Ph.D. Preliminary Examination Spring 2015 Solution ... · Ph.D. Preliminary Examination Spring 2015 Solution ... The examination is closed book and closed notes. ... are inside the

Ph.D. Preliminary ExaminationSpring 2015 Solution

INSTRUCTIONS

1. Please check to ensure that you have a complete exam booklet. There are 22 numberedproblems. Note that Problem 1 occupies 3 pages, Problem 10 occupies 2 pages, Problem22 occupies 2 pages. Including the cover sheet, you should have 58 pages. There should beno blank pages in the booklet.

2. The examination is closed book and closed notes. No reference material is allowed at yourdesk. A calculator is permitted.

3. All wireless devices must be turned off for the entire duration of the exam.

4. You may work a problem directly on the problem statement (if there is room) or on blanksheets of paper available from the exam proctor. Do not write on the back side of any sheet.

5. Your examination code number MUST APPEAR ON EVERY SHEET. This includes thiscover sheet, the problem statement sheets, and any additional work sheets you turn in. DONOT write your name on any of these sheets. Use the preprinted numbers whenever possible,or WRITE LEGIBLY!!!

6. Under the rules of the examination, you must choose 8 problems to be handed in for grading.Each problem to be graded should be separated from the rest of the materials, stapled to theassociated worksheets, and placed on the top of the appropriate envelope in the front of theexam room. DO NOT TURN IN ANY SHEETS FOR THE OTHER 14 PROBLEMS!!

7. The examination lasts 4 hours, from 9:30 AM to 1:30 PM EST.

8. When you hand in the exam:

(a) Separate the 8 problems to be graded as explained above.

(b) Check to see that your Code Number is in EVERY sheet you are turning in.(c) On the section at the bottom of this page, CIRCLE the problem numbers that you are

turning in for grading.

(d) Turn in this cover sheet (containing your code number) and the 8 problems to be graded.

(e) All other material is to be placed in the discard box at the front of the room. You arenot allowed to take any of the exam booklet pages from the room!

1 2 3 4 5 6 7 8 910 11 12 13 14 15 16 17 1819 20 21 22

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Problem 1 (Core: VLSI - ECE 2020) Solution

PROBLEMCMPE 2020 PRELIM PROBLEM The state transition diagram of a 3-state Mealy FSM (2 flip flops) is given below.

(a) First fill in the next state transition table for the above FSM in the table below. Assume that all the FSM flip flops are reset to logic 0 after power-up.

Present State = A(t),B(t)

Input = I Next State = A(t+1), B(t+1)

Output=Z

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

(b) You are to design a finite state machine that realizes the above FSM with D flip flops. Below, fill in the K-maps for D(A), D(B) and Z (see figure on next page) and write the minimal Boolean expressions for the same.

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D(A) =

D(B) =

Z =

A

BI

0

1

00 01 11 10

A

BI

0

1

00 01 11 10

A

BI

0

1

00 01 11 10

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(c) Draw a circuit diagram for the finite state machine showing all logic, FSM input and output and the two D flip flops corresponding to A(t) and B(t).

D(A)

D(B)

A(t)

B(t)

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SOLUTION

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Problem 2 (Core: DSP - ECE 2026) Solution

PROBLEM

Problem XX (AREA) Code Number:

PROBLEM

Problem (10 points): The following three parts are independent of each other.

(a) (3 points) Consider a causal discrete system with the system functionH(z) = 2z113z1 and

the corresponding impulse response h[n]. Define a new system with impulseresponse h0[n] = (0.1)nh[n]. Determine the frequency response of the newsystem.

(b) (4 points) Consider a discrete system with input x[n] and output y[n]. LetH(z) = 11 4

9z2

be the z transform of the causal LTI system. Find the output y[n] for the inputx[n] = cos(

2n +

6) where < n < .

(c) (3 points) LetH(z) have two real poles at p1 > p2 > 1, and two real zeros at z1 > z2 > 1.Assume that H(z) is causal with impulse response h[n]. Define g[n] = rnh[n],where r is a positive real number. Find the largest value of r such that the polesof G(z) are inside the unit circle.

3

Problem XX (AREA) Code Number:

PROBLEM

Problem (10 points): The following three parts are independent of each other.

(a) (3 points) Consider a causal discrete system with the system functionH(z) = 2z113z1 and

the corresponding impulse response h[n]. Define a new system with impulseresponse h0[n] = (0.1)nh[n]. Determine the frequency response of the newsystem.

(b) (4 points) Consider a discrete system with input x[n] and output y[n]. LetH(z) = 11 4

9z2

be the z transform of the causal LTI system. Find the output y[n] for the inputx[n] = cos(

2n +

6) where < n < .

(c) (3 points) LetH(z) have two real poles at p1 > p2 > 1, and two real zeros at z1 > z2 > 1.Assume that H(z) is causal with impulse response h[n]. Define g[n] = rnh[n],where r is a positive real number. Find the largest value of r such that the polesof G(z) are inside the unit circle.

3

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Problem 2 (Core: DSP - ECE 2026) Solution

SOLUTION

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Problem 3 (Core: CSS - ECE 2035) Solution

PROBLEM

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SOLUTION

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Problem 4 (Core: EDA - ECE 2040) Solution

PROBLEM

Problem XX (AREA) Code Number: ___________

1

PROBLEM For the circuit below, suppose that the input current source is sinusoidal, Ii = Acos(t). Find the expression for the transfer function H() that relates the phasor of the input current source to the phasor of the output current Io. Use a systematic and efficient solution method when solving. Credit will be given based on the appropriateness of the method chosen as well as the accuracy of the answer. There is no need to simplify your answer.

