5 two port networks

8/18/2019 5 Two Port Networks

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Network Theory

Two-Port NetworksECE - 20061. In the two port network shown in the figure below, Z 12 and Z 21are, respectively

(A) r and βr (B) 0 and βr

(C) 0 and βr (D) r and βr

2. A two-port network is represented by ABCD parameters given by

VI A BC DVI If port-2 is terminated by R L, then input impedance seen at port-1 is given by

(A) ++ (B) + +

(C) ++ (D) ++

ECE - 2008

Statement for linked Answer Questions 3 and 4A two port network shown below is excited by external dc sources. The voltages andcurrents are measured with voltmeters V,V and ammeters A,A (All assumed to beideal) as indicated. Under following switch conditions, the readings obtained are:(i) S –open , S - closed A 0A,V 4.5V,V1.5V,A1A (ii) S –closed , S – open A 4A,V 6V,V6V,A0A

3. The z-parameter matrix for this network is

(A) 1.5 1.54.5 1.5 (B) 1.5 4.51.5 4.5

(C) 1.5 4.51.5 1.5 (D) 4.5 1.51.5 4.5

4.

The h-parameter matrix for this network is(A) 3 31 0.67 (C) 3 31 0.67

A A V V

1 2

2′ 1′ Two port network

S S

6V 1.5V

r βI r

I I


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Network Theory

(B) 3 13 0.67 (D) 3 13 0.67 5. The driving point impedance of the following network is given by

Z(s) .+ . +. The component values are

(A) L = 5H, R = 0.5Ω, C = 0.1F (B) L = 0.1H, R = 0.5Ω, C = 5F

(C) L = 0.1H, R = 2 Ω, C = 0.1F (D) L = 0.1H, R = 2Ω, C = 5F

ECE - 20106. For the two-port network shown below, the short circuit admittance parameter matrix is

(A) 4 22 4S(B)

1 0.50.5 1S(C) 1 0.50.5 1S(D)

4 22 4SECE - 20117. In the circuit shown below, the network N is described by the following Y matrix:

Y0.1 S 0.01 S0.01 S 0.1 S.The voltage gain is

(A) 1/90(B) 1/90

(C) 1/99(D) 1/11

ECE/EE/IN - 2012Common Data for questions 8 and 9With 10 V dc connected at port A in the linear nonreciprocal two-port network shown

below, the following were observed:(i) 1Ω connected at port B draws a current of 3 A(ii) 2.5Ω connected at port B draws a current of 2 A

100Ω

25Ω

100 V V

N

V

I I

20.5Ω0.5Ω 0.5Ω

1

2′ 1′

R C L Z s


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Network Theory

8. With 10 V dc connected at port A, the current drawn by 7 Ω. connected at port B is(A) 3/7 A(B) 5/7 A

(C) 1 A(D) 9/7 A

9. For the same network, with 6 V dc connected at port A, 1 Ω connected at port B draws 7/3A. If 8 V dc is connected to port A, the open circuit voltage at port B is(A) 6 V(B) 7 V

(C) 8 V(D) 9 V

ECE - 2014

10. A two-port network has scattering parameters given by S s s s s . If the port-2 of thetwo- port is short circuited, the s parameter for the resultant one-port network isA s ss ss1 s B s ss ss1 s C s ss ss1 s

D s ss ss1 s 11. In the h-parameter model of the 2-port network given in the figure shown, the value of h

(in S) is ______ .

12. Consider the building block called ‘Network N’ shown in the figure.

Let C 100 μF and R 10 kΩ.

Two such blocks are connected in cascade, as shown in the figure.

