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## Ethernet

Read this section. MAC addresses are the datalink services found in an Ethernet environment.

##### 6.3.2 Ethernet

Ethernet was designed in the 1970s at the Palo Alto Research Center [Metcalfe1976]. The first prototype 4 used a coaxial cable as the shared medium and 3 Mbps of bandwidth. Ethernet was improved during the late 1970s and in the 1980s, Digital Equipment, Intel and Xerox published the first official Ethernet specification [DIX]. This specification defines several important parameters for Ethernet networks. The first decision was to standardise the commercial Ethernet at 10 Mbps. The second decision was the duration of the slot time. In Ethernet, a long slot time enables networks to span a long distance but forces the host to use a larger minimum frame size. The compromise was a slot time of 51.2 microseconds, which corresponds to a minimum frame size of 64 bytes.

The third decision was the frame format. The experimental 3 Mbps Ethernet network built at Xerox used short frames containing 8 bit source and destination addresses fields, a 16 bit type indication, up to 554 bytes of payload and a 16 bit CRC. Using 8 bit addresses was suitable for an experimental network, but it was clearly too small for commercial deployments. Although the initial Ethernet specification [DIX] only allowed up to 1024 hosts on an Ethernet network, it also recommended three important changes compared to the networking technologies that were available at that time. The first change was to require each host attached to an Ethernet network to have a globally unique datalink layer address. Until then, datalink layer addresses were manually configured on each host. [DP1981] went against that state of the art and noted “Suitable installation-specific administrative procedures are also needed for assigning numbers to hosts on a network. If a host is moved from one network to another it may be necessary to change its host number if its former number is in use on the new network. This is easier said than done, as each network must have an administrator who must record the continuously changing state of the system (often on a piece of paper tacked to the wall !). It is anticipated that in future office environments, hosts locations will change as often as telephones are changed in present-day offices.” The second change introduced by Ethernet was to encode each address as a 48 bits field [DP1981]. 48 bit addresses were huge compared to the networking technologies available in the 1980s, but the huge address space had several advantages [DP1981] including the ability to allocate large blocks of addresses to manufacturers. Eventually, other LAN technologies opted for 48 bits addresses as well [802]_. The third change introduced by Ethernet was the definition of broadcast and multicast addresses. The need for multicast Ethernet was foreseen in [DP1981] and thanks to the size of the addressing space it was possible to reserve a large block of multicast addresses for each manufacturer.

The datalink layer addresses used in Ethernet networks are often called MAC addresses. They are structured as shown in the figure below. The first bit of the address indicates whether the address identifies a network adapter or a multicast group. The upper 24 bits are used to encode an Organisation Unique Identifier (OUI). This OUI identifies a block of addresses that has been allocated by the secretariat 5 who is responsible for the uniqueness of Ethernet addresses to a manufacturer. Once a manufacturer has received an OUI, it can build and sell products with one of the 16 million addresses in this block.

Figure 6.18: 48 bits Ethernet address format

The original 10 Mbps Ethernet specification [DIX] defined a simple frame format where each frame is composed of five fields. The Ethernet frame starts with a preamble (not shown in the figure below) that is used by the physical layer of the receiver to synchronise its clock with the sender’s clock. The first field of the frame is the destination address. As this address is placed at the beginning of the frame, an Ethernet interface can quickly verify whether it is the frame recipient and if not, cancel the processing of the arriving frame. The second field is the source address. While the destination address can be either a unicast or a multicast/broadcast address, the source address must always be a unicast address. The third field is a 16 bits integer that indicates which type of network layer packet is carried inside the frame. This field is often called the EtherType. Frequently used EtherType values 6 include 0x0800 for IPv4, 0x86DD for IPv6 and 0x806 for the Address Resolution Protocol (ARP).

The fourth part of the Ethernet frame is the payload. The minimum length of the payload is 46 bytes to ensure a minimum frame size, including the header of 512 bits. The Ethernet payload cannot be longer than 1500 bytes. This size was found reasonable when the first Ethernet specification was written. At that time, Xerox had been using its experimental 3 Mbps Ethernet that offered 554 bytes of payload and RFC 1122 required a minimum MTU of 572 bytes for IPv4. 1500 bytes was large enough to support these needs without forcing the network adapters to contain overly large memories. Furthermore, simulations and measurement studies performed in Ethernet networks revealed that CSMA/CD was able to achieve a very high utilization. This is illustrated in the figure below based on [SH1980], which shows the channel utilization achieved in Ethernet networks containing different numbers of hosts that are sending frames of different sizes.

Figure 6.19: Impact of the frame length on the maximum channel utilisation [SH1980]

The last field of the Ethernet frame is a 32 bit Cyclical Redundancy Check (CRC). This CRC is able to catch a much larger number of transmission errors than the Internet checksum used by IP, UDP and TCP [SGP98]. The format of the Ethernet frame is shown below.

Note: Where should the CRC be located in a frame ?

The transport and datalink layers usually chose different strategies to place their CRCs or checksums. Transport layer protocols usually place their CRCs or checksums in the segment header. Datalink layer protocols sometimes place their CRC in the frame header, but often in a trailer at the end of the frame. This choice reflects implementa- tion assumptions, but also influences performance RFC 893. When the CRC is placed in the trailer, as in Ethernet, the datalink layer can compute it while transmitting the frame and insert it at the end of the transmission. All Eth- ernet interfaces use this optimisation today. When the checksum is placed in the header, as in a TCP segment, it is impossible for the network interface to compute it while transmitting the segment. Some network interfaces provide hardware assistance to compute the TCP checksum, but this is more complex than if the TCP checksum were placed in the trailer 8.

Figure 6.20: Ethernet DIX frame format

The Ethernet frame format shown above is specified in [DIX]. This is the format used to send both IPv4 RFC 894 and IPv6 packets RFC 2464. After the publication of [DIX], the Institute of Electrical and Electronic Engineers (IEEE) began to standardise several Local Area Network technologies. IEEE worked on several LAN technolo- gies, starting with Ethernet, Token Ring and Token Bus. These three technologies were completely different, but they all agreed to use the 48 bits MAC addresses specified initially for Ethernet [802]_. While developing its Ethernet standard [802.3], the IEEE 802.3 working group was confronted with a problem. Ethernet mandated a minimum payload size of 46 bytes, while some companies were looking for a LAN technology that could trans- parently transport short frames containing only a few bytes of payload. Such a frame can be sent by an Ethernet host by padding it to ensure that the payload is at least 46 bytes long. However since the Ethernet header [DIX] does not contain a length field, it is impossible for the receiver to determine how many useful bytes were placed inside the payload field. To solve this problem, the IEEE decided to replace the Type field of the Ethernet [DIX] header with a length field 9. This Length field contains the number of useful bytes in the frame payload. The pay- load must still contain at least 46 bytes, but padding bytes are added by the sender and removed by the receiver. In order to add the Length field without significantly changing the frame format, IEEE had to remove the Type field. Without this field, it is impossible for a receiving host to identify the type of network layer packet inside a received frame. To solve this new problem, IEEE developed a completely new sublayer called the Logical Link Control [802.2]. Several protocols were defined in this sublayer. One of them provided a slightly different version of the Type field of the original Ethernet frame format. Another contained acknowledgements and retransmissions to provide a reliable service... In practice, [802.2] is never used to support IP in Ethernet networks. The figure below shows the official [802.3] frame format.

Figure 6.21: Ethernet 802.3 frame format

Note: What is the Ethernet service ?

An Ethernet network provides an unreliable connectionless service. It supports three different transmission modes [unicast, multicast and broadcast. While the Ethernet service is unreliable in theory, a good Ethernet network should, in practice, provide a service that:]

• delivers frames to their destination with a very high probability of successful delivery
• does not reorder the transmitted frames

The first property is a consequence of the utilisation of CSMA/CD. The second property is a consequence of the physical organisation of the Ethernet network as a shared bus. These two properties are important and all evolutions of the Ethernet technology have preserved them.

Several physical layers have been defined for Ethernet networks. The first physical layer, usually called 10Base5, provided 10 Mbps over a thick coaxial cable. The characteristics of the cable and the transceivers that were used then enabled the utilisation of 500 meter long segments. A 10Base5 network can also include repeaters between segments.

