This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Discrete Fourier Transform”.

1. Given that S_{1} and S_{2} are two discrete time systems. Consider the following statements:

i) If S_{1} and S_{2} are linear, then S is linear

ii) If S_{1} and S_{2} are non-linear, then S is non-linear

iii) If S_{1} and S_{2} are causal, then S is causal

iv) If S_{1} and S_{2} are time invariant, then S is time invariant

The true statements from the above are ____________

a) i, ii, iii

b) ii, iii, iv

c) I, iii, iv

d) I, ii, iii, iv

View Answer

Explanation: Only statement ii is false.

For example, S

_{1}: y[n] = x[n] +b and S

_{2}: y[n] = x[n]-b, where, b≠0.

S{x[n]} = S2 {S

_{1}{x[n]}} = S

_{2}{x[n] +b} = x[n]

Hence, S is linear.

2. For two discrete time systems, consider the following statements:

i) If S_{1} and S_{2} are linear and time invariant, then interchanging their order does not change the system.

ii) If S_{1} and S_{2} are linear and time variant, then interchanging their order does not change the system

The correct statement from the above is __________

a) Both i & ii

b) Only i

c) Only ii

d) Neither i, nor ii

View Answer

Explanation: S

_{1}: y[n] = n x[n]

And S

_{2}: y[n] = n x [n+1]

If x[n] = δ[n], then S

_{2}{S1 {δ[n]}} = S

_{2}[0] = 0

S

_{1}{S

_{2}{δ[n]}} = S

_{1}{δ[n+1]} = – δ[n+1]≠0.

3. The following input-output pairs have been observed during the operation of a time invariant system

i) x_{1}[n] = {1, 0, 2} (Laplace transform) y_{1} [n] = {0, 1,2}

ii) x_{2}[n] = {0,0, 3} (Laplace transform) y_{2}[n] = {0,1,0,2}

iii) x_{3}[n] = {0,0,0,1} (Laplace transform) y_{3}[n] = {1,2,1}

The conclusion regarding the linearity of the system is _____________

a) Linear

b) Non-linear

c) One more observation is required

d) Conclusion cannot be drawn from observation

View Answer

Explanation: System is not linear. This is evident from the observation of the pairs, x

_{3}[n] – y

_{3}[n] and x

_{2}[n] and y

_{2}[n]. If the system were linear y

_{2}[n] would be of the form y

_{2}[n] = {3, 6, 3}.

4. S_{1} and S_{2} are two DT systems which are connected together to form a new system. Consider the following statements:

i) If S_{1} and S_{2} are non-causal, then S is non-causal

ii) If S_{1} and/or S_{2} are unstable, then S is unstable

The correct statement from the above is ____________

a) Both i and ii

b) Only i

c) Only ii

d) Neither i nor ii

View Answer

Explanation: S

_{1}: y[n] = x [n+1] …… (Non-causal)

S

_{2}: y[n] = x [n-2] ……… (Causal)

S: y[n] = x [n-1] which is causal ………. (False)

S

_{1}: y[n] = e

^{x[n]}stable, S

_{2}: y[n] = ln(x[n]) ……… (Unstable)

But S: y[n] = x[n] ………. (Stable, false)

5. Given a signal x[n] = δ[n] + 0.9 δ [n − 6]. The Discrete Time Fourier Transform for 8 points is __________

a) 1 – 0.9 \(e^{-j \frac{2π}{8} k6}\)

b) 1 + 0.9 \(e^{-j \frac{2π}{8} k6}\)

c) 1 + 0.9 \(e^{j \frac{2π}{8} k6}\)

d) 1 – 0.9 \(e^{j \frac{2π}{8} k6}\)

View Answer

Explanation: Given N = 8.

Now, x[k] = \(∑_0^{N-1} x[n] e^{-j \frac{2π}{N} kn}\)

= \(∑_0^7 x[n] e^{-j \frac{2π}{8} kn}\)

= \(∑_{N=0}^7 (δ[n] + 0.9 δ [n – 6])e^{-j \frac{2π}{8} kn}\)

= 1 + 0.9 \(e^{-j \frac{2π}{8} kn}\)

Here, n = 6, from given question.

