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ECE-202 FINAL May 3, 2013
Name: __________________________________
(Please print clearly)
Student ID: _________________
CIRCLE YOUR DIVISION
DeCarlo-Morning DeCarlo-Afternoon
Meyer—Tu-Thur
INSTRUCTIONS
There are 34 multiple choice worth 6 points each.
This is a closed book, closed notes exam. No scrap paper or calculators are permitted. A transform table will be handed out separately.
Carefully mark your multiple choice answers on the scantron form and also on the text booklet as you will need to put the scantron inside the test booklet and turn both in at the end of the exam.
Nothing is to be on the seat beside you. When the exam ends, all writing is to stop. No writing while turning in the
exam/scantron or risk an F in the exam. Again, turn in the exam booklet with the scantron inside the exam. There are two exam forms and this is necessary for the proper grading of your scantron. All students are expected to abide by the customary ethical standards of the university, i.e., your answers must reflect only your own knowledge and reasoning ability. As a reminder, at the very minimum, cheating will result in a zero on the exam and possibly an F in the course. Communicating with any of your classmates, in any language, by any means, for any reason, at any time between the official start of the exam and the official end of the exam is grounds for immediate ejection from the exam site and loss of all credit for this exercise. The professor reserves the right to move students around during the exam. Do not open, begin, or peek inside this exam until you are instructed to do so.
EE-202, Final Sp 13 page 2
MULTIPLE CHOICE. 1. A circuit characterizing the decay of the ordinary attention span of a student in a DeCarlo circuit’s
class is H (s) = 30(s +1)(s + 3)
. The step response of the associated transfer function has a term of the
form Ae−tu(t) where A = : (1) -10 (2) 15 (3) –15 (4) 10 (5) 5 (6) 6 (7) –5 (8) -6 (9) None of above Solution 1. (Transfer function, step response, partial fraction expansion distinct poles, inverse transform)
L[StepResponse]= H (s)s
= 30s(s +1)(s + 3)
= 10s− 15s +1
+ 5s + 3
.
3StepResponse 10 15 5 ( )t te e u t− −⎡ ⎤= − +⎣ ⎦ . ANSWER: (3).
2. Consider the circuit below in which L = 2 H, iL (0− ) = −2 A, and Z(s) = 4s + 2
s. Note that we have
used a resistor symbol to denote a general impedance. Then iL (t) has a term of the form Ke−atu(t) where (K , a) = : (1) (2, 2) (2) (2, 2)− (3) (2,1) (4) ( 2,1)− (5) ( 2, 2)− (6) ( 2, 2)− − (7) (2, 1)− (8) ( 2, 1)− − (9) None of above
Solution 2. (Initial inductor current model, parallel impedance, ohm’s law, inductor impedance) Using the series voltage source model of the inductor, we have that
IL (s) =LiL (0
− )Ls + Z(s)
= −2 × 2
2s + 4s + 2s
= −2ss2 + 2s +1
= −2s(s +1)2
= −2s +1
+ 2(s +1)2
. Hence
iL (t) = −2e−tu(t)+ 2te−tu(t) A. ANSWER: (4).
EE-202, Final Sp 13 page 3
3. Consider the circuit below in which C = 0.25 F and L = 0.25 H. Suppose vin (t) = 10e−2tu(t) and all
initial conditions are ZERO. Then vC (t) has a term of the form K sin(ωt) where (K ,ω ) = (K in volts, ω in rad/s)): (1) (4, 2) (2) (10,4) (3) ( 10,4)− (4) (4, 4) (5) (16,4) (6) (10,2) (7) ( 4, 2)− (8) ( 16,4)− (9) none of above
Solution 3. (Impedance, Voltage Division, Response calculation, partial fraction expansion with complex poles)
VC (s) =
1Cs
Ls + 1Cs
Vin (s) =
1LC
s2 + 1LC
Vin (s) =10
(s + 2)× 16s2 +16
= 10 ×16(s + 2)(s2 +16)
2 28 8 16
2 216 16A Bs C ss ss s
+ − += + = ++ ++ +
. Thus 2( ) 8 ( ) 8cos(4 ) 4sin(4 )tCv t e u t t t−= − + V.
