# Bilinear Transformation MCQ [Free PDF] – Objective Question Answer for Bilinear Transformation Quiz

1. Bilinear Transformation is used for transforming an analog filter to a digital filter.

A. True
B. False

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

The bilinear transformation uses the 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

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. $$[\frac{x(nT)+x[(n+1)T]}{2}]T$$

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. $$[\frac{x(n)+x(n+1)}{2}]T$$

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. $$\frac{T}{2}[\frac{1-z^{-1}}{1+z^{-1}}]$$

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

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

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

s = $$\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]$$
=>σ = $$\frac{2}{T}[\frac{r^2-1}{r^2+1+2rcosω}]$$