastic Cooper-pair tunnelling through a d.c.-biased Josephson junction. The key to the low noise of this approach is a well-defined auxiliary idler mode, which allows for operation analogous to parametric amplifiers.

## Access options

\$32.00

All prices are NET prices.

## References

1. Caves, C. M. Quantum limits on noise in linear amplifiers. Phys. Rev. D 26, 1817–1839 (1982).

2. Yurke, B. et al. Observation of 4.2-K equilibrium-noise squeezing via a Josephson-parametric amplifier. Phys. Rev. Lett. 60, 764–767 (1988).

3. Castellanos-Beltran, M. A. & Lehnert, K. W. Widely tunable parametric amplifier based on a superconducting quantum interference device array resonator. Appl. Phys. Lett. 91, 083509 (2007).

4. Bergeal, N. et al. Phase-preserving amplification near the quantum limit with a Josephson ring modulator. Nature 465, 64–68 (2010).

5. Roch, N. et al. Widely tunable, nondegenerate three-wave mixing microwave device operating near the quantum limit. Phys. Rev. Lett. 108, 147701 (2012).

6. Mutus, J. Y. et al. Strong environmental coupling in a Josephson parametric amplifier. Appl. Phys. Lett. 104, 263513 (2014).

7. Roy, T. et al. Broadband parametric amplification with impedance engineering: beyond the gain–bandwidth product. Appl. Phys. Lett. 107, 262601 (2015).

8. Simoen, M. et al. Characterization of a multimode coplanar waveguide parametric amplifier. J. Appl. Phys. 118, 154501 (2015).

9. Macklin, C. et al. A near-quantum-limited Josephson traveling-wave parametric amplifier. Science 350, 307–310 (2015).

10. Hover, D. et al. Superconducting low-inductance undulatory galvanometer microwave amplifier. Appl. Phys. Lett. 100, 063503 (2012).

11. Lähteenmäki, P., Vesterinen, V., Hassel, J., Seppä, H. & Hakonen, P. Josephson junction microwave amplifier in self-organized noise compression mode. Sci. Rep. 2, 276 (2012).

12. Lähteenmäki, P. et al. Advanced concepts in Josephson junction reflection amplifiers. J. Low Temp. Phys. 175, 868–876 (2014).

13. Holst, T., Esteve, D., Urbina, C. & Devoret, M. H. Effect of a transmission line resonator on a small capacitance tunnel junction. Phys. Rev. Lett. 73, 3455–3458 (1994).

14. Hofheinz, M. et al. Bright side of the Coulomb blockade. Phys. Rev. Lett. 106, 217005 (2011).

15. Leppäkangas, J., Johansson, G., Marthaler, M. & Fogelström, M. Nonclassical photon pair production in a voltage-biased Josephson junction. Phys. Rev. Lett. 110, 267004 (2013).

16. Westig, M. et al. Emission of nonclassical radiation by inelastic Cooper pair tunneling. Phys. Rev. Lett. 119, 137001 (2017).

17. Russer, P. Parametric amplification with Josephson junctions. Archiv für Elektronik und Übertragungstechnik 23, 417–420 (1969).

18. Russer, P. & Russer, J. A. Nanoelectronic RF Josephson devices. IEEE Trans. Microw. Theory Techn. 59, 2685–2701 (2011).

19. Jebari, S. The Inelastic Cooper Pair Tunneling Amplifier (ICTA). PhD thesis, Université Grenoble Alpes (2017).

20. Ingold, G.-L. & Nazarov, Y. V. in Single Charge Tunneling: Coulomb Blockade Phenomena in Nanostructures (eds Grabert, H. & Devoret, M. H.) 21–107 (Plenum, New York, 1992).

21. Basset, J., Bouchiat, H. & Deblock, R. Emission and absorption quantum noise measurement with an on-chip resonant circuit. Phys. Rev. Lett. 105, 166801 (2010).

22. Safi, I. & Joyez, P. Time-dependent theory of nonlinear response and current fluctuations. Phys. Rev. B 84, 205129 (2011).

23. Roussel, B., Degiovanni, P. & Safi, I. Perturbative fluctuation dissipation relation for nonequilibrium finite-frequency noise in quantum circuits. Phys. Rev. B 93, 045102 (2016).

24. Chen, F. et al. Realization of a single-Cooper-pair Josephson laser. Phys. Rev. B 90, 020506 (2014).

25. Cassidy, M. C. et al. Demonstration of an AC Josephson junction laser. Science 355, 939–942 (2017).

26. Tien, P. K. & Gordon, J. P. Multiphoton process observed in the interaction of microwave fields with the tunneling between superconductor films. Phys. Rev. 129, 647–651 (1963).

