All India Electronics and Communication Engineering (ECE) Group

Which of the component performs integration operation?
  • a)
    Resistor
  • b)
    Diode
  • c)
    Capacitor
  • d)
    Inductor
Correct answer is option 'C'. Can you explain this answer?

Capacitor performs integration. V (t) developed across capacitor is given by
v (t) = (1/C)* ∫t-∞ i (∂).d∂, I (t) is the current flowing through a capacitor of capacitance C.

AM radio signal is an example for __________
  • a)
    y (t) = a x (t)
  • b)
    y (t) = x1 (t) + x2 (t)
  • c)
    y (t) = x1 (t) * x2 (t)
  • d)
    y (t) = -x(t)
Correct answer is option 'C'. Can you explain this answer?

Telecom Tuners answered
AM radio signal is an example for y (t) = x1 (t) * x2 (t) where, x1 (t) consists of an audio signal plus a dc component and x2 (t) is a sinusoidal signal called carrier wave.

The Nyquist frequency for the signal x (t) = 3 cos 50πt + 10 sin 300πt – cos 100t is ___________
  • a)
    50 Hz
  • b)
    100 Hz
  • c)
    200 Hz
  • d)
    300 Hz
Correct answer is option 'D'. Can you explain this answer?

Telecom Tuners answered
We know that Nyquist frequency is twice the maximum frequency, i.e. fs = 2 fm.
The maximum frequency present in the signal is ωm = 300 π or fm = 150 Hz. Therefore the Nyquist frequency fs = 2 fm = 300 Hz.

The time system which operates with a continuous time signal and produces a continuous time output signal is _________
  • a)
    CTF system
  • b)
    DTF System
  • c)
    Time invariant System
  • d)
    Time variant System
Correct answer is option 'A'. Can you explain this answer?

Telecom Tuners answered
DTF System operates with a discrete signal, on the other hand time invariant system is a system whose output does not depend explicitly on time. For continuous time system, the inputs as well as output both are CT signals.

The system described by the difference equation y(n) – 2y(n-1) + y(n-2) = X(n) – X(n-1) has y(n) = 0 and n<0. If x (n) = δ(n), then y (z) will be?
  • a)
    2
  • b)
    1
  • c)
    0
  • d)
    -1
Correct answer is option 'C'. Can you explain this answer?

Telecom Tuners answered
Given equation = y (n) – 2y (n-1) + y (n-2) = X (n) – X (n-1) has y (n) = 0
For n = 0, y (0)2y (-1) + y (-2) = x (0) – x (-1)
∴ y(0) = x(0) – x(-1)
Or, y (n) = 0 for n<0
For n=1, y (1) = -2y (0) + y (-1) = x (1) – x (0)
Or, y (1) = x (1) – x (0) + 2x (0) – 2x (-1)
Or, y (1) = x (1) +x (0) – 2x (-1)
For n=2, y (2) = x (2) – x (1) + 2y (1) – y (0)
Or, y(2) = x(2) – x(1) + 2x(1) + 2x(0) – 4x(-1) – x(0) + x(-1)
∴y (2) = d (2) + d (1) + d (0) – 3d (-1).

A discrete time signal is as given below X [n] = cos (n/8) cos (πn/8)
The period of the signal X [n] is _____________
  • a)
    16 π
  • b)
    16(π+1)
  • c)
    8
  • d)
    Non-periodic
Correct answer is option 'D'. Can you explain this answer?

Telecom Tuners answered
We know that for X [n] = X1 [n] × X2 [n] to be periodic, both X1 [n] and X2 [n] should be periodic with finite periods.
Here X2 [n] = cos (πn/8), is periodic with fundamental period as 8/n
But X1 [n] = cos (n/8) is non periodic.
∴ X [n] is a non-periodic signal.

F(t) and G(t) are the one-sided z-transforms of discrete time functions f(nt) and g(nt), the z-transform of ∑f(kt)g(nt-kt) is given by _____________
  • a)
    ∑f(nt)g(nt)z-n
  • b)
    ∑f(nt)g(nt)zn
  • c)
    ∑f(kt)g(nt-kt) z-n
  • d)
    ∑f(nt-kt)g(nt)z-n
Correct answer is option 'A'. Can you explain this answer?

Telecom Tuners answered
Given that F (t) and G (t) are the one-sided z-transforms.
Also, f (nt) and g (nt) are discrete time functions, which means that property of Linearity, time shifting and time scaling will be similar to that of continuous Fourier transform. Since, for a continuous Fourier transform, the value of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z-n.
∴ z-transform of ∑f(kt)g(nt-kt) is given by∑f(nt)g(nt)z-n.

A Discrete signal is said to be even or symmetric if X(-n) is equal to __________
  • a)
    X(n)
  • b)
    0
  • c)
    –X(n)
  • d)
    –X(-n)
Correct answer is option 'A'. Can you explain this answer?

