Digital Signal Processing Questions and Answers – Bilinear Transformations

This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Bilinear Transformations”.

1. Bilinear Transformation is used for transforming an analog filter to a digital filter.
a) True
b) False
View Answer

Answer: a
Explanation: The bilinear transformation can be regarded as a correction of the backward difference method. The bilinear transformation is used for transforming an analog filter to a digital filter.

2. Which of the following rule is used in the bilinear transformation?
a) Simpson’s rule
b) Backward difference
c) Forward difference
d) Trapezoidal rule
View Answer

Answer: d
Explanation: Bilinear transformation uses trapezoidal rule for integrating a continuous time function.

3. Which of the following substitution is done in Bilinear transformations?
a) s = \(\frac{2}{T}[\frac{1+z^{-1}}{1-z^1}]\)
b) s = \(\frac{2}{T}[\frac{1+z^{-1}}{1+}]\)
c) s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\)
d) None of the mentioned
View Answer

Answer: c
Explanation: In bilinear transformation of an analog filter to digital filter, using the trapezoidal rule, the substitution for ‘s’ is given as
s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\).
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4. What is the value of \(\int_{(n-1)T}^{nT} x(t)dt\) according to trapezoidal rule?
a) \([\frac{x(nT)-x[(n-1)T]}{2}]T\)
b) \([\frac{x(nT)+x[(n-1)T]}{2}]T\)
c) \([\frac{x(nT)-x[(n+1)T]}{2}]T\)
d) \([\frac{x(nT)+x[(n+1)T]}{2}]T\)
View Answer

Answer: b
Explanation: The given integral is approximated by the trapezoidal rule. This rule states that if T is small, the area (integral) can be approximated by the mean height of x(t) between the two limits and then multiplying by the width. That is
\(\int_{(n-1)T}^{nT} x(t)dt=[\frac{x(nT)+x[(n-1)T]}{2}]T\)

5. What is the value of y(n)-y(n-1) in terms of input x(n)?
a) \([\frac{x(n)+x(n-1)}{2}]T\)
b) \([\frac{x(n)-x(n-1)}{2}]T\)
c) \([\frac{x(n)-x(n+1)}{2}]T\)
d) \([\frac{x(n)+x(n+1)}{2}]T\)
View Answer

Answer: a
Explanation: We know that the derivative equation is
dy(t)/dt=x(t)
On applying integrals both sides, we get
\(\int_{(n-1)T}^{nT}dy(t)=\int_{(n-1)T}^{nT} x(t)dt\)
=> y(nT)-y[(n-1)T]=\(\int_{(n-1)T}^{nT} x(t)dt\)
On applying trapezoidal rule on the right hand integral, we get
y(nT)-y[(n-1)T]=\([\frac{x(nT)+x[(n-1)T]}{2}]T\)
Since x(n) and y(n) are approximately equal to x(nT) and y(nT) respectively, the above equation can be written as
y(n)-y(n-1)=\([\frac{x(n)+x(n-1)}{2}]T\)
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6. What is the expression for system function in z-domain?
a) \(\frac{2}{T}[\frac{1+z^{-1}}{1-z^1}]\)
b) \(\frac{2}{T}[\frac{1+z^{-1}}{1-z^1}]\)
c) \(\frac{T}{2}[\frac{1+z^{-1}}{1-z^1}]\)
d) \(\frac{T}{2}[\frac{1-z^{-1}}{1+z^{-1}}]\)
View Answer

Answer: c
Explanation: We know that
y(n)-y(n-1)= \([\frac{x(n)+x(n-1)}{2}]T\)
Taking z-transform of the above equation gives
=>Y(z)[1-z-1]=([1+z-1]/2).TX(z)
=>H(z)=Y(z)/X(z)=\(\frac{T}{2}[\frac{1+z^{-1}}{1-z^1}]\).

7. In bilinear transformation, the left-half s-plane is mapped to which of the following in the z-domain?
a) Entirely outside the unit circle |z|=1
b) Partially outside the unit circle |z|=1
c) Partially inside the unit circle |z|=1
d) Entirely inside the unit circle |z|=1
View Answer

Answer: d
Explanation: In bilinear transformation, the z to s transformation is given by the expression
z=[1+(T/2)s]/[1-(T/2)s].
Thus unlike the backward difference method, the left-half s-plane is now mapped entirely inside the unit circle, |z|=1, rather than to a part of it.
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8. The equation s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\) is a true frequency-to-frequency transformation.
a) True
b) False
View Answer

Answer: a
Explanation: Unlike the backward difference method, the left-half s-plane is now mapped entirely inside the unit circle, |z|=1, rather than to a part of it. Also, the imaginary axis is mapped to the unit circle. Therefore, equation s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\) is a true frequency-to-frequency transformation.

9. If s=σ+jΩ and z=re, then what is the condition on σ if r<1?
a) σ > 0
b) σ < 0
c) σ > 1
d) σ < 1
View Answer

Answer: b
Explanation: We know that if = σ+jΩ and z=re, then by substituting the values in the below expression
s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\)
=>σ = \(\frac{2}{T}[\frac{r^2-1}{r^2+1+2rcosω}]\)
When r<1 => σ < 0.
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10. If s=σ+jΩ and z=re and r=1, then which of the following inference is correct?
a) LHS of the s-plane is mapped inside the circle, |z|=1
b) RHS of the s-plane is mapped outside the circle, |z|=1
c) Imaginary axis in the s-plane is mapped to the circle, |z|=1
d) None of the mentioned
View Answer

Answer: c
Explanation: We know that if =σ+jΩ and z=re, then by substituting the values in the below expression
s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\)
=>σ = \(\frac{2}{T}[\frac{r^2-1}{r^2+1+2rcosω}]\)
When r=1 => σ = 0.
This shows that the imaginary axis in the s-domain is mapped to the circle of unit radius centered at z=0 in the z-domain.

11. If s=σ+jΩ and z=re, then what is the condition on σ if r>1?
a) σ > 0
b) σ < 0
c) σ > 1
d) σ < 1
View Answer

Answer: a
Explanation: We know that if = σ+jΩ and z=rejω, then by substituting the values in the below expression
s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\)
=>σ = \(\frac{2}{T}[\frac{r^2-1}{r^2+1+2rcosω}]\)
When r>1 => σ > 0.

12. What is the expression for the digital frequency when r=1?
a) \(\frac{1}{T} tan⁡(\frac{ΩT}{2})\)
b) \(\frac{2}{T} tan⁡(\frac{ΩT}{2})\)
c) \(\frac{1}{T} tan^{-1}(\frac{ΩT}{2})\)
d) \(\frac{2}{T} tan^{-1}⁡(\frac{ΩT}{2})\)
View Answer

Answer: d
Explanation: When r=1, we get σ=0 and
Ω = \(\frac{2}{T} [\frac{2 sin⁡ω}{1+1+2 cos⁡ω}]\)
=>ω=\(\frac{2}{T} tan^{-1}⁡(\frac{ΩT}{2})\).

13. What is the kind of relationship between Ω and ω?
a) Many-to-one
b) One-to-many
c) One-to-one
d) Many-to-many
View Answer

Answer: c
Explanation: The analog frequencies Ω=±∞ are mapped to digital frequencies ω=±π. The frequency mapping is not aliased; that is, the relationship between Ω and ω is one-to-one. As a consequence of this, there are no major restrictions on the use of bilinear transformation.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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