The second physical layer was 10Base2. This physical layer used a thin coaxial cable that was easier to install than the 10Base5 cable, but could not be longer than 185 meters. A 10BaseF physical layer was also defined to transport Ethernet over point-to-point optical links. The major change to the physical layer was the support of twisted pairs in the 10BaseT specification. Twisted pair cables are traditionally used to support the telephone service in office buildings. Most office buildings today are equipped with structured cabling. Several twisted pair cables are installed between any room and a central telecom closet per building or per floor in large buildings. These telecom closets act as concentration points for the telephone service but also for LANs.

The introduction of the twisted pairs led to two major changes to Ethernet. The first change concerns the physical topology of the network. 10Base2 and 10Base5 networks are shared buses, the coaxial cable typically passes through each room that contains a connected computer. A 10BaseT network is a star-shaped network. All the devices connected to the network are attached to a twisted pair cable that ends in the telecom closet. From a maintenance perspective, this is a major improvement. The cable is a weak point in 10Base2 and 10Base5 networks. Any physical damage on the cable broke the entire network and when such a failure occurred, the network administrator had to manually check the entire cable to detect where it was damaged. With 10BaseT, when one twisted pair is damaged, only the device connected to this twisted pair is affected and this does not affect the other devices. The second major change introduced by 10BaseT was that is was impossible to build a 10BaseT network by simply connecting all the twisted pairs together. All the twisted pairs must be connected to a relay that operates in the physical layer. This relay is called an Ethernet hub. A hub is thus a physical layer relay that receives an electrical signal on one of its interfaces, regenerates the signal and transmits it over all its other interfaces. Some hubs are also able to convert the electrical signal from one physical layer to another (e.g. 10BaseT to 10Base2 conversion).

Figure 6.22: Ethernet hubs in the reference model

Computers can directly be attached to Ethernet hubs. Ethernet hubs themselves can be attached to other Ethernet hubs to build a larger network. However, some important guidelines must be followed when building a complex network with hubs. First, the network topology must be a tree. As hubs are relays in the physical layer, adding a link between Hub2 and Hub3 in the network below would create an electrical shortcut that would completely disrupt the network. This implies that there cannot be any redundancy in a hub-based network. A failure of a hub or of a link between two hubs would partition the network into two isolated networks. Second, as hubs are relays in the physical layer, collisions can happen and must be handled by CSMA/CD as in a 10Base5 network. This implies that the maximum delay between any pair of devices in the network cannot be longer than the 51.2 microseconds slot time. If the delay is longer, collisions between short frames may not be correctly detected. This constraint limits the geographical spread of 10BaseT networks containing hubs.

Figure 6.23: A hierarchical Ethernet network composed of hubs

In the late 1980s, 10 Mbps became too slow for some applications and network manufacturers developed several LAN technologies that offered higher bandwidth, such as the 100 Mbps FDDI LAN that used optical fibers. As the development of 10Base5, 10Base2 and 10BaseT had shown that Ethernet could be adapted to different physical layers, several manufacturers started to work on 100 Mbps Ethernet and convinced IEEE to standardise this new technology that was initially called Fast Ethernet. Fast Ethernet was designed under two constraints. First, Fast Ethernet had to support twisted pairs. Although it was easier from a physical layer perspective to support higher bandwidth on coaxial cables than on twisted pairs, coaxial cables were a nightmare from deployment and maintenance perspectives. Second, Fast Ethernet had to be perfectly compatible with the existing 10 Mbps Ethernets to allow Fast Ethernet technology to be used initially as a backbone technology to interconnect 10 Mbps Ethernet networks. This forced Fast Ethernet to use exactly the same frame format as 10 Mbps Ethernet. This implied that the minimum Fast Ethernet frame size remained at 512 bits. To preserve CSMA/CD with this minimum frame size and 100 Mbps instead of 10 Mbps, the duration of the slot time was decreased to 5.12 microseconds.

The evolution of Ethernet did not stop. In 1998, the IEEE published a first standard to provide Gigabit Ethernet over optical fibers. Several other types of physical layers were added afterwards. The 10 Gigabit Ethernet standard appeared in 2002. Work is ongoing to develop standards for 40 Gigabit and 100 Gigabit Ethernet and some are thinking about Terabit Ethernet. The table below lists the main Ethernet standards. A more detailed list may be found at http://en.wikipedia.org/wiki/Ethernet_physical_layer

 Standard Comments 10Base510Base210BaseT10Base-F100Base-Tx100Base-FX1000Base-CX1000Base-SX10 Gbps40-100 Gbps Thick coaxial cable, 500mTwo pairs of category 3+ UTP10 Mb/s over optical fiberCategory 5 UTP or STP, 100 m maximumTwo multimode optical fiber, 2 km maximumTwo pairs shielded twisted pair, 25m maximumTwo multimode or single mode optical fibers with lasersOptical fiber but also Category 6 UTPBeing developed, standard expected in 2010Thin coaxial cable, 185m

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# structure of atom class 11 mcq with answers