Telecom Tuners answered
We know that any signal be it discrete or continuous is said to be even or symmetric when that signal f(x) = f (-x). Here given signal is X (n). It is a discrete time signal. So, the signal will be even symmetric if X (n) = X (-n).

or the system, y (t) = x (t - 5) – x (3 - t) which of the following holds true?
  • a)
    System is Linear, time-invariant, causal and stable
  • b)
    System is time-invariant, causal and stable
  • c)
    System is Linear, time-invariant and stable
  • d)
    System is Linear, time-invariant and causal
Correct answer is option 'C'. Can you explain this answer?

Telecom Tuners answered
y1 (t) = v (t - 5) – v (3 - t)
y2 (t) = k v (t - 5) – k v (3 - t) = ky1 (t)
Let x1 (t) = v (t), then y1 (t) = v (t - 5) – v (3 - t)
Let x2 (t) = 2w (t), then y2 (t) = w (t - 5)-w (3 - t)
Let x3 (t) = x (t) + w (t)
Then, y3 (t) = y1 (t) + y2 (t)
Hence it is linear.
Again, y1 (t) = v (t - 5) – v (3 - t)
∴ y2 (t) = y1 (t-t0)
Hence, system is time-invariant
If x (t) is bounded, then, x (t-5) and x (3-t) are also bounded, so stable system.
At t = 0, y (0) = x (-5)-x (3), therefore, the response at t=0 depends on the excitation at a later time t = 3.
Therefore Non-Causal.

Consider an LTI system is such that
h1(n) = δ(n) – 5 δ (n – 1)
h2(n) = 5n u(n)
Find the output of signal at n = 3 for applied input is 
  • a)
    0.111
  • b)
    0.333
  • c)
    0.037
  • d)
    0.537
Correct answer is option 'C'. Can you explain this answer?

Telecom Tuners answered
For an LTI system the output is given by
y(n) = x(n) ⨂ [h1(n) ⨂ h2(n)]
given
h1(n) = δ(n)-5δ(n­-1)
h2(n) = 5n u(n)
h1(n) ⨂ h2(n) = [δ(n) - 5δ(n-1) ⨂ 5nu(n)
= 5nu(n) – 5 (5n-1 u(n-1))
= 5nu(n) – 5nu(n-1)
= 5n [u(n) – u(n-1)]
A = -5
B = 9
Poles of the H(z) is at |z| = 2, |z| = 3. So, these are 3 possible ROC.
Case (1) 2 < |z|<3
Case (2) |z| ≤ 2
Case (3) |z| ≥ 3
Only in case (2) signal is anti-causal and stable for stable ROC must cuts the unity circle so for ROC |z| ≤ 2
h(n)=-5(-2nu(-n-1) + 9(-3nu(-n-1)
h(n)=5×2nu(-n-1) – 9 × 3nu(-n-1)
Note: u(n) – u(n-1) = δ(n)
so h1(n) ⨂ h2(n) = 5nδ(n) = δ(n)
y(n) = x(n) ⨂ [h1(n) ⨂ h2(n)]
= (1/3)nu(n) ⨂ [δ(n)]

Given f(t) and g(t) as shown below
g(t) can be expressed as
  • a)
    g(t) = f (2t - 3)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Telecom Tuners answered
Concept:
Time-shifting property: When a signal is shifted in time domain it is said to be delayed or advanced based on whether the signal is shifted to the right or left. 
For example:
Time scaling property: A signal is scaled in the time domain with the scaling factor 'a'. 
If a > 1, then the signal is contracted by a factor of 'a' along the time axis. 
if a < 1, then the signal is expanded by a factor of 'a' along the time axis. 
For example:
Analysis:
We can observe that if we scale f(t) by a factor of ½ and then shift, we will get g(t).
First scale f(t) by a factor of ½ ,
g1(t) = f (t/2)
Shift g1(t) by 3
 

For a signal u(sint); Fourier series is having
  • a)
    Odd harmonics of sine only.
  • b)
    Zero DC.
  • c)
    DC and odd Harmonics of sine.
  • d)
    All Harmonics of sine.
Correct answer is option 'C'. Can you explain this answer?

Telecom Tuners answered
Given x(t) = u(sint)
Drawing waveform

We know for waveform like y(t) shown
DC component is zero and have sine odd harmonics.
Performing upward shift in y(t) by adding dc value of 0.5, then
x(t) = y(t) + 0.5
So, x(t) → DC + sine odd harmonics

The Fourier series coefficient of a periodic signal, x(t) is defined as

Which of the following statement is correct.
  • a)
    x(t) is real
  • b)
    dx/dt is even 
  • c)
    x(t) is even
  • d)
    All of the above
Correct answer is option 'C'. Can you explain this answer?

Telecom Tuners answered
CASE 1: Let us assume that x(t) is real, then x(t) = x*(t). This implies that for x(t) to be real, ak = a*-k. Since this is not true. So, x(t) is not real.
CASE 2: Let us assume that x(t) is even, then x(t) = x(-t) and ak = a–k. Since this is true for this case, hence x(t) is even.
CASE 3: Let us assume

Therefore,

Since bk is not even, So, g(t) is not even.
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