Electronics and Communication Engineering - Exam Questions Papers

31.
If a signal f(t) has energy E, the energy of the signal f(2t) is equal to __________ .
E
E/2
E/4
2E
Answer: Option
Explanation:

Energy content of a signal x(t), E = |f(t)|2 dt

Now, E' = |f(2t)|2 dz for signal f(2t)

Putting 2t = z, we get

E' = |f(t)|2 dz = .


32.
For a npn BJT transistor fβ is 1.64 x 108 Hz. Cμ = 10-14 F; Cp = 4 x 10-13 F and DC current gain is 90. Find fT and gm (fβ = cut off frequency, Cμ = capacitance, Cp = parasitic capacitance, gm = transconductance, fT = gain BW product)
fT = 1.47 x 1010 Hz ; gm = 38 milli mho
fT = 1.64 x 108 Hz ; gm = 30 milli mho
fT = 1.47 x 109 Hz ; gm = 38 mho
fT = 1.33 x 1012 Hz; gm = 0.37 m-mho
Answer: Option
Explanation:

fT = 90 x 1.64 x 108 = 1.47 x 1010 Hz

gm = 2pfT(Cμ + Cp) = 2 x p x 1.47 x 1010

= (10-14 + 4 x 10-13)

gm= 38 mμ.


33.
The impulse response h(t) of a linear time-invariant continuous time system is described by h(t) = exp (at) u(t) + exp (βt) u (- t), where u(t) denotes the unit step function, and a and β are constants. This system is stable if
a is positive and β is positive
a is negative and β is negative
a is positive and β is negative
a is negative and β is positive
Answer: Option
Explanation:

h(t) = e+atu(t) + eβtu(- t)

For h(t) to be stableh(t) dt < ∞

It will happen when a is negative and β is positive.


34.
The probability density function of a random variable x is as shown.

The value of A is:
Answer: Option
Explanation:

= 1

= 1

Solve this to get value of A = 1/5.


35.

ωL =
108 rad/sec
104 rad/sec
102 rad/sec
10 rad/sec
Answer: Option
Explanation:

R1 = (10 + 10) = 20 kΩ

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