ANSWER: (4). 4. Consider the circuit below, driven by the input voltage vin (t) = f (t) V where C = 4 F, G = 2 mho (or S), K = 10 , T = 2 , and the initial capacitor voltage is zero. Then vout (t) = (in V): (1) 10 ( 2) 10 ( )u t u t− − (2) 5 ( ) 5 ( 2)u t u t− − (3) 10 ( ) 10 ( 2)r t r t− − (4) 5 ( ) 5 ( 2)r t r t− − (5) 5 ( 2) 5 ( )r t r t− − (6) 10 ( 2) 10 ( )r t r t− − (7) 5 ( 2) 5 ( )u t u t− − (8) 10 ( ) 10 ( 2)u t u t− − (9) None of above
EE-202, Final Sp 13 page 4
Solution 4. (op amp transfer function, input decomposition in terms of ramps, inverse transform)
The transfer function of the op amp circuit isH (s) = −CsG
= −2s . The input vin (t) = 5r(t)− 5r(t − 2) .
Vin (s) = 51− e−2s
s2. Vout (s) = −10 (1− e
−2s )s
. vout (t) = 10u(t − 2)−10u(t) .
ANSWER: (1). 5. Suppose R = 2 Ω, L = 2 H, and r0 = 16 Ω. The Thevenin equivalent impedance seen by the circuit below is:
(1)2 92ss+ (2)
2 2ss +
(3) 72ss− (4)
27s
s −
(5) 23s
s + (6) 9
2ss+ (7) 2
9s
s + (8) 2 2s
s+
(9) none of above
Solution 5. (Thevenin equivalent, node analysis, impedance, admittance)
Iin (s) = 0.5Vin (s)+Vin (s)+16IR(s)
2s= 0.5Vin (s)+
Vin (s)+ 8Vin (s)2s
= 0.5 + 12s
+ 82s
⎡⎣⎢
⎤⎦⎥Vin (s) =
s + 92s
Vin (s)
Hence Zin (s) =2ss + 9
.
ANSWER: (7).
6. Consider the circuit below in which R = 4 Ω and L = 4 H. Suppose vin (t) = 10−t
∫ 6e−q cos(q)u(q)dq
V and all initial conditions are ZERO. Then iL (t) has a term of the form Ke−at sin(ωt)u(t) where (K ,a,ω ) = :
EE-202, Final Sp 13 page 5
(1) ( 2, 2, 2)− (2) (2, 1,1)− (3) (2,1,1) (4) (2,2,2) (5) ( 2, 1,1)− − (6) (1,1,1) (7) ( 1,1,1)− (8) ( 2,1,1)− (9) None of above Solution 6. (Integration property of Laplace transform, Ohm’s Law, partial fraction expansion)
Vin (s) =1s
16(s +1)(s +1)2 +12
⎛⎝⎜
⎞⎠⎟
, IL (s) =1s
16(s +1)(s +1)2 +12
⎛⎝⎜
⎞⎠⎟· 14(s +1)
= 1s
4(s +1)2 +12
⎛⎝⎜
⎞⎠⎟
,
IL (s) =2s+ −2(s +1)(s +1)2 +12
+ −2(s +1)2 +12
. Thus iL (t) = [2 − 2e−t cos(t)− 2e−t sin(t)]u(t)
ANSWER: (8). 7. Consider the circuit below in which R1 = 1 Ω, R2 = 0.25 Ω, C = 1 F , vs2 (t) = 5u(t) , is1(t) = 2δ (t) , and all initial conditions are ZERO. Let vC (t) denote the node voltage at the top of the capacitor. Then vC (t)= (in V):
(1) 54 )2 ] ([ t u te−− (2) 5[2 ] ( )t u te−− (3) 10 ]2 ( )[4 te u t−−
(4) 2 ( ) 20 ( )t u t+δ
(5) 51 )2 ] ([ t u te−− (6) 54 )2 ] ([ t u te−+ (7)
52 )4 ] ([ t u te−− (8) 10 ]4 ( )[2 te u t−−
(9) None of above
Solution 7. (node analysis, Ohm’s Law, partial fraction expansion)
Is1(s)+Vs2 (s)R2
= 2 + 20s= 2 s +10
s= Cs + 1
R1+ 1R2
⎛⎝⎜
⎞⎠⎟VC (s) = (s + 5)VC (s)
Hence VC (s) = 2s +10s(s + 5)
= 4s− 2s + 5
and vC (t) = 4u(t)− 2e−5tu(t) V.
