Ph.D. Preliminary Examination
Spring 2015
Solution
I NSTRUCTIONS
1. Please check to ensure that you have a complete exam booklet. There are 22 numbered
problems. Note that Problem 1 occupies 3 pages, Problem 10 occupies 2 pages, Problem
22 occupies 2 pages. Including the cover sheet, you should have 58 pages. There should be
no blank pages in the booklet.
2. The examination is closed book and closed notes. No reference material is allowed at your
desk. 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 blank
sheets 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 this
cover sheet, the problem statement sheets, and any additional work sheets you turn in. DO
NOT 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 the
associated worksheets, and placed on the top of the appropriate envelope in the front of the
exam 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 are
not allowed to take any of the exam booklet pages from the room!
1
10
19
2
11
20
3
12
21
4
13
22
5
14
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6
15
7
16
8
17
9
18
Problem 1 (Core: VLSI - ECE 2020)
CMPE 2020 PRELIM PROBLEM
Solution
P ROBLEM
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)
0
0
0
0
0
1
0
1
1
0
1
0
1
1
1
1
Input = I
Next State = A(t+1),
B(t+1)
Output=Z
0
1
0
1
0
1
0
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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Problem 1 (Core: VLSI - ECE 2020)
Solution
BI
A
00
01
11
10
01
11
10
11
10
0
1
D(A) =
BI
A
00
0
1
D(B) =
BI
A
01
00
0
1
Z =
3 of 58
Problem 1 (Core: VLSI - ECE 2020)
Solution
(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).
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D(A)
A(t)
D(B)
B(t)
Problem 1 (Core: VLSI - ECE 2020)
S OLUTION
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Solution
Problem 1 (Core: VLSI - ECE 2020)
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Solution
Problem 1 (Core: VLSI - ECE 2020)
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Solution
Problem 2 (Core:
DSP - ECE 2026)
Problem XXProblem
(AREA)XX (AREA)
Code Number: Solution
Code Number:
P ROBLEM
P ROBLEM
P ROBLEM
Problem (10 points): The following three parts are independent of each other.
Problem (10 points): The following three parts are independent of each other.
−1
2z
(a) (3 points) Consider a causal discrete system with the system function
2z −1 H(z) = 1−3z −1 and
(a) (3 points) Consider a causal discrete system with the system function H(z) = 1−3z
−1 and
the corresponding impulse response h[n]. Define a new system with impulse
the corresponding impulse response h[n].
Define a new system with impulse
responsenh0 [n] = (0.1)n h[n]. Determine the frequency response of the new
response h0 [n] = (0.1) h[n]. Determine the frequency response of the new
system.
system.
(b) (4 points) Consider a discrete system with input x[n] and output y[n]. Let H(z) = 1− 41z −2
(b) (4 points) Consider a discrete system with input x[n] and output y[n]. Let H(z) = 1− 41z −2
9
9
be the z transform of the causal LTI system. Find the output
y[n] for the input
be the z transform of the causal
LTI system. Find the output y[n] for the input
x[n] = cos( π2 n + π6 ) where −∞ < n < ∞.
x[n] = cos( π2 n + π6 ) where −∞
< n < ∞.
(c) (3 points) Let H(z) have two real poles at p > p2 > 1, and two real zeros at z1 > z2 > 1.
(c) (3 points) Let H(z) have two real poles at p1 > p2 > 1, and1two real
zeros at z1 > z2 > 1.
Assume that H(z) is causal with impulse response h[n]. Define
g[n] = r n h[n],
Assume that H(z) is causal with impulse response h[n]. Define g[n] = r n h[n],
where r is a positive real number. Find the largest value of r such that the poles
where r is a positive real number. Find the largest value of r such that the poles
of G(z) are inside the unit circle.
of G(z) are inside the unit circle.
