Digital Signal Processing Questions and Answers Part-16

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

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

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) \(Ae^{j(\frac{nπ}{2}+26.6°)}\)

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)=\(\frac{2}{\sqrt{5}} Ae^{j(\frac{nπ}{2}-26.6°)}\)

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

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) \(\sum_{k=-∞}^∞ h(k) cos⁡ωk\)

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)=-\(\sum_{k=-∞}^∞h(k) sin⁡ωk\)

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

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|

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|

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

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) \(\frac{|b|}{\sqrt{1-2acosω+a^2}}\)

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

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

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).