Digital Signal Processing Multiple Choice Questions on “Bilinear Transformations”.
1. Bilinear Transformation is used for transforming an analog filter to a digital filter.
A. True
B. False
Answer: A
Clarification: 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
Answer: D
Clarification: 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
Answer: C
Clarification: 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}}]).
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)
Answer: B
Clarification: 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)
Answer: A
Clarification: 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)
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}}])
Answer: C
Clarification: 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
Answer: D
Clarification: 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.
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
Answer: A
Clarification: 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=rejω, then what is the condition on σ if r<1?
A. σ > 0
B. σ < 0
C. σ > 1
D. σ < 1
Answer: B
Clarification: 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.
10. If s=σ+jΩ and z=rejω 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
Answer: C
Clarification: 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.
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=rejω, then what is the condition on σ if r>1?
A. σ > 0
B. σ < 0
C. σ > 1
D. σ < 1
Answer: A
Clarification: 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}))
Answer: D
Clarification: When r=1, we get σ=0 and
Ω = (frac{2}{T} [frac{2 sinω}{1+1+2 cosω}])
=>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
Answer: C
Clarification: 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.