3
3
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Problem 2 (Core: DSP - ECE 2026)
S OLUTION
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Solution
Problem 3 (Core: CSS - ECE 2035)
P ROBLEM
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Solution
Problem 3 (Core: CSS - ECE 2035)
S OLUTION
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Solution
Problem
(Core:
EDA - ECE 2040)
Problem 4XX
(AREA)
Solution
Code Number: ___________
P
PROBLEM
ROBLEM
For the circuit below, suppose that the input current source is sinusoidal, Ii = Acos(ωt).
Find the expression for the transfer function H(ω) 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
Io
R1
C
L
R2
1
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Problem 4 (Core: EDA - ECE 2040)
S OLUTION
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Solution
Problem
(Core:
Problem 5
3025
(EMEMAG
AREA)- ECE 3025)
Solution
Code Number: ___________
P ROBLEM
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, ε1, ε2, a, b, c, and the spherical coordinates and unit vectors.
Hint: use Gauss’ Law.
D(r , θ , φ ) = ___________________________ (a ≤ r ≤ c)
b) Derive an expression for the capacitance. Express your answer in terms of ε1, ε2, a, b,
and c.
C = ____________________________________
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Problem 5 (Core: EMAG - ECE 3025)
S OLUTION
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Solution
Problem 6 (Core: VLSI - ECE 3030)
ECE#3030#
Solution
P ROBLEM
!
!"
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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ECE#3030#6 (Core: VLSI - ECE 3030)
Problem
!
!
S OLUTION
Solution#
Delay!definition:!
!"
!=
!
!
Drain!current!at!saturation:!
1 !
!! = !′ (!!" − !! )! !
2 !
Substituting!power!supply!voltage:!
!
!!!!
1 !
!=
, ! = !′ !
!
!(!!! − !! )
2 !
!
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Solution
Problem 7 (Core: MICROSYS - ECE 3040)
P ROBLEM
$
0.55eV
Ec
Ef
0 eV
Ei
−0.1eV
−0.55eV
Ev
0
1μm
μm
x
)$ .(
.)&$
*$ $
+$ $
.(.)&$
,$ $
.(.)&$
#
" ni = 1010 cm −3, # kT ≈ 26 meV @300K
0/)+-(*%
18 of 58
Solution
Problem 7 (Core: MICROSYS - ECE 3040)
S OLUTION
Solutions:*
$
1. The$electron$density$is$given$as$$
(E −E )/kT
n = ni e f i $
$
@$x=0,$ n = 1010 e 0.1/0.026 ≈ 4.7 ×1011 cm −3 $
@$x=1um,$ n = 1010 e 0 = 1010 cm −3 $
$
$
-3
n (cm )
4.7×10
11
$
10
10
0
1µm
$
$
$
$
$
2. The$electric$field$is$given$as:$$
$
0.1(V )
C
ε = 1q dE
=
= 0.1 V / µ m = 1000 V / cm $
dx 1( µ m)
$
$
Ε(V/µm)
0.15
0
$
$
$
$
$
$
1µm
$
19 of 58
Solution
Problem 7 (Core: MICROSYS - ECE 3040)
Solution
3. The$drift$current$density$is$given$as$$
$
$
!
!
!
(E − E )/ kT
J n,drift = q n µn ε = q ni e i f µn ε $
$
2
Jn (A/cm )
0.1
$
0.002
0
$
$
!
J n−diff
1µm
$
1. The$net$current$density$due$to$electrons$has$to$be$zero$since$the$Fermi$
energy$is$constant.$Therefore,$$
!$
= − J n−drift $
$
$
2
Jn (A/cm )
−0.002
1µm
$
−0.1
$
20 of 58
Problem 8XX
(AREA)
Problem
(Core:
CSS - ECE 3056)
Code Number: ___________
Solution
3056 PROBLEM: EVALUATING
COMPUTER PERFORMANCE
P ROBLEM
Your company wants to optimize a key kernel for the stock options pricing application. Profiling this kernel
on your current set up revealed:
1.
