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.

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)

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).

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)

Answer: B

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

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).

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

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.

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

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.

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

Answer: A

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 a 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

Answer: B

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 in a 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

Answer: C

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 a 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

Answer: A

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

Answer: D

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.