EE-202, Final Sp 13 page 6
ANSWER: (1). 8. Reconsider the circuit of problem 7. Suppose both sources are set to zero. Suppose further that the initial voltage on the capacitor (top to bottom is plus to minus) is vC (0
− ) = 10 V. Then vC (t) = (in V): (1) 205 ( )te u t− (2) 2010 ( )te u t− (3) 105 ( )te u t− (4) 1010 ( )te u t−
(5) 55 ( )te u t− (6) 510 ( )te u t− (7) 520 ( )te u t− (8) 2020 ( )te u t−
(9) None of above Solution 8. (Equivalent Circuit for initialized capacitor in s-domain, node analysis, Ohm’s Law, partial
fraction expansion, zero input response) CvC (0− ) = 10 = Cs + 1
R1+ 1R2
⎛⎝⎜
⎞⎠⎟VC (s) = (s + 5)VC (s) . Thus
vC (t) = 10e−5tu(t) V.
ANSWER: (6).
9. (step response and impulse response) The step response of a circuit is vout,step(t) = 20 1− e−2t( )u(t)
V. Then the impulse response is: (1) 22 ( )te u t− (2) 220 ( )te u t− (3) 210 ( )te u t−
(4) 240 ( )te u t−− (5) ( )220 1 ( )te t−− δ (6) 220 ( )te u t−−
(7) 240 ( )te u t− (8) 22 ( )te u t−− (9) none of these
Solution 9. dvout,step
dt= 20 2e−2t( )u(t)+ 20(1− e−2t )δ (t) = 40e−2tu(t) . ANSWER: (7).
10. (Switching circuits, LC Circuit Response computation with and without initialized elements) In the circuit below, L = 0.25 H, C = 0.25 F, R = 2 Ω, and iin(t) = 10 A (dc) for all time. Suppose that switch S1 has been in position B for a long time and switch S2 has been open for all past time. At time t = 0 ,
switch S1 moves to position A and switch S2 closes on position B instantaneously. If vout (0− ) = 0 , then
at t = 0.5 seconds, vout (t) is closest to (in volts): (1) –1.5 (2) – 2 (3) –3 (4) –4.5 (5) –4 (6) 1.5 (7) 3 (8) –15 (9) none of these
EE-202, Final Sp 13 page 7
x -1 -2 -3 -4
ex 0.4 0.15 0.05 0.02
Solution 10. iL (0− ) = 10 A. Using the voltage source equivalent circuit for the inductor for t > 0 we
have by voltage division Vout (s) = −
1Cs
1Cs
+ Ls + RLiL (0
− ) =− 1C
s2 + RLs + 1
LC
iL (0− ) . Hence,
Vout (s) =−40
s2 + 8s +16= −40(s + 4)2
. Thus vout (t) = −40te−4tu(t)⇒ vout (0.5) = −20e−2 ≅ −3 V.
ANSWER: (3). 11. (Sinusoidal steady state analysis) A linear circuit with transfer function
H (s) = Vout (s)Vin(s)
= 10 3− s0.5s2 + s +11
has voltage input vin(t) = 5cos(4t) V. Then vout,ss (t) has magnitude
and phase (in rads) equal to:
(1) 50, 1 42 tan3
− ⎛ ⎞⎜ ⎟⎝ ⎠
(2) 50, 1 42 tan3
− ⎛ ⎞− ⎜ ⎟⎝ ⎠ (3) 50, 1 44 tan
3− −⎛ ⎞⎜ ⎟⎝ ⎠
(4) 25, 1 44 tan3
− −⎛ ⎞⎜ ⎟⎝ ⎠
(5) 25, 1 42 tan3
− ⎛ ⎞− ⎜ ⎟⎝ ⎠ (6) 25, 1 42 tan
3− ⎛ ⎞⎜ ⎟⎝ ⎠
(7) 10, 1 42 tan3
− ⎛ ⎞⎜ ⎟⎝ ⎠
(8) 10, 1 42 tan3
− ⎛ ⎞− ⎜ ⎟⎝ ⎠ (9) none of these
Solution 11. H ( j4) = 10 3− 4 j−8 + 4 j +11
= 10 3− 4 j3+ 4 j
. H (4 j) = 10 and
∠H (4 j) = tan−1 −43
⎛⎝⎜
⎞⎠⎟ − tan
−1 43
⎛⎝⎜
⎞⎠⎟ = −2 tan−1 4
3⎛⎝⎜
⎞⎠⎟ . vout,ss (t) = 50cos 4t − 2 tan
−1 43
⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
V.
