text:
# Reviews

RICHARD MATEOSIAN

Pages: pp. 128

This time I write about two books by economists. The first looks at the economics and winning business strategies of industries that require software platforms. The second addresses globalization.

Invisible Engines: How Software Platforms Drive Innovation and Transform Industries, David S. Evans, Andrei Hagiu and Richard Schmalensee MIT Press, 2006, 395 pp., ISBN 0-262- 05085-4, http://www.mitpress.mit.edu, US$34.95) The authors are experts who consult regularly with leading platform-based businesses. Evans and Schmalensee have advised Microsoft for many years, particularly about Microsoft's antitrust difficulties. Hagiu wrote a doctoral dissertation on two-sided economies—situations in which two groups must participate (for example, auctions, which need both buyers and sellers). In this book, the authors lay out the underlying principles and parallels of personal computers, mobile phones, video games, online auctions, search engines, and similar industries. All of these have in common that they have more than one "side." For example, the Microsoft Windows operating system creates an ecosystem of hardware manufacturers, application developers, and end users. The various sides in the ecosystem that a software platform creates depend on each other in various ways. Hardware manufacturers need software support to maximize their performance/price ratios. Hardware manufacturers and developers need buyers for their products, while the existence of desirable products on a platform leads to more buyers. These are the well-known network effects that cause some platform ecosystems to grow rapidly while others shrink and die. The interdependence of the sides in a platform-based ecosystem is commonplace knowledge. This book, however, goes into the details. It provides brief histories of various platforms and shows how their business and pricing strategies evolved. The authors find that in each platform ecosystem, one side gets a better deal than the others. For example, in the Windows ecosystem, developers receive valuable tools and support for minimal cost, while buyers of Windows-based computers and software provide most of Microsoft's profits. In the video game business, on the other hand, users receive the game platforms at highly favorable prices, and the platform makers receive most of their profit from game developers. Media players (for example, Real Player or Apple QuickTime) are always free to end users, while the developers of the player software obtain their profits from content developers and distributors. The book delves into pricing issues such as price discrimination, bundling, and charging for access versus charging for usage. The authors find, sadly, that it almost always pays the manufacturer to bundle features, even though most buyers won't use most of what they buy. This leads to a kind of Moore's Law of software bloat: Software platforms double in size roughly every two years. The access versus usage question arises most notably in communication platforms such as phones or personal digital assistants (PDAs). In theory, platform providers should charge for both, but in practice they usually charge for just one. Interestingly, an industry rarely changes its pricing strategy as it matures. It tends to stick with the model that made it successful. An industry does, however, change its integration strategy. New platforms usually provide all sides of the ecosystem, because the network effects have not yet occurred. Later, as the ecosystem grows, the provider defines application programming interfaces (APIs) to allow outside vendors to supply alternatives to the platform provider's offerings. Microsoft Windows, for example, defines many standards to allow computer and peripheral makers, application developers, and even browser makers to complement each other's offerings. The Apple media delivery system, on the other hand, works best when all of the components (iTunes, iPod, the Apple music store, and the associated digital rights management software) come from Apple. Apple buys and resells content for this system. As the authors discuss one platform after another, they repeatedly stress the importance of APIs and other standards that allow different sides of the platform ecosystem to innovate without entanglement with other players. We all know this, but it is interesting and useful to see the principle at work in many different platform ecologies. The best thing about this book is the concrete detail. Although the authors work hard to show the similarities between different platform-based businesses, their theories are not a procrustean bed. They discuss the quirks and odd corners of each industry and show how those quirks lead to the unique business strategies of that industry. This is a valuable book for anyone who wants to understand or engage in a business based on a software platform. I recommend it highly. Making Globalization Work, Joseph E. Stiglitz (W.W. Norton, 2006, 384 pp., ISBN 0-393-06122-1, http://www. wwnorton.com, US$26.95)

In The World Is Flat (Farrar, Straus and Giroux, 2005; Micro Review, May/June 2005), New York Times columnist Tom Friedman emphasized the bright side of globalization. In Globalization and Its Discontents (Norton, 2002), Nobel Prize winning economist Joseph Stiglitz addressed the dark side of globalization. In his latest work, Stiglitz fuses these threads. He proposes ways to correct the problems he described in his earlier work, so everyone can benefit from globalization.

Joseph Stiglitz shared the 2001 Nobel Prize in Economic Sciences for his analysis of markets with asymmetric information. His research disproved many cherished assumptions that had persisted from the time of Adam Smith to the late twentieth century. In 1776, Adam Smith argued that free markets lead to efficient outcomes as if by an invisible hand. Starting with his field work in Kenya in 1969, Stiglitz accumulated the insights that led him finally to the conclusion that the invisible hand, if it ever existed, had disappeared.

In its place, Stiglitz argues, government has a role in keeping free markets working. For example, free markets naturally lead to rapid flows of speculative capital into and out of developing countries. Institutions and regulations must protect those countries from the resulting instability.

By the time he received his Nobel Prize, Stiglitz had put his ideas into practice. He served as chair of President Clinton's Council of Economic Advisors, then as Chief Economist of the World Bank. While at the World Bank, Stiglitz went to war against the policies of the International Monetary Fund (IMF) and its chief supporter and shareholder, the United States Treasury Department.

Stiglitz's 2002 book Globalization and Its Discontents contends that IMF policies favored United States financial interests, while short-changing the poor nations that the IMF is supposed to help. That book was widely influential, especially after huge protests against globalization like that at the World Trade Organization meeting in Seattle, in November 1999. The book identified many problems and suggested some changes, but it didn't offer a comprehensive solution. Four years later, Stiglitz continues the story of the adverse effects of globalization, but also offers solutions to many of the problems he sees.

Stiglitz summarizes the main complaints against the current implementation of globalization:

• The rules favor the industrialized countries. The rules are so unfair that they make the poorest countries worse off.
• Material values are more important than the environment or even life itself.
• Democracy suffers when developing countries lose sovereignty over decisions that affect their citizens' welfare.
• There are many losers in both developed and developing countries.
• Developing countries must accept Americanization of their economies and their culture.

An issue that exemplifies several of these points is especially relevant to technology-based industries. The protection of intellectual property ensures that those who invest their time and money in writing books or developing inventions receive a reward for their efforts. Granting a time-limited monopoly to creators encourages innovation and helps to expand the body of knowledge.