V

V R C

Network N

3Ω 3Ω

2Ω 2Ω

2Ω

2

2′

1′

1 3Ω

B A


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Network Theory

The transfer functionof the cascaded network is(A) + (B) + +

(C) + (D) +

13. For the two-port network shown in the figure, the impedance (Z) matrix (in Ω) is

(A) 6 2442 9 (B) 9 88 24

(C) 9 66 24 (D) 42 66 60 EE - 2006

1. The parameters of the circuit shown in the figure are R = 1 MΩ, R = 10 Ω,A =

10 V/V.If

V = 1

μ V, then output voltage, input impedance and output impedance

respectively are

(A) 1V,∞, 10 Ω (B) 1 V,0, 10 Ω

(C) 1 V, 0,∞ (D) 10 V,∞,10Ω

2. The parameter type and the matrix representation of the relevant two port parametersthat describe the circuit shown are

(A) z parameters, 0 00 0 (B) h parameters, 1 0

0 1

(C) h parameters, 0 00 0 (D) z parameters, 1 00 1

V I

V I

R R

AV V +

+

_

_

2

2′ 1′ 1

60Ω 10Ω 30Ω

Network N Network N Vs Vs


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Network Theory

EE - 20093. The equivalent capacitance of the input loop of the circuit shown is

(A) 2μ F (B) 100μ F

(C) 200μ F (D) 4μFs

EE - 20104. The two-port network P shown in the figure has ports 1 and 2, denoted by terminals (a, b)

and (c, d), respectively. It has an impedance matrix Z with parameters denoted by z. A 1Ω resistor is connected in series with the network at port 1 as shown in the figure. Theimpedance matrix of the modified two-port network (shown as a dashed box) is

(A) z 1 z1z z 1 (B) z 1 zz z 1 (C) z 1 zz z (D) z 1 z1z 1 z

IN - 2007

1. The DC voltage gain in the following circuit is given by.

(A) AV + (B) AV +

(C) AV + + R (D) AV

IN - 2008

2. For the circuit shown below the input resistance R 11 = = is

V R R

R V

V AV

1 Ω ae

bf

P

d

c

I1

1kΩ100μF100μF

49i 1 Input loop

1kΩ 1kΩ


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Network Theory

(A) 3Ω (B) 2 Ω

(C) 3 Ω(D) 13 Ω

IN - 20133. Considering the transformer to be ideal, the transmission parameter ‘A’ of the

2 port network shown in the figure below is

(A) 1.3(B) 1.4

(C) 0.5(D) 2.0

Answer Keys and Explanations

ECE1. [Ans. B]

Z I 0 ∵current source will be open)Z I 0 =

− . βr 2. [Ans .D]

The ABCD parameter equations are given by,V AVBI I CVDI when the network is terminal by R fig.1,V−I R Z −− AI R BICI R DIA RBC RD

Fig. 1

I I V

V

RL

1 2

5 5

2 22

I1 I2 V2 V1

1′ 2′

1:2

V 2 V3

+1Ω

I1 I2

V1

2Ω

+

+

2Ω

2V3

3I 2 +


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Network Theory

3. [Ans. C]

V = ZI ZI V

=

ZI ZI

case (1)⇒I = 0∴Z = 4.5 Z VI = 1.5case (2)⇒I = 0∴Z = 1.5 Z = = 1.5Z= 1.5 4.51.5 1.5

4. [Ans. A]

V = hI hV I = hI hV From given Z parameters,I I23V V = 1.5I 4.5 IV = 3I 3V ∴ H = 3 31 0.67 5. [Ans. D]ZS R|| SL||1SC RSL.1SC⁄(RSL R.1SC⁄ LC⁄ )

RSLSC⁄SRCL R SLSC⁄

⇒ZS SRLSRCL SL R

S.1C⁄(SSRC⁄ 1LC⁄ )……… 1 0.2SS 0.1s 2…………… 2

Comparing (1) and (2)

1C⁄ 0.2⇒C 5F 1RC⁄ 0.1⇒R 2Ω 1LC⁄ 2⇒L 0.1 H 6. [Ans. A]

Y = . 4Y = . 4


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Network Theory

Y = 2 Y

∴y4 22 4S

7. [Ans. D]

Given Y0.1S 0.01S0.01S 0.1S ∴I 0.1 V0.01 V…..1 I 0.01 V0.1 V…….. 2 From the circuit shown in Fig. 1

I V100 From eqn (2)0.01 V0.1 V0.01 V 0.11 V0.01 V VV

0.010.11

111

Note: is Independent of I 8. [Ans. C]

As per the given conditions, we can draw the following two figures.