Hence, x[k] = 1 + 0.9 \(e^{-j \frac{2π}{8} k6}\).

6. The Z transform of δ (n − m) is ___________

a) z^{-n}

b) z^{-m}

c) \(\frac{1}{z-n}\)

d) \(\frac{1}{z-m}\)

View Answer

Explanation: δ (n − m) is a delayed impulse function which is delayed by m units. We know that the Z-transform of a delayed function f (n-m) is (z-m) times the Z-transform of the function f (n). So, the Z transform of δ (n − m) is z

^{-m}.

7. A 10 V is connected across a load whose V-I characteristics is given by 7I = V^{2} + 2V. The internal resistance of the battery is of magnitude 1Ω. The current delivered by the battery is ____________

a) 6 A

b) 5 A

c) 7 A

d) 8 A

View Answer

Explanation: 7I = V

^{2}+ 2V …………………. (1)

Now, V = 10 – 1 × I

Putting the value of V in eqt (1), we get,

&I = (10 – I) 2 + 2(10 – I) …………………. (2)

Or, I = 100 + I2 – 20I + 20 – 2I

Or, I2 – 29I + 120 = 0

∴ \(I = \frac{+29 ± \sqrt{29^2 – 4(120)}}{2} = \frac{29 ± 19}{2}\)

I = 5 A, 24 A

Now, I = 24 A is not possible because V will be negative from eqt (2)

∴ I = 5 A.

8. The period of the signal x(t) = 10 sin 12 π t + 4 cos18 π t is ____________

a) \(\frac{π}{4}\)

b) \(\frac{1}{6}\)

c) \(\frac{1}{9}\)

d) \(\frac{1}{3}\)

View Answer

Explanation: There are two waveforms of frequencies 6 and 9, respectively. Hence the combined frequency is the highest common factor between 6 and 9 which is 3. Therefore the period is \(\frac{1}{3}\).

9. Given a series RLC circuit with V = 5V, R = 200 kΩ, C = 10µF. Sampling frequency of the circuit is 10 Hz. The expression and the ROC of the z-transform of the sampled signal are ____________

a) \(\frac{5z}{z-e^{-5′}}\), |z|<e^{-5}

b) \(\frac{5z}{z-e^{-0.05′}}\), |z|<e^{-0.05}

c) \(\frac{5z}{z-e^{-0.05′}}\), |z|>e^{-0.05}

d) \(\frac{5z}{z-e^{-5′}}\), |z|>e^{-5}

View Answer

Explanation: I (t) = \(\frac{V}{R} e^{-t/RC}\)

Voltage across resistor = R I (t)

= V e

^{-t/RC}= 5 e

^{-t/RC}

= 5 \(e{\frac{-t}{200 × 10 × 10^{-6} × 10^3}}\) = 5 e

^{-t/2}

Given that, the Sampling frequency of the circuit = 10 Hz

Hence, x (n) = 5e

^{-n/2 X 10}= 5e

^{-0.05n}

Now, X (z) = \(∑_{n=-∞}^∞ x[n]z^{-n}\)

= 5 \(∑_{n=-∞}^∞ (e^{-0.05} z^{-n})^n\)

= 5. \(\frac{1}{1-e^{-0.05} Z^{-1′}}\), ROC |z|>e

^{-0.05}

= \(\frac{5z}{z-e^{-0.05′}}\), |z|>e

^{-0.05}.

10. Given a series RLC circuit with V = 5V, R = 200 kΩ, C = 10µF. Sampling frequency of the circuit is 10 Hz. The samples x (n), where n=0,1,2,…., is ___________

a) 5(1-e^{-0.05n})

b) 5e^{-0.05n}

c) 5(1-e^{-5n})

d) 5e^{-5n}

View Answer

Explanation: The charging current in circuit I (t) = I (0

^{+}) e

^{-t/RC}

Since the capacitor acts as short circuit, I (0

^{+}) = \(\frac{V}{R}\)

∴ I (t) = \(\frac{V}{R}\) e

^{-t/RC}

Voltage across resistor = R I (t)

= V e

^{-t/RC}= 5 e

^{-t/RC}

= 5 \(e{\frac{-t}{200 × 10 × 10^{-6} × 10^3}}\) = 5 e

^{-t/2}

Given that, the Sampling frequency of the circuit is 10 Hz

∴ x (n) = 5e

^{-n/2 X 10}= 5e

^{-0.05n}.