27. Safi, I. & Sukhorukov, E. V. Determination of tunneling charge via current measurements. Eur. Phys. Lett. 91, 67008 (2010).

28. Parlavecchio, O. et al. Fluctuation–dissipation relations of a tunnel junction driven by a quantum circuit. Phys. Rev. Lett. 114, 126801 (2015).

## Acknowledgements

The authors acknowledge discussions with J. Leppäkangas, C. Altimiras, M. Devoret, B. Kubala and J. Ankerhold, as well as financial support from the Grenoble Nanosciences Foundation (grant WiQOJo), the European Union (ERC starting grant 278203 WiQOJo and ICT grant 218783 SCoPE) and the ANR (grants Masquelspec, AnPhoTEQ, JosePhSCharLi).

## Author information

Authors

### Contributions

S.J., F.B. and A.G. performed the measurements and analysed the data. M.H., F.P. and D.V. designed and fabricated the sample. A.G., M.H., F.B., R.A., S.J. and D.H. built the set-up and wrote software. M.H. and S.J. wrote the manuscript with input from all authors.

### Corresponding author

Correspondence to M. Hofheinz.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Supplementary information

### Supplementary Information

Supplementary Note 1, Supplementary Figures 1–2

## Rights and permissions

Reprints and Permissions

Jebari, S., Blanchet, F., Grimm, A. et al. Near-quantum-limited amplification from inelastic Cooper-pair tunnelling. Nat Electron 1, 223–227 (2018). https://doi.org/10.1038/s41928-018-0055-7

• Accepted:

• Published:

• Issue Date:

• DOI: https://doi.org/10.1038/s41928-018-0055-7

• ### Cooper-Pair Tunneling in Small Josephson Junction Arrays Under Radio-Frequency Irradiation

• Godwill Mbiti Kanyolo
• Kouichi Takeda

Journal of Low Temperature Physics (2020)

• ### Microwave amplifiers keep the noise down

• Nicolas Bergeal

Nature Electronics (2018)


text:
Browse Questions

# Four candidates A,B,C,D have applied for the assignment to coach a school cricket team.If A is twice as likely to be selected as B,and B and C are given about the same chance of being selected,while C is twice as likely to be selected as D,what are the probabilities that A will not be selected?

$\begin{array}{1 1}(A)\;\large\frac{5}{9}\\(B)\;\large\frac{2}{9}\\(C)\;\large\frac{1}{9}\\(D)\;\text{None of these}\end{array}$

Step 1:
Let the probability of D selected be $'x'$
$\therefore P(D)=x$
$P(C)=2x$
$P(B)=2x$
$P(A)=4x$
$P(A)+P(B)+P(C)+P(D)=1$
Sum of all probabilities =1
$\therefore 4x+2x+2x+x=1$
$9x=1$
$x=\large\frac{1}{9}$
Step 2:
A will not be selected
$\therefore P(A$ will not selected)=$1-P(A)=1-4x$
$\Rightarrow 1-\large\frac{4}{9}=\frac{5}{9}$
Hence (A) is the correct answer.


text:
# Thread: limit for a sequence...

1. ## limit for a sequence...

Hey all,

Having trouble trying to find the limit for the following sequence
$\displaystyle \frac{(\frac{1}{2})^{2n+1}-(\frac{1}{3})^{3n-1}}{(\frac{1}{4})^{n}+(\frac{1}{3})^{2n-1}}$ as the sequence goes from $\displaystyle n = 1$ to $\displaystyle \infty$

Thanks for all the help....

2. ## Re: limit for a sequence...

$\displaystyle \lim_{n \rightarrow \infty}\frac{(\frac{1}{2})^{2n+1}-(\frac{1}{3})^{3n-1}}{(\frac{1}{4})^{n}+(\frac{1}{3})^{2n-1}}$ aquires the indeterminate form, $\displaystyle \frac{\infty}{\infty}$. So I would suggest the use of L' Hopital's rule.

Also check what wolfram alpha has to say:chick here

Thanks :-)

4. ## Re: limit for a sequence...

Not to dissuade you from trying this, but I don't think L'Hopital's will help here. First off, it is of indeterminate form $\displaystyle \frac 0 0$. Furthermore, since these are all exponential functions, they will all reappear when applying L'Hopital's rule, and will only be multiplied by constants, so you will still have an indeterminate form. WolframAlpha doesn't show the steps, unfortunately.

5. ## Re: limit for a sequence...

Yes, I am sorry. It is of the indeterminate form $\displaystyle \frac{0}{0}$.