Let Vth and R th be the Thevenin voltage & resistance as seen from part B.

N ± 2.5 Ω 10 V AB

2A

N

± 1Ω 10 A B3A

V

V

N

25Ω I I

100 Ω

100 V

Fig.1


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Network Theory

Vth = 3Rth + 3 (1)&

Vth = 2Rth + 5 (2)Solving (1) & (2)Rth = 2Ω So, Vth = 3 x 2 + 3 = 9VNow,

i = +

9. [Ans. B]

So, Vth = 7/3 x 2 + x = = 7V.

The open circuit voltage at port B is 7V.

10. [Ans. B]

b Sa sa→ ① b sa sa→ ② b

b

a a Port 1 Port 2 S parametersLoad

1Ω R 2Ω

V 7/3A

7Ω 2Ω

9V B.i

1Ω

R

V

3A

2.5Ω R V

2A


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Network Theory

Port 2 is short circuited ⇒a b The new sb/a From ②

a sa sa ⇒a a1 ss ⇒a sa1 s From ① b sa s.sa1 s ⇒s ba s ss1 s ⇒s s ss s.s1 s

11. [Ans. ] Range 1.24 to 1.26In the figure two port networks in parallel

The y parameter of the parallel network is equal to the sum of the individual network yparameterFor network A

y23

y 13 y 13 y 23 For network B

y 1 2Ω

2Ω

2Ω

3Ω 3Ω 3Ω

N/WA

N/WB V→


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Network Theory

y 12

y1

2

y 1 ∴y| 53 y| 56 y| 56 y| 53

∴hΔyy

7536×

35

151254 1.25 12. [Ans. B]

Apply mesh analysis to determine the current Is R1scI s RIs Vs …① 2R1scIs RI s 0 …② ⇒Is 2R Is…③ [from ②] Combining ① and③

R1sc1R 2R1sc R I s Vs⇒2R R RIs RVs Vs Is.R RVs2R R R ⇒VsVs Rsc2Rsc 1Rsc 1Rsc Rsc1 3src sRc asR 10kΩ

C 100μF}RC 1

∴VsVs s1 3s s

Is

R R

Is

Vs Vs

1sc 1sc


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Network Theory

13. [Ans. C]

y 30− 10− 30−30−

60−

30−

0.133 0.0330.033 0.05 z y− 9 66 24 EE1. [Ans. A]

V 10−×10 = 1V Z = →∞ Z = →R = 10

2. [Ans. C]

I hV hI V hV hI

Since port – 1 is open – circuit , I 1=0Port – 2 is short – circuit, V 2 =0

h IV= 0V 0

h II = 0I 0 h VV= 0V 0 h VI = 0I 0 So,h – parametersh hh h 0 00 0

3. [Ans. A]

Assume a 1A current source at input terminals,

∴I = 1A

Applying KVL

V i1 150 ijX0⇒ V i2 j50X Input impedance 2 j50X As imaginary part is negative, input impedance has equivalent capacitive reactance X.

100μF 1kΩ 1kΩ

1kΩ i

~ 100μF 50i V

49 i

V1 V2

I1 I2


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Network Theory

X .50X 1ωC.

50ωC

50ω×10012ω C 2μF 4. [Ans. C]

The impedance matrix Z of the modified network is calculated from fig. given below:VV zII V Z I Z I V Z I Z I V 1×I VI ZI Z I 1 ZI ZI VV z II Z Z 1 ZZ Z Z 1 00 0

IN1. [Ans. A]

V =V+ V = AV A+ 2. [Ans. D]

Apply KVL on abcdef

V I 3I2I I 2V2I I 0 V 5II 4V …① Where V2I I→put it in equation ①

V 5II 8I8I V 13I9I

V3

+1Ω I1 I2 2Ω

+

+

2Ω 2V3

3I 2 +

a

I I

+

I I 2V c

bd

ef

V1 V

I 2V

I 1 Ω I I I

V V V

Fig.


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5 two port networks

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