(iii) Proton = ($$\frac{1}{1}$$) = 1 Therefore, the kinetic energy of same orbit of hydrogen atom is 3.4 eV. (d) Iron. States Of Matter . For l = 2, total number of electrons = 5 [3d5], Question 13. (d) 9.84 × 105 J mol-1, Answer: (d) 9.84 × 105 J mol-1 ⇒ 3s23p63d5 Atoms further consist of smaller charged particles that are either positively or negatively charged. (a) 518 nm So I am trying to upload chapter wise and Topic wise question practice of all chapter. Therefore max 10 electrons can be accommodated as each orbital can have maximum of 2 electrons. Question Bank for NEET Chemistry Structure of Atom Topic Test - Structure of Atom. ($$\frac{1}{355}$$) = ($$\frac{1}{680}$$) + ($$\frac{1}{λ_2}$$) (d) The relative number and kind of atoms are constant in a given compound. (b) Na+, Ca2+ , Sc3+, F– n = 6, ℓ = 2 means 6d → will have 5 orbitals. Order of energy is: (b) Azimuthal quantum number (c) -13.6 eV (a) 3.4 eV The solved questions answers in this Structure Of Atom - Practice Test, Class 9 Science quiz give you a … MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. Explanation: JEE Main 2020 & NEET 2020: Get important MCQs (with answers) from NCERT book of CBSE Class 11 Chemistry (Chapter 2 - Structure of Atom). n = 3, l = 0 represents 3s orbital n = 3, l = 1 represents 3p orbital n = 3, l = 2 represents 3d orbital n = 4, l = 0 represents 4s orbital The order of increasing energy of the orbitals is 3s < 3p < 4s < 3d. Answer. Answer: Yes, you can find important questions for CBSE Class 11 Chemistry Chapter 2- Structure of Atom online on Vedantu. (b) The electron in the orbit nearest the nucleus has the lowest energy structure of atom class 11 question bank mcq. (c) Spin quantum number Here you will find Chapter Wise NCERT MCQ Questions for Class 11 Chemistry with Answers PDF Free Download based on the important concepts and topics given in the textbook. Multiple Choice Questions (MCQ) for Structure of Atom - CBSE Class 11-science Chemistry on Topperlearning. This contains 20 Multiple Choice Questions for Class 9 Structure Of Atom - Practice Test, Class 9 Science (mcq) to study with solutions a complete question bank. (a) 1 You will get all level (easy, medium and hard) of question from Structure of Atom Check the below NCERT MCQ Questions for Class 9 Science Chapter 4 Structure of the Atom with Answers Pdf free download. λ = $$\frac {(h)}{(mv)}$$ (c) Electrons revolve in different orbits around the nucleus Free Online MCQ Questions for Class 9 Science with Answers was Prepared Based on Latest Exam Pattern. (b) 6.8 eV Newton Law Of Motion || Class 11 physics || Physics Chapter 5. The four atomic species can be represented as follows. =−3.28 × 105−(−13.2 × 106) (b) Mn ET = E1 + E2 ….. (1) The energy required to excite the electron in the atom from n = 1 to n = 2 is (a) s < p < d < f (a) nucleons Structure of Atom Class 11 MCQs Questions with Answers. We have Provided Atoms and Molecules Class 9 Science MCQs Questions with Answers to help students understand the concept very well. s < p < d < f, Question 4. Or $$\frac{(1)}{(λ_T)}$$ = ($$\frac{1}{355}$$) − ($$\frac{1}{680}$$) Answer: (d) 3s, 3p and 3d orbitals all have the same energy Consider the ground state of Cr atom (Z = 24). Explanation: Which one of the following sets of ions represents a collection of isoelectronic species? (b) n = 3, l = 1, m = 1, s = + $$\frac {1}{2}$$ (a) 6.63 × 10-37 m Class 11 Atomic Structure MCQs JEE/NEET Multiple Choice Questions with One Correct Answer. Question 7. Physics MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. (c) n, p, α, e (1) 0.1518 Å (2) 6.1518 Å (3) 2.3518 Å (4) 4.1218 Å Atoms and Molecules Class 9 MCQs Questions with Answers. (Atomic nos. = 10 × 10-6 kg (d) n = 4, l = 0, m = 0, s = + $$\frac {1}{2}$$. (b) 3p orbital is lower in energy than 3d orbital (ii) α− particle = ($$\frac{2}{4}$$) = 0.5 eVidyarthi. (d) Sc, Answer: (a) K ROTATE YOUR DEVICE FOR BEST VIEWING EXPERIENCE. The kinetic energy is equal to the negative of the total energy. Thus all these ions have 18 electrons in them. For hydrogen atom, MCQs Test on Structure of Atom, Atomic Structure with answers, Test: 1, Total Questions: 15 Chemical Bonding And Molecular Structure . (a) Energy of the electrons in the orbit is quantised And the potential energy is equal to the twice of the total energy. (b) 6.56 × 105 J mol-1 Solve free mcq on structure of atom class 11 chemistry chapter 2 useful for CBSE, NEET/ AIPMT, IIT JEE and CET exam preparation. Then find out KE of same orbit of Hydrogen atom Total energy of electron in an orbit = -($$\frac {e^2}{2r_n}$$) E2=−(1.312 × 106 × (1)²)/(2²) (a) 3s orbital is lower in energy than 3p orbital (b) Shape of orbitals (b) Chromium Answer: (c) Spin quantum number The numbers of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively: Number of electrons in M shell = 13 (a) e, p, n, α 24Cr → 1s2 2s22p6 3s2 3p6 3d5 4s1 Thermodynamics . (c) 2n2. Answer: (a) K+, Ca2+, Sc3+, Cl– Explanation: These important questions will play significant role in clearing concepts of Chemistry. (b) s > p > d > f (d) -2, Answer: (c) -1 Explanation: Explanation: RD Sharma Solutions , RS Aggarwal Solutions and NCERT Solutions. Which of the following sets of quantum numbers represents the highest energy of an atom? Multiple Choice Questions (Type-I) Explanation: (b) 6.63 × 10-31 m Which of the following statements in relation to the hydrogen atom is correct? ΔE = 9.84×105 J mol-1, Question 3. More.. (b) 6.56 × 105 J mol-1 = −3.28 × 105 J mol-1 Practice Now. (d) 6.63 × 10-35 m, Answer: (b) 6.63 × 10-31 m eVidyarthi; School. So the electron which are available in the same orbitals, must have opposite spins. Here we have covered Important Questions on Structure of Atom for Class 11 Chemistry subject.. Chemistry Important Questions Class 11 are given below.. (a) n = 3, l = 0, m = 0, s = + $$\frac {1}{2}$$ (a) K+, Ca2+, Sc3+, Cl– Find Tutor. (a) K Explanation: Explanation: ⇒ ($$\frac{1}{λ_2}$$) = (680 − 355)/ (355 × 680) (c) K+, Cl–, Mg2+, Sc3+ (b) 12 and 5 Question 12. so obviously there are only 2 possible directions. Question 11. (a) 12 electrons Electrons occupying the same orbital are distinguished by Spin quantum number. The first excited state energy of orbital = -3.4 eV Equilibrium . These questions will test students knowledge of this chapter and help in analysing their exam preparation. (i) (e/m) for (i) neutron = ($$\frac{0}{1}$$) = 0 Answer: (c) 10 electrons M shell means it is third shell ⇒ n = 3 (iv) electron = ($$\frac{1}{1837}$$) = 1837. Explanation: X and Y are: (a) isobars (b) isotones (c) isomers (d) isotopes Q7: In an atom, the mass number of an atom is equal to the number of _____. v = 100 ms-1 (a) 8.51 × 105 J mol-1 2. (a) Copper (d) The position and velocity of the electrons in the orbit cannot be determined simultaneously, Answer: (d) The position and velocity of the electrons in the orbit cannot be determined simultaneously The increasing order (lowest first) for the values of e/m (charge/mass) for (a) e, p, n, α (b) n, p, e, α (c) n, p, α, e (d) n, α, p, e. Answer. Which of the following element has least number of electrons in its M-shell? Hence the value of n, l , m are same for the two electrons occupying in the same orbitals, but only the is different, which is Answer: (c) Orientation of orbitals Question 2. In the Bohrs model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in a quantum state n is: Question 20. Therefore, $$\frac {(KE)}{(TE)}$$ = -1, Question 18. (d) +13.6 eV. and The kinetic energy of same orbital = -(-3.4 eV) = 3.4 eV Isoelectronic species are those which have same number of electrons. These important questions are curated and prepared by our subject matter experts, with close reference to the NCERT Chemistry textbook for Class 11 and various other reference study materials. (b) n, p, e, α (a) 3s orbital is lower in energy than 3p orbital (b) 2 If you want to practice full question from science to score maximum mark in all Exam then this website will help you to score maximum mark in all exam like NEET, AIIMS, as well as for Class 11. The energy required to excite the electron in the atom from n = 1 to n = 2 is = −3.28 × 105−(−13.2 × 106) (d) n, α, p, e. Answer: (d) n, α, p, e Hydrogen atom is … Explanation: (c) 16 and 4 The wavelength of de-Broglie wave associated with it would be (Note: h = 6.63 × 10-34 Js) We have Provided Structure of the Atom Class 9 Science MCQs Questions with Answers to help students understand the concept very well. The ionization enthalpy of hydrogen atom is 1.312 × 106 J mol-1. MCQ quiz on Atomic Structure multiple choice questions and answers on atomic structure MCQ questions quiz on atomic structure objectives questions with answer test pdf. Copyright (c) 2020 Sora Templates All Right Reseved, Question Bank for NEET Chemistry Structure of Atom. Answer: (c) n = 3, l = 2, m = 1, s = + $$\frac {1}{2}$$ Answer of this question will be given in the end of question. (d) 3s, 3p and 3d orbitals all have the same energy. The element is_____. (c) there is a lot of empty space in an atom (d) the nucleus carries the most of the mass Q6: Two elements X and Y have the same atomic mass but their atomic numbers are 20 and 21 respectively. These MCQ's are extremely critical for all CBSE students to score better marks. (d) f < p < d < s, Answer: (a) s < p < d < f (hc/λT) = (hc)/ (λ1) + (hc)/ (λ2) Which of the following statements in relation to the hydrogen atom is correct? MCQs from CBSE Class 9 Science Chapter 4: Structure of The Atom 1. Who discovered the electron? (d) Magnetic quantum number. Online Test for Class 11 Chemistry Structure of Atom. (c) Nickel Explanation: (d) Nuclear Stability. (d) 2l + 1. (c) 6.63 × 10-34 m Maximum number of electrons in a subshell can be. (b) 3p orbital is lower in energy than 3d orbital Main Menu. The electronic configuration is: (1s2) (2s2 2p6) (3s2 3p6 3d5) (4s1) MCQ Questions for Class 9 Science with Answers were prepared based on the latest exam pattern. View the Important Question bank for Class 11 & 12 Chemistry developed by expert faculties from Kota. Explanation: (C) 2n2. So I am trying to upload chapter wise and Topic wise question practice of all chapter. Free PDF Download of CBSE Physics Multiple Choice Questions for Class 12 with Answers Chapter 13 Nuclei. Students can solve NCERT Class 12 Physics Nuclei MCQs Pdf with Answers to know their preparation level. Download CBSE Class 11 Chemistry Structure of Atom MCQs Set A in pdf, Chemistry chapter wise Multiple Choice Questions free. 3s23p6 = m-shell (c) Orientation of orbitals We hope the given NCERT MCQ Questions for Class 11 Chemistry Chapter 2 Structure of Atom with Answers Pdf free download will help you. Multiple Choice Questions (MCQs) on CBSE Class 9 Science Chapter- Structure of the Atom are provided here with answers. A hydrogen atom has 1st configuration and these its, 3p and 3d orbitals will have same energy wrt 1s orbital. (A) 4l + 2. (b) 4l – 2. 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text:
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 typo and remark

The key to do this is to realize the difference between computation and verification. Although computing value A(a,b) of the Ackermann function cannot be done primitive recursively, verifying whether a proposed number c is the correct value of A(a,b) can be done primitive recursively. (Note that computation and verification is also what distinguishes P and NP.)