ANSWER: (2).
EE-202, Final Sp 13 page 8
12. Consider the functions ( )f t and ( )h t (both shown below). Suppose y(t) = f (t)∗h(−t) . Note here that f (t) is convolved with h(−t) NOT h(t) . Then y(2)=: (1) 6 (2) 24 (3) 8 (4) 4 (5) 12 (6) 9 (7) 16 (8) none of these
01−
( )f t
3
4
0
( )h t
1−
3
t t3
Solution 12. (Graphical Convolution) Do a flip and shift on h(−t) to produce h(τ − 2) on the τ − axis . This would overlap f (τ ) for 1≤ τ ≤ 3 . The product h(τ − 2) f (τ ) is nonzero only for 1≤ τ ≤ 3 and for this interval is a right triangle of height 12 and a base of 2. Hence, AREA = 12= y(2) . ANSWER: (5).
Questions 13 and 14 are in regards to the circuit below in which RL = 5 Ω
and Rs = 10 Ω. It is known that ( ) 10cos(10 ) ( )inv t t u t= V. Questions 13 and
14 require that you compute the values of L and C that lead to maximum power transfer to the load at the frequency determined by the input cosine.
13. (Maximum power transfer, LC Coupling Networks, Resonance) From the information given above, the value of L that leads to maximum power transfer is (in H):
(1) 1 (2) 2 (3) 0.75 (4) 0.25 (5) 0.1 (6) 0.2 (7) 0.5 (8) none of above
EE-202, Final Sp 13 page 9
Solution 13: Yin( jω ) = jωC + 1
RL + jωL=
RL
RL2 +ω 2L2 + jω C − L
RL2 +ω 2L2
⎡
⎣⎢
⎤
⎦⎥ . For maximum power
transfer the real part has to equal 1 / sR and the imaginary part is zero. For Problem 13,
RL
RL2 +ω 2L2 = 1
10⇒ 50 = 25+100L2 → L = 25
100= 0.5 H . ANSWER: (7).
14. (Maximum power transfer, LC Coupling Networks, Resonance) From the information given above and the value of L computed in question 13, the value of C that leads to maximum power transfer is (in F):
(1) 0.1 (2) 1200
(3) 3200
(4) 2100
(5) 5200
(6) 6100
(7) 0.01 (8) none of above
Solution 14: From Problem 13 solution work, set imaginary part to zero now.
C − L
RL2 +ω 2L2 = 0⇒C = L
RL2 +ω 2L2 = 0.5
25+100·0.25= 1
100. ANSWER: (7).
15. (Op amp circuit design from transfer function) If C1 = 2 F, the value of R2 (in Ω) for which the
transfer function for the op amp circuit below is H (s) = − 4s + 0.4s +1
is R2 = (in Ω):
(1) 1 (2) 4 (3) 0.25 (4) 2 (5) 0.5 (6) 0.2 (7) 10 (8) None of above
Solution 15: H (s) = −Yin(s)Yf (s)
= − C1s +G1C2s +G2
= −
C1C2
s + G1C2
s + G2C2
= −
2C2
s + G1C2
s + G2C2
= − 4s + 0.4s +1
. Thus
G2 = C2 = 0.5⇒ R2 = 2 Ω . ANSWER: (4).
EE-202, Final Sp 13 page 10
16. (Op amp circuit design from transfer function) In the circuit below C1 = C2 = 1 F and R2 = 0.25
Ω and R4 = 0.5 Ω. If the transfer function H (s) = Vout (s)Vs (s)
= H1(s)H2(s) =−8s + 4
× −4s + 2
. We assume
that H1(s) =−8s + 4
corresponds to the first stage (first op amp circuit). The values for R1 and R3 (in Ω)
respectively are: (1) 8, 4 (2) 0.125, 0.5 (3) 0.125, 0.125 (4) 0.25, 0.5 (5) 4, 8 (6) 0.125, 0.25 (7) two of above (8) none of above
Solution 16. For stage 1 H1(s) =−G1
C1s +G2= −G1s + 4
and for stage 2, H2 (s) =−G3
C2s +G4= −G3s + 2
. Hence
G1 = 8 mhos and G3 = 4 mhos. R1 = 0.125 Ω and R3 = 0.25 Ω. ANSWER: (6). 17. (Op amp circuit design from transfer function) Referring again to problem 16, the impulse response of the circuit is: (1) 4 28( ) ( )t te e u t− −− (2) 2 48( ) ( )t te e u t− −+ (3) 2 48( ) ( )t te e u t− −−
(4) 4 24( ) ( )t te e u t− −− (5) 2 44( ) ( )t te e u t− −− (6) 2 416( ) ( )t te e u t− −−
(7) 2 416( ) ( )t te e u t− −+ (8) 4 216( ) ( )t te e u t− −− (9) none of these
Solution 17. H (s) = 32(s + 2)(s + 4)
= 16s + 2
− 16s + 4
. Hence vimpulse(t) = 16(e−2t − e−4t )u(t) V.
