Signal and System Frequency Analysis MCQ Quiz – Objective Question with Answer for Signal and System Frequency Analysis

61. If the Eigenfunction of an LTI system is x(n) = Aejnπ and the impulse response of the system is h(n) = (1/2)nu(n), then what is the Eigenvalue of the system?

A. 3/2
B. -3/2
C. -2/3
D. 2/3

Answer: D
First, we evaluate the Fourier transform of the impulse response of the system h(n)

H(ω) = \(\sum_{n = -∞}^∞ h(n) e^{-jωn} = \frac{1}{1-\frac{1}{2} e^{-jω}}\)

At ω = π, the above equation yields,

H(π) = \(\frac{1}{1+\frac{1}{2}}\) = 2/3

If the input signal is a complex exponential signal, then the input is known as the Eigen function and H(ω) is called the Eigenvalue of the system. So, the Eigenvalue of the system mentioned above is 2/3.

62. If h(n) is the real-valued impulse response sequence of an LTI system, then what is the imaginary part of the Fourier transform of the impulse response?

A. –\(\sum_{k = -∞}^∞ h(k) sin⁡ωk\)

B. \(\sum_{k = -∞}^∞ h(k) sin⁡ωk\)

C. –\(\sum_{k = -∞}^∞ h(k) cos⁡ωk\)

D. \(\sum_{k = -∞}^∞ h(k) cos⁡ωk\)

Answer: A

From the definition of H(ω), we have

H(ω) = \(\sum_{k = -∞}^∞h(k) e^{-jωk}\)

= \(\sum_{k = -∞}^∞h(k) cos⁡ωk-j\sum_{k = -∞}^∞h(k) sin⁡ωk\)

= HR(ω)+j HI(ω)

= > HI(ω) = -\(\sum_{k = -∞}^∞h(k) sin⁡ωk\)

63. If h(n) is the real-valued impulse response sequence of an LTI system, then what is the phase of H(ω) in terms of HR(ω) and HI(ω)?

A. \(tan^{-1}\frac{H_R (ω)}{H_I (ω)}\)

B. –\(tan^{-1}\frac{H_R (ω)}{H_I (ω)}\)

C. \(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)

D. –\(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)

Answer: C

If h(n) is the real-valued impulse response sequence of an LTI system, then H(ω) can be represented as HR(ω)+j HI(ω).

= > tanθ = \(\frac{H_I (ω)}{H_R (ω)}\) = > Phase of H(ω) = \(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)

64. What is the magnitude of H(ω) for the three point moving average system whose output is given by
y(n) = \(\frac{1}{3}[x(n+1)+x(n)+x(n-1)]\)?

A. \(\frac{1}{3}|1-2cosω|\)

B. \(\frac{1}{3}|1+2cosω|\)

C. |1-2cosω|

D. |1+2cosω|

Answer: B

For a three point moving average system, we can define the output of the system as

y(n) = \(\frac{1}{3}[x(n+1)+x(n)+x(n-1)] = >h(n) = \{\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\}\)

it follows that H(ω) = \(\frac{1}{3}(e^{jω}+1+e^{-jω}) = \frac{1}{3}(1+2cosω)\)

= >| H(ω)| = \(\frac{1}{3}\)|1+2cosω|

65. What is the response of the system with impulse response
h(n) = (1/2)nu(n) and the input signal x(n) = 10-5sinπn/2+20cosπn?

A. 20-\(\frac{10}{\sqrt{5}} sin(π/2n-26.60)+ \frac{40}{3}cosπn\)

B. 20-\(\frac{10}{\sqrt{5}} sin(π/2n-26.60)+ 40cosπn\)

C. 20-\(\frac{10}{\sqrt{5}} sin(π/2n+26.60)+ \frac{40}{3cosπn}\)

D. None of the mentioned

Answer: A

The frequency response of the system is

H(ω) = \(\sum_{n = -∞}^∞ h(n) e^{-jωn} = \frac{1}{1-\frac{1}{2} e^{-jω}}\)

For first term, ω = 0 = >H(0) = 2

For second term, ω = π/2 = >H(π/2) = \(\frac{1}{1+j\frac{1}{2}} = \frac{2}{\sqrt{5}} e^{-j26.6°}\)

For third term, ω = π = > H(π) = \(\frac{1}{1+\frac{1}{2}}\) = 2/3

Hence the response of the system to x(n) is

y(n) = 20-\(\frac{10}{\sqrt{5}} sin(π/2n-26.6^0)+ \frac{40}{3}cosπn\)

66. What is the magnitude of the frequency response of the system described by the difference equation y(n) = ay(n-1)+bx(n), 0<a<1?