In this case, the fact that this can be done hinges on the strong monotonicity properties of Ackermann's function. Indeed, if A(a,b) = c then all the previous values of A needed to compute A(a,b) are bounded by c. Therefore, the search for a valid computation verifying that A(a,b) = c can be bounded by a primitive recursive function B(a,b,c). Knowing this B(a,b,c) function B we can take a proposed value c for A(a,b), compute the bound B(a,b,c) and search up to this bound for a valid computation verifying that c is indeed equal to A(a,b). If no such computation is found we return 1, otherwise we return 0.

To make this a little more specific, let's say that computation for A(a,b) = c is a (coded) finite sequence of triples (ai,bi,ci) which ends with the triple (a,b,c). Each such triple codes the fact that A(ai,bi) = ci. The computation is valid when each such triple follows from previous triples and the rules for computing the Ackermann function. (Checking this is obviously primitive recursive.) For example a valid computation for A(1,1) = 3 is the sequence (0,1,2), (0,2,3), (1,0,2), (1,1,3).

Using the rules for computing Ackermann function and our specific method for coding finite sequences, we can compute a precise an explicit bound B(a,b,c) on a valid computation verifying A(a,b) = c. The verification that B is primitive recursive should be straightforward if our coding of finite sequences is reasonable. Therefore, verifying A(a,b) = c can be done by a primitive recursive computation.

(The details of the last paragraph are best carried out in the privacy of one's office.)

2 typo and detail

The key to do this is to realize the difference between computation and verification. Although computing value A(a,b) of the Ackermann function cannot be done primitive recursively, verifying whether a proposed number c is the correct value of A(a,b) can be done primitive recursively.

In this case, the fact that this can be done hinges on the strong monotonicity properties of Ackermann's function. Indeed, if A(a,b) = c then all the previous values of A needed to compute A(a,b) are bounded by c. Therefore, the search for a valid computation verifying that A(a,b) = c can be bounded by a primitive recursive function B(a,b,c). Knowing this B(a,b,c) we can take a proposed value c for A(a,b), compute the bound B(a,b,c) and search up to this bound for a valid computation verifying that c is indeed equal to A(a,b). If no such computation is found we return 1, otherwise we return 0.

To make this a little more specific, let's say that computation for A(a,b) = c is a (coded) finite sequence of triples (ai,bi,ci) which ends with the triple . (a,b,c). Each such triple codes the fact that A(ai,bi) = ci. The computation is valid when each such triple follows from previous triples and the rules for computing the Ackermann function. (Checking this is obviously primitive recursive.) For example a valid computation for A(1,1) = 3 is the sequence (0,1,2), (0,2,3), (1,0,2), (1,1,3).

Using the rules for computing Ackermann function and our specific method for coding finite sequences, we can compute a precise bound B(a,b,c) on a valid computation verifying A(a,b) = c. The verification that B is primitive recursive should be straightforward if our coding of finite sequences is reasonable. Therefore, verifying A(a,b) = c can be done by a primitive recursive computation.

(The details of the last paragraph are best carried out in the privacy of one's office.)

1

The key to do this is to realize the difference between computation and verification. Although computing value A(a,b) of the Ackermann function cannot be done primitive recursively, verifying whether a proposed number c is the correct value of A(a,b) can be done primitive recursively.

In this case, the fact that this can be done hinges on the strong monotonicity properties of Ackermann's function. Indeed, if A(a,b) = c then all the previous values of A needed to compute A(a,b) are bounded by c. Therefore, the search for a valid computation verifying that A(a,b) = c can be bounded by a primitive recursive function B(a,b,c). Knowing this B(a,b,c) we can take a proposed value c for A(a,b), compute the bound B(a,b,c) and search up to this bound for a valid computation verifying that c is indeed equal to A(a,b). If no such computation is found we return 1, otherwise we return 0.

To make this a little more specific, let's say that computation for A(a,b) = c is a (coded) finite sequence of triples (ai,bi,ci) which ends with the triple . Each such triple codes the fact that A(ai,bi) = ci. The computation is valid when each such triple follows from previous triples and the rules for computing the Ackermann function. For example a valid computation for A(1,1) = 3 is the sequence (0,1,2), (0,2,3), (1,0,2), (1,1,3).

Using the rules for computing Ackermann function and our specific method for coding finite sequences, we can compute a precise bound B(a,b,c) on a valid computation verifying A(a,b) = c. The verification that B is primitive recursive should be straightforward if our coding of finite sequences is reasonable. Therefore, verifying A(a,b) = c can be done by a primitive recursive computation.

(The details of the last paragraph are best carried out in the privacy of one's office.)


text:
# How to find the sum of the sides of a polygon whose one vertex goes from the north of a circle and the other comes from the east in its perimeter?

The problem is as follows:

In figure 1. there is a circle as shown. The radius is equal to 10 inches and its center is labeled with the letter O. If $\measuredangle PC=30^{\circ}$. $\textrm{Find AB+BC}$.

The existing alternatives in my book are:

• $3\left( \sqrt{2}+\sqrt{6}\right)$
• $4\left( \sqrt{6}-\sqrt{2}\right)$
• $5\left( \sqrt{3}-\sqrt{2}\right)$
• $5\left( \sqrt{3}+\sqrt{2}\right)$
• $5\left( \sqrt{2}+\sqrt{6}\right)$

After analyzing the drawing the figure from below shows all all the relationships which I could found and it is summarized as follows:

The triangle $\textrm{COP}$ is isosceles since it shares the same side from the radius of the circle and since $\measuredangle PC=30^{\circ}$, then all is left to do is to apply the identity which it says that the sum of inner angles in a triangle must equate to $180^{\circ}$.

$$2x+30^{\circ}=180^{\circ}$$ $$x=\frac{150^{\circ}}{2}=75^{\circ}$$

Since $\measuredangle OCP = \measuredangle OPC$, its supplementary angle would become:

$$180^{\circ}-75^{\circ}=105^{\circ}$$

Since it is given from the problem:

$$\measuredangle COA = 90^{\circ}$$

therefore its complementary angle with $\measuredangle COP = 30^{\circ}$ would become into:

$$\measuredangle POA = 60^{\circ}$$

Since $PO = OA$ this would also make another isosceles triangle and by recurring to the previous identity:

$$2x+60^{\circ}=180^{\circ}$$ $$x=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ}$$

Therefore the triangle POA is an equilateral one so,

$$\textrm{PA=10 inches}$$

As $\measuredangle OPB = 105 ^{\circ}$ and $\measuredangle OPA = 60^{\circ}$ then its difference is: $\measuredangle APB = 45^{\circ}$.

From this its easy to note that $\measuredangle PAB = 45^{\circ}$.

Since the vertex $\textrm{B}$ of the triangle $\textrm{ABP}$ is $\measuredangle = 90 ^{\circ}$. I did identified a special right triangle with the form $45^{\circ}-45^{\circ}-90^{\circ}$ or $\textrm{k, k,}\,k\sqrt{2}$.

By equating the newly found side $\textrm{PA = 10 inches}$ to $k\sqrt{2}$ this is transformed into:

$$k\sqrt{2} = 10$$

$$k = \frac{10}{\sqrt{2}}$$

From this is established that:

$$AB = \frac{10}{\sqrt{2}}$$

Since we have $\textrm{AB}$ we also know $\textrm{PB}$ as $AB = PB = \frac{10}{\sqrt{2}}$

Therefore all that is left to do is to find $\textrm{CP}$ as $CP+PB = BC$

To find $CP$ I used cosines law as follows:

$$a^{2}=b^{2}+c^{2}-2bc\,\cos A$$

Being a, b and c the sides of a triangle ABC and A the opposing angle from the side taken as a reference in the left side of the equation.