Ii

R1 LR2CIo

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Problem 4 (Core: EDA - ECE 2040) Solution

SOLUTION

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Problem 5 (Core: EMAG - ECE 3025) Solution

PROBLEM

Problem 3025 (EM AREA) Code Number: ___________

PROBLEM Consider a spherical capacitor containing two materials as shown below. The total charge on the surface on the inner conductor is Q, and on the outer conductor is -Q.

a) Derive an expression for the electric flux density in the capacitor. Express your

answer in terms of Q, 1, 2, a, b, c, and the spherical coordinates and unit vectors. Hint: use Gauss Law.

),,( rD = ___________________________ )( cra b) Derive an expression for the capacitance. Express your answer in terms of 1, 2, a, b, and c. C = ____________________________________

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Problem 5 (Core: EMAG - ECE 3025) Solution

SOLUTION

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PROBLEMECE#3030#!Use!the!relationship!! = !"! !derive!a!formula!to!describe!the!relationship!between!inverter!delay!and!power!supply!voltage!!!! .!!You!can!assume!a!classical!linear/!saturation!characteristic!for!transistor!voltage/current!behavior.!!!

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SOLUTION

ECE#3030#!!Solution#Delay!definition:!! = !"! !Drain!current!at!saturation:!!! =

12 !

!! (!!" !!)

!!Substituting!power!supply!voltage:!!! = !!!!!(!!! !!)!

,! = 12 !!! !

!

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Problem 7 (Core: MICROSYS - ECE 3040) Solution

PROBLEM

ni =1010cm3,

kT 26meV@300K

Ec

Ev

Ef

x 1m m0

0 eV

0.55eV

0.55eV

0.1eV

Ei

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SOLUTIONSolutions:*$1. The$electron$density$is$given$as$$

n = nie(E f Ei )/kT $

$@$x=0,$n =1010e0.1/0.026 4.71011cm3 $@$x=1um,$n =1010e0 =1010cm3 $$$

$$$$

$2. The$electric$field$is$given$as:$$

$ = 1

qdECdx

= 0.1(V )1(m)

= 0.1V / m =1000 V / cm $$$

$$$$$$$

1m 0

$

4.71011

1010

n (cm-3

)

1m 0

0.15

(V/m)

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3. The$drift$current$density$is$given$as$$$$

!Jn,drift = q n n

! = q nie(Ei Ef )/kTn

! $

$

$$$

1. The$net$current$density$due$to$electrons$has$to$be$zero$since$the$Fermi$energy$is$constant.$Therefore,$$$!

Jndiff = !Jndrift $

$$

$

1m 0

$

0.1

0.002

Jn (A/cm2)

1m

$0.1

0.002

Jn (A/cm2)

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PROBLEM


1

3056 PROBLEM: EVALUATING COMPUTER PERFORMANCE Your company wants to optimize a key kernel for the stock options pricing application. Profiling this kernel on your current set up revealed:

1. The number of instructions (N) in the kernel is: 2 Billion 2. The Cycles per Instruction (CPI) of the current machine for this kernel is: 2 3. The Operating frequency of the current machine is: 2 GHz

(A) The execution time for this kernel for your current set up is ________ seconds

You hire three companies: Brainiacs, which tries to make the microarchitecture more efficient with the goal of improving the CPI; SpeedDemons, which believe in deep pipelines to improve frequency; CompiLinx, which work on optimizing the compiler to produce more efficient code. The three companies gave you the following choices:

Metric Brainiacs SpeedDemons CompiLinx Instructions 2 Billion 2 Billion 1.5 Billion

CPI 1.7 2.5 2 Frequency 2 GHz 3 GHz 2 GHz

B) Compute the MIPS (Million Instructions per Second) for these options C) Compute the Execution time (in seconds) for these options D) The option that has the highest performance is _____________, and it provides a speedup of _____ compared to the current machine (from part (A)).

Brainiacs SpeedDemons CompiLinx

Brainiacs SpeedDemons CompiLinx

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SOLUTION


2

SOLUTIONS

Your company wants to optimize a key kernel for the stock options pricing application. Profiling this kernel on your current set up revealed:

1. The number of instructions (N) in the kernel is: 2 Billion 2. The Cycles per Instruction (CPI) of the current machine for this kernel is: 2 3. The Operating frequency of the current machine is: 2 GHz

(A) The execution time for this kernel for your current set up is ____2____ seconds

You hire three companies: Brainiacs, which tries to make the microarchitecture more efficient with the goal of improving the CPI; SpeedDemons, which believe in deep pipelines to improve frequency; CompiLinx, which work on optimizing the compiler to produce more efficient code. The three companies gave you the following choices:

Metric Brainiacs SpeedDemons CompiLinx Instructions 2 Billion 2 Billion 1.5 Billion

CPI 1.7 2.5 2 Frequency 2 GHz 3 GHz 2 GHz

B) Compute the MIPS (Million Instructions per Second) for these machines: C) Compute the Execution time (in seconds) for these options D) The option that has the highest performance is ______CompiLinx_______, and it provides a speedup of ___1.33x__ compared to the current machine (from part (A)).

Brainiacs 1176 SpeedDemons 1200 CompiLinx 1000

Brainiacs 1.7 seconds SpeedDemons 1.66 seconds CompiLinx 1.5 seconds

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Problem 9 (Core: POWER - ECE 3072) Solution

PROBLEM PROBLEM for ECE3072 A 15 kVA, 2300/230 volt single phase transformer is tested as follows. The open circuit and short circuit tests are carried out with the instruments on the high voltage side for both tests. Open circuit test Voc = 2300 V; Ioc = 0.21 A; Poc =50 W Short circuit test Vsc =47 V; Isc = 6.0 A Psc = 160 W

a) Calculate the four impedances in the following equivalent circuit referred to the high voltage side. b) Calculate the voltage regulation when the transformer is supplying 15 kVA at unity power factor

and 230 V to a load.