ANSWER: (6).
EE-202, Final Sp 13 page 11
18. (Poles and zeros, transfer functions) The pole-zero plot of a transfer function of an active circuit is
A bounded input which will make the zero-state response unbounded (increasing with time) is: (1) δ (t) (2) u(t) (3) cos(t − 30o )u(t)
(4) sin(t − 30o )u(t) (5) cos(2t − 30o )u(t) (6) sin(2t − 30o )u(t) (7) two of above (8) none of above ANSWER: (7). 19. (Coupled Inductors, Stored Energy) If iin (t) = 5cos(2t)u(t)V, then the maximum instantaneous energy stored in the inductors is (in J): (1) 150 (2) 175 (3) 162.5 (4) 50 (5) 25 (6) 60 (7) 125 (8) none of above
Solution 19. Wmax = 0.5 × 2 × 25 + 0.5 × 2 × 25 +1× 25 + 0.5 × 8 × 25 = 175 J. ANSWER: (2). 20. (Approximate BP Circuit Analysis) Consider the circuit below. The approximate bandwidth in rad/s is: (1) 1 (2) 2 (3) 3 (4) 5 (5) 6 (6) 8 (7) 40 (8) 0.2 (9) none of above
EE-202, Final Sp 13 page 12
Solution 20. ω0 = 1LC
=1
25 ×10−3= 40 rad/s. Qcoil ( j40) =
40 × 0.0250.1
= 10 . Req =40 ×1050
= 8
Ω. Bw =1
ReqC=10.2
= 5 rad/s. Answer: (4).
21. (Approximate BP Circuit Analysis) Referring again to the circuit of problem 20, the maximum value (at least approximately) of the steady state response, vout (t) in V, when iin(t) = 5sin(ω0t)u(t)
where ω0 = 1LC
:
(1) 1 (2) 2 (3) 10 (4) 4 (5) 5 (6) 40 (7) 20 (8) 100 (9) none of above Solution 21: Hmax ≅ Req = 8 Ω. Hence, vout,max = 40 V. Answer: (6).
22. (Coupled Inductors, Impedance, Equivalent Circuits, Equivalent Inductance) In the circuit below, all initial conditions are zero. Suppose L1 = 0.8 H, L2 = 0.6 H, M = 0.4 H, L3 = 0.4 H, and L4 = 0.2 H. Then the equivalent inductance of the inductors is Leq = (in H): (1) 0.1 (2) 0.2 (3) 1.6 (4) 0.4 (5) 1 (6) 0.6 (7) 0.8 (8) none of above
Solution 22. Using the T-model of the coupled inductors, we have (L3 in series with L1-M) in series with ( M in parallel with (L2-M in series with L4) ). Equivalently, 0.4 + 0.4 + (0.4 // (0.2 + 0.2)) = 1. ANSWER: (5).
EE-202, Final Sp 13 page 13
23. (Coupled Inductors, Impedance) Consider the mutually coupled inductor circuit below. The 2 H
inductor is short circuited. The currents in the coupled inductors are i1(t) = 2e−tu(t) A and
i2 (t) = e−tu(t) A. The value of the mutual inductance M (in H) is:
(1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 9 (8) 8 (9) none of above
Solution 23. 0 = L1sI1(s)−MsI2 (s)⇒M = L1I1(s)I2 (s)
= 4 H. ANSWER: (4).
