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Electronics and Communication Engineering - Exam Questions Papers

Exercise : Exam Questions Papers - Exam Paper 8
26.
The electric field of uniform plane wave is given by E = 20 sin (2 p x 108t- pz) x + 20 cos (2 p x 108t - pz) y. The polarization of wave is
Elliptical, left handed
Circular, left handed
Elliptical, right handed
Circular, right handed
Answer: Option
Explanation:

The given two components are equal in amplitude, perpendicular and out of phase by 90°.

∴ The wave is circularly polarized.

Now we consider z = 0 and put t = 0 and

∴ At t = 0, E = 10ay

At t

∴ Left circularly polarized.

27.
A two port network shown below is external dc sources. The voltages and the currents are measured with voltmeters V1 V2 and ammeter A1 A2 (all assumed to be ideal), as indicated. Under following switch conditions, the readings obtained are:
(i) S1 - Open, S2 - Closed A1 = 0A, V1 = 4.5 V, V2 = 1.5 V, A2 = 1 A
(ii) S1 - Closed, S2 - Open A1 = 4A, V1 = 6 V, V2 = 6 V, A2 = 0 A

The z-parameter matrix for this network is
Answer: Option
Explanation:

V1 = Z11I1 + Z12I2

V2 = Z22I + Z22I2

Using the given information; S1-open, S2-closed

4.5 = Z121 ⇒Z12 = 4.5

1.5 = Z221 ⇒ Z22 = 1.5

S1-closed, S2-open

6 = Z111 ⇒ Z11 = 1.5

6 = Z214 ⇒ Z21 = 1.5

Thus Z matrix =

28.
The Z-transform of a particular signal is given as X(Z) = where
This system after practical implementation will be
Non Causal and Stable
Causal and Stable
Non Causal and Unstable
Causal and Unstable
Answer: Option
Explanation:

Let x =

zx + x2 = 1

Substitute in the equation given, we get

Hence we have pole at Z = 0.8 for stability the ROC should be away from Z = 0.8 towards infinity.

29.
The JFET in the circuit shown in figure has an IDSS = 10 mA and Vp = -5 V. The value of the resistance Rs for a drain current IDS = 6.4 mA is (select the nearest value)
150 Ω
150 Ω
560 Ω
1 kΩ
Answer: Option
Explanation:

6.4 x 10-3 = 10 x 10-3

= 0.8

= 0.8 - 1 = - 0.2

VGS = -5 x 0.2 = -1 V

IDSRS = -VGS = - (-1) = 1

= 0.156 x 103 = 156 Ω.

30.
In the circuit s closed for a long time and steady state is reached. s is opened at t = 0+ Determine current through resistor.
A
A
- A
- A
Answer: Option
Explanation:

Steady state with s closed.

C is open Is = 4.

This source current is divided between two resistors Ω and 1 Ω.

Across 1 Ω, C is open, I in 1 Ω is = 1 A.

Hence voltage across C is 1 V. The initial charge on C is V0 = 1 V.

At t = 0+, s is opened bringing R = 1 Ω into the circuit.

C is replaced by 1 V source with positive polarity left.

The circuit has two sources Is = 4 and V0 = 1.

The currents in the resistor R = and R = 1 can be determined by superposition.

For R =

Due to Is only (short V0)

I = Is x= 3

Due to V0 only (open Is)

I == Downwards

Adding, current through R = is 3= .

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