2.
3.
The number of instructions (N) in the kernel is: 2 Billion
The Cycles per Instruction (CPI) of the current machine for this kernel is: 2
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
Instructions
CPI
Frequency
Brainiacs
2 Billion
1.7
2 GHz
SpeedDemons
2 Billion
2.5
3 GHz
CompiLinx
1.5 Billion
2
2 GHz
B) Compute the MIPS (Million Instructions per Second) for these options
Brainiacs
SpeedDemons
CompiLinx
C) Compute the Execution time (in seconds) for these options
Brainiacs
SpeedDemons
CompiLinx
D) The option that has the highest performance is _____________, and it provides a speedup of _____
compared to the current machine (from part (A)).
1
21 of 58
Problem
(Core:
Problem
XX8(AREA)
CSS - ECE 3056)
Code Number: ___________
S OLUTION
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.
2.
3.
The number of instructions (N) in the kernel is: 2 Billion
The Cycles per Instruction (CPI) of the current machine for this kernel is: 2
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
Instructions
CPI
Frequency
Brainiacs
2 Billion
1.7
2 GHz
SpeedDemons
2 Billion
2.5
3 GHz
CompiLinx
1.5 Billion
2
2 GHz
B) Compute the MIPS (Million Instructions per Second) for these machines:
Brainiacs
SpeedDemons
CompiLinx
1176
1200
1000
C) Compute the Execution time (in seconds) for these options
Brainiacs
SpeedDemons
CompiLinx
1.7 seconds
1.66 seconds
1.5 seconds
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)).
2
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Solution
Problem 9 (Core: POWER - ECE 3072)
Solution
P ROBLEM
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
Rc
Xm
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Xequivalent
Problem 9 (Core: POWER - ECE 3072)
Solution for ECE3072
Solution
S OLUTION
(a) From the open circuit test: arc cos Θoc = 50W/[(2300V)(0.21A)] and Θoc = 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 Θsc = 160/[(47V)(6A)] and Θsc = 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)
Problem XX (AREA)
P ROBLEM
P ROBLEM
Code Number:
Solution
Let 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:
1
rate
Recall that we use the notation X ⇠ Exp( ) to denote that X is an exponential random variable with
> 0, i.e., a random variable with probability density function given by
(
e x x 0,
fX (x) =
0
x < 0.
Recall also that if X ⇠ Exp( ) then E[X] = 1 .
3
25 of 58
Problem 10 (Core: DSP/TLCOM - ECE 3077)
Problem XX (AREA)
Code Number:
Solution
3. (4 points) Let Z be the amount of time until both Merlin and Arthur are finished with
the 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)
Problem XX (AREA)
S OLUTION
Solution
Code Number:
S OLUTIONS
Let 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) =
(
1
e a/3 m
3
0
a, m 0
otherwise
2. (4 points) What is the probability that Merlin finishes the exam before Arthur?
Answer: 0.75
P[M < A] =
Z
1
Za=0
1
Z
a
m=0
1
e
3
a/3 m
dm da
1 a/3
e
(1 ea ) da
a=0 3
Z 1
Z
1 a/3
1 1
=
e
da
e
3 a=0
a=0 3
1 3
=1
·
3 4
3
= .
4
=
1
rate
4a/3
da
Recall that we use the notation X ⇠ Exp( ) to denote that X is an exponential random variable with
> 0, i.e., a random variable with probability density function given by
(
e x x 0,
fX (x) =
0
x < 0.
Recall also that if X ⇠ Exp( ) then E[X] = 1 .
3
27 of 58
Problem
DSP/TLCOM - ECE 3077) Code Number:
Problem10
XX(Core:
(AREA)
3. (4 points) Let Z be the amount of time until both Arthur and Merlin are finished with
the 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
◆ ✓Z z
1 a/3
=
e
da
e
0 3
0
= (1
=1
z/3
e
e
e
dm
e z)
)(1
z/3
m
z
+e
◆
4z/3
for z
0. Thus the probability density function for Z is given by the derivative of
the cumulative distribution function:
1
fZ (z) = e
3
z/3
+e
z
4
e
3
4z/3
.
