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
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
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
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)=\(\frac{2}{\sqrt{5}} Ae^{j(\frac{nπ}{2}-26.6°)}\)
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
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) –\(\sum_{k=-∞}^∞ h(k) sinωk\)
b) \(\sum_{k=-∞}^∞ h(k) sinωk\)
c) –\(\sum_{k=-∞}^∞ h(k) cosωk\)
d) \(\sum_{k=-∞}^∞ h(k) cosωk\)
View Answer
Explanation: From the definition of H(ω), we have
H(ω)=\(\sum_{k=-∞}^∞h(k) e^{-jωk}\)
=\(\sum_{k=-∞}^∞h(k) cosωk-j\sum_{k=-∞}^∞h(k) sinωk\)
= HR(ω)+j HI(ω)
=> HI(ω)=-\(\sum_{k=-∞}^∞h(k) sinωk\)
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
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
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|
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
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
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)|=\(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)
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
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
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
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
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
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.
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