In this case

$$(CP)^{2}= 10^{2}+10^{2}-2(10)(10)\cos30^{\circ}$$ $$(CP)^{2}= 10^{2} \left(1+1-2\left(\frac{\sqrt{3}}{2}\right)\right)$$ $$CP = 10 \sqrt{ 2-\sqrt{3}}$$

Therefore $CP = 10 \sqrt{ 2-\sqrt{3}}$ and we have all the parts so the rest is just adding them up.

$$CP+PB= BC = 10 \sqrt{ 2-\sqrt{3}} + \frac{10}{\sqrt{2}}$$

$$AB= \frac{10}{\sqrt{2}}$$

$$AB + BC = \frac{10}{\sqrt{2}} + 10 \sqrt{ 2-\sqrt{3}} + \frac{10}{\sqrt{2}}$$

And that's how far I went, but from then on I don't know if what I did was correct or did I missed something? as my answer doesn't appear within the alternatives.

The best I could come up with by simplifying was:

$$\frac{10\sqrt{2}}{2}+10\sqrt{2-\sqrt{3}}+\frac{10\sqrt{2}}{2}$$

$$10\sqrt{2}+10\sqrt{2-\sqrt{3}}$$

$$10\left(\sqrt{2}+\sqrt{2-\sqrt{3}}\right)$$

and, that's it. But it doesn't seem to be in the choices given. Can somebody help me to find if did I do something wrong?. If a drawing is necessary please include one as I'm not savvy enough to notice these things easily.

• If $∡PC=30º$ then I think $\angle AOP=60º$ since the angle to the center is twice as the angle to the side. Sep 13, 2018 at 4:41

Notice that $$\angle ACB=\angle OCB-\angle OCA=75^o-45^o=30^o,$$and $$AC=10\sqrt{2}.$$Thus $$AB+BC=AC\cdot(\sin \angle ACB+\cos \angle ACB)=5(\sqrt{2}+\sqrt{6}).$$

Assuming your work is correct so far, I think we may be able to simplify further.

$$\sqrt{2-\sqrt3}=\sqrt{\frac{4-2\sqrt3}{2}}=\frac{\sqrt{\sqrt3^2-2(1)\sqrt3+1^2}}{\sqrt2}=\frac{\sqrt{(\sqrt3-1)^2}}{\sqrt2}=\frac{\sqrt3-1}{\sqrt2}$$.

Now let's see if that helps.

$$10\left(\sqrt2+\sqrt{2-\sqrt3}\right)=5\sqrt2\left(2+\sqrt3-1\right)=5(\sqrt2+\sqrt6)$$

Again, assuming everything you've done is correct, the last answer is the solution.

• I'm still stuck at why $\sqrt{\frac{{4-2\sqrt{3}}}{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}$ ? Perhaps can you expand with steps how this reverse rationalization works?. Because I tried all the methods I know and I am still unable to reach what you have found. Does it exist an identity am I unaware of?. Sep 14, 2018 at 3:48
• @ChrisSteinbeckBell The numerator is simply $\sqrt3^2-2(1)\sqrt3+1^2=(\sqrt3-1)^2$.
– Mike
Sep 14, 2018 at 4:37
• Thanks! but I voice my opinion that what have you just added in the comments should be added as part of the answer since, well maybe is not too obvious. Other than that is great to know that I was on the right track. Sep 14, 2018 at 4:51
• @ChrisSteinbeckBell All right, I've added the requested intermediate steps.
– Mike
Sep 14, 2018 at 5:20

I have checked your work and it is correct.

Your answer is correct and $$10\left(\sqrt2+\sqrt{2-\sqrt3} \right)=5\left( \sqrt{2}+\sqrt{6}\right)=19.31851653$$


text:
# Field Trial of Duplex, 10 Gbps $\,\times\,$8-User DPSK-OCDMA System Using a Single 16$\,\times\,$ 16 Multi-Port Encoder/Decoder and 16-Level Phase-Shifted SSFBG Encoder/Decoders

#### Abstract

An optical code-division multiple access (OCDMA) over wavelength-division multiplexing (WDM) system could be one promising solution to the symmetric gigabit access network with high spectral efficiency, cost effective, good flexibility and enhanced security. A cost-effective OCDMA/WDM system using a single multi-port en/decoder at an optical line terminal… (More)


text:
# isomorphism of Chern character in kk-theory

Suppose we work with Fréchet algebras. Cuntz defined kk-theory for those algebras and hence we have the notions of K-theory and K-homology for those algebras. Now suppose Chern character is isomorphism in K-theory for those algebras (after tensoring with complex numbers). Does that mean that Chern character is also isomorphism in K-homology (after tensoring with complex numbers)?

• Usually, what you probably refer to, Chern character is an isomorphism (after tensoring with complex numbers) with one side being K-homology (- not K-theory). To answer your question: unlikely. K-homology = Hom(A,C). K-theory = Hom(C,A). Why should they coincide? Chern character is also defined as a functor from the category $kk_0$ to $HP^0$. So your question seems unprecise. – hänsel Mar 12 '16 at 17:13


text:
# Differential equations and Vector spaces

I was reading Cohn's book on Lie Groups.In introduction part he has given the motivation behind Lie Groups.It is like this

If solution of the differential equation $\frac{dx_{i}}{dt}=u(t)$ is $x_{i}=f_{i}(x,t)$ then we can think of this solution as a linear transformation $x'=S_{t}(x)$ such that each point $x$ can be associated with a new point $x'$ reached after time $t$ .Then family of transformations $(S_{t})$ on the whole space forms a group,with the operation $S_{t}.S_{t'}(x)=S_{t+t'}$.

I could not understand the whole concept of constructing transformations from differential equations and the defined operation

• The group property that you mention fails in general, unless $u=u(x_i)$. Please clarify this point first. – John B Jan 25 '16 at 2:12
• Can you explain this with a simple example of $\frac{dx}{dt}=2t$ in a single variable case ? – Madhu Jan 25 '16 at 11:04


text:
# 10. Auxiliary tools

LAMMPS is designed to be a computational kernel for performing molecular dynamics computations. Additional pre- and post-processing steps are often necessary to setup and analyze a simulation. A list of such tools can be found on the LAMMPS webpage at these links:

The last link for Pizza.py is a Python-based tool developed at Sandia which provides tools for doing setup, analysis, plotting, and visualization for LAMMPS simulations.

Note that many users write their own setup or analysis tools or use other existing codes and convert their output to a LAMMPS input format or vice versa. The tools listed here are included in the LAMMPS distribution as examples of auxiliary tools. Some of them are not actively supported by the LAMMPS developers, as they were contributed by LAMMPS users. If you have problems using them, we can direct you to the authors.

The source code for each of these codes is in the tools sub-directory of the LAMMPS distribution. There is a Makefile (which you may need to edit for your platform) which will build several of the tools which reside in that directory. Most of them are larger packages in their own sub-directories with their own Makefiles and/or README files.

## 10.4. Tool descriptions

### 10.4.1. amber2lmp tool

The amber2lmp sub-directory contains two Python scripts for converting files back-and-forth between the AMBER MD code and LAMMPS. See the README file in amber2lmp for more information.

These tools were written by Keir Novik while he was at Queen Mary University of London. Keir is no longer there and cannot support these tools which are out-of-date with respect to the current LAMMPS version (and maybe with respect to AMBER as well). Since we don’t use these tools at Sandia, you’ll need to experiment with them and make necessary modifications yourself.

### 10.4.2. binary2txt tool

The file binary2txt.cpp converts one or more binary LAMMPS dump file into ASCII text files. The syntax for running the tool is

binary2txt file1 file2 ...

which creates file1.txt, file2.txt, etc. This tool must be compiled on a platform that can read the binary file created by a LAMMPS run, since binary files are not compatible across all platforms.

### 10.4.3. ch2lmp tool

The ch2lmp sub-directory contains tools for converting files back-and-forth between the CHARMM MD code and LAMMPS.

They are intended to make it easy to use CHARMM as a builder and as a post-processor for LAMMPS. Using charmm2lammps.pl, you can convert a PDB file with associated CHARMM info, including CHARMM force field data, into its LAMMPS equivalent. Support for the CMAP correction of CHARMM22 and later is available as an option. This tool can also add solvent water molecules and Na+ or Cl- ions to the system. Using lammps2pdb.pl you can convert LAMMPS atom dumps into PDB files.

These tools were created by Pieter in’t Veld (pjintve at sandia.gov) and Paul Crozier (pscrozi at sandia.gov) at Sandia.