Recognizing that each term in this sum looks like the probability density function of
a simple exponential random variable, we have that
E[Z] = 3 + 1
4
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3
= 3.25.
4
Solution
Problem XX
11 (Core:
- ECE 3084)
Problem
(Core: CONTROLS
S&C-ECE3084)
Solution
Code Number: ___________
ROBLEM
PPROBLEM
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(jω) and H2(jω).
The input is x(t) and the output is y(t).
x(t)
y(t)
LTI System #1
LTI System #2
(a) Suppose that h1(t) = tu(t) – 2(t – t0)u(t – t0) + (t – 2t0)u(t – 2t0) and h2(t) = δ(t−td).
Sketch h(t), the impulse response of the overall system for 2t0 < td.
(b) Find an expression for H(jω), 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 µs but you do not know td. You perform an
experiment where you set x(t) = cos(2πft) and sweep the frequency from f = 100 kHz
to 1.1 MHz. You observe nulls (locations where H(jω) = 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?
1
29 of 58
Problem
11(Core:
(Core:S&C-ECE3084)
CONTROLS - ECE 3084) Code Number: ___________
Solution
Problem
XX
S OLUTION
SOLUTION
(a) Sketch of the impulse response of the overall system:
h(t)
t
t0
0
2t0
td
td+t0
td+2t0
(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.
h(t )
h1 (t ) h1 (t td )
hˆ(t ) hˆ(t )
h1 (t )
hˆ(t ) u (t ) u (t t0 )
Hˆ ( j )
H1 ( j )
sin( t0 / 2) j
e
/2
Hˆ ( j ) Hˆ ( j )
sin( t0 / 2)
e
/2
4sin 2 ( t0 / 2)
2
H( j )
t0 / 2
2
j t0 /2
e
j t0
H1 ( j ) H1 ( j )e
4sin 2 ( t0 / 2)
2
4sin 2 ( t0 / 2)
2
8sin 2 ( t0 / 2)
2
j td
e
j t0
(1 e
e
j t0
e
e
j ( t0 td /2)
j td
j td /2
)
(e j
td /2
e
j td /2
)
cos( td / 2)
(c) The sine term causes zeros in the frequency response at
t
2
n 0
n ;
n
2 n
; fn
t0
n
t0
n
1 s
n 1 MHz, n 1, 2,3,...
The null at 1 MHz can be explained by the sine term, which does not give us any
information about td.
2
30 of 58
Problem 11 (Core: CONTROLS - ECE 3084)
Problem XX (Core: S&C-ECE3084)
Solution
Code Number: ___________
The nulls at 200 kHz and 600 kHz must be due to the cosine term:
t
2
3 5
, ...
2 2 2
,
k d
(2k 1)
,k
2
1, 2,3,...
If we assume that the null at 200 kHz is the lowest one, then
t
2
1 d
2
and td
2 f1
1
2 f1
1
2(200 kHz)
2.5 μs
Assuming that this value is correct, the first four nulls can be calculated as:
2 f k td
2
f1
(2k 1)
; fk
2
200 kHz, f 2
(2k 1)
2t d
(2k 1)(200 kHz)
600 kHz, f 3 1 MHz, f 4
1.4 MHz
These values are consistent with the measurements, so we can conclude that td = 2.5 µs 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:
(2m 1)
2(200 kHz)
td
(2m 1) 2.5 μs.
The (m + 1)st null is,
fm
1
[2( m 1) 1]
2t d
[2( m 1) 1]
2(2m 1)(2.5 μs)
2m 1
200 kHz.
2m 1
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 µs is the only one that
explains the measurements and is therefore unique.