11. For the circuit given below, if the frequency of the source is 50 Hz, then a value of to which results in a transient free response is _________________

a) 0

b) 1.78 ms

c) 7.23 ms

d) 9.21 ms

View Answer

Explanation: T = \(\frac{L}{R} \)

Or, T = \(\frac{0.01}{5}\) = 0.002 s = 2 ms

For the ideal case, transient response will die out with time constant.

Practically, T will be less than 2 ms.

12. If G(f) represents the Fourier Transform of a signal g (t) which is real and odd symmetric in time, then G (f) is ____________

a) Complex

b) Imaginary

c) Real

d) Real and non- negative

View Answer

Explanation: Fourier transform of g (t) is G (f)

Given that, g (t) is real, odd and symmetric with respect to time.

∴G*(jm) = – G(jm); G(jm) purely imaginary.

13. If R_{1} is the region of convergence of x (n) and R_{2} is the region of convergence of y(n), then the region of convergence of x (n) convoluted y (n) is ___________

a) R_{1} + R_{2}

b) R_{1} – R_{2}

c) R_{1} ∩ R_{2}

d) R_{1} ∪ R_{2}

View Answer

Explanation: The z-transform of x (n) = X (z). Let the region of convergence be R

_{1}

The z-transform of y (n) = y (z). Let the region of convergence be R

_{2}

The z-transform of x (n) * y (n) is X (z).Y (z) [from property]

So, the region of convergence is R

_{1}∩ R

_{2}.

14. The system under consideration is an RC low-pass filter with R = 1 kΩ and C = 1 µF. Let H (f) denotes the frequency response of the RC, low-pass filter. Let f1 be the highest frequency, such that 0≤|f|≤f_{1}, \(\frac{|H(f1)|}{H(0)}\)≥0.95 Then f_{1} is ___________

a) 327.8

b) 163.9

c) 52.2

d) 104.4

View Answer

Explanation: H (ω) = \(\frac{\frac{1}{jωC}}{R+(\frac{1}{jωC})} = \frac{1}{1+jωRC}\)

H (f) = \(\frac{1}{1+j2πfRC}\)

|H (f)| = \(\frac{1}{\sqrt{1+4π^2 f_1^2 R^2 C^2}}\)

H (0) = 1

Given that \(\frac{|H(f1)|}{H(0)}\)≥0.95

Or, 1 + 4π

^{2}f

_{1}

^{2}R

^{2}C

^{2}≤ 1.108

Simplifying, f

_{1}≤ \(\frac{0.329}{2πRC}\)

∴f

_{1}≤ 52.2 Hz.

15. The response of the LTI system for \(\frac{d^2 y(t)}{dt^2} + \frac{dy(t)}{dt} + 5y(t) = \frac{dx(t)}{dt}\). Given that y(0^{–}) = 2, \(\frac{dx(t)}{dt}\) (at t=0) = 0, x(t) = u(t) is __________

a) 2e^{-t} cos t u(t)

b) 0.5 e^{-t} sin t u(t)

c) 2e^{-t} cos t u(t) + 0.5 e^{-t} sin t u(t)

d) 0.5 e^{-t} cos t u(t-1) + 2e^{-t} sin t u(t-1)

View Answer

Explanation: s

^{2}Y(s) – 2s + 2sY(s) – 2 + 5Y(s) = 1

∴ (s

^{2}+2s+5) Y(s) = 3+2s

Or, Y(s) = \(\frac{2s+3}{s^2+2s+5}\)

= \(\frac{2(s+1)}{(s+1)^2 + 2^2} + \frac{1}{(s+1)^2 + 2^2}\)

Hence, y (t) = 2e

^{-t}cos t u(t) + 0.5 e

^{-t}sin t u(t).

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