Requivalent Xequivalent

Rc Xm

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Problem 9 (Core: POWER - ECE 3072) Solution

SOLUTION Solution for ECE3072 (a) From the open circuit test: arc cos oc = 50W/[(2300V)(0.21A)] and oc = 840

Excitation admittance is Ye = (Ioc/Voc) /_-840 = (0.21/2300) /_-840 = 9.5 j 90.8 micro S

Hence Rc = 1/9.5microS = 105 k ohm And Xm = 1/90.8 microS = 11 k ohm From the s/c test: arc cos sc = 160/[(47V)(6A)] and sc = 55.40 Series impedance Zeq = (47V/6A) /_55.40 = 4.45 + j6.45 Hence Req = 4.45 ohm and Xeq = 6.45 ohm

(b) Load current I referred to HV side = 15000/2300 = 6.52 A Hence input voltage to the circuit is Vin = 2300/_00 + (4.45)(6.52/_00 ) + j(6.45)(6.52/_00 ) = 2329.4/_1.040

Voltage regulation = [(2329.4-2300)/2300] 100 = 1.28%

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Problem 10 (Core: DSP/TLCOM - ECE 3077) Solution

PROBLEMProblem XX (AREA) Code Number:

PROBLEMLet A and M be the amount of time that Arthur and Merlin (respectively) take to finish

the preliminary exam. Assume that A Exp(13), that M Exp(1), and that A and M are

independent of each other.1

1. (2 points) What is the joint probability density function of A and M?

fA,M(a, m) =

2. (4 points) What is the probability that Merlin finishes the exam before Arthur?

Answer:

1Recall that we use the notation X Exp() to denote that X is an exponential random variable withrate > 0, i.e., a random variable with probability density function given by

fX(x) =

(ex x 0,0 x < 0.

Recall also that if X Exp() then E[X] = 1 .

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Problem 10 (Core: DSP/TLCOM - ECE 3077) SolutionProblem XX (AREA) Code Number:

3. (4 points) Let Z be the amount of time until both Merlin and Arthur are finished withthe exam. (In other words, let Z = max(A, M).) Calculate E[Z].

Answer:

4

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Problem 10 (Core: DSP/TLCOM - ECE 3077) Solution

SOLUTIONProblem XX (AREA) Code Number:

SOLUTIONSLet A and M be the amount of time that Arthur and Merlin (respectively) take to finish

the preliminary exam. Assume that A Exp(13), that M Exp(1), and that A and M are

independent of each other.1

1. (2 points) What is the joint probability density function of A and M?

fA,M(a, m) =

(13ea/3m a, m 0

0 otherwise

2. (4 points) What is the probability that Merlin finishes the exam before Arthur?

Answer: 0.75

P[M < A] =Z 1

a=0

Z a

m=0

1

3ea/3m dm da

=

Z 1

a=0

1

3ea/3(1 ea) da

=

Z 1

a=0

1

3ea/3 da 1

3

Z 1

a=0

e4a/3 da

= 1 13

34

=3

4.

1Recall that we use the notation X Exp() to denote that X is an exponential random variable withrate > 0, i.e., a random variable with probability density function given by

fX(x) =

(ex x 0,0 x < 0.

Recall also that if X Exp() then E[X] = 1 .

3

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Problem 10 (Core: DSP/TLCOM - ECE 3077) SolutionProblem XX (AREA) Code Number:

3. (4 points) Let Z be the amount of time until both Arthur and Merlin are finished withthe exam. (In other words, let Z = max(A, M).) Calculate E[Z].

Answer: 3.25

The random variable Z = max(A, M) has cumulative distribution function given by

FZ(z) = P [A z and M z]= P [A z] P [M z]

=

Z z

0

1

3ea/3 da

Z z

0

em dm

= (1 ez/3)(1 ez)= 1 ez/3 ez + e4z/3

for z 0. Thus the probability density function for Z is given by the derivative ofthe cumulative distribution function:

fZ(z) =1

3ez/3 + ez 4

3e4z/3.

Recognizing that each term in this sum looks like the probability density function ofa simple exponential random variable, we have that

E[Z] = 3 + 1 34

= 3.25.

4

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Problem 11 (Core: CONTROLS - ECE 3084) Solution

PROBLEM

Problem XX (Core: S&C-ECE3084) Code Number: ___________

1

PROBLEM Consider the system shown below that is composed of two LTI systems with impulse responses of h1(t) and h2(t) and corresponding frequency responses of H1(j) and H2(j). The input is x(t) and the output is y(t).

(a) Suppose that h1(t) = tu(t) 2(t t0)u(t t0) + (t 2t0)u(t 2t0) and h2(t) = (ttd).

Sketch h(t), the impulse response of the overall system for 2t0 < td.

(b) Find an expression for H(j), the frequency response of the overall system, in terms of t0 and td. Simplify your answer as much as possible.

(c) Suppose that you know that t0 = 1 s but you do not know td. You perform an experiment where you set x(t) = cos(2ft) and sweep the frequency from f = 100 kHz to 1.1 MHz. You observe nulls (locations where H(j) = 0) at f = 200 kHz, 600 kHz, and 1 MHz. Assuming that you have found all nulls in this frequency range, can you uniquely determine td from these measurements? If so, what is it? If not, what is the set of all possible values?

LTI System #1x(t) y(t)

LTI System #2

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Problem 11 (Core: CONTROLS - ECE 3084) Solution

SOLUTION

Problem XX (Core: S&C-ECE3084) Code Number: ___________

2

SOLUTION (a) Sketch of the impulse response of the overall system:

(b) Approach: Break h(t) down into the sum of two triangular pulses, where each one is expressed as the convolution of two rectangular pulses. (c) The sine term causes zeros in the frequency response at

00 0

2; ; 1 MHz, 1,2,3,...2 1 sn

n nt n n nn f n n

t t

The null at 1 MHz can be explained by the sine term, which does not give us any information about td.