In recent years, however, United States laws about intellectual property have become a drag on innovation in the software industry. There are far too many software patents, and most of them are unwarranted. In any event, nobody can keep track of them all. Stiglitz quotes Tim Berners-Lee on this point: "Developers are stalling their efforts in a given direction when they hear rumors that some company may have a patent that may involve the technology."

In the drug industry, patent laws favor the development of minor variations of high-volume drugs rather than cures for "orphan" illnesses with low potential sales. They also keep life-saving drugs out of the hands of people who need them but can't afford to pay.

In 1993, bypassing the existing World Intellectual Property Organization, the World Trade Organization adopted the TRIPs agreement (for "trade-related" intellectual property). This subjected developing countries to even more onerous intellectual property restrictions. The most well-known example is the fight over the right of countries with high incidences of HIV-AIDS to manufacture generic versions of expensive drugs for that disease.

More humorous examples are attempts to enforce western patents on traditional Chinese medicine or on the medicinal use of turmeric in India.

I never advocate any particular political point of view in this column. Globalization, however, is inherently political. In laying out his program to address the problems of globalization, Stiglitz makes many statements that others might regard as politically biased. Nonetheless, Stiglitz presents a compelling, comprehensive case for his program. Like Thomas Jefferson, he believes that an informed electorate that is willing to hold its leaders accountable can solve problems like those of globalization. He hopes to help today's voters and leaders understand the issues that globalization has raised.

I won't go further into Stiglitz's views here. I recommend that you read his book for the details of his analysis and his solutions.


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A numerical solution to the problem of hydrodynamic lubrication of rigid point contacts with an isoviscous, incompressible lubricant has been obtained. The hydrodynamic load-carrying capacity under unsteady (or dynamic) conditions arising from the combined effects of squeeze motion superposed upon the entraining motion has been determined for both normal approach and separation. Superposed normal motion considerably increases net load-carrying capacity during normal approach and substantially reduces net load-carrying capacity during separation. Geometry has also been found to have a significant influence on the dynamic load-carrying capacity. The ratio of dynamic to steady state load-carrying capacity increases with increasing geometry parameter for normal approach and decreases during separation. The cavitation (film rupture) boundary is also influenced significantly by the normal motion, moving downstream during approach and upstream during separation. For sufficiently high normal separation velocity the rupture boundary may even move upstream of the minimum-film-thickness position. Sixty-three cases were used to derive a functional relationship for the ratio of the dynamic to steady state load-carrying capacity β in terms of the dimensionless normal velocity parameter q (incorporating normal velocity, entraining velocity, and film thickness) and the geometry parameter α. The result is expressed in the form
$β={α−0.028sech(1.68q)}1/q$
The ratio of the dynamic to steady state peak pressures in the contact ξ increases considerably with increasing normal velocity parameter during normal approach, with a similar decrease during separation. The ratio is expressed as a function of q and α by
$ξ={α−0.032sech(2q)}1/q$
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### #6 The optimal concentration of nectar-I

/*-------------Divider-------------*/
There are nectar-feeding species in bees, butterflies, moths, ants, hummingbirds, honeyeaters, sunbirds, and bats. The more concentrated the nectar is, the more energy per unit volume it contains. However, the nectar would also become more viscous and therefore compromise the maximum flow rate. Does this phenomenon place any limitation on nectar-feeding species? How would animals utilizing different nectar-feeding mechanisms be affected differently? That’s what we are going to discuss in these serial episodes and also the research problem that Kim, Gilet and Bush were interested in.
/*-------------Divider-------------*/

The authors of the suggested reading consider 3 different nectar-feeding mechanisms. They are
(1)  Active suction utilized by butterflies, moths, and some bees and ants. These animals produce negative pressure in their oral cavity to suck in the nectar.
(2)  Capillary suction employed by hummingbirds, honeyeaters, and sunbirds. These birds intake nectar by capillary flow in the hollow tube formed by their tongue.
(3)  Viscous dipping exploited by bats, some bees and ants. They place their tongue into the nectar and rapidly retract it back to their mouth.
In these series, we will talk about active suction in episode I, capillary suction in episode II, viscous dipping in episode III, and have extra explanation about LLD theory in episode IV.

/*-------------Divider-------------*/
Let’s consider the physics mechanism of active suction. Assume the mouthpiece of butterfly a hollow tube with inner radius a, tube length L. Assume nectar density ρ, viscosity μ. Assume mean flow velocity in the tube u. Assume the negative pressure produced by butterflies ΔP. According to the Newton’s second law, we could write down the following for the nectar in the tube:
Inertial force = Pressure difference X area + Gravitational force + Viscous force
or
$(\pi&space;a^2L\rho)&space;\textup{d}u/\textup{d}t=&space;\Delta&space;P(\pi&space;a^2)&space;-&space;(\pi&space;a^2L\rho)g&space;–&space;8\pi\mu&space;uL$

The viscous force might be a little bit unfamiliar. Please refer to the end of this episode. There are 4 terms in the above equation and it looks terrible, so we are going to remove some terms by physics reasoning.

How much negative pressure could be produced by butterflies and moths? Biophysicists believe the work producible by unit volume of muscles is a constant and therefore the force producible by muscle is proportional to its cross-section area. (This is deducible from scaling laws. Please refer to our previous articles: Scaling law of locomotion-Ihttps://goo.gl/2olyQH & Scaling law of locomotion-IIhttps://goo.gl/61l7vv) Consequently, if it is possible to define a characteristic length L for a creature, the force producible by it is F ~ L^2. The pressure it could achieve is therefore
$P&space;\sim&space;F/L^2&space;\sim&space;L^0$
That is to say all creatures could produce approximately the same negative suction pressure and that was validate experimentally. The maximum negative pressure produced by human and mosquitos are both around 10kPa.

What about the gravitational force? The length of their mouthpiece is about 1cm. Consider the ratio of gravitational force and pressure force:
It is small enough compared with the pressure force, so we discard it from our equation.

And finally let’s consider the ratio between the inertial force and viscous force (This is actually the well-known Reynolds number introduced by Stokes in 1851):
$\frac{(\pi&space;a^2L\rho)\textup{d}&space;u/&space;\textup{d}t}{8\pi\mu&space;uL}&space;=&space;\frac{(\rho&space;a^2)\textup{d}u/\textup{d}t&space;}{8\mu&space;u&space;}=&space;\textup{Re}$
To estimate the term du/dt, assume the suction frequency of butterflies f, which is experimentally observed to be 10 Hz. The acceleration of nectar in the mouthpiece could then be estimated as
$\textup{d}&space;u/&space;\textup{d}t&space;\sim&space;u/\Delta&space;t\sim&space;uf$
So the Reynold number could now be formulated as
$\textup{Re}&space;=&space;\frac{\rho&space;a^2uf}{8\mu&space;u&space;}\sim\frac{\rho&space;a^2&space;f}{\mu}$
The inner radius of the mouthpiece of butterflies is about 100μm. Substitute the real scale of these variables back and we will get Re ~ 0.1. The inertial force is rather small so we discard it, too.