31 of 58 3
Problem 12 (Breadth: VLSI - ECE 3150)
Solution
P ROBLEM
Consider the following Boolean function F = a(bc + d ′) . 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 τ 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 τ = R·CINV. 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.
32 of 58
Problem 12 (Breadth: VLSI - ECE 3150)
S OLUTION
33 of 58
Solution
Problem 12 (Breadth: VLSI - ECE 3150)
34 of 58
Solution
Problem 13 (Breadth: POWER - ECE 3300)
Solution
P ROBLEM
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 = 0.25Ω, R2 = 0.2Ω, X1 = X2 = 0.5Ω. 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,
s=
nsync − nrated
nrated
=
3600 − 3500
= 0.0277
3600
Neglecting the magnetizing branch implies that the stator and rotor current are the same.
Therefore the torque at this slip is,
Tem = 3
( 460 / 3 )
2
p 1 2 R2
2 1
0.2
I2
=3
= 71.69 N ⋅ m
2
2 ωs
s
2 2π 60 ⎛
0.2 ⎞
2 0.0277
⎜ 0.25 +
⎟ + ( 0.5 + 0.5 )
0.0277 ⎠
⎝
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.
Pg = 3I 22
(
460 / 3
)
2
R2
0.2
=3
= 27.03 kW
2
s
0.0277
0.2
2
⎛
⎞
⎜ 0.25 +
⎟ + ( 0.5 + 0.5 )
.0277 ⎠
⎝
35 of 58
a) Determine the torque at the rated speed for this machine.
b) Determine the power produced by this machine at zero speed.
c) Determine
power that crosses
the air gap
from the
stator to the rotor at rated speed.
Problem
13 the
(Breadth:
POWER
- ECE
3300)
S OLUTION
SOLUTION
a) At the rated speed, the slip, s, is,
s=
nsync − nrated
nrated
=
3600 − 3500
= 0.0277
3600
Neglecting the magnetizing branch implies that the stator and rotor current are the same.
Therefore the torque at this slip is,
(
)
2
460 / 3
p 1 2 R2
2 1
0.2
Tem = 3
I2
=3
= 71.69 N ⋅ m
2
2 ωs
s
2 2π 60 ⎛
0.2 ⎞
2 0.0277
⎜ 0.25 +
⎟ + ( 0.5 + 0.5 )
0.0277 ⎠
⎝
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
460 / 3
R
0.2
Pg = 3I 22 2 = 3
= 27.03 kW
2
s
0.2 ⎞
2 0.0277
⎛
⎜ 0.25 +
⎟ + ( 0.5 + 0.5 )
.0277 ⎠
⎝
36 of 58
Solution
Problem
EDA - ECE 3400)
Problem14
XX(Breadth:
(Core: EDA-ECE3400)
Solution
Code Number: ___________
PPROBLEM
ROBLEM
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 saturationmode, 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 triodemode? Please answer YES or NO.
If your answer is YES, find the minimum Vin value for triode-operation to happen.
37 of 58
1
Problem
(Breadth:
EDA - ECE 3400)
Problem
XX 14
(Core:
EDA-ECE3400)
Code Number: ___________ Solution
S OLUTION
SOLUTIONS
2
38 of 58
Problem XX (Core: EDA-ECE3400)
Code Number: ___________
Problem 14 (Breadth: EDA - ECE
3400)
SOLUTIONS
3
39 of 58
Solution
Problem 15
XX(Breadth:
(AREA) MICROSYS - ECE 3450) Code Number: ___________
Problem
Solution
PPROBLEM
ROBLEM
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 εr = 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):
1
40 of 58
Problem
(Breadth: MICROSYS - ECE 3450)Code Number: ___________
Solution
Problem
XX15(AREA)
S OLUTION
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, εr = 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):
1
41 of 58
Problem 16 (Breadth: CONTROLS - ECE 3550)
Solution
P ROBLEM
!