0 t0 td2t0 td+t0 td+2t0

h(t)

t

0

0

0

0

1 1

1

0

/20

1

2/20

20

2

1 1

20

2

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )sin( / 2) ( )

/ 2 ( ) ( ) ( )

sin( / 2)/ 2

4sin ( / 2)

( ) ( ) ( )

4sin ( / 2)(1

d

d

j t

j t

j t

j t

j t

h t h t h t t

h t h t h t

h t u t u t ttH j e

H j H j H j

te

t e

H j H j H j e

te

0

0

2/2 /2 /20

2

2( /2)0

2

)

4sin ( / 2) ( )

8sin ( / 2) cos( / 2)

d

d d d

d

j t

j t j t j t j t

j t td

e

t e e e e

t e t

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Problem 11 (Core: CONTROLS - ECE 3084) SolutionProblem XX (Core: S&C-ECE3084) Code Number: ___________

3

The nulls at 200 kHz and 600 kHz must be due to the cosine term:

3 5 (2 1), , ... , 1,2,3,...

2 2 2 2 2k dt k k

If we assume that the null at 200 kHz is the lowest one, then

1

1 1

1 1 and 2.5 s

2 2 2 2 2(200 kHz)d

dt t

f f

Assuming that this value is correct, the first four nulls can be calculated as:

1 2 3 4

2 (2 1) (2 1); (2 1)(200 kHz)

2 2 2

200 kHz, 600 kHz, 1 MHz, 1.4 MHz

k dk

d

f t k kf kt

f f f f

These values are consistent with the measurements, so we can conclude that td = 2.5 s is a possible solution. This value is unique, which can be seen by assuming that 200 kHz is the mth null (i.e., there are m 1 nulls below 100 kHz). For that case, td is:

(2 1)(2 1) 2.5 s.

2(200 kHz)dmt m

The (m + 1)st null is,

1

[2( 1) 1] [2( 1) 1] 2 1 200 kHz.

2 2(2 1)(2.5 s) 2 1m d

m m mft m m

This null assumes its maximum value of 600 kHz for m = 1, and is smaller than 600 kHz for all other values of m. Therefore the solution of td = 2.5 s is the only one that explains the measurements and is therefore unique.

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Problem 12 (Breadth: VLSI - ECE 3150) Solution

PROBLEM

Consider the following Boolean function )( dbcaF += . Assume that the complemented inputs are available.

a) (2pts) Draw the CMOS transistor-level schematic of F using a single pair of pull-up and pull-down network. Minimize the number of transistors used.

b) (2pts) Assume that the width of all transistors in part a) is 3w. Compute the worst-case RC delay in terms of when F is driving a load of 10Cinv. w and R are the width and the resistance of the minimum-size nFET, respectively. CINV is the input capacitance of the minimum-sized inverter, and = RCINV. Assume that the hole mobility is 2x slower than that of electron.

c) (3pts) Draw the stick diagram of part (a) such that there is no break on the diffusion strips.

d) (3pts) Draw the stick diagram of F using the minimum number of NAND2s and INVs only. You are to use a single Vdd/Gnd rail.

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SOLUTION

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Problem 13 (Breadth: POWER - ECE 3300) Solution

PROBLEMPROBLEM Induction Machines A 460 V line to line, 60 Hz, 2 pole, 3500 rpm, Y-connected, three phase induction motor has the following equivalent circuit parameters per phase: R1 = 0.25, R2 = 0.2, X1 = X2 = 0.5. The core loss resistance and magnetizing inductance can be neglected.

a) Determine the torque at the rated speed for this machine.

b) Determine the power produced by this machine at zero speed.

c) Determine the power that crosses the air gap from the stator to the rotor at rated speed.

SOLUTION

a) At the rated speed, the slip, s, is, 3600 3500 0.02773600

sync rated

rated

n ns

n = = =

Neglecting the magnetizing branch implies that the stator and rotor current are the same. Therefore the torque at this slip is,

( )( )

2

2 22 2

2

460 / 31 2 1 0.23 3 71.69 N m2 2 2 60 0.02770.20.25 0.5 0.5

0.0277

ems

RpT Is

= = = + + +

b) At zero speed, there is no output power since any power is always torque times speed. c) The power that crosses the air gap, Pg, is the power in R2/s.

( )( )

2

2 22 2

2

460 / 3 0.23 3 27.03 kW0.02770.20.25 0.5 0.5

.0277

gRP Is

= = = + + +

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Problem 13 (Breadth: POWER - ECE 3300) Solution

SOLUTION

PROBLEM Induction Machines A 460 V line to line, 60 Hz, 2 pole, 3500 rpm, Y-connected, three phase induction motor has the following equivalent circuit parameters per phase: R1 = 0.25, R2 = 0.2, X1 = X2 = 0.5. The core loss resistance and magnetizing inductance can be neglected.

a) Determine the torque at the rated speed for this machine.

b) Determine the power produced by this machine at zero speed.

c) Determine the power that crosses the air gap from the stator to the rotor at rated speed.

SOLUTION

a) At the rated speed, the slip, s, is, 3600 3500 0.02773600

sync rated

rated

n ns

n = = =

Neglecting the magnetizing branch implies that the stator and rotor current are the same. Therefore the torque at this slip is,

( )( )

2

2 22 2

2

460 / 31 2 1 0.23 3 71.69 N m2 2 2 60 0.02770.20.25 0.5 0.5

0.0277

ems

RpT Is

= = = + + +

b) At zero speed, there is no output power since any power is always torque times speed. c) The power that crosses the air gap, Pg, is the power in R2/s.

