Inverse Systems and Deconvolution Questions and Answers

This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Inverse Systems and Deconvolution”.

1. 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
View Answer

Answer: a
Explanation: 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 one-to-one correspondence between the input and output signals.

2. 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)
View Answer

Answer: c
Explanation: 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).

3. 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)
View Answer

Answer: b
Explanation: 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).

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4. 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
View Answer

Answer: c
Explanation: 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.

5. 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
View Answer

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

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6. If the frequency response of an FIR system is given as H(z)=6+z^-1-z^-2, then the system is ___________
a) Minimum phase
b) Maximum phase
c) Mixed phase
d) None of the mentioned
View Answer

Answer: a
Explanation: Given H(z)=6+z^-1-z^-2
By factoring the system function we find the zeros for the system.
The zeros of the given system are at z=-1/2, 1/3
So, the system is minimum phase.

7. If the frequency response of an FIR system is given as H(z)=1-z^-1-6z^-2, then the system is ___________
a) Minimum phase
b) Maximum phase
c) Mixed phase
d) None of the mentioned
View Answer

Answer: b
Explanation: Given H(z)=1-z^-1-6z^-2
By factoring the system function we find the zeros for the system.
The zeros of the given system are at z=-2,3
So, the system is maximum phase.

8. If the frequency response of an FIR system is given as H(z)=1-5/2z^-1-3/2z^-2, then the system is ___________
a) Minimum phase
b) Maximum phase
c) Mixed phase
d) None of the mentioned
View Answer

Answer: c
Explanation: Given H(z)= 1-5/2z^-1-3/2z^-2
By factoring the system function we find the zeros for the system.
The zeros of the given system are at z=-1/2, 3
So, the system is mixed phase.

9. An IIR system with system function H(z)=\(\frac{B(z)}{A(z)}\) is called a minimum phase if ___________
a) All poles and zeros are inside the unit circle
b) All zeros are outside the unit circle
c) All poles are outside the unit circle
d) All poles and zeros are outside the unit circle
View Answer

Answer: a
Explanation: For an IIR filter whose system function is defined as H(z)=\(\frac{B(z)}{A(z)}\) to be said a minimum phase,
then both the poles and zeros of the system should fall inside the unit circle.

10. An IIR system with system function H(z)=\(\frac{B(z)}{A(z)}\) is called a mixed phase if ___________
a) All poles and zeros are inside the unit circle
b) All zeros are outside the unit circle
c) All poles are outside the unit circle
d) Some, but not all of the zeros are outside the unit circle
View Answer

Answer: d
Explanation: For an IIR filter whose system function is defined as H(z)=\(\frac{B(z)}{A(z)}\) to be said a mixed phase and if the system is stable and causal, then the poles are inside the unit circle and some, but not all of the zeros are outside the unit circle.

11. A causal system produces the output sequence y(n)={1,0.7} when excited by the input sequence x(n)={1,-0.7,0.1}, then what is the impulse response of the system function?
a) [3(0.5)ⁿ+4(0.2)ⁿ]u(n)
b) [4(0.5)ⁿ-3(0.2)ⁿ]u(n)
c) [4(0.5)ⁿ+3(0.2)ⁿ]u(n)
d) None of the mentioned
View Answer

Answer: b
Explanation: The system function is easily determined by taking the z-transforms of x(n) and y(n). Thus we have
H(z)=\(\frac{Y(z)}{X(z)} = \frac{1+0.7z^{-1}}{1-0.7z^{-1}+0.1z^{-2}} = \frac{1+0.7z^{-1}}{(1-0.2z^{-1})(1-0.5z^{-1})}\)
Upon applying partial fractions and applying the inverse z-transform, we get
[4(0.5)ⁿ-3(0.2)ⁿ]u(n).

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