Backward Difference Method MCQ Quiz – Objective Question with Answer for Backward Difference Method

Answer: A

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

Answer: D
The bilinear transformation uses the trapezoidal rule for integrating a continuous-time function.

Answer: C

In bilinear transformation of an analog filter to digital filter, using the trapezoidal rule, the substitution for ‘s’ is given as

s = \(\frac{2}{T}[\frac{1-z^{-1}}{1+z^{-1}}]\).

Answer: B

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

\(\int_{(n-1)T}^{nT} x(t)dt=[\frac{x(nT)+x[(n-1)T]}{2}]T\)

Answer: A

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)=\([\frac{x(n)+x(n-1)}{2}]T\)

Answer: C

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)=\(\frac{T}{2}[\frac{1+z^{-1}}{1-z^1}]\).

Answer: D

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.

Answer: A

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.

Answer: B

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.

Answer: C

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.

Answer: A

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.

Answer: D

When r=1, we get σ=0 and

Ω = \(\frac{2}{T} [\frac{2 sin⁡ω}{1+1+2 cos⁡ω}]\)

=>ω=\(\frac{2}{T} tan^{-1}⁡(\frac{ΩT}{2})\).

Answer: C

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.

Answer: A

If ak and bk are the filter coefficients, then the transfer function of a general IIR filter is given by the expression

H(z)=\(\frac{\sum_{k=0}^M b_k z^{-k}}{1+\sum_{k=1}^N a_k z^{-k}}\)