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