Finally the equation becomes
$0&space;=&space;+\Delta&space;P(\pi&space;a^2)&space;-&space;8\pi\mu&space;uL$
That is just Poiseuille flow.

Ultimately we would like to know the relations between the flow rate and viscosity, for they are most easily observed experimentally. A naïve way to deduce the relations would be done by assuming the suction pressure ΔP a constant. Then
$Q&space;=&space;(\pi&space;a^2)u&space;=&space;\Delta&space;P(\pi&space;a^4)/8\mu&space;L&space;\propto&space;\mu^{-1}$
Nevertheless, the experimentally observed power law is
$Q&space;\propto&space;\mu^{-0.5}$
What shall we do? Another possible hypothesis proposed by Pivnick is that the power during active suction is a constant. That is to say
$\dot{W}=&space;Q\Delta&space;P&space;=\Delta&space;P^2(\pi&space;a^4)/8\mu&space;L&space;=&space;\textup{Constant}$
Because
$\Delta&space;P&space;=8\mu&space;LQ/(\pi&space;a^4)$
We now arrive at
$\dot{W}&space;=&space;Q^2(8\mu&space;L)/(\pi&space;a^4)&space;=&space;\textup{Constant}$
So
$Q&space;\propto&space;\mu^{-0.5}$
And that is the correct power law observed experimentally.

/*Supplementary: Poiseuille flow */
In 17th century, Newton quantitatively discuss the viscosity of fluid first. He believed that there would be viscous force whenever there is a velocity gradient between fluid layers. And the viscous stress τ (the viscous force per unit area) was “assumed” to be
$\tau&space;=&space;\mu\frac{\textup{d}u}{\textup{d}y}$
The fluid following this assumption is said to be Newtonian. Those violate this assumption is termed non-Newtonian instead.

That’s then consider a hollow tube streaming Newtonian fluid. Assume the flow is driven by the pressure difference ΔP, the tube length L, inner radius R, fluid density ρ, viscosity μ. Consider the “fluid column” with radial distance less than r. At steady state, the pressure force will balance out the viscous force. We could then write down that:
$-\Delta&space;P(\pi&space;r^2)&space;=&space;\tau&space;(2\pi&space;rL)&space;=\mu&space;\frac{\textup{d}u(r)}{\textup{d}r}(2\pi&space;rL)$
or
$\frac{\textup{d}u(r)}{\textup{d}r}=&space;\frac{-\Delta&space;P&space;r}{2&space;\mu&space;L}$
Integrate it with the no-slip boundary condition (flow velocity = 0 at r = R) and we will get
$u(r)&space;=&space;\Delta&space;P(R^2&space;-&space;r^2)/(4\mu&space;L)$
The velocity profile is quadratic and it has its maximum value at the central of the tube.

After knowing the velocity profile, it is easy to get the volumetric flow rate Q: $Q=\int_{r=0}^{R}u(r)(2\pi&space;r)\textup{d}r&space;=&space;\int_{r=0}^{R}\frac{\Delta&space;P(R^2&space;-&space;r^2)}{4\mu&space;L}(2\pi&space;r)\textup{d}r&space;=\frac{\pi\Delta&space;PR^4}{8\mu&space;L}$
And this is the well-known Poiseuille’s law for incompressible fluid.

Often we would love to express the flow rate in terms of mean velocity u by assuming
$Q=\pi&space;R^2&space;u$
Substitute it into Poiseuille law and we will get

$u&space;=\frac{\Delta&space;P&space;R^2}{8\mu&space;L}$
Because the above equation comes from the balance between viscous force and pressure force, the viscous force could be estimated as
$F_{\textup{viscous}}&space;=&space;-\Delta&space;P(\pi&space;R^2)&space;=&space;-(8u\mu&space;L/R^2)(\pi&space;R^2)&space;=&space;-8\pi&space;u\mu&space;L$

As what we stated previously.
/*-------------Divider-------------*/


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# Print the frequency of characters as a horizontal histogram

Well, I've used some of the feedback from my last posting to improve how I wrote this program. Mainly using the void parameter in main, initializing my int array to zero without using a for-loop, and better naming of variables. If there are some issues that are going on with my code, or there's some things I can do better, please let me know.

Any feedback is welcome!

#include <stdio.h>

/* print histogram of the frequencies of different characters in its input */
int main(void)
{
int histogram[95] = {0}; /* initialize 95 spaces for ASCII characters 32 - 127 */
int c, i, j;

while ((c = getchar()) != EOF) {
++histogram[c-' '];
}

for (i = 32; i < 127; ++i) {
printf("%c ", i);
++histogram[c-' '];
for (j = 0; j < histogram[i-' ']; ++j) {
putchar('x');
}
putchar('\n');
}
}

When dealing with offset arrrays (using position 0 to indicate character 32), it is important for readability to use symmetrical access patterns for the process. In your case, you load the values as c-' ', but what you use to read them, is accessed by i = 32; i < 127 .... and this asymmetrical access is inconvenient.

You also have no handling for out-of-bound characters.... can you trust your input sources to not have a carriage return, or a tab? Even esoteric null characters, or other binary values. I would have a catch-all 'bucket' for invalid values. That will prevent segfaults or other undefined actions.

You have a small bug in your code. In the print-loop, you increment the last character received.... what is this line for in the 'print' loops:

++histogram[c-' '];

Finally, for code reuse, it is often important to extract your functionality to a single reentrant method, other than the main method. Consider a signature like:

void histogram(const int offset, const int range)

where your histogram can be configured to take an offset position, and range, like:

histogram('a', 'z' - 'a' + 1);

which will count just the lower-case letters...