( )( )

2

2 22 2

2

460 / 3 0.23 3 27.03 kW0.02770.20.25 0.5 0.5

.0277

gRP Is

= = = + + +

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Problem 14 (Breadth: EDA - ECE 3400) Solution

PROBLEM

Problem XX (Core: EDA-ECE3400) Code Number: ___________

1

PROBLEM Please analyze the following feedback circuit. Assume the op-amp is ideal with infinite voltage gain, infinite bandwidth, infinite input impedance, and zero output impedance. The op-amp is powered by 20V. Neglect the channel-length-modulation and body-effect of the MOSFET transistor. Please answer the following questions. 1. Label the op-amp Plus + and Minus - inputs to ensure a negative feedback. 2. Based on the configuration at the Input and Output, what type of feedback is this circuit? Please choose ONLY ONE feedback from the following four feedback types.

(a) Input Series Output Series; (b) Input Series Output Shunt; (c) Input Shunt Output Series; (d) Input Shunt Output Shunt 3. Assume the input voltage Vin is 1V DC. Assume the MOSFET is in its saturation-mode, solve all the node voltages (V1, V2, V3, V4 and Vout) and branch currents (I1, I2, I3 and I4). Recall that for a MOSFET in saturation

, where , and . 4. Assume the input voltage Vin is 1 Volt DC and all the other circuit parameters stay the same except R4. If one keeps increasing the resistance value of R4, will the MOSFET enter its triode-mode? Recall that a MOSFET enters triode when or equivalently . Please answer YES or NO.

If your answer is YES, find the minimum R4 value for triode-operation to happen. 5. Assume R4 is 1kohm and all the other circuit parameters stay the same except the input DC voltage Vin. If one keeps increasing the Vin, will the MOSFET enter its triode-mode? Please answer YES or NO.

If your answer is YES, find the minimum Vin value for triode-operation to happen.

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SOLUTION


2

SOLUTIONS

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3

SOLUTIONS

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Problem 15 (Breadth: MICROSYS - ECE 3450) Solution

PROBLEM


1

PROBLEM Below are listed the doping levels of uniform samples of silicon at equilibrium at room temperature. For silicon you may take Eg = 1.1242 eV, NC = 2.86 1019 cm-3, NV = 3.10 1019 cm-3, and r = 11.8. The Boltzmann constant is k = 8.617 10-5 eV/K, and the Planck constant is h = 4.136 10-15 eV s. (a) ND = 0.0, NA = 2.0e18 cm-3. Calculate: (i) Hole concentration (p): (ii) Fermi level energy minus valence band edge energy (EF EV): (b) ND = 5.0e17 cm-3; NA = 2.0e18 cm-3. Calculate: (i) Conduction band electron concentration (n): (ii) Hole concentration (p): (iii) Conduction band edge energy minus valence band edge energy (EC EV):

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Problem 15 (Breadth: MICROSYS - ECE 3450) Solution

SOLUTION


1

PROBLEM Below are listed the doping levels of uniform samples of silicon at equilibrium at room temperature. For silicon you may assume Eg = 1.1242 eV, NC = 2.86e19 cm-3, NV = 3.10e19 cm-3, r = 11.8. (a) ND = 0.0, NA = 2.0e18 cm-3. Calculate: (ii) Hole concentration (p): (iii) Fermi level energy minus valence band edge energy (EF EV): (b) ND = 5.0e17 cm-3; NA = 2.0e18 cm-3. Calculate: (i) Conduction band electron concentration (n): (ii) Hole concentration (p): (iii) Conduction band edge energy minus valence band edge energy (EC EV):

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Problem 16 (Breadth: CONTROLS - ECE 3550) Solution

PROBLEM

!

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Problem 16 (Breadth: CONTROLS - ECE 3550) Solution

SOLUTION

!

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Problem 17 (Breadth: TELECOM - ECE 3600) Solution

PROBLEMECE3600 Spring 2015 Prelim Question Between hosts A in Atlanta and B in Las Vegas there are 2 routers (X,Y). The link between routers (---) is 10 Mbps. The access links (LANs, ===) are 1000 Mbps. The total distance from A to B is 2500 km. Host A starts to send a large file using TCP, sending a 1500 byte packet to B. There is no other traffic on this network. When the TCP segment is received, B sends back a short ACK. Be accurate to 0.1 ms. A ===X---Y===B 1. What is the time required (for X) to transmit a 1500 Byte datagram at 10 Mbit/s? _______________ ms 2. What is the propagation delay in milliseconds (speed in cable is 2e10 m/s): ____________________ ms

3. If the router buffers are empty, what is the round trip time (neglect processing delay)? __________ ms

4. What is an organization that has been assigned a block of IP addresses called: _________________

5. This type of organization must maintain an authoritative: _________________________________

6. Name a common type of wired Local Area Network: _______________________________

7. Name a common type of wireless Local Area Network: _______________________________

8. What is the secure (encrypted) protocol uses by Web browsers and servers: ______________

How do (9) CIDR and (10) NAT contribute to much more efficient use of the available IP address space?