42 of 58
Problem 16 (Breadth: CONTROLS - ECE 3550)
Solution
S OLUTION
!
43 of 58
Problem 17 (Breadth: TELECOM - ECE 3600)
ECE3600 Spring 2015 Prelim Question
Solution
P ROBLEM
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: ___________________________________________________________________________
44 of 58
Problem 17 (Breadth: TELECOM - ECE 3600)
S OLUTION
ECE3600 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
45 of 58
Solution
Problem 18 (Breadth: EMAG - ECE 4350)
Solution
P ROBLEM
Problem for ECE 4350
You are to design a two-element dipole antenna array (arranged along the x axis, and with dipoles
oriented along the z direction) that will exhibit radiation maxima in the H-plane pattern that
occur at ⌅ = ⇤/4, 3⇤/4, 5⇤/4, and 7⇤/4, where ⌅ is the angle in the xy plane, measured from the
x axis. All maxima carry equal intensities at wavelength . There is to be zero power radiated in
the 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 the
dipoles, and k = 2⇤/ .
a) Specify the smallest relative current phase in radians, ⇥, that will yield zero intensity in the
broadside directions.
b) With the current phasing thus established as per part a, find the smallest antenna spacing d
in 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 endfire
directions, compared to the ⇤/4 direction?
46 of 58
Problem 18 (Breadth: EMAG - ECE 4350)
Solution
S OLUTION
Problem Solution for ECE 4350
You are to design a two-element dipole antenna array (arranged along the x axis, and with dipoles
oriented along the z direction) that will exhibit radiation maxima in the H-plane pattern that
occur at ⌅ = ⇤/4, 3⇤/4, 5⇤/4, and 7⇤/4, where ⌅ is the angle in the xy plane, measured from the
x axis. All maxima carry equal intensities at wavelength . There is to be zero power radiated in
the broadside (⌅ = ±⇤/2) directions. The array function magnitude in the H-plane is given by
⇤ ⌃ ⌥⇤
⇤
⇧ ⇤⇤
|A(⇧)| = ⇤⇤cos
with ⇧ = ⇥ + kd cos ⌅
2 ⇤
where ⇥ is the relative phase between the two driving currents, d is the spacing along x between
the dipoles, and k = 2⇤/ .
a) Specify the smallest relative current phase, ⇥, that will yield zero intensity in the broadside
directions.
When ⌅ = ⇤/2, ⇧ = ⇥, and we have
⇤ ⌃ ⌥⇤
⇤
⇥ ⇤⇤
|A(⇧)| = ⇤⇤cos
= 0 when ⇥ = ±⇤
2 ⇤
b) With the current phasing thus established as per part a, find the smallest antenna spacing d
in wavelengths that will yield a maximum in the direction ⌅ = ⇤/4.
Consider the primary maximum of |A|, occurring when ⇧ = 0. Choosing ⇥ =
⌅⇤ ⇧
2⇤d
⇧= ⇤+
cos
=0
4
so that
⇤, we have
⌥
1
⌥
=1 ⇧ d= ⌥
2
2
With this result, the array function is now
⇤
⇧ ⇤
⌥
⇤⌅
⇤
⇤
|A(⇧)| = ⇤cos
1
2 cos ⌅ ⇤
2
2d
⌃
This will maximize at all four required angles as stated in the beginning, while zeroing at
broadside.
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 endfire
directions, compared to the ⇤/4 direction?
In the endfire (⌅ = 0, ⇤) directions, we have
⇤
⌥ ⇧ ⇤⇤
⇤⌅
⇤
|A(⇧)| = ⇤cos
1 ± 2 ⇤ = ⇤0.796
2
Radiation intensity is proportional to the square of the array function, so the decibel drop
in intensity between the endfire and ⇤/4 direction will be
⇥
10 log10 0.7962 = 1.985 or ⌅ 2 dB
47 of 58
Problem19
XX(Specialized:
(AREA)
Problem
OPTICS - ECE 4500)
Solution
Code Number: ___________
PPROBLEM
ROBLEM
Mach–Zehnder Interferometric Gas Sensor
The Mach–Zehnder 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 Mach–Zehnder 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 (λ0 = 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.