Something like:

#include <stdio.h>
#include <memory.h>

void histogram(const int offset, const int range) {
int histogram[range];
memset(histogram, 0, sizeof(histogram));

int special = 0;

int c;
while ((c = getchar()) != EOF) {
if (c < offset || c >= (offset + range)) {
special++;
} else {
++histogram[c - offset];
}
}

for (int i = 0; i < range; ++i) {
c = i + offset;
printf("%c ", c);
for (int j = 0; j < histogram[i]; ++j) {
putchar('x');
}
putchar('\n');
}

printf("- ");
for (int j = 0; j < special; j++) {
putchar('x');
}
putchar('\n');
}

int main(void) {
histogram(' ', 95);
}

• Ok, so you ensure that the input is always going to be plotted in the histogram by assigning the characters which aren't in the range of the offset to special, then plotting those after the printf("- "); at the end? I'm sure that this is an easier method than say, trying to get the input of every possible ASCII value... but is it possible then to catch all the ASCII values if I didn't want to use special to plot a histogram? Is there a function that can do that in stdio.h? Thanks! Sep 7, 2015 at 22:32
• @cody.codes: look at the various isfoo functions in <ctypes.h>, e.g. isalpha. (All 7-bit values are ASCII, including the control codes and non-printable chars.)
– Mat
Sep 8, 2015 at 4:46


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# Partial Fractions Integration Question

$$\int\frac{x^5+x-1}{x^3 +1} dx$$

Have tried everything ... polynomial long division, partial fractions, trig substitution etc...

Not for an assignment, so if a complete solution could be provided that'd be much appreaciated

Hint:

$$\frac{x^5+x-1}{x^3 +1} = \frac{(x^3+x^2-1)(x^2-x+1)}{(x+1)(x^2-x+1)}$$

• Would the first step not be to use polynomial long division? – Guest May 27 '14 at 0:10
• @Guest Do you want the first step to be polynomial division? – David H May 27 '14 at 0:13
• Well, generally partial fractions is utilised when degree of the numerator is less than the degree of the denominator – Guest May 27 '14 at 0:28
• @Guest Sure, just like how you were taught to use long division to divide two numbers and find the remainder when the given some fraction like $\frac{1210}{55}$. How would you go about simplifying that fraction? – David H May 27 '14 at 0:36
• Using the division algorithm .... – Guest May 27 '14 at 0:39

First divide the denominator into the numerator to get

$$\frac{x^5+x-1}{x^3+1}=x^2 -\frac{x^2-x+1}{x^3+1}$$

Now we factor the denominator, $$x^3+1=(x+1)(x^2-x+1)$$

Note that we have cancellation so we get $$\int x^2-\frac{1}{x+1} dx=\frac{x^3}{3}-\ln |x+1|+C$$

• Thanks a lot ... can't believe I missed such a simple detail on cancelling out like terms – Guest May 27 '14 at 0:12
• I didnt see it either, you just need to follow the method. – Rene Schipperus May 27 '14 at 0:30


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How do you simplify x/(2x^0)^2 and write it using only positive exponents?

Apr 27, 2017

$\frac{x}{4}$
${x}^{0}$ is simply $1$.
So the denominator simplifies to just ${2}^{2}$, or $4$.
There are no common factors between the numerator, $x$, and the denominator, $4$, so the answer is just $\frac{x}{4}$.


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# 3.5 Fourier transform of common signals

 Page 1 / 1
Fourier transform of common signals are derived.

## Fourier transform of common signals

Next, we'll derive the FT of some basic continuous-time signals. [link] summarizes these transform pairs.

## Rectangular pulse

Let's begin with the rectangular pulse

$\text{rect}\left(t,\tau \right)\equiv \left\{\begin{array}{cc}1,\hfill & t\le \tau /2\hfill \\ 0,\hfill & t>\tau /2\hfill \end{array}\right)$

The pulse function, $\text{rect}\left(t,\tau \right)$ is shown in [link] . Evaluating the Fourier transform integral with $x\left(t\right)=\text{rect}\left(t,\tau \right)$ gives

$\begin{array}{ccc}\hfill X\left(j\Omega \right)& =& {\int }_{-\tau /2}^{\tau /2}{e}^{-j\Omega t}dt\hfill \\ & =& {\left(\frac{-1}{j\Omega },{e}^{-j\Omega t}|}_{-\tau /2}^{\tau /2}\hfill \\ & =& \frac{1}{j\Omega }\left[{e}^{j\Omega \tau /2},-,{e}^{-j\Omega \tau /2}\right]\hfill \\ & =& \tau \frac{sin\left(\Omega \tau /2\right)}{\Omega \tau /2}\hfill \\ & =& \tau \text{sinc}\left(\Omega \tau /2\right)\hfill \end{array}$

A plot of $X\left(j\Omega \right)$ is shown in [link] .

Note that when $\Omega =0$ , $X\left(j\Omega \right)=\tau$ . We now have the following transform pair:

$\text{rect}\left(t,\tau \right)↔\tau \frac{sin\left(\Omega \tau /2\right)}{\Omega \tau /2}$

## Impulse

The unit impulse function was described in a previous section. From the sifting property of the impulse function we find that

$\begin{array}{ccc}\hfill X\left(j\Omega \right)& =& {\int }_{-\infty }^{\infty }\delta \left(t\right){e}^{-j\Omega t}dt\hfill \\ & =& 1\hfill \end{array}$

or

$\delta \left(t\right)↔1$

## Complex exponential

The complex exponential function, $x\left(t\right)={e}^{j{\Omega }_{0}t}$ , has a Fourier transform which is difficult to evaluate directly. It is easier to start with the Fourier transform itself and work backwards using the inverse Fourier transform. Suppose we want to find the time-domain signal which has Fourier transform $X\left(j\Omega \right)=\delta \left(\Omega -{\Omega }_{0}\right)$ . We can begin by using the inverse Fourier transform [link]

$\begin{array}{ccc}\hfill x\left(t\right)& =& \frac{1}{2\pi }{\int }_{-\infty }^{\infty }\delta \left(\Omega -{\Omega }_{0}\right){e}^{j\Omega t}d\Omega \hfill \\ & =& \frac{1}{2\pi }{e}^{\Omega t}\hfill \end{array}$

This result follows from the sifting property of the impulse function. By linearity, we can then write

${e}^{j\Omega t}↔2\pi \delta \left(\Omega -{\Omega }_{0}\right)$

## Cosine

The cosine signal can be expressed in terms of complex exponentials using Euler's Identity

$cos\left({\Omega }_{0}t\right)=\frac{1}{2}\left({e}^{j{\Omega }_{0}t},+,{e}^{-j{\Omega }_{0}t}\right)$

Applying linearity and the Fourier transform of complex exponentials to the right side of [link] , we quickly get:

$cos\left({\Omega }_{0}t\right)↔\pi \delta \left(\Omega -{\Omega }_{0}\right)+\pi \delta \left(\Omega +{\Omega }_{0}\right)$

## Real exponential

The real exponential function is given by $x\left(t\right)={e}^{-\alpha t}u\left(t\right)$ , where $\alpha >0$ . To find its FT, we start with the definition

$\begin{array}{ccc}\hfill X\left(j\Omega \right)& =& {\int }_{0}^{\infty }{e}^{-\alpha t}{e}^{-j\Omega t}dt\hfill \\ & =& {\int }_{0}^{\infty }{e}^{-\left(\alpha +j\Omega \right)t}dt\hfill \\ & =& \frac{-1}{\alpha +j\Omega }{\left({e}^{-\left(\alpha +j\Omega \right)t}|}_{0}^{\infty }\hfill \\ & =& \frac{-1}{\alpha +j\Omega }\left(0-1\right)\hfill \\ & =& \frac{1}{\alpha +j\Omega }\hfill \end{array}$

therefore,

${e}^{-\alpha t}u\left(t\right)↔\frac{1}{\alpha +j\Omega }$

## The unit step function

In a previous section, we looked at the unit step function,

$u\left(t\right)=\left\{\begin{array}{cc}1,\hfill & t\ge 0\hfill \\ 0,\hfill & t<0\hfill \end{array}\right)$

A closely related signal is the signum function, defined by

$sgn\left(t\right)=\left\{\begin{array}{cc}1,\hfill & t\ge 0\hfill \\ -1,\hfill & t<0\hfill \end{array}\right)$

from which it follows that

$u\left(t\right)=\left(sgn\left(t\right)+1\right)/2$

The signum function can be described as follows:

$sgn\left(t\right)=\underset{a\to 0}{lim}\left[{e}^{-at},u,\left(t\right),-,{e}^{at},u,\left(-t\right)\right]$

Since we already have the Fourier transform of the exponential signal,

$\begin{array}{cc}\hfill SGN\left(j\Omega \right)& =\underset{a\to 0}{lim}\left[\frac{1}{a+j\Omega },-,\frac{1}{a-j\Omega }\right]\hfill \\ & =\underset{a\to 0}{lim}\frac{-2j\Omega }{{a}^{2}+{\Omega }^{2}}\hfill \\ & =\frac{2}{j\Omega }\hfill \end{array}$

Using [link] and linearity then leads to

$u\left(t\right)↔\pi \delta \left(\Omega \right)+\frac{1}{j\Omega }$
 $x\left(t\right)$ $X\left(j\Omega \right)$ $\text{rect}\left(t,\tau \right)$ $\tau \frac{sin\left(\Omega \tau /2\right)}{\Omega \tau /2}$ $\delta \left(t\right)$ 1 ${e}^{j{\Omega }_{0}t}$ $2\pi \delta \left(\Omega -{\Omega }_{0}\right)$ $cos\left({\Omega }_{0}t\right)$ $\pi \delta \left(\Omega -{\Omega }_{0}\right)+\pi \delta \left(\Omega +{\Omega }_{0}\right)$ ${e}^{-\alpha t}u\left(t\right)$ $\frac{1}{\alpha +j\Omega }$ $u\left(t\right)$ $\pi \delta \left(\Omega \right)+\frac{1}{j\Omega }$

When working problems involving finding the Fourier transform, it is often preferable to use a table of transform pairs rather than to recalculate the Fourier transform from scratch. Often, transform pairs in can be combined with known Fourier transform properties to find new Fourier transforms.

Example 3.1 Find the Fourier transform of: $y\left(t\right)=2{e}^{5t}u\left(-t\right)$ . Clearly, we can write $y\left(t\right)=x\left(-t\right)$ where $x\left(t\right)=2{e}^{-5t}u\left(t\right)$ . Therefore, we can combine the known transform of $x\left(t\right)$ from [link] , namely,

$X\left(j\Omega \right)=\frac{2}{5+j\Omega }$

with the time reversal property:

$x\left(-t\right)↔X{\left(j\Omega \right)}^{*}$

to get the answer:

$Y\left(j\Omega \right)=\frac{2}{5-j\Omega }$

#### Questions & Answers

Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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Berger describes sociologists as concerned with
Got questions? Join the online conversation and get instant answers!


text:
# Question #710f4

$4 N {H}_{3} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 4 N O \left(g\right) + 6 {H}_{2} O \left(l\right)$
Nitrous oxide is commonly an oxidation product of ammonia. If this is the reaction you propose, then $3$ $\text{equiv}$ of water result from $2$ $\text{equiv}$ of ammonia.
You react $40 \cdot m o l$ of $N {H}_{3}$, then clearly you get $60 \cdot m o l$ of $O {H}_{2}$. I acknowledge that this might not have been the reaction you had in mind.


text:
The Kyoto Journal of Mathematics, formerly The Journal of Mathematics of Kyoto University, has a long and distinguished history of publishing high-quality and original mathematical research. A quarterly, it publishes original research papers at the forefront of pure mathematics, including surveys that contribute to advances in pure mathematics.

### Volume 32, Number 2

#### Publication Date: 1992

Vector valued Fourier hyperfunctions

Yoshifumi Ito; 259-285

A necessary condition for $H^\infty$-wellposed Cauchy problem of Schrödinger type equations with variable coefficients

Shigeomi Hara; 287-305

Mixed problems or Cauchy problems for semi-degenerate hyperbolic equations of 2-nd order with a parameter

Reiko Sakamoto; 307-331

Estimates for degenerate Schrödinger operators and hypoellipticity for infinitely degenerate elliptic operators

Yoshinori Morimoto; 333-372

Hypoellipticity for some infinitely degenerate elliptic operators of second order

Tatsushi Morioka; 373-386

Range characterization of Radon transforms on complex projective spaces

Tomoyuki Kakehi; 387-399

Stability properties of regular flows of heat-conducting compressible fluids


text:
# Slope of the tangent lines?

The slope of the tangent line to the parabola $y = x^2 + 6x + 7$ at the point $( 3 , 61 )$ is: The equation of this tangent line can be written in the form $y = mx+b$ where $m$ is: __ and where $b$ is: __?

and also:

Let $f(x) = \sqrt{10-x}$ The slope of the tangent line to the graph of $f(x)$ at the point $(6,2)$ is . The equation of the tangent line to the graph of $f(x)$ at $(6,2)$ is $y=mx+b$ for $m=$ __ and $b=$ __ Hint: the slope is given by the derivative at $x=6$, ie. $\lim_{x\to6} \frac{(f(6+h)-f(6))}{h}$

I'm absolutely stumped.... :( Help?!