9. CIDR:___________________________________________________________________________

10. NAT: ___________________________________________________________________________

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Problem 17 (Breadth: TELECOM - ECE 3600) Solution

SOLUTIONECE3600 Spring 2015 Prelim Question - Answers Between hosts A in Atlanta and B in Las Vegas there are 2 routers (X,Y). The bitrate of the link between routers (---) is 10 Mbps. The access links (LANs, ===) are 1000 Mbps. The total distance from A to B is 2500 km. Host A starts to send a large file using TCP, sending a 1500 byte packet to B. There is no other traffic on this network. When the TCP segment is received, B sends back a short ACK. Be accurate to 0.1 ms. A ===X---Y===B 1. What is the time required (for X) to transmit a 1500 Byte datagram at 10 Mbit/s? _____1.2______ ms 2. What is the propagation delay in milliseconds (speed in cable is 2e10 m/s): ________12.5_______ ms

3. If the router buffers are empty, what is the round trip time (neglect processing delay)? __26.5____ ms

4. What is an organization that has been assigned a block of IP addresses, called: _AS, or autonomous system_ (half-credit for ISP)

5. This type of organization must maintain an authoritative: ______DNS server (or DNS)_________

6. Name a common type of wired Local Area Network: ____Ethernet, or IEEE 802.3_________

7. Name a common type of wireless Local Area Network: ___ WiFi, or IEEE 802.11________

8. What is the secure (encrypted) protocol uses by Web browsers and servers: ___HTTPS____

How do (9) CIDR and (10) NAT contribute to much more efficient use of the available IP address space?

9. CIDR:_Allows IP address blocks to be appropriately sized (not just 256, 65,536, or 16,000,000)._

10. NAT: Allows private networks to exist behind a NAT router with a single routable IP address. John Copeland 404 894-5177

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Problem 18 (Breadth: EMAG - ECE 4350) Solution

PROBLEMProblem for ECE 4350

You are to design a two-element dipole antenna array (arranged along the x axis, and with dipolesoriented along the z direction) that will exhibit radiation maxima in the H-plane pattern thatoccur at = /4, 3/4, 5/4, and 7/4, where is the angle in the xy plane, measured from thex axis. All maxima carry equal intensities at wavelength . There is to be zero power radiated inthe broadside ( = /2) directions. The array function magnitude in the H-plane is given by

|A()| =cos

2

where = + kd cos

is the relative phase between the two driving currents, d is the spacing along x between thedipoles, and k = 2/.

a) Specify the smallest relative current phase in radians, , that will yield zero intensity in thebroadside directions.

b) With the current phasing thus established as per part a, find the smallest antenna spacing din wavelengths that will yield a maximum in the direction = /4.

c) With the spacing as found in part b, determine the value of |A()| in the endfire ( = 0,)directions. What decibel reduction in radiation intensity is therefore seen along the endfiredirections, compared to the /4 direction?

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Problem 18 (Breadth: EMAG - ECE 4350) Solution

SOLUTIONProblem Solution for ECE 4350

You are to design a two-element dipole antenna array (arranged along the x axis, and with dipolesoriented along the z direction) that will exhibit radiation maxima in the H-plane pattern thatoccur at = /4, 3/4, 5/4, and 7/4, where is the angle in the xy plane, measured from thex axis. All maxima carry equal intensities at wavelength . There is to be zero power radiated inthe broadside ( = /2) directions. The array function magnitude in the H-plane is given by

|A()| =cos

2

with = + kd cos

where is the relative phase between the two driving currents, d is the spacing along x betweenthe dipoles, and k = 2/.

a) Specify the smallest relative current phase, , that will yield zero intensity in the broadsidedirections.

When = /2, = , and we have

|A()| =cos

2

= 0 when =

b) With the current phasing thus established as per part a, find the smallest antenna spacing din wavelengths that will yield a maximum in the direction = /4.

Consider the primary maximum of |A|, occurring when = 0. Choosing = , we have

= + 2d

cos

4

= 0

so that2d

12

= 1 d =

2

With this result, the array function is now

|A()| =cos

2

1

2 cos

This will maximize at all four required angles as stated in the beginning, while zeroing atbroadside.

c) With the spacing as found in part b, determine the value of |A()| in the endfire ( = 0,)directions. What decibel reduction in radiation intensity is therefore seen along the endfiredirections, compared to the /4 direction?

In the endfire ( = 0,) directions, we have

|A()| =cos

2

1

2 = 0.796

Radiation intensity is proportional to the square of the array function, so the decibel dropin intensity between the endfire and /4 direction will be

10 log100.7962

= 1.985 or 2 dB

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Problem 19 (Specialized: OPTICS - ECE 4500) Solution

PROBLEM


1

PROBLEM MachZehnder Interferometric Gas Sensor The MachZehnder interferometer is a device used to determine the relative phase difference between two collimated beams derived by splitting light from a single source. A typical MachZehnder interferometer, as shown below, consists of two 50:50 beamsplitters, two flat mirrors, and a light source emitting a collimated beam of stable intensity. A gas cell of length 10 cm is placed in one arm, and a photodetector is used to monitor the interferometric output power.

(1) The photodetector will have a zero reading when destructive interference occurs in the output channel. Where does the optical energy go in this case?

(2) A collimated He-Ne laser (0 = 632.8 nm) is used as the light source. The cell is initially evacuated (i.e., vacuum), and a zero reading is obtained in the photodetector at the beginning. As the cell is gradually pressurized, we observe that the detected power goes high and becomes zero again, periodically with a total of 100 complete cycles. Determine the refractive index of the gas at the final pressure.

(3) Now, a collimated sodium lamp is used as the light source. The yellow light from a sodium lamp contains two closely spaced spectral lines of wavelengths 589.0 nm and 589.6 nm. The cell is initially evacuated (i.e., vacuum), and a zero reading is obtained in the photodetector at the beginning. As the cell is gradually pressurized, the detected power oscillates in time, and becomes absolutely zero again after many quasi-periodically cycles.

(3a) Roughly sketch the detected power as a function of time during this period.

(3b) Determine the refractive index of the gas at the final pressure.

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SOLUTION


1

SOLUTIONS

MachZehnder Interferometric Gas Sensor

(1) The power goes to the other interferometric channel. When a Mach-Zehnder

interferometer is constructed, two interferometric channels are formed simultaneously

(one output beam as shown in the figure, and the other goes upward from the beam

combiner). Law of conservation of energy always holds, and the input power is

distributed between the two output channels.