$\displaystyle \lim_{n%20\rightarrow%20\infty}\frac{(\frac{1}{2}) ^{2n+1}-(\frac{1}{3})^{3n-1}}{(\frac{1}{4})^{n}+(\frac{1}{3})^{2n-1}}$

$\displaystyle =\lim_{n%20\rightarrow%20\infty}\frac{2^{-2n-1}-3^{1-3n}}{3^{1-2n}+4^{-n}}$

Tried L' Hopital's. It is also of no use here.

Wolfram Alpha shows (1/2) but there seems no way to solve it.

6. ## Re: limit for a sequence...

Originally Posted by Oiler
Hey all,

Having trouble trying to find the limit for the following sequence
$\displaystyle \frac{(\frac{1}{2})^{2n+1}-(\frac{1}{3})^{3n-1}}{(\frac{1}{4})^{n}+(\frac{1}{3})^{2n-1}}$ as the sequence goes from $\displaystyle n = 1$ to [TEX]\infty[TEX]
Note
$\displaystyle \frac{(\frac{1}{2})^{2n+1}-(\frac{1}{3})^{3n-1}}{(\frac{1}{4})^{n}+(\frac{1}{3})^{2n-1}}=$$\displaystyle \frac{(\frac{1}{4})^{n}(\frac{1}{2})-(\frac{1}{27})^{n}(3)}{(\frac{1}{4})^{n}+(\frac{1} {9})^{n}(3)}$

Dividing by $\displaystyle \left(\tfrac{1}{4}\right)^n$ it is clear that the limit is $\displaystyle \frac{1}{2}$.


text:
# How do I add a method to a class without editing the source file?

I am trying to add a method to a Minecraft source file, but I have to figure out how to do it without actually editing the source files, as it would be illegal to redistribute Minecraft's source files with the mod I am creating.

I need to add the method setInPortalZub() to the EntityPlayer class located in net.minecraft.entity.player. I am using the MCP/Minecaft Forge API. I have tried creating an EntityPlayer instance, but I'm not exactly sure how to make this work.

• So you mean you can only make instances of EntityPlayer? If that's the case, you could make a wrapper and add your own functions from there.. – Krienie Sep 2 '13 at 22:53
• Nice to see software licensing terms are being studied before coding these days. I know that sounds sarcastic, but I genuinely mean it. Often on mod. projects people dive into the coding before realizing whether what they're doing is legal or not. That said, why do you have to insert something into the source code? Cannot you create a specialized subclass? The only way to really insert something into EntityPlayer without modifying the source code would be to disassemble, insert some Java bytecode and then re-assemble, and that's even less likely to be legal :) – Andon M. Coleman Sep 2 '13 at 23:48
• Welcome to Minecraft modding, where basically you have to work around the crappy modloader/forge apis, and not to mention Minecraft itself. What I usually found myself doing was creating hash maps with player ids and flags/data to store arbitrary data rather than relying on the actual entity player class. It's pretty dodgy but hey it's a mod – Luke Nov 5 '13 at 19:36
• Wait, is the source file available to the legal mod users? Cause if so I have an answer for this question. :) – wolfdawn Mar 15 '14 at 8:33
• Would reflection work? (I've used it sparingly in Java coursework when told "don't change the given source".) – Anko Apr 8 '14 at 19:23

Only way to change a class without extending it (or without access to the source) is by using reflection. For java there is the rather comprehensive Java Reflection API that is made for this task. It might be a bit daunting at first.

Anyway the operation you are looking for is called injection. Here is an example of injecting a method into the class at runtime.

Actually, if you don't need to change any of the minecraft core functionality, I would recommend using a mod API such as Minecraft Forge. It allows you to listen to player events and add blocks/entities. However, if you do need to edit core minecraft functionality, use the Mincraft Coder Pack, or MCP, it allows you to decompile minecraft, as well as recompile and obfusicate the code, and it will give you the compiled files that were modified, if you do that, then you can just, give them those files to put in the minecraft.jar

Extend the class, then make your own method on the extension

MyClass extends TargetClass
{
void myMethod();
}

• For this to work, all existing instantiations of TargetClass (EntityPlayer in the OP's case) would need to be replaced with instantiations of MyClass unless the OP does not need to use his new method on anything except objects he also manually creates (which seems unlikely). How is that accomplished without modifying the code, violating the original constraint? – Josh Apr 23 '14 at 16:38
• @JoshPetrie That's a very good point. For some reason, I took that as 'without modifying the class' rather than 'without modifying the source'. I stand very much corrected. Thanks! – wes Apr 23 '14 at 19:21