-
Have you tried using the hint given? –  J. M. is back. Oct 8 '11 at 5:55
Do you know how to take the derivative of $4x^2 + 6x + 7$? –  Unreasonable Sin Oct 8 '11 at 6:06

Equation of tangent line at point $(a,f(a))$ is $y = f(a) + f'(a)(x - a)$, so we have to find $f'(x)$ and than plug in value $a$ into the result.

$f'(x)=(4x^2+6x+7)'=8x+6 \Rightarrow f'(3)=30$

Since $f(a)=f(3)=61$, we may write next tangent line equation:

$y=61+30(x-3) \Rightarrow y=30x-29$

For the second tangent line we have that $f'(x)=\frac{-1}{2\sqrt{10-x}}\Rightarrow f'(6)=\frac{-1}{4}$ ,and $f(6)=2$, so the second tangent line is:

$y=2-\frac{1}{4}(x-6) \Rightarrow y=\frac{-1}{4}x+\frac{7}{2}$

-
pedja, for things like these, it's usually not a good idea to give full solutions... –  J. M. is back. Oct 8 '11 at 6:30
@J.M,Do you think that full solution is counterproductive ? –  pedja Oct 8 '11 at 6:37
Ideally we want to leave students something they have to figure out on their own... –  J. M. is back. Oct 8 '11 at 6:39

Let find $m_{\text{tangent}}=f'(6)$ when $f'(x)=\sqrt{10-x}$ by using the definition not applying rules. We know that $$f'(6)=\lim_{x\to 6}\frac{f(x)-f(6)}{x-6}=\lim_{x\to 6}\frac{\sqrt{10-x}-2}{x-6}$$ which is $\frac{0}{0}$. So we do it as follows: $$\lim_{x\to 6}\frac{\sqrt{10-x}-2}{x-6}=\lim_{x\to 6}\frac{(\sqrt{10-x}-2)\color{blue}{(\sqrt{10-x}+2)}}{(x-6)\color{blue}{(\sqrt{10-x}+2)}}=\lim_{x\to 6}\frac{(10-x)-4}{(x-6)\color{blue}{(\sqrt{10-x}+2)}}\\ =\lim_{x\to 6}\frac{6-x}{(x-6)\color{blue}{(\sqrt{10-x}+2)}}=\frac{-1}{4}$$ Now the tangent line at $(6,2)$ is $y-2=\frac{-1}4(x-2)$.

-

If you do not know how to take the derivative, you could try computing for the slope using $m = \lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}$

All you have to do is to plug in the values: $$lim_{\Delta x \to 0}\frac{4(x + \Delta x)^2 + 6(x + \Delta x) + 7 - (4x^2 + 6x + 7)}{\Delta x}$$ and you should get $\lim_{\Delta x\to0} 8x + 4\Delta x + 6$

Also, review the equation of a line: $y-y_1 = m(x - x_1)$

Voila! You should now have a tangent line.

-


text:
# 23.9 Inductance  (Page 4/10)

 Page 4 / 10

## Section summary

• Inductance is the property of a device that tells how effectively it induces an emf in another device.
• Mutual inductance is the effect of two devices in inducing emfs in each other.
• A change in current $\Delta {I}_{1}/\Delta t$ in one induces an emf ${\text{emf}}_{2}$ in the second:
${\text{emf}}_{2}=-M\frac{\Delta {I}_{1}}{\Delta t}\text{,}$
where $M$ is defined to be the mutual inductance between the two devices, and the minus sign is due to Lenz’s law.
• Symmetrically, a change in current $\Delta {I}_{2}/\Delta t$ through the second device induces an emf ${\text{emf}}_{1}$ in the first:
${\text{emf}}_{1}=-M\frac{\Delta {I}_{2}}{\Delta t}\text{,}$
where $M$ is the same mutual inductance as in the reverse process.
• Current changes in a device induce an emf in the device itself.
• Self-inductance is the effect of the device inducing emf in itself.
• The device is called an inductor, and the emf induced in it by a change in current through it is
$\text{emf}=-L\frac{\Delta I}{\Delta t}\text{,}$
where $L$ is the self-inductance of the inductor, and $\Delta I/\Delta t$ is the rate of change of current through it. The minus sign indicates that emf opposes the change in current, as required by Lenz’s law.
• The unit of self- and mutual inductance is the henry (H), where $1 H=1 \Omega \cdot \text{s}$ .
• The self-inductance $L$ of an inductor is proportional to how much flux changes with current. For an $N$ -turn inductor,
$L=N\frac{\Delta \Phi }{\Delta I}\text{.}$
• The self-inductance of a solenoid is
$L=\frac{{\mu }_{0}{N}^{2}A}{\ell }\text{(solenoid),}$
where $N$ is its number of turns in the solenoid, $A$ is its cross-sectional area, $\ell$ is its length, and ${\text{μ}}_{0}=4\pi ×{\text{10}}^{\text{−7}}\phantom{\rule{0.25em}{0ex}}\text{T}\cdot \text{m/A}\phantom{\rule{0.10em}{0ex}}$ is the permeability of free space.
• The energy stored in an inductor ${E}_{\text{ind}}$ is
${E}_{\text{ind}}=\frac{1}{2}{\text{LI}}^{2}\text{.}$

## Conceptual questions

How would you place two identical flat coils in contact so that they had the greatest mutual inductance? The least?

How would you shape a given length of wire to give it the greatest self-inductance? The least?

Verify, as was concluded without proof in [link] , that units of $\text{T}\cdot {\text{m}}^{2}/A=\Omega \cdot \text{s}=\text{H}$ .

## Problems&Exercises

Two coils are placed close together in a physics lab to demonstrate Faraday’s law of induction. A current of 5.00 A in one is switched off in 1.00 ms, inducing a 9.00 V emf in the other. What is their mutual inductance?

1.80 mH

If two coils placed next to one another have a mutual inductance of 5.00 mH, what voltage is induced in one when the 2.00 A current in the other is switched off in 30.0 ms?

The 4.00 A current through a 7.50 mH inductor is switched off in 8.33 ms. What is the emf induced opposing this?

3.60 V

A device is turned on and 3.00 A flows through it 0.100 ms later. What is the self-inductance of the device if an induced 150 V emf opposes this?

