LTI System Frequency Selective Filter MCQ Quiz – Objective Question with Answer for Frequency Selective Filter

11. A digital resonator is a special two-pole bandpass filter with a pair of complex conjugate poles located near the unit circle.
A. True
B. False

Answer: A

The magnitude response of a bandpass filter with two complex poles located near the unit circle is as shown below.
The filtered gas has a large magnitude response at the poles and hence it is called a digital resonator.

 

12. Which of the following filters have a frequency response as shown below?

A. Bandpass filter
B. Bandstop filter
C. All pass filter
D. Notch filter

Answer: D

The given figure represents the frequency response characteristic of a notch filter with nulls at frequencies at ω0 and ω1.

 

13. A comb filter is a special case of a notch filter in which the nulls occur periodically across the frequency band.

A. True
B. False

Answer: A

A comb filter can be viewed as a notch filter in which the nulls occur periodically across the frequency band, hence the analogy to an ordinary comb that has periodically spaced teeth.

 

14. The filter with the system function H(z)=z-k is a _________

A. Notch filter
B. Bandpass filter
C. All pass filter
D. None of the mentioned

Answer: C

The system with the system function given as H(z)=z-k is a pure delay system. It has a constant gain for all frequencies and is hence called the All pass filter.

 

15. If the system has an impulse response as h(n)=Asin(n+1)ω0u(n), then the system is known as a digital frequency synthesizer.

A. True
B. False

Answer: A

The given impulse response is h(n)=Asin(n+1)ω0u(n).
According to the above equation, the second-order system with complex conjugate poles on the unit circle is a sinusoid and the system is called a digital sinusoidal oscillator or a Digital frequency synthesizer.

 

16. If a system is said to be invertible, then?

A. One-to-one correspondence between its input and output signals
B. One-to-many correspondence between its input and output signals
C. Many-to-one correspondence between its input and output signals
D. None of the mentioned

Answer: A

If we know the output of a system y(n) of a system and if we can determine the input x(n) of the system uniquely, then the system is said to be invertible. That is there should be a one-to-one correspondence between the input and output signals.

 

17. If h(n) is the impulse response of an LTI system T and h1(n) is the impulse response of the inverse system T-1, then which of the following is true?

A. [h(n)*h1(n)].x(n)=x(n)
B. [h(n).h1(n)].x(n)=x(n)
C. [h(n)*h1(n)]*x(n)=x(n)
D. [h(n).h1(n)]*x(n)=x(n)

Answer: C

If h(n) is the impulse response of an LTI system T and h1(n) is the impulse response of the inverse system T-1, then we know that h(n)*h1(n)=δ(n)=>[h(n)*h1(n)]*x(n)=x(n).

 

18. What is the inverse of the system with impulse response h(n)=(1/2)nu(n)?

A. δ(n)+1/2 δ(n-1)
B. δ(n)-1/2 δ(n-1)
C. δ(n)-1/2 δ(n+1)
D. δ(n)+1/2 δ(n+1)

Answer: B

Given impulse response is h(n)=(1/2)nu(n)
The system function corresponding to h(n) is

H(z)=\(\frac{1}{1-\frac{1}{2} z^{-1}}\) ROC:|z|>1/2

This system is both stable and causal. Since H(z) is all pole system, its inverse is FIR and is given by the system function

HI(z)=\(1-\frac{1}{2} z^{-1}\)

Hence its impulse response is δ(n)-1/2 δ(n-1).

 

19. What is the inverse of the system with impulse response h(n)=δ(n)-1/2δ(n-1)?

A. (1/2)nu(n)
B. -(1/2)nu(-n-1)
C. (1/2)nu(n) & -(1/2)nu(-n-1)
D. None of the mentioned

Answer: C

The system function of given system is H(z)=\(1-\frac{1}{2} z^{-1}\)

The inverse of the system has a system function as H(z)=\(\frac{1}{1-\frac{1}{2} z^{-1}}\)

Thus it has a zero at origin and a pole at z=1/2.So, two possible cases are |z|>1/2 and |z|<1/2
So, h(n)= (1/2)nu(n) for causal and stable(|z|>1/2)
and h(n)= -(1/2)nu(-n-1) for anti causal and unstable for |z|<1/2.

 

20. What is the causal inverse of the FIR system with impulse response h(n)=δ(n)-aδ(n-1)?

A. δ(n)-aδ(n-1)
B. δ(n)+aδ(n-1)
C. a-n
D. an

Answer: D

Given h(n)= δ(n)-aδ(n-1)
Since h(0)=1, h(1)=-a and h(n)=0 for n≥a, we have
hI(0)=1/h(0)=1.
and
hI(n)=-ahI(n-1) for n≥1
Consequently, hI(1)=a, hI(2)=a2,….hI(n)=an
Which corresponds to a causal IIR system as expected.

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