(2) The output power completes one dark-bright-dark circle when the relative path

difference varies by one 0 (i.e., the phase difference varies by 2). When 100 cycles are

completed,

! 1 ! = 100!! (or, !!!! ! 1 ! = 1002!). So ! = 1+!""!!! = 1.0006328

(3) The two lights with distinct wavelengths are not coherent, and the two sets of

interference should be considered separately. For each wavelength, the intensity varies as !!! (1+ cos !), where the phase difference ! =

!!! ! 1 !. So the total intensity that

reaches the detector is:

!!"# = !!! 1+ cos !! +!!! 1+ cos !! =

!!! 2+ cos !! + cos !!

= !!! 1+ cos!!!!!! ! cos

!!!!!! !

= !!2 1+ cos !"(! 1)!! !!!!!!

! cos !"(! 1) !! + !!!!!!!

Slowly varying envelope high frequency oscillation

(3a) The detected signal follows a beating pattern, where a high frequency oscillation is

modulated by a slowly varying envelop.

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PROBLEMProblem for ECE 4502

Over certain wavelength ranges, it is possible to approximate nearly linear behavior of single modefiber dispersion with wavelength, and we may use the two-term Taylor series form:

D() D(m) + ( m)dD

d

m

where m is an operating wavelength (at which a measured dispersion value is quoted), and wheredD/d is the dispersion slope (also quoted at the operating wavelength). In OFS Truewave fiber,and at m = 1550 nm, the dispersion is D(1550) = +2.8 ps/nm-km, and the slope at thatwavelength is dD/d = +0.07 ps/nm2-km.

a) What is the approximate zero-dispersion wavelength, 0?

b) Suppose a light pulse of rms width T = 10.0 ps, center wavelength = 1530nm, and withrms spectral width = 0.50 nm is input to a length L = 100 km of this fiber. Find theoutput rms width, T .

c) A length of inverse-slope dispersion-compensating fiber (ISDCF) is now attached to the outputend of the Truewave fiber, and is to be used to fully restore the pulse to its original width.The DCF fiber specifications are D(1550) = 100 ps/nm-km, and dD/d = 2.5 ps/nm2-kmat 1550 nm. What length of DCF is needed?

d) Briefly comment on the wavelength robustness of the two-fiber link. Specifically, as thewavelength is tuned away from 1530 nm, would good dispersion compensation be achieved?....and what requirement rests on the ISDCF, relative to the Truewave fiber, such that perfectcompensation would be achieved in principle within the linear regime, regardless of wave-length?

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SOLUTION

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Problem 20 (Specialized: OPTICS - ECE 4502) SolutionProblem Solution for ECE 4502

Over certain wavelength ranges, it is possible to approximate nearly linear behavior of single modefiber dispersion with wavelength, and we may write:

D() D(m) + ( m)dD

d

m

where m is an operating wavelength (at which a measured dispersion value is quoted), and wheredD/d is the dispersion slope (also quoted at the operating wavelength). OFS Truewave fiber isdispersion-shifted, and at = 1550 nm, dispersion data are D(1550) = +2.8 ps/nm-km, and theslope at that wavelength is dD/d|1550 = +0.07 ps/nm2-km.a) What is the approximate zero-dispersion wavelength, 0?

Set up2.8 + 0.07(0 1550) = 0 0 = 1510nm

b) Suppose a light pulse of rms width T = 10.0 ps, center wavelength = 1530nm, and withrms spectral width = 0.50 nm is input to a length L = 100 km of this fiber. Find theoutput pulse width, T :

The pulse spread will be

= D(m)L = 0.5 [2.8 + (1530 1550)(0.7)] (100) = 14.0 psThe output pulsewidth will then be:

T =2T +

2 =

(10)2 + (14)2 = 17.2 ps

c) A length of inverse-slope dispersion-compensating fiber (ISDCF) is now attached to the outputend of the Truewave fiber, and is to be used to fully restore the pulse to its original width.The DCF fiber specifications are D(1550) = 100 ps/nm-km, and dD/d = 2.5 ps/nm2-kmat 1550 nm. What length of DCF is needed?

We need to achieve the equal and opposite pulse spread as in part a. So

(DCF ) = 0.5 [100 (1530 1550)(2.5)]L2 = +14.0 psSolve for L2 to get 0.56m.

d) Briefly comment on the wavelength robustness of the two-fiber link. Specifically, as thewavelength is tuned away from 1530 nm, would good dispersion compensation be achieved,and what requirement rests on the ISDCF, relative to the Truewave fiber, such that perfectcompensation would be achieved in principle within the linear regime, regardless of wave-length?

The requirement is that both the dispersion and the slope are compensated. Specifically,we need

D1|mL1 = D2|mL2 anddD1d

|mL1 = dD2d

|mL2This will occur for all wavelengths within the linear regime if the ratio of the first andsecond terms in the expansion is the same for both fibers.

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Problem 21 (Specialized: BIO - ECE 4782) Solution

PROBLEM

Problem 4782 (AREA) Code Number: ___________

1

PROBLEM (3 pts) Cardiovascular Anatomy / Physiology. Fill in the blanks with the best possible answer. The large artery which outputs oxygenated blood from the left ventricle to most of the body is the __________________. The large artery which outputs deoxygenated blood from the right ventricle to the lungs is the ___________________. The volume of blood pumped by the heart in one heartbeat is referred to as ____________________. (4 pts) Draw an electrocardiogram (ECG) signal for a healthy subject. Label the important peaks. For each peak, explain the underlying phenomenon that is associated with the peak (e.g. the Y peak is associated with the contraction of the atria). (3 pts) Draw a pressure-volume loop for the left ventricle, labeling each segment and each corner point. Show which segments are associated with filling (diastole) and which are associated with ejection (systole).