Starting with ${\text{emf}}_{2}=-M\frac{\Delta {I}_{1}}{\Delta t}$ , show that the units of inductance are $\left(\text{V}\cdot \text{s}\right)\text{/A}=\Omega \cdot \text{s}$ .

Camera flashes charge a capacitor to high voltage by switching the current through an inductor on and off rapidly. In what time must the 0.100 A current through a 2.00 mH inductor be switched on or off to induce a 500 V emf?

A large research solenoid has a self-inductance of 25.0 H. (a) What induced emf opposes shutting it off when 100 A of current through it is switched off in 80.0 ms? (b) How much energy is stored in the inductor at full current? (c) At what rate in watts must energy be dissipated to switch the current off in 80.0 ms? (d) In view of the answer to the last part, is it surprising that shutting it down this quickly is difficult?

(a) 31.3 kV

(b) 125 kJ

(c) 1.56 MW

(d) No, it is not surprising since this power is very high.

(a) Calculate the self-inductance of a 50.0 cm long, 10.0 cm diameter solenoid having 1000 loops. (b) How much energy is stored in this inductor when 20.0 A of current flows through it? (c) How fast can it be turned off if the induced emf cannot exceed 3.00 V?

A precision laboratory resistor is made of a coil of wire 1.50 cm in diameter and 4.00 cm long, and it has 500 turns. (a) What is its self-inductance? (b) What average emf is induced if the 12.0 A current through it is turned on in 5.00 ms (one-fourth of a cycle for 50 Hz AC)? (c) What is its inductance if it is shortened to half its length and counter-wound (two layers of 250 turns in opposite directions)?

(a) 1.39 mH

(b) 3.33 V

(c) Zero

The heating coils in a hair dryer are 0.800 cm in diameter, have a combined length of 1.00 m, and a total of 400 turns. (a) What is their total self-inductance assuming they act like a single solenoid? (b) How much energy is stored in them when 6.00 A flows? (c) What average emf opposes shutting them off if this is done in 5.00 ms (one-fourth of a cycle for 50 Hz AC)?

When the 20.0 A current through an inductor is turned off in 1.50 ms, an 800 V emf is induced, opposing the change. What is the value of the self-inductance?

60.0 mH

How fast can the 150 A current through a 0.250 H inductor be shut off if the induced emf cannot exceed 75.0 V?

Integrated Concepts

A very large, superconducting solenoid such as one used in MRI scans, stores 1.00 MJ of energy in its magnetic field when 100 A flows. (a) Find its self-inductance. (b) If the coils “go normal,” they gain resistance and start to dissipate thermal energy. What temperature increase is produced if all the stored energy goes into heating the 1000 kg magnet, given its average specific heat is $\text{200 J/kg·ºC}$ ?

(a) 200 H

(b) $\text{5.00ºC}$

Unreasonable Results

A 25.0 H inductor has 100 A of current turned off in 1.00 ms. (a) What voltage is induced to oppose this? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible?

Physics is a physical science that deals with the study of matter in relation to energy
what is physics
a15kg powerexerted by the foresafter 3second
what is displacement
movement in a direction
Jason
hello
Hosea
Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
Hi
saeid
hi
Yimam
What is thê principle behind movement of thê taps control
while
Hosea
what is atomic mass
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
show me how to get the accuracies of the values of the resistors for the two circuits i.e for series and parallel sides
Explain why it is difficult to have an ideal machine in real life situations.
tell me
Promise
what's the s . i unit for couple?
Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
Oluwakayode
İt iş diffucult to have idêal machine because of FRİCTİON definitely reduce thê efficiency
Oluwakayode
if the classica theory of specific heat is valid,what would be the thermal energy of one kmol of copper at the debye temperature (for copper is 340k)
can i get all formulas of physics
yes
haider
what affects fluid
pressure
Oluwakayode
Dimension for force MLT-2
what is the dimensions of Force?
how do you calculate the 5% uncertainty of 4cm?
4cm/100×5= 0.2cm
haider


text:
Home > CCG > Chapter 7 > Lesson 7.1.1 > Problem7-7

7-7.

Use the relationships in the diagram at right to find the values of each variable. Name which geometric relationships you used.

Refer to the Math Notes box in Lesson 2.1.4 if you need help establishing any angle pair relationship.

Since the highlighted angles are vertical angles, and vertical angles are equal, $37°+a=63°$.

$∠c$ is an alternate interior angle to $∠a$. Recall that alternate interior angles are equal.

Use the Triangle Angle Sum Theorem to calculate $b$: $37°+c+52+b=180°$.

To find $m∠d$, notice that it is the exterior angle in the small triangle that contains $∠b$ and $52°$. What relationship do those angles have?


text:
# How to print \unit with "quantity-product" in siunitx?

siunitx allows us to print numbers, units and quantities, the later understood as a combination of numbers and units. For that we have, respectively, \num, \unit and \qty.

Using \qty{<number>}{<unit>} is not the same as just juxtaposing \num{<number>}\unit{<unit>}, since \qty takes care of proper spacing between number and unit, as per the option quantity-product. So far, so good.

However, I'd like to be able to split the number from the unit and still get the spacing from quantity-product. The use case is a table, where I'd like to use an S column, and pass the unit in the table's preamble to all numbers in the column.

Now, I know most everyone here will tell me I should not add the unit to each number but should rather just give it once in the table's header. I agree this is usually best practice. But I have a case where the header is already crammed, and the numbers and units are short, so I think my table would look better if I did it this way. So, regardless of table formatting best practices and taste, I'm looking for a technical solution for the situation.

The MWE below summarizes the situation:

\documentclass{article}

\usepackage{siunitx}
\usepackage{array}

\begin{document}

\begin{tabular}{
S[table-format = 4.0, group-digits = false]
S[table-format = 1.1{\unit{\percent}}]<{{\unit{\percent}}}
}
\hline
& \multicolumn{1}{c}{$\frac{\text{imports}}{\text{apparent consumption}}$} \\
\hline
1924 & 3,7 \\
1925 & 3,4 \\
1926 & 3,6 \\
1927 & 4,0 \\
1928 & 3,8 \\
1929 & 4,0 \\
1930 & 4,5 \\
1935 & 4,7 \\
1936 & 4,4 \\
1937 & 3,3 \\
\hline
\end{tabular}

\end{document}

The above example, of course, is missing the space between the numbers and the units in the second column. So I'm trying to get a, let's say, \qtyunit macro to use instead of \unit there.

