Digital Signal Processing Questions and Answers – Frequency Domain Characteristics of LTI System

This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Frequency Domain Characteristics of LTI System”.

1. If x(n)=Aejωn is the input of an LTI system and h(n) is the response of the system, then what is the output y(n) of the system?
a) H(-ω)x(n)
b) -H(ω)x(n)
c) H(ω)x(n)
d) None of the mentioned
View Answer

Answer: c
Explanation: If x(n)= Aejωn is the input and h(n) is the response o the system, then we know that
y(n)=\(\sum_{k=-∞}^∞ h(k)x(n-k)\)
=>y(n)=\(\sum_{k=-∞}^∞ h(k)Ae^{jω(n-k)}\)
= A \([\sum_{k=-∞}^∞ h(k) e^{-jωk}] e^{jωn}\)
= A. H(ω). ejωn
= H(ω)x(n)

2. If the system gives an output y(n)=H(ω)x(n) with x(n) = Aejωnas input signal, then x(n) is said to be Eigen function of the system.
a) True
b) False
View Answer

Answer: a
Explanation: An Eigen function of a system is an input signal that produces an output that differs from the input by a constant multiplicative factor known as Eigen value of the system.

3. What is the output sequence of the system with impulse response h(n)=(1/2)nu(n) when the input of the system is the complex exponential sequence x(n)=Aejnπ/2?
a) \(Ae^{j(\frac{nπ}{2}-26.6°)}\)
b) \(\frac{2}{\sqrt{5}} Ae^{j(\frac{nπ}{2}-26.6°)}\)
c) \(\frac{2}{\sqrt{5}} Ae^{j({nπ}{2}+26.6°)}\)
d) \(Ae^{j(\frac{nπ}{2}+26.6°)}\)
View Answer

Answer: b
Explanation: First we evaluate the Fourier transform of the impulse response of the system h(n)
H(ω)=\(\sum_{n=-∞}^∞ h(n) e^{-jωn} = \frac{1}{1-1/2 e^{-jω}}\)
At ω=π/2, the above equation yields,
H(π/2)=\(\frac{1}{1+j 1/2}=\frac{2}{\sqrt{5}} e^{-j26.6°}\)
We know that if the input signal is a complex exponential signal, then y(n)=x(n) . H(ω)
=>y(n)=\(\frac{2}{\sqrt{5}} Ae^{j(\frac{nπ}{2}-26.6°)}\)
advertisement
advertisement

4. If the Eigen function 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 Eigen value of the system?
a) 3/2
b) -3/2
c) -2/3
d) 2/3
View Answer

Answer: d
Explanation: 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 Eigen function and H(ω) is called the Eigen value of the system. So, the Eigen value of the system mentioned above is 2/3.

5. If h(n) is the real valued impulse response sequence of an LTI system, then what is the imaginary part of 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) \(\sum_{k=-∞}^∞ h(k) cos⁡ωk\)
View Answer

Answer: a
Explanation: 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)=-\(\sum_{k=-∞}^∞h(k) sin⁡ωk\)
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. 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) –\(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)
View Answer

Answer: c
Explanation: 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)=\(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)

7. 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|
View Answer

Answer: b
Explanation: 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|
advertisement

8. 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
View Answer

Answer: a
Explanation: 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-\(\frac{10}{\sqrt{5}} sin(π/2n-26.6^0)+ \frac{40}{3}cosπn\)

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

Answer: d
Explanation: 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)|=\(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)
advertisement

10. 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) 1-a
c) 1+a
d) none of the mentioned
View Answer

Answer: b
Explanation: 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).

11. 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) \(5+0.888sin(\frac{π}{2}n+420)-1.06cos(πn+\frac{π}{4})\)
View Answer

Answer: c
Explanation: 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)=\(5+0.888sin(\frac{π}{2}n-42^0)-1.06cos(πn+\frac{π}{4})\)

12. 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
View Answer

Answer: a
Explanation: 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.

13. 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) \(\frac{1}{(1+\frac{1}{2} e^{-jω})(1+\frac{1}{4} e^{-jω})}\)
View Answer

Answer: b
Explanation: 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)
=\(\frac{1}{(1-\frac{1}{2} e^{-jω})(1-\frac{1}{4} e^{-jω})}\).

14. What is the frequency response of the system described by the system function H(z)=\(\frac{1}{1-0.8z^{-1}}\)?
a) \(\frac{e^{jω}}{e^{jω}-0.8}\)
b) \(\frac{e^{jω}}{e^{jω}+0.8}\)
c) \(\frac{e^{-jω}}{e^{-jω}-0.8}\)
d) None of the mentioned
View Answer

Answer: a
Explanation: Given H(z)=\(\frac{1}{1-0.8z^{-1}}\)=z/(z-0.8)
Clearly, H(z) has a zero at z=0 and a pole at p=0.8. Hence the frequency response of the system is given as
H(ω)=\(\frac{e^{jω}}{e^{jω}-0.8}\).

Sanfoundry Global Education & Learning Series – Digital Signal Processing.

To practice all areas of Digital Signal Processing, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.