48 of 58
1
Code Number: ___________
Problem XX (AREA)
Problem 19 (Specialized: OPTICS
- ECE 4500)
SOLUTIONS
Solution
S OLUTION
Mach–Zehnder 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 λ0 (i.e., the phase difference varies by 2π). When 100 cycles are
completed,
! − 1 ! = 100!! (or,
!!
! − 1 ! = 100×2!). 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
!! !!!
!
! cos
!!
1 + cos !! =
!
!! !!!
!
!!
2 + cos !! + cos !!
!
!
!!
!! − !!
!! + !!
1 + cos !"(! − 1)
! cos !"(! − 1)
!
2
!! !!
!! !!
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.
1
49 of 58
Problem 20 (Specialized: OPTICS - ECE 4502)
Solution
P ROBLEM
Problem for ECE 4502
Over certain wavelength ranges, it is possible to approximate nearly linear behavior of single mode
fiber 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 where
dD/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 that
wavelength 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 with
rms spectral width ⇥ = 0.50 nm is input to a length L = 100 km of this fiber. Find the
output rms width, ⇥T .
c) A length of inverse-slope dispersion-compensating fiber (ISDCF) is now attached to the output
end 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 -km
at 1550 nm. What length of DCF is needed?
d) Briefly comment on the wavelength “robustness” of the two-fiber link. Specifically, as the
wavelength 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 perfect
compensation would be achieved in principle within the linear regime, regardless of wavelength?
50 of 58
Problem 20 (Specialized: OPTICS - ECE 4502)
S OLUTION
51 of 58
Solution
Problem 20 (Specialized: OPTICS - ECE 4502)
Solution
Problem Solution for ECE 4502
Over certain wavelength ranges, it is possible to approximate nearly linear behavior of single mode
fiber 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 where
dD/d is the dispersion slope (also quoted at the operating wavelength). OFS Truewave fiber is
dispersion-shifted, and at = 1550 nm, dispersion data are D(1550) = +2.8 ps/nm-km, and the
slope at that wavelength is dD/d |1550 = +0.07 ps/nm2 -km.
a) What is the approximate zero-dispersion wavelength,
0?
Set up
2.8 + 0.07(
0
1550) = 0
⇤
0
= 1510nm
b) Suppose a light pulse of rms width ⇥T = 10.0 ps, center wavelength = 1530nm, and with
rms spectral width ⇥ = 0.50 nm is input to a length L = 100 km of this fiber. Find the
output pulse width, ⇥T :
The pulse spread will be
⇥⇥ =
⇥ D(
m )L
=
0.5 [2.8 + (1530
1550)(0.7)] (100) =
14.0 ps
The output pulsewidth will then be:
⇤
⇥
⇥T = ⇥T2 + ⇥⇥2 = (10)2 + (14)2 = 17.2 ps
c) A length of inverse-slope dispersion-compensating fiber (ISDCF) is now attached to the output
end 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 -km
at 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 ps
Solve for L2 to get 0.56 m.
d) Briefly comment on the wavelength “robustness” of the two-fiber link. Specifically, as the
wavelength 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 perfect
compensation would be achieved in principle within the linear regime, regardless of wavelength?
The requirement is that both the dispersion and the slope are compensated. Specifically,
we need
dD1
dD2
D1 | m L1 = D2 | m L2 and
| m L1 =
| m L2
d
d
This will occur for all wavelengths within the linear regime if the ratio of the first and
second terms in the expansion is the same for both fibers.
52 of 58
Problem
(Specialized:
BIO - ECE 4782)
Problem 21
4782
(AREA)
Solution
Code Number: ___________
P
ROBLEM
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).