24. (Coupled Inductors, Impedance, Response Calculation) Reconsider problem 23. The correct expression for v2 (t) for t ≥ 0 is (in V):
(1) ( )te u t− (2) ( )tδ (3) ( )te u t−−
(4) ( )t−δ (5) ( ) ( )tt e u t−+δ (6) ( ) ( )tt e u t−−δ
(7) ( )17 ( ) ( )tt e u t−−δ (8) ( )17 ( ) ( )tt e u t−+δ (9) none of above
Solution 24. V2 (s) = −MsI1(s)+ L2sI2 (s) =−8ss +1
+ 9ss +1
= ss +1
= 1− 1s +1
. ANSWER: (6).
25. (Ideal Transformers, Maximum Power Transfer) In the transformer circuit below, the load is RL
and the turns ratio n1n2
= 1a= 101
. Suppose R1 = 400 Ω and R2 = ∞ (open circuit). The value of RL for
maximum power transfer (in Ω) is: (1) 200 (2) 2 (3) 4 (4) 40 (5) 0.4 (6) 40,000 (7) 400 (8) 20 (9) none of above
EE-202, Final Sp 13 page 14
Solution 25. Z1 =RLa2
= 100RL = 400 Ω implies that RL = 4 Ω. ANSWER: (3).
26. (Ideal Transformers, Maximum Power Transfer) Reconsider problem 25 with vin (t) = 40e
−tu(t) V. The maximum power transfer to the load as a function of time is Pmax,load (t)= (in watts):
(1) 2 ( )te u t− (2) 20.05 ( )te u t− (3) 0.05 ( )te u t− (4) 0.5 ( )te u t−
(6) ( )te u t− (7) 22 ( )te u t− (8) 24 ( )te u t− (9) none of above
Solution 26. Z1 = 400 Ω. v1(t) =40e−tu(t)
2= 20e−tu(t) V. v2 (t) = 2e
−tu(t) V. Hence
Pmax,load (t) =4e−2tu(t)
4= e−2tu(t) watts. ANSWER: (1).
27. (Ideal Transformer Circuits, Impedance Calculation) For the transformer circuit shown below, a = 0.5 , C = 4 F, Zs (s) = 1 Ω. Then Z1(s) = :
(1) 16s (2) 2s
(3) 116s
(4) 14s
(5) 0.5s (6) 1s
(7) 18s
(8) 8s
(9) none of above
EE-202, Final Sp 13 page 15
Solution 27. Z1(s) =1a2
× 1Cs
= 44s
= 1s
. ANSWER: (6).
28. (Ideal Transformer Circuits, Impedance Calculation) Reconsider problem 27. Additionally, suppose that vC (0) = 0 and vin (t) = 10e
−tu(t) V. v2 (t) = (in V):
(1) ( )tte u t− (2) 10 ( )tte u t− (3) 20 ( )tte u t− (4) 5 ( )tte u t−
(5) 20 ( )tte u t−− (6) ( )tte u t−− (7) 5 ( )tte u t−− (8) 10 ( )tte u t−− (9) none of above
Solution 28. By V-division, V1(s) =1s +1
Vin (s) =10
(s +1)2. V2 (s) =
5(s +1)2
and v2 (t) = 5te−tu(t) V.
ANSWER: (4). 29. (Two ports) The z-parameter, z22 , of the two port below is: (1) 0.25 (2) 0.75 (3) 1 (4) 1.25 (5) 1.5 (6) 1.75 (7) –0.25 (8) 2 (9) none of above
Solution 29.
EE-202, Final Sp 13 page 16
V1V20
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥=
2 0 −10 1 1−1 1 4
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
I1I2I3
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
. By Matrix-Partitioning,
V1V2
⎡
⎣⎢⎢
⎤
⎦⎥⎥= 2 0
0 1⎡
⎣⎢
⎤
⎦⎥ +
1−1
⎡
⎣⎢
⎤
⎦⎥14 −1 1⎡⎣ ⎤⎦
⎛
⎝⎜⎞
⎠⎟I1I2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
Thus [zij ]=1.75 0.250.25 0.75
⎡
⎣⎢
⎤
⎦⎥ . ANSWER: (2).
30. (Two Ports) In the cascade of 2-ports below, R = 0.5 Ω and tij⎡⎣ ⎤⎦ =1 10 1
⎡
⎣⎢
⎤
⎦⎥ are the t-parameters
of the 2-port N1. The t-parameter, t22 , of the cascade is: (1) 1 (2) 2 (3) 3 (4) 4 (5) 0 (6) –1 (7) –3 (8) –2 (9) none of above
Solution 30. tij,overall⎡⎣ ⎤⎦ =1 02 1
⎡
⎣⎢
⎤
⎦⎥
1 10 1
⎡
⎣⎢
⎤
⎦⎥ =
1 12 3
⎡
⎣⎢
⎤
⎦⎥ . ANSWER: (3).
