Test: Inverse Systems And Deconvolution

# Test: Inverse Systems And Deconvolution

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## 10 Questions MCQ Test Digital Signal Processing | Test: Inverse Systems And Deconvolution

Test: Inverse Systems And Deconvolution for Electrical Engineering (EE) 2022 is part of Digital Signal Processing preparation. The Test: Inverse Systems And Deconvolution questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Inverse Systems And Deconvolution MCQs are made for Electrical Engineering (EE) 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Inverse Systems And Deconvolution below.
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Test: Inverse Systems And Deconvolution - Question 1

### If a system is said to be invertible, then:

Detailed Solution for Test: Inverse Systems And Deconvolution - Question 1

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.

Test: Inverse Systems And Deconvolution - Question 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?

Detailed Solution for Test: Inverse Systems And Deconvolution - Question 2

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).

Test: Inverse Systems And Deconvolution - Question 3

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

Detailed Solution for Test: Inverse Systems And Deconvolution - Question 3

Explanation: Given impulse response is h(n)=(1/2)nu(n)
The system function corresponding to h(n) is
H(z)=1/(1-1/2 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- 1/2 z-1
Hence its impulse response is δ(n)-1/2 δ(n-1).

Test: Inverse Systems And Deconvolution - Question 4

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

Detailed Solution for Test: Inverse Systems And Deconvolution - Question 4

Explanation: The system function of given system is H(z)= 1- 1/2 z-1
The inverse of the system has a system function as H(z)= 1/(1-1/2 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.

Test: Inverse Systems And Deconvolution - Question 5

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

Detailed Solution for Test: Inverse Systems And Deconvolution - Question 5

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
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.

Test: Inverse Systems And Deconvolution - Question 6

If the frequency response of an FIR system is given as H(z)=6+z-1-z-2, then the system is:

Detailed Solution for Test: Inverse Systems And Deconvolution - Question 6

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.

Test: Inverse Systems And Deconvolution - Question 7

If the frequency response of an FIR system is given as H(z)=1-z-1-z-2, then the system is:

Detailed Solution for Test: Inverse Systems And Deconvolution - Question 7

Explanation: Given H(z)= 1-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=-2,3
So, the system is maximum phase.

Test: Inverse Systems And Deconvolution - Question 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:

Detailed Solution for Test: Inverse Systems And Deconvolution - Question 8

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.

Test: Inverse Systems And Deconvolution - Question 9

An IIR system with system function H(z)=(B(z))/(A(z)) is called a minimum phase if:

Detailed Solution for Test: Inverse Systems And Deconvolution - Question 9

Explanation: For an IIR filter whose system function is defined as H(z)=(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.

Test: Inverse Systems And Deconvolution - Question 10

An IIR system with system function H(z)=(B(z))/(A(z)) is called a mixed phase if:

Detailed Solution for Test: Inverse Systems And Deconvolution - Question 10

Explanation: For an IIR filter whose system function is defined as H(z)=(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.

## Digital Signal Processing

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## Digital Signal Processing

3 videos|50 docs|54 tests