I have tried a couple of alternatives, but I'm not really confident in or satisfied with either. The first one was:

\NewDocumentCommand\qtyunit{m}{\qty[print-unity-mantissa=false]{1}{#1}}

This looks nice, and gets the job done. But I'm not really sure if I should rely on passing 1 as number with print-unity-mantissa=false actually printing the quantity-product. This seems very much as a side-effect, which might change (one could argue it should not be printed), and the effect is not documented.

(siunitx does allow to pass an empty number argument to \qty, but the case is treated specially, and quantity-product is skipped).

So I went to siunitx.sty and cooked (based on the definition of \unit):

\ExplSyntaxOn
\NewDocumentCommand \qtyunit { O { } > { \TrimSpaces } m }
{
\mode_leave_vertical:
\group_begin:
\siunitx_unit_options_apply:n {#2}
\keys_set:nn { siunitx } {#1}
% print quantity-product spacing, from ‘\siunitx_quantity_print’
\tl_use:N \l__siunitx_quantity_product_tl
\bool_if:NTF \l__siunitx_quantity_break_bool
{ \penalty \binoppenalty }
{ \nobreak }
% now print the unit proper
\siunitx_unit_format:nN {#2} \l__siunitx_tmp_tl
\siunitx_print_unit:V \l__siunitx_tmp_tl
\group_end:
}
\ExplSyntaxOff

But this, while it works, leaves me the sensation of being much overkill for this purpose.

So, the question is, am I missing something in siunitx from which I could get this in a more straightforward and robust way? If not, does anyone see an alternative which would be better than either of the two above?

Edit: I came up with a third alternative:

\NewDocumentCommand\qtyunit{m}{\qty[parse-numbers=false]{\relax}{#1}}

Looks promising, but it still feels like cheating somewhat.

• If all numbers in a column share the same unit, wouldn't it be better to add this unit to the column header instead of repeating it in each column? Jun 16, 2021 at 4:56
• @leandriis Please see my comment about this in the OP. Paragraph starting with "Now, I know most everyone here will tell me I should not add the unit to each number but should rather just give it once in the table's header." Jun 16, 2021 at 10:40

In v3 there are a number of code-level interfaces that are used for communication between sub-modules, but which are not directly exposed as document commands. Here, you are really after \siunitx_quantity_print:nn, which prints a pre-formatted quantity and which will accept a missing numerical part. (This is needed for example in printing complex quantities.)

We first need to format the unit, then pass it to the printing routine.

\documentclass{article}

\usepackage{siunitx}
\usepackage{array}
\ExplSyntaxOn
\NewDocumentCommand \qtyunit { m }
{
\siunitx_unit_format:nN {#1} \l_tmpa_tl
\siunitx_quantity_print:nV { } \l_tmpa_tl
}
\ExplSyntaxOff

\begin{document}

\begin{tabular}{
S[table-format = 4]
S[table-format = 1.1{\qtyunit{\percent}}]<{{\qtyunit{\percent}}}
}
\hline
& \multicolumn{1}{c}{$\frac{\text{imports}}{\text{apparent consumption}}$} \\
\hline
1924 & 3,7 \\
1925 & 3,4 \\
1926 & 3,6 \\
1927 & 4,0 \\
1928 & 3,8 \\
1929 & 4,0 \\
1930 & 4,5 \\
1935 & 4,7 \\
1936 & 4,4 \\
1937 & 3,3 \\
\hline
\end{tabular}

\end{document}

• Hi Joseph, thank you! I'm starting to have some fun with v3, as you see, and it's looking damn good. I particularly like the new names. They do read better. So, thank you very much for v3 too! Jun 16, 2021 at 10:50

In the example below, the contents of each cell in the second column is collected, using collcell package, then passed to \mynum as its argument (like \mynum{<cell contents>}).

That is, the spec for the second column

S[table-format = 1.1{\unit{\percent}}]<{{\unit{\percent}}}
% is replace with
>{\collectcell\mynum}c<{\endcollectcell}

\documentclass{article}

\usepackage{siunitx}
\usepackage{array}
\usepackage{collcell}

\newcommand\mynum[1]{%
\qty{#1}{\percent}%
}

\begin{document}

\begin{tabular}{
S[table-format = 4.0, group-digits = false]
>{\collectcell\mynum}c<{\endcollectcell}
}
\hline
& \multicolumn{1}{c}{$\frac{\text{imports}}{\text{apparent consumption}}$} \\
\hline
1924 & 3,7 \\
1925 & 3,4 \\
1926 & 3,6 \\
1927 & 4,0 \\
1928 & 3,8 \\
1929 & 4,0 \\
1930 & 4,5 \\
1935 & 4,7 \\
1936 & 4,4 \\
1937 & 3,3 \\
\hline
\end{tabular}

\end{document}

See more collcell examples in package manuals of both collcell and siunitx.

• Hi @muzimuzhi Z, thank you very much for your stab at it! It is indeed an interesting and clever approach. However, I am looking for something more general. In particular, I'd like to use the S column type, for its number alignment capabilities (true the MWE was boring in that regard). Also, this approach would require to define one such macro for each unit I'd need to do this, which is not quite convenient. Besides, this definition would probably have to be kept distant from the table, and would make for the document to become a little opaque in this regard. Jun 15, 2021 at 23:17
• I see. So I guess you won't satisfied with S[table-format = 1.1{\siprod\unit{\percent}}]<{{\siprod\unit{\percent}}} where \def\siprod{\l__siunitx_quantity_product_tl}, will you? Or one can hack the l3keys to provide an expandable \keys_get_tl_set:nn{<module>}{<key>} to retrieve \l__siunitx_quantity_product_tl from \keys_get_tl_set:nn{siunitx}{quantity-product}. Jun 16, 2021 at 0:21
• That'd be much better already. But I sort of been there ;-). Don't get me wrong I appreciate your chiming in too. Thanks. I can point a shortcoming of \siprod in that it does not get the units specific options. So, if you'd use it, for example, in \siprod\unit{\degree}, you'd get the wrong spacing. The solution of \qty[parse-numbers=false]{\relax}{#1} with which I came up later and added in an edit, is good, but I guess I was really hoping for a more "institutionalized" way to do this. I'd expect this to be something people might want to do, I'm sort of surprised I didn't just "find" it. Jun 16, 2021 at 0:38
• Btw, feel free to edit with \siprod, if you also think it is better. I have already given my upvote. Jun 16, 2021 at 0:42