31. (Two Ports) Consider the following terminated 2-port having general h-parameters
V1I2
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
h11 h12h21 h22
⎡
⎣⎢⎢
⎤
⎦⎥⎥
I1V2
⎡
⎣⎢⎢
⎤
⎦⎥⎥
EE-202, Final Sp 13 page 17
Of course Admittance = 1Impedance
. The formula for the gain GV2 =V2V1
is:
(1) 11
12
inZ hh− (2) 11
21
1 inh Yh
− (3) 11
12
in
in
Z hZ h− (4) 11
12
hh
(5) 22
21
loadh Yh
(6) 11
12
1 inh Yh
− (7) 21
22( )in L
hZ h Y
−+
(8) 21 22 loadh h Y+
(9) none of above
Solution 31. I2 = h21I1 + h22V2 and I2 = −YLV2 . Thus −(h22 +YL )V2 = h21I1 =h21V1Zin
.
GV 2 =V2V1
= − h21Zin (h22 +YL )
. Similarly, V1 = h11I1 + h12V2 =h11V1Zin
+ h12V2 . Hence, 1− h11Zin
⎛⎝⎜
⎞⎠⎟V1 = h12V2 . A
second gain formula then is: GV 2 =1− h11
Zin
⎛⎝⎜
⎞⎠⎟
h12= Zin − h11
h12.
ANSWER: (1) and/or (7).
32. (Two Ports) Reconsider problem 31. Suppose YL (s) = s and h11 h12h21 h22
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
2s
2
0.5 s
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
. Then
Zin (s) = (in Ω):
(1) 12s
(2) 32s
(3) 52s
(4) 23s (5) 2
5s
(6) 2s (7) 1s
(8) s (9) none of above
Solution 32. Zin (s) = h11 −h12h21h22 +YL
= 2s− 12s
= 32s
. ANSWER: (2).
33. (Series interconnection of two ports) Suppose the unspecified y-parameter matrix in the
interconnected 2-port below is Y1 =3 21 1
⎡
⎣⎢
⎤
⎦⎥ . The resistances are: R1 = 1 Ω, R2 = 2 Ω, and R3 = 3 Ω.
The overall z-parameter matrix is:
(1) 6 44 10⎡ ⎤⎢ ⎥⎣ ⎦
(2) 4 43 8⎡ ⎤⎢ ⎥⎣ ⎦
(3) 6 42 2⎡ ⎤⎢ ⎥⎣ ⎦
(4) 6 33 6⎡ ⎤⎢ ⎥⎣ ⎦
EE-202, Final Sp 13 page 18
(5) 2 42 6
−⎡ ⎤⎢ ⎥−⎣ ⎦
(6) 4 01 8⎡ ⎤⎢ ⎥⎣ ⎦
(7) 0 00 0⎡ ⎤⎢ ⎥⎣ ⎦
(8) Cannot be determined
Solution 33. Z1 = Y1−1 = 1 −2
−1 3⎡
⎣⎢
⎤
⎦⎥ . Z2 =
3 22 5
⎡
⎣⎢
⎤
⎦⎥ . Zeq = Z1 + Z2 =
4 01 8
⎡
⎣⎢
⎤
⎦⎥ . ANSWER: (6).
34. (Series interconnection of two ports) For the interconnected 2-port below, Y1 =1 −2−1 3
⎡
⎣⎢
⎤
⎦⎥ and
Z2 =3 12 1
⎡
⎣⎢
⎤
⎦⎥ . Assume the standard 2-port labeling. The z-parameters of the interconnected 2-port are
(1) 6 24 2⎡ ⎤⎢ ⎥⎣ ⎦
(2) 4 11 4
−⎡ ⎤⎢ ⎥⎣ ⎦
(3) 6 33 2⎡ ⎤⎢ ⎥⎣ ⎦
(4) 2 42 6
−⎡ ⎤⎢ ⎥−⎣ ⎦
(5) 0 00 0⎡ ⎤⎢ ⎥⎣ ⎦
(6) 2 33 2⎡ ⎤⎢ ⎥−⎣ ⎦
(7) Cannot be determined
ANSWER: (7).
EE-202, Final Sp 13 page 19