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SOLUTION

Problem 4782 (AREA) Code Number: ___________

1

SOLUTION (3 pts) Cardiovascular Anatomy / Physiology. Fill in the blanks with the best possible answer. The large artery which outputs oxygenated blood from the left ventricle to most of the body is the ___aorta__________. The large artery which outputs deoxygenated blood from the right ventricle to the lungs is the _______pulmonary artery_. The volume of blood pumped by the heart in one heartbeat is referred to as ___stroke volume____. (4 pts) Draw an electrocardiogram (ECG) signal for a healthy subject. Label the important peaks, and show the typical amplitude (in millivolts) and time scale (in milliseconds). For each peak / complex, explain the underlying phenomenon that is associated with the peak / complex (e.g. the Y peak is associated with the contraction of the atria). Students should show the ECG for a beat. They should show that it is approximately 1-5 mV in amplitude, and the entire beat should be approximately 1 sec in width. They should note the p, q, r, s, and t waves, and that the p wave is associated with atrial contraction; the qrs with ventricular contraction, and t wave with ventricular repolarization. (3 pts) Draw a pressure-volume loop for the left ventricle, labeling each segment and each corner point. Show which segments are associated with filling (diastole) and which are associated with ejection (systole). Students should show the classic PV loop, labeling isovolumetric contraction, ejection, isovolumetric relaxation, and filling. IVC and ejection are systole, IVR and filling are diastole.

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PROBLEM

Problem 4784 (Bio) Code Number:

PROBLEM

(4 points) You are measuring the electrical membrane potential of neurons in a solution thatis prepared by dissolving 0.5g CaCl2 and 6.5g NaCl in one liter of water. You determinethrough imaging experiments that the intracellular concentrations of Cl- and Ca2+ are 4.0mM and 100 M, respectively. Calculate ECl and ECa, the Nernst potentials of Cl- andCa2+.

Physical constants: R=8.314 J/(mol K), T=37C, F=96485 Coulombs/molAtomic weights: Na (22.99), Ca (40.08), Cl (35.45)

(2 points) A neuron is sitting at rest (V = resting potential) and a brief 0.5 ms stimuluscurrent elicits the firing of an action potential. Draw the time course of 1) the membranepotential and 2) the membrane whole-cell conductance. On the plot of membrane potential,draw a horizontal line indicating the resting potential. On the plot of membrane conduc-tance, draw a horizontal line indicating the resting membrane conductance. Exact valuesare not important - focus on qualitative features.

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Problem 22 (Specialized: BIO - ECE 4784) SolutionProblem 4784 (Bio) Code Number:

(4 points) Consider an excitable nerve cell whose membrane potential is described by thefollowing set of equations:

CmdV

dt= INa(V, m) IK(V ) + Istim

dm

dt=

mss(V ) m

INa(V, m) = gNam(V ENa)IK(V ) = gK(V EK)

mss(V ) =1

1 + eV 15

4

100 50 0 50 1000

0.2

0.4

0.6

0.8

1

V (mV)

mss

(V)

where Cm = 1 F/cm2, gNa = 3 S/cm2, gK = 1 S/cm2, = 1 sec, EK = -70 mV, ENa=+50mV, and mss(V ) vs. V is plotted in the figure at the top right.

Draw a steady-state current voltage curve (Istim vs. V ) over a voltage range from -80mV to +60 mV. On the x-axis (V ), label -80 mV, 0 mV, +60 mV, ENa and EK . It is notnecessary to draw the exact shape of Istim) vs. V . What is important is 1) the sign of Istim(positive or negative) over specific ranges of V and 2) qualitative features of the plot, suchas slope and location of zero crossings.

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SOLUTION

Problem 4784 (Bio) Code Number:

SOLUTIONS

(4 points). You are measuring the electrical membrane potential of neurons in a solutionthat is prepared by dissolving 0.5g CaCl2 and 6.5g NaCl in one liter of water. You determinethrough imaging experiments that the intracellular concentrations of Cl- and Ca2+ are 4.0mM and 100 M, respectively. Calculate ECl and ECa, the Nernst potentials of Cl- andCa2+.

Physical constants: R=8.314 J/(mol K), T=37C, F=96485 Coulombs/molAtomic weights: Na (22.99), Ca (40.08), Cl (35.45)

(2 points) A neuron is sitting at rest (V = resting potential) and a brief 100 ms stimuluscurrent elicits the firing of an action potential. Draw the time course of 1) the mem-brane potential and 2) the membrane whole-cell conductance. On the plot of membranepotential, draw a horizontal line indicating the resting potential. On the plot of mem-brane conductance, draw a horizontal line indicating the resting membrane conductance.

1

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Problem 22 (Specialized: BIO - ECE 4784) SolutionProblem 4784 (Bio) Code Number:

(4 points) Consider an excitable nerve cell whose membrane potential is described by thefollowing set of equations:

CmdV

dt= INa(V, m) IK(V ) + Istim

dm

dt=

mss(V ) m

INa(V, m) = gNam(V ENa)IK(V ) = gK(V EK)

mss(V ) =1

1 + eV 15

4

100 50 0 50 1000

0.2

0.4

0.6

0.8

1

V (mV)

mss

(V)

where Cm = 1 F/cm2, gNa = 3 S/cm2, gK = 1 S/cm2, = 1 sec, EK = -70 mV, ENa=+50mV, and mss(V ) vs. V is plotted in the figure at the top right.

Draw a steady-state current voltage curve (Istim vs. V ) over a voltage range from -80mV to +60 mV. On the x-axis (V ), label -80 mV, 0 mV, +60 mV, ENa and EK . It is notnecessary to draw the exact shape of Istim) vs. V . What is important is 1) the sign of Istim(positive or negative) over specific ranges of V and 2) qualitative features of the plot, suchas slope and location of zero crossings.

2

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