A. \(\frac{|b|}{\sqrt{1+2acosω+a^2}}\)

B. \(\frac{|b|}{1-2acosω+a^2}\)

C. \(\frac{|b|}{1+2acosω+a^2}\)

D. \(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)

Answer: D

Given y(n) = ay(n-1)+bx(n)

= >H(ω) = \(\frac{|b|}{1-ae^{-jω}}\)

By calculating the magnitude of the above equation we get

|H(ω)| = \(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)

67. If an LTI system is described by the difference equation y(n) = ay(n-1)+bx(n), 0 < a < 1, then what is the parameter ‘b’ so that the maximum value of |H(ω)| is unity?

A. a
B.

C. 1+a
D. none of the mentioned

Answer: B

We know that,

|H(ω)| = \(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)
Since the parameter ‘a’ is positive, the denominator of |H(ω)| becomes minimum at ω = 0. So, |H(ω)| attains its maximum value at ω = 0. At this frequency we have,

\(\frac{|b|}{1-a}\) = 1 = > b = ±(1-A..

68. If an LTI system is described by the difference equation y(n) = ay(n-1)+bx(n), 0<a<1, then what is the output of the system when input
x(n) = \(5+12sin\frac{π}{2}n-20cos(πn+\frac{π}{4})\)?(Given a = 0.9 and b = 0.1)

A. \(5+0.888sin(\frac{π}{2}n-420)-1.06cos(πn-\frac{π}{4})\)

B. \(5+0.888sin(\frac{π}{2}n-420)+1.06cos(πn+\frac{π}{4})\)

C. \(5+0.888sin(\frac{π}{2}n-420)-1.06cos(πn+\frac{π}{4})\)

D. \(5+0.888sin(\frac{π}{2}n+420)-1.06cos(πn+\frac{π}{4})\)

Answer: C

From the given difference equation, we obtain

|H(ω)| = \(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)

We get |H(0)| = 1, |H(π/2)| = 0.074 and |H(π)| = 0.053

θ(0) = 0, θ(π/2) = -420 and θ(π) = 0 and we know that y(n) = H(ω)x(n)

= >y(n) = \(5+0.888sin(\frac{π}{2}n-42^0)-1.06cos(πn+\frac{π}{4})\)

69. The output of the Linear time-invariant system cannot contain the frequency components that are not contained in the input signal.

A. True
B. False

Answer: A

If x(n) is the input of an LTI system, then we know that the output of the system y(n) is y(n) = H(ω)x(n) which means the frequency components are just amplified but no new frequency components are added.

70. An LTI system is characterized by its impulse response h(n) = (1/2)nu(n). What is the spectrum of the output signal when the system is excited by the signal x(n) = (1/4)nu(n)?

A. \(\frac{1}{(1-\frac{1}{2} e^{-jω})(1+\frac{1}{4} e^{-jω})}\)

B. \(\frac{1}{(1-\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\)

C. \(\frac{1}{(1+\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\)

D. \(\frac{1}{(1+\frac{1}{2} e^{-jω})(1+\frac{1}{4} e^{-jω})}\)

Answer: B

The frequency response function of the system is

H(ω) = \(\sum_{n = 0}^∞ (\frac{1}{2})^n e^{-jωn}\)
= \(\frac{1}{1-\frac{1}{2} e^{-jω}}\)

Similarly, the input sequence x(n) has a Fourier transform

X(ω) = \(\sum_{n = 0}^∞ (\frac{1}{4})^n e^{-jωn}\)

= \(\frac{1}{1-\frac{1}{4} e^{-jω}}\)

Hence the spectrum of the signal at the output of the system is

Y(ω) = X(ω)H(ω)

= \(\frac{1}{(1-\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\).

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