CMAP support added and tested by Xiaohu Hu (hux2 at ornl.gov) and Robert A. Latour (latourr at clemson.edu), David Hyde-Volpe, and Tigran Abramyan, (Clemson University) and Chris Lorenz (chris.lorenz at kcl.ac.uk), King’s College London.

### 10.4.4. chain tool

The file chain.f creates a LAMMPS data file containing bead-spring polymer chains and/or monomer solvent atoms. It uses a text file containing chain definition parameters as an input. The created chains and solvent atoms can strongly overlap, so LAMMPS needs to run the system initially with a “soft” pair potential to un-overlap it. The syntax for running the tool is

chain < def.chain > data.file

See the def.chain or def.chain.ab files in the tools directory for examples of definition files. This tool was used to create the system for the chain benchmark.

### 10.4.5. colvars tools

The colvars directory contains a collection of tools for post-processing data produced by the colvars collective variable library. To compile the tools, edit the makefile for your system and run “make”.

Please report problems and issues the colvars library and its tools at: https://github.com/colvars/colvars/issues

abf_integrate:

MC-based integration of multidimensional free energy gradient Version 20110511

Syntax: ./abf_integrate < filename > [-n < nsteps >] [-t < temp >] [-m [0|1] (metadynamics)] [-h < hill_height >] [-f < variable_hill_factor >]

The LAMMPS interface to the colvars collective variable library, as well as these tools, were created by Axel Kohlmeyer (akohlmey at gmail.com) at ICTP, Italy.

### 10.4.6. createatoms tool

The tools/createatoms directory contains a Fortran program called createAtoms.f which can generate a variety of interesting crystal structures and geometries and output the resulting list of atom coordinates in LAMMPS or other formats.

See the included Manual.pdf for details.

The tool is authored by Xiaowang Zhou (Sandia), xzhou at sandia.gov.

### 10.4.7. doxygen tool

The tools/doxygen directory contains a shell script called doxygen.sh which can generate a call graph and API lists using the Doxygen software.

See the included README file for details.

### 10.4.8. drude tool

The tools/drude directory contains a Python script called polarizer.py which can add Drude oscillators to a LAMMPS data file in the required format.

See the header of the polarizer.py file for details.

The tool is authored by Agilio Padua and Alain Dequidt: agilio.padua at univ-bpclermont.fr, alain.dequidt at univ-bpclermont.fr

### 10.4.9. eam database tool

The tools/eam_database directory contains a Fortran program that will generate EAM alloy setfl potential files for any combination of 16 elements: Cu, Ag, Au, Ni, Pd, Pt, Al, Pb, Fe, Mo, Ta, W, Mg, Co, Ti, Zr. The files can then be used with the pair_style eam/alloy command.

The tool is authored by Xiaowang Zhou (Sandia), xzhou at sandia.gov, and is based on his paper:

X. W. Zhou, R. A. Johnson, and H. N. G. Wadley, Phys. Rev. B, 69, 144113 (2004).

### 10.4.10. eam generate tool

The tools/eam_generate directory contains several one-file C programs that convert an analytic formula into a tabulated embedded atom method (EAM) setfl potential file. The potentials they produce are in the potentials directory, and can be used with the pair_style eam/alloy command.

The source files and potentials were provided by Gerolf Ziegenhain (gerolf at ziegenhain.com).

### 10.4.11. eff tool

The tools/eff directory contains various scripts for generating structures and post-processing output for simulations using the electron force field (eFF).

These tools were provided by Andres Jaramillo-Botero at CalTech (ajaramil at wag.caltech.edu).

### 10.4.12. emacs tool

The tools/emacs directory contains an Emacs Lisp add-on file for GNU Emacs that enables a lammps-mode for editing input scripts when using GNU Emacs, with various highlighting options set up.

These tools were provided by Aidan Thompson at Sandia (athomps at sandia.gov).

### 10.4.13. fep tool

The tools/fep directory contains Python scripts useful for post-processing results from performing free-energy perturbation simulations using the USER-FEP package.

The scripts were contributed by Agilio Padua (Universite Blaise Pascal Clermont-Ferrand), agilio.padua at univ-bpclermont.fr.

See README file in the tools/fep directory.

### 10.4.14. i-pi tool

The tools/i-pi directory contains a version of the i-PI package, with all the LAMMPS-unrelated files removed. It is provided so that it can be used with the fix ipi command to perform path-integral molecular dynamics (PIMD).

The i-PI package was created and is maintained by Michele Ceriotti, michele.ceriotti at gmail.com, to interface to a variety of molecular dynamics codes.

See the tools/i-pi/manual.pdf file for an overview of i-PI, and the fix ipi doc page for further details on running PIMD calculations with LAMMPS.

### 10.4.15. ipp tool

The tools/ipp directory contains a Perl script ipp which can be used to facilitate the creation of a complicated file (say, a lammps input script or tools/createatoms input file) using a template file.

ipp was created and is maintained by Reese Jones (Sandia), rjones at sandia.gov.

See two examples in the tools/ipp directory. One of them is for the tools/createatoms tool’s input file.

### 10.4.16. kate tool

The file in the tools/kate directory is an add-on to the Kate editor in the KDE suite that allow syntax highlighting of LAMMPS input scripts. See the README.txt file for details.

The file was provided by Alessandro Luigi Sellerio (alessandro.sellerio at ieni.cnr.it).

### 10.4.17. lmp2arc tool

The lmp2arc sub-directory contains a tool for converting LAMMPS output files to the format for Accelrys’ Insight MD code (formerly MSI/Biosym and its Discover MD code). See the README file for more information.

This tool was written by John Carpenter (Cray), Michael Peachey (Cray), and Steve Lustig (Dupont). John is now at the Mayo Clinic (jec at mayo.edu), but still fields questions about the tool.

This tool was updated for the current LAMMPS C++ version by Jeff Greathouse at Sandia (jagreat at sandia.gov).

### 10.4.18. lmp2cfg tool

The lmp2cfg sub-directory contains a tool for converting LAMMPS output files into a series of *.cfg files which can be read into the AtomEye visualizer. See the README file for more information.

This tool was written by Ara Kooser at Sandia (askoose at sandia.gov).

### 10.4.19. matlab tool

The matlab sub-directory contains several MATLAB scripts for post-processing LAMMPS output. The scripts include readers for log and dump files, a reader for EAM potential files, and a converter that reads LAMMPS dump files and produces CFG files that can be visualized with the AtomEye visualizer.

These scripts were written by Arun Subramaniyan at Purdue Univ (asubrama at purdue.edu).

### 10.4.20. micelle2d tool

The file micelle2d.f creates a LAMMPS data file containing short lipid chains in a monomer solution. It uses a text file containing lipid definition parameters as an input. The created molecules and solvent atoms can strongly overlap, so LAMMPS needs to run the system initially with a “soft” pair potential to un-overlap it. The syntax for running the tool is

micelle2d < def.micelle2d > data.file

See the def.micelle2d file in the tools directory for an example of a definition file. This tool was used to create the system for the micelle example.

### 10.4.21. moltemplate tool

The moltemplate sub-directory contains instructions for installing moltemplate, a Python-based tool for building molecular systems based on a text-file description, and creating LAMMPS data files that encode their molecular topology as lists of bonds, angles, dihedrals, etc. See the README.txt file for more information.

This tool was written by Andrew Jewett (jewett.aij at gmail.com), who supports it. It has its own WWW page at http://moltemplate.org. The latest sources can be found on its GitHub page

### 10.4.22. msi2lmp tool

The msi2lmp sub-directory contains a tool for creating LAMMPS template input and data files from BIOVIA’s Materias Studio files (formerly Accelrys’ Insight MD code, formerly MSI/Biosym and its Discover MD code).

This tool was written by John Carpenter (Cray), Michael Peachey (Cray), and Steve Lustig (Dupont). Several people contributed changes to remove bugs and adapt its output to changes in LAMMPS.

This tool has several known limitations and is no longer under active development, so there are no changes except for the occasional bug fix.

### 10.4.23. phonon tool

The phonon sub-directory contains a post-processing tool useful for analyzing the output of the fix phonon command in the USER-PHONON package.

See the README file for instruction on building the tool and what library it needs. And see the examples/USER/phonon directory for example problems that can be post-processed with this tool.