text:
Problem using UART0 in TM4C123 Tiva C Launch Pad

moahmedatef

New Member
I'm trying to interface UART0 with the my laptop using the ICDI
the code I used is from Mazidi's book :
C:
#include <stdint.h>
#include "tm4c123gh6pm.h"
void UART0Tx(char c);
void delayMs(int n);
int main(void)
{
SYSCTL->RCGCUART |= 1; /* provide clock to UART0 */
SYSCTL->RCGCGPIO |= 1; /* enable clock to PORTA */
delayMs(20); /* wait for output line to stabilize */
/* UART0 initialization */
UART0->CTL = 0; /* disable UART0 */
UART0->IBRD =  104; /* 16MHz/16=1MHz, 1MHz/104=9600 baud rate */
UART0->FBRD =  11; /* fraction part, see Example 4-4 */
UART0->CC = 0; /* use system clock */
UART0->LCRH = 0x60; /* 8-bit, no parity, 1-stop bit, no FIFO */
UART0->CTL = 0x301; /* enable UART0, TXE, RXE */
/* UART0 TX0 and RX0 use PA0 and PA1. Set them up. */
GPIOA->DEN = 0x03; /* Make PA0 and PA1 as digital */
GPIOA->AFSEL = 0x03; /* Use PA0,PA1 alternate function */
GPIOA->PCTL = 0x11; /* configure PA0 and PA1 for UART */
delayMs(1); /* wait for output line to stabilize */
for(;;)
{
UART0Tx('M');
delayMs(1000) ;

}
}
/* UART0 Transmit */
/* This function waits until the transmit buffer is available then */
/* writes the character in the transmit buffer. It does not wait */
/* for transmission to complete. */
void UART0Tx(char c)
{
while((UART0->FR & 0x20) != 0); /* wait until Tx buffer not full */
UART0->DR = c; /* before giving it another byte */
}
void delayMs(int n)
{
int i = 0 , j = 0 ;
for(i = 0 ; i < n ; i++)
{
for(j = 0 ; j < 3180 ; j++)
{

}
}
}
and there are no problems with the COM port I've set it up in a correct way and I made sure to modify options for target to get 16Mhz clock frequency and to debug using ICDI ( I'm using keil uvision IDE)

and All I get is this symbol

What should I do ?

rjenkinsgb

Well-Known Member
Are you getting a new character each second or so?
Or just those then nothing?

If you are getting new characters at regular intervals, it's most likely a baud rate problem.

moahmedatef

New Member
a new character each second and the baudrate is is the same in both the code and puTTY

rjenkinsgb

Well-Known Member
The small Y character appears to be UTF-8; C3BF

For it to receive more than one character for each one sent, the baud rate must be too slow by some amount, so a zero bit in the sent character is seens as the start of a second character.

Looking at the bit patterns, possibly it's sending at 4800 ? Or it could be some intermediate speed.

moahmedatef

New Member
The small Y character appears to be UTF-8; C3BF

For it to receive more than one character for each one sent, the baud rate must be too slow by some amount, so a zero bit in the sent character is seens as the start of a second character.

Looking at the bit patterns, possibly it's sending at 4800 ? Or it could be some intermediate speed.
• I have changed the baud rate in the virtual terminal installed on my pc to 28800 and the baud rate in the code is still 9600 and the character was transmitted correctly what does that mean ?

• The baud rate in this code is 9600 and I got the divisor using the data sheet and the reference book When the baud rate in the pc terminal was 9600 the transmitted character was not the desired When I changed it in the pc terminal to 28800 and in my code it is still 9600

rjenkinsgb

Well-Known Member
28800 is three times 9600.
That implies the UART clock is 3x faster than expected for the calculations you have used?

What is the main oscillator frequency you are using, vs. the osc frequency of the device in the example code?

moahmedatef

New Member
28800 is three times 9600.
That implies the UART clock is 3x faster than expected for the calculations you have used?

What is the main oscillator frequency you are using, vs. the osc frequency of the device in the example code?
16 MHz external oscillator which is by default the oscillator of the MCU

rjenkinsgb

Well-Known Member
Very strange... I have no idea why!

Try tripling the UART clock divider, from 104 to 312 and see if that gives 9600 baud??

moahmedatef

New Member
Very strange... I have no idea why!

Try tripling the UART clock divider, from 104 to 312 and see if that gives 9600 baud??
Yes
Does this mean I have a clock frequency of 48Mz ?


text:
## Basic College Mathematics (10th Edition)

Compute how many cups of milk are needed, then convert to quarts and gallons. Multiply amount per toddler by number of toddlers by number of days. $\frac{2}{3}\times15\times5=\frac{30}{3}\times5=10\times5=50$ cups of milk are needed each week. Convert to quarts. $50 c\times\frac{1 qt}{4 c} = \frac{50}{4} qt = 12.5 qt$ Convert to gallons. $12.5 qt\times\frac{1 gal}{4 qt}=\frac{12.5}{4} gal=3.125 gal$