1
53 of 58
Problem
21 (Specialized:
BIO - ECE 4782)
Problem
4782
(AREA)
Solution
Code Number: ___________
S OLUTION
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.
54 of158
Problem 22 (Specialized: BIO - ECE 4784)
Problem 4784 (Bio)
Code Number:
Solution
P ROBLEM
P ROBLEM
(4 points) You are measuring the electrical membrane potential of neurons in a solution that
is prepared by dissolving 0.5g CaCl2 and 6.5g NaCl in one liter of water. You determine
through imaging experiments that the intracellular concentrations of Cl- and Ca2+ are 4.0
mM and 100 µM, respectively. Calculate ECl and ECa , the Nernst potentials of Cl- and
Ca2+ .
Physical constants: R=8.314 J/(mol K), T=37 C, F=96485 Coulombs/mol
Atomic 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 stimulus
current elicits the firing of an action potential. Draw the time course of 1) the membrane
potential 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 conductance, draw a horizontal line indicating the resting membrane conductance. Exact values
are not important - focus on qualitative features.
1
55 of 58
Problem 22 (Specialized: BIO - ECE 4784)
Problem 4784 (Bio)
Code Number:
Solution
(4 points) Consider an excitable nerve cell whose membrane potential is described by the
following set of equations:
1
dV
dt
dm
dt
IN a (V, m)
IK (V )
Cm
=
IN a (V, m)
0.8
IK (V ) + Istim
mss(V)
mss (V ) m
⌧
= ḡN a m(V EN a )
= ḡK (V EK )
1
mss (V ) =
V 15
1+e 4
=
0.6
0.4
0.2
0
−100
−50
0
V (mV)
50
where Cm = 1 µF/cm2 , ḡN a = 3 µS/cm2 , ḡK = 1 µS/cm2 , ⌧ = 1 sec, EK = -70 mV, EN a =+50
mV, 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 -80
mV to +60 mV. On the x-axis (V ), label -80 mV, 0 mV, +60 mV, EN a and EK . It is not
necessary 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, such
as slope and location of zero crossings.
2
56 of 58
100
Problem 22 (Specialized: BIO - ECE 4784)
Problem 4784 (Bio)
Code Number:
S OLUTION
S OLUTIONS
(4 points). You are measuring the electrical membrane potential of neurons in a solution
that is prepared by dissolving 0.5g CaCl2 and 6.5g NaCl in one liter of water. You determine
through imaging experiments that the intracellular concentrations of Cl- and Ca2+ are 4.0
mM and 100 µM, respectively. Calculate ECl and ECa , the Nernst potentials of Cl- and
Ca2+ .
Physical constants: R=8.314 J/(mol K), T=37 C, F=96485 Coulombs/mol
Atomic 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 stimulus
current elicits the firing of an action potential. Draw the time course of 1) the membrane potential 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 conductance, draw a horizontal line indicating the resting membrane conductance.
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Solution
Problem
22 (Specialized:
BIO - ECE 4784)
Problem
4784 (Bio)
Solution
Code Number:
(4 points) Consider an excitable nerve cell whose membrane potential is described by the
following set of equations:
1
dV
dt
dm
dt
IN a (V, m)
IK (V )
Cm
=
IN a (V, m)
0.8
IK (V ) + Istim
mss(V)
mss (V ) m
⌧
= ḡN a m(V EN a )
= ḡK (V EK )
1
mss (V ) =
V 15
1+e 4
=
0.6
0.4
0.2
0
−100
−50
0
V (mV)
50
where Cm = 1 µF/cm2 , ḡN a = 3 µS/cm2 , ḡK = 1 µS/cm2 , ⌧ = 1 sec, EK = -70 mV, EN a =+50
mV, 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 -80
mV to +60 mV. On the x-axis (V ), label -80 mV, 0 mV, +60 mV, EN a and EK . It is not
necessary 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, such
as slope and location of zero crossings.
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100