This tool was written by Ling-Ti Kong at Shanghai Jiao Tong University.

### 10.4.24. polybond tool

The polybond sub-directory contains a Python-based tool useful for performing “programmable polymer bonding”. The Python file lmpsdata.py provides a “Lmpsdata” class with various methods which can be invoked by a user-written Python script to create data files with complex bonding topologies.

See the Manual.pdf for details and example scripts.

This tool was written by Zachary Kraus at Georgia Tech.

### 10.4.25. pymol_asphere tool

The pymol_asphere sub-directory contains a tool for converting a LAMMPS dump file that contains orientation info for ellipsoidal particles into an input file for the PyMol visualization package or its open source variant.

Specifically, the tool triangulates the ellipsoids so they can be viewed as true ellipsoidal particles within PyMol. See the README and examples directory within pymol_asphere for more information.

This tool was written by Mike Brown at Sandia.

### 10.4.26. python tool

The python sub-directory contains several Python scripts that perform common LAMMPS post-processing tasks, such as:

• extract thermodynamic info from a log file as columns of numbers

• plot two columns of thermodynamic info from a log file using GnuPlot

• sort the snapshots in a dump file by atom ID

• convert multiple NEB dump files into one dump file for viz

• convert dump files into XYZ, CFG, or PDB format for viz by other packages

These are simple scripts built on Pizza.py modules. See the README for more info on Pizza.py and how to use these scripts.

### 10.4.27. reax tool

The reax sub-directory contains stand-alone codes that can post-process the output of the fix reax/c/bonds command from a LAMMPS simulation using ReaxFF. See the README.txt file for more info.

These tools were written by Aidan Thompson at Sandia.

### 10.4.28. smd tool

The smd sub-directory contains a C++ file dump2vtk_tris.cpp and Makefile which can be compiled and used to convert triangle output files created by the Smooth-Mach Dynamics (USER-SMD) package into a VTK-compatible unstructured grid file. It could then be read in and visualized by VTK.

See the header of dump2vtk.cpp for more details.

This tool was written by the USER-SMD package author, Georg Ganzenmuller at the Fraunhofer-Institute for High-Speed Dynamics, Ernst Mach Institute in Germany (georg.ganzenmueller at emi.fhg.de).

### 10.4.29. spin tool

The spin sub-directory contains a C file interpolate.c which can be compiled and used to perform a cubic polynomial interpolation of the MEP following a GNEB calculation.

See the README file in tools/spin/interpolate_gneb for more details.

This tool was written by the SPIN package author, Julien Tranchida at Sandia National Labs (jtranch at sandia.gov, and by Aleksei Ivanov, at University of Iceland (ali5 at hi.is).

### 10.4.30. vim tool

The files in the tools/vim directory are add-ons to the VIM editor that allow easier editing of LAMMPS input scripts. See the README.txt file for details.

These files were provided by Gerolf Ziegenhain (gerolf at ziegenhain.com)

### 10.4.31. xmgrace tool

The files in the tools/xmgrace directory can be used to plot the thermodynamic data in LAMMPS log files via the xmgrace plotting package. There are several tools in the directory that can be used in post-processing mode. The lammpsplot.cpp file can be compiled and used to create plots from the current state of a running LAMMPS simulation.

See the README file for details.

These files were provided by Vikas Varshney (vv0210 at gmail.com)


text: /*PINs configuration, interrupts, timers initialization etc*/, /*program's execution will never reach here*/, Run-length encoding (find/print frequency of letters in a string), Sort an array of 0's, 1's and 2's in linear time complexity, Checking Anagrams (check whether two string is anagrams or not), Find the level in a binary tree with given sum K, Check whether a Binary Tree is BST (Binary Search Tree) or not, Capitalize first and last letter of each word in a line, Greedy Strategy to solve major algorithm problems. & ans. This is certainly standard practice in both languages and almost unavoidable in C++. There is only one difference between 'Super Loop' and 'Infinite Loop': There may only one 'Super Loop' but the 'Infinite Loop' may be infinite (i.e. » PHP » C w This page will talk about a common program architecture called the Super-Loop Architecture, that is very useful in meeting these requirements. A Special Function Register (or Special Purpose Register, or simply Special Register) is a register within a microprocessor that controls or monitors the various functions of a microprocessor. There the loop is called (tight) (main) game loop. They are also designed to be efficient and cheap when performing their task. » Certificates e Embedded C language is used to develop microcontroller-based applications. Barr Group's Embedded C Coding Standard was developed to minimize bugs in firmware by focusing on practical rules that keep bugs out, while also improving the maintainability and portability of embedded software. » Feedback Ad: A » Data Structure » Networks » C C For loop. Embedded C, even if it’s similar to C, and embedded languages in general requires a different kind of thought process to use. Previous Page. :   99.9 ( Step 1: First initialization happens and the counter variable gets initialized. Home Embedded. » Internship » Puzzles : A In a previous article on structures in embedded C, we observed that rearranging the order of the members in a structure can change the amount of memory required to store a structure. % ) r That mean at this section before ‘Super Loop’ you can place initialization related codes (like initialisations of interrupts, times, pins configuration, memory and other attached devices). In an Embedded C Application, there are set of statements which need to be executed forever. % Because there are no operating system to return to or an embedded device is running until the power supply is removed. Choice of a good data structure or algorithm for a given problem comes after a deep understanding of the underlying theory. A » Cloud Computing The count is initialized to 1 and the test expression is evaluated. o This approach is well suited for small systems but has limitations for more complex applications. These are some of the benefits of using super loop in an Embedded Application. Solved programs: 0.1 A super loop is a program structure comprised of an infinite loop, with all the tasks of the system contained in that loop. This is especially important because few embedded applications will require 100% of processor resources. RTOS Advantages. Join our Blogging forum. Notice how we can cut down our power consumption by adding in a substantial delay? When a C program enters an endless loop, it either spews output over and over without end or it sits there tight and does nothing. 100 » SQL 20 » C++ STL Advertisements. Syntax of for loop: for (initialization; condition test; increment or decrement) { //Statements to be executed repeatedly } Flow Diagram of For loop. % Create new STM32 project in Keil uVision 5 tutorial. Embedded systems, like cameras or TV boxes, are simple computers that are designed to perform a single specific task. × » Java » Java Embedded Systems - SFR Registers. v » Java Usually, the loop continually checks for events, and takes appropriate action when an event is signalled. One functions is used to flash the LED by changing the state of glow and the other function is used to … . Support new debugger Support for … and how it is different from infinite loop?. Well, it’s doing what you ordered it to do, which is to sit and spin forever. This page was last edited on 22 May 2019, at 19:10. » Embedded Systems In this tutorial we see how to create project in KEIL MDK uVision 5 for STM32 ARM Cortex-M based MCUs. The ability to write code that gets close to the hardware is essential and C provides this facility. » SEO Game of Snake in C: Snake game was popular in old mobile phones which can be very easily devolped using c program. In an Embedded C Application, there are set of statements which need to be executed forever. To build this project you require basic understanding of c syntax. In this section we will go over some of the key aspects and problems of a circular buffer implementation. Are you a blogger? You aren't alone. For embedded - … and how it is different from infinite loop? » Java » Linux = » Privacy policy, STUDENT'S SECTION We also saw that a compiler has certain constraints when allocating memory for the members of a structure. And a developer using an RTOS still has to understand how the startup code works and get to main() just like a super-loop developer. [Proper use of volatile is part of the bug-killing Embedded C Coding Standard.] So, embedded systems software is not the only type of software which uses this kind of architecture. r In the previous tutorial we learned for loop. Most cross-compilers for embedded systems include an assembly language file called startup.asm, crt0.s (short for C runtime), or something similar. It is very simple to use, edit, debug and understand. » C++ g 1) The Super Loop. In this example, we used while (1) as 'Super Loop', here while is a looping statement and 1 is a non zero value that will also true and programming will run forever. » News/Updates, ABOUT SECTION Because there are no operating system to return to or an embedded device is running until the power supply is removed. Creative Commons Attribution-ShareAlike License. If the average loop time of the program is 1ms, and it requires only few instructions to be checked every second the program will save the state and build a delay that will be caused to read the input on every loop and it saves lot of energy or the power that needs to be used. In this situation, the program will loop 1000 times before it needs to read the input, and the other 999 loops of the program will just be a countdown to the next read. ) Note: Since, program's execution will not reach to end of the program, hence return 0 will never be executed, we can also use void main() instead of int main() and then there is no need to use return 0. Output. See IAR Embedded Workbench IDE Project Management and Building Guide for more information. One of the key benefits of the C language, which is the reason it is so popular for embedded applications, is that it is a high-level, structured programming language, but has low-level capabilities. » Facebook It is just an endless loop, and the actions of the system all happen continually within the loop. C language is not an extension to any programming language, but a general-purpose programming language: Embedded C is an extension to the C programming language including different features such as addressing I/O, fixed-point arithmetic, multiple-memory addressing, etc.   » DBMS » Machine learning » Articles A loop is used for executing a block of statements repeatedly until a given condition returns false. Do it until you run out of input data "cards". Embedded C Programming Language, which is widely used in the development of Embedded Systems, is an extension of C Program Language. For desktop applications, where memory is freely available, these difficulties can be ignored. » C Posted By Umang Gajera Posted date: July 25, 2018 in: Embedded No Comments. » Content Writers of the Month, SUBSCRIBE m & ans. Super loop is an infinite loop which is suitable only in embedded c programming because there you have to run your code for very very long time and wants explicitly terminate when the behavior is change for your robot or whatever. » DS A super loop is a program structure comprised of an infinite loop, with all the tasks of the system contained in that loop. CS Subjects: » Ajax Aptitude que. More: The embedded system is defined as the combination of embedded C programming software and hardware part majorly consist of microcontrollers and it is intended to perform the specific task. + » Python ホットウィール マテル ミニカー ホットウイール BGJ55 【送料無料】Hot Wheels Super Loop Chase Race Trackset (Discontinued by manufacturer)ホットウィール マテル ミニカー ホットウイール BGJ55 7hyOlM4t 32901円-0%-32901円 Newer version of C-STAT The static analysis tool C-STAT has been updated with 20 additional rules, a number of bugfixes and better reporting function. » C++ That is just one of the responsibilities assigned to startup code for C programs. » C#.Net If we build this delay to delay for 999ms, we don't need to loop 1000 times, we can read the input on every loop. The loop is in fact a variant of the classic "batch processing" control flow… These types of embedded systems are being used in our daily life such as washing machines and video recorders, refrigerators and so on. This is one of the most frequently used loop in C programming. We will now implement an expanded superloop to build in a delay: Notice how we added a delay at the end of the super loop? However, the handling of such dynamic memory can be problematic and inefficient. The value entered by the user is stored in the variable num.Suppose, the user entered 10. Syntax of while loop: » DBMS A In C and C++, it can be very convenient to allocate and de-allocate blocks of memory as and when needed. In this case, it is sorely inefficient to have the processor chugging away at 100% capacity all the time. » Contact us e 5 Interview que. 0000010286 00000 n Please confirm the information below before signing in. » C Unfortunately, most books about the C programming language dismiss volatile in a sentence or two. It seems a waste to continue looping the program, especially when we don't need to do anything most of the time. » Web programming/HTML Let's also say that we are using the example superloop above, which is in "Low-Power Mode" 99.9% of the time (1ms of calculations every second), and is only in normal mode 0.1% of the time: P From Wikibooks, open books for an open world, https://en.wikibooks.org/w/index.php?title=Embedded_Systems/Super_Loop_Architecture&oldid=3547395. For example, computer games often use a similar loop. I dislike the term "super loop" as it implies that there is something special about the loop - but there isn't. Beware the endless loop! » Kotlin Most embedded systems are able to just sit and wait in a low-power state until needed. However, once you get to main you start the RTOS instead of starting the super-loop. During a project for developing a light JPEG library which is enough to run on a mobile device without compromising quality graphics on a mobile device, I have seen and worked out a number of ways in which a given computer program can be made to run faster. ( » Node.js » HR Here is a general pseudocode for a superloop implementation: We perform the initialization routines before we enter the super loop, because we only want to initialize the system once. The loop is in fact a variant of the classic "batch processing" control flow: Read input, calculate some values, write out values. = m A for loop is a repetition control structure that allows you to efficiently write a loop that needs to execute a specific number of times.. Syntax. Web Technologies: The location and contents of this file are usually described in the documentation supplied with the compiler. Also, it is important to note that many microcontrollers have power-save modes, where they will require less electrical power, which can be especially good if the system is running off a battery. Enter a positive integer: 10 Sum = 55. The syntax of a for loop in C programming language is −. Hopefully, this will allow you to make informed decisions on the choice of data structure. m A developer using an RTOS still needs to understand the hardware registers in order to write a peripheral driver. there is no limit of infinite loops in a program). C's volatile keyword is a qualifier that is applied to a variable when it is declared. Let's say we have an embedded system which has an average loop time of 1ms, and needs only to check a certain input once per second. Simple embedded systems typically use a Super-Loop concept where the application executes each function in a fixed order. Sometimes, this setup is done on purpose, but mostly it […] Once the infinite loop begins, we don't want to reset the values, because we need to maintain persistent state in the embedded system. The use of volatile is poorly understood by many programmers. Let's say that we have a microcontroller that uses 20mA of current in "normal mode", but only needs 5mA of power in "Low-Power Mode". Here is a general pseudocode for a superloop implementation: We perform the initialization routines before we enter the super loop, because we only want to initialize the system once. © https://www.includehelp.com some rights reserved. » CS Basics Here, anything() is also not a standard library function, at this section you can actual code that you want to execute again and again to keep running the device. 'Super Loop' runs give statements within the scope forever. a » Android In this guide we will learn while loop in C. C – while loop. Learn: What is 'Super Loop' in Embedded C programming language? To check the loop time of the program the power-save super loop is used. Here, initialize() is not a standard library function, we just wrote this function as an example. Once the infinite loop begins, we don't want to reset the values, because we need to maintain persistent state in the embedded system. For example, they aren’t supposed to use a lot of power to operate and they are supposed to be as cheap as po… » O.S. {\displaystyle \mathrm {Power} ={\frac {(99.9\%\times 5\ \mathrm {mA} )+(0.1\%\times 20\ \mathrm {mA} )}{100\%}}=5.015\ \mathrm {mA} \quad \mathrm {Average} }. When programming an embedded system, it is important to meet the time deadlines of the system, and to perform all the tasks of the system in a reasonable amount of time, but also in a good order. The Embedded C Programming Language uses the same syntax and semantics of the C Programming Language like main function, declaration of datatypes, defining variables, loops, functions, statements, etc. Example:for loop,while loop,etc.With building this type of game project your programmin… » CS Organizations » Embedded C » About us Languages: In this function, we use a super loop in which we alternately call two functions LED_FLASH_Change_State(), and DELAY_ LOOP_Wait(). » CSS » JavaScript So, to run set of statements, we need a loop that must not be finished, such kind of loops are known as 'Super Loop' or 'Infinite Loop'. » C++ 5.015 » Subscribe through email. × Next Page . Interrupt Service Routines (ISR) are used for time-critical program portions. » C# for ( init; condition; increment ) { statement(s); } Here is the flow of control in a 'for' loop − The init step is executed first, and only once. Its for beginners who want to get started in programming STM32 with Keil. » LinkedIn e   Learn: What is 'Super Loop' in Embedded C programming language? Non-Confidential PDF versionARM DUI0375H ARM® Compiler v5.06 for µVision® armcc User GuideVersion 5Home > Using the Inline and Embedded Assemblers of the ARM Compiler > Inline assembler and register access in C and C++ code 6.14 Inline assembler and register access in C and C++ code The inline assembler provides no direct access to the physical registers of an ARM processor. » DOS Because SST doesn't rely in any way on the interrupt stack frame layout, with most embedded C compilers, the ISRs can be written entirely in C.One notable difference between a simple “super-loop” and SST ISRs is that SST requires the programmer to insert some simple actions at each ISR entry and exit. Registers in order to write a peripheral driver using super loop in an embedded C programming language is to. 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# Gradient for the max function

• October 8th 2009, 12:10 AM
noschemer
Suppose $f(\vec{x})=\max_ix_i$. What is the gradient $\bigtriangledown f(\vec{x})=?$
If you were to graph max(x,y) on, say $-1\le x\le 1$, $-1\le y\le 1$ you would see sharp corners along the lines y= x and y= -x. Generally speaking, only "smooth" functions have derivatives.