Transfer Function of a System MCQ Quiz - Objective Question with Answer for Transfer Function of a System - Download Free PDF

Concept:

To determine Pole-zero plot, we have to first find the transfer function of LTI System, i.e.

$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}$

Calculation:

Input-output difference equation is given as:

$y\left(n\right)=\sum _{K=0}^3{\left(-1\right)}^{k}x\left(n-K\right)$

y[n] can be represented as:

y(n) = x(n) – x(n - 1) + x(n - 2) – x(n - 3)

Taking the Z-transform, we get:

Y(z) = X(z) – z^-1 X(z) + z^-2 X(z) – z^-3 X(z)

$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}=1-z^{-1}+z^{-2}-z^{-3}$

$H\left(z\right)=\frac{z^3-z^2+z-1}{z^3}=\frac{\left(z-1\right)\left(z^2+1\right)}{z^3}$

Hence, Pole-zero plot is:

Concept:

The transfer function is given by:

$H\left(z\right)=\frac{b_0+b_1{z}^{-1}+\dots +b_m{z}^{-m}}{a_0+a_1{z}^{-1}+\dots +a_N{z}^{-N}}$

a₀ = b₀ = 1

a₁ is to be determined and all other coefficients are zero.

Calculations:

$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}=\frac{1}{1+a_1{z}^{-1}}$

Y(z) [1 + a₁z^-1] = X(z)

Y(z) + a₁z^-1Y(z) = X(z)

y[n] + a₁y[n - 1] = x[n]

y[n] = x[n] - a₁y[n - 1]

at n = 3

y[3] = x[3] - a₁y[2]

0 = 1/2 -a₁(1)

Concept:

Causality and Stability:

For a system with rational transfer function H(z) to be causal, the ROC should lie outside the outermost pole, and for BIBO stability, the ROC should include the unit circle |z| = 1. 

∴ An LTI discrete-time causal system with the rational system function H(z) is said to be stable if all the poles of H(z) lie inside the unit circle.

Calculation:

The transfer function can be written as:

$\mathrm{H}\left(\mathrm{z}\right)=\frac{\mathrm{k}{\mathrm{z}}^4}{\left(\mathrm{z}-0.5-\mathrm{j}0.5\right)\left(\mathrm{z}-0.5+\mathrm{j}0.5\right)\left(\mathrm{z}+0.5+\mathrm{j}0.5\right)\left(\mathrm{z}+0.5-\mathrm{j}0.5\right)}$

$\mathrm{H}\left(\mathrm{z}\right)=\frac{\mathrm{k}{\mathrm{z}}^4}{\left({\mathrm{z}}^2-\mathrm{z}+\frac{1}{2}\right)\left({\mathrm{z}}^2+\mathrm{z}+\frac{1}{2}\right)}$

Now, since the system is causal and stable we have,

$\mathrm{h}\left[0\right]=\underset{\mathrm{z}\to \mathrm{\infty }}{lim}\mathrm{H}\left(\mathrm{z}\right)=\mathrm{k}=1$

Thus, the transfer function will be:

$\mathrm{H}\left(\mathrm{z}\right)=\frac{{\mathrm{z}}^4}{\left({\mathrm{z}}^2-\mathrm{z}+\frac{1}{2}\right)\left({\mathrm{z}}^2+\mathrm{z}+\frac{1}{2}\right)}$

H(z) = 1 - 0.25z^-4.......

h[n] = [1,0,0,0,-0.25,.....]

Now,

h[0] = 1

so,h[n] is real for all n

Note:

Causality:

A linear time-invariant discrete-time system is said to be causal if the impulse response h[n] = 0, for n < 0 and it is therefore right-sided.

The ROC of such a system H(z) is the exterior of a circle. If H(z) is rational, then the system is said to be causal if:

1) The ROC is the exterior of a circle outside the outermost pole; and

2) The degree of the numerator polynomial of H(z) should be less than or equal to the degree of the denominator polynomial.

Stability:

A discrete-time LTI system is said to be BIBO stable if the impulse response h[n] is summable, i.e.

${\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left|h\left[n\right]\right|<\mathrm{\infty }$

z-transform of h[h] is given as:

$H\left(z\right)={\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}h\left[h\right]z^{-n}$

Let z = ejΩ (which describes a unit circle in the z-plane), then

$\left|H\left[e^{j\mathrm{\Omega }}\right]\right|=\left|{\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}h\left[n\right]e^{-j\mathrm{\Omega }n}\right|$

The above equality can be written as:

$\le {\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left|h\left[n\right]e^{-j\mathrm{\Omega }n}\right|$

$={\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left|h\left[n\right]\right|<\mathrm{\infty }$

This is the condition for stability. Thus we can conclude that an LTI system is stable if the ROC of the system function H(z) contains the unit circle |z| = 1

Concept:

Poles and Zeros of a transfer function are the frequencies for which the value of the denominator and numerator of transfer function becomes zero respectively.

The values of the poles and the zeros of a system determine whether the system is stable, and how well the system performs.

Calculation:

$\begin{array}{l}\frac{\mathrm{Y}\left(\mathrm{z}\right)}{\mathrm{X}\left(\mathrm{z}\right)}=\mathrm{H}\left(\mathrm{z}\right)=\frac{1}{1-\frac{5}{6}{\mathrm{z}}^{-1}+\frac{{\mathrm{z}}^{-2}}{6}}=\frac{{\mathrm{z}}^2}{{\mathrm{z}}^2-\frac{5}{6}\mathrm{z}+\frac{1}{6}}\\ =\frac{{\mathrm{z}}^2}{\left(\mathrm{z}-\frac{1}{2}\right)\left(\mathrm{z}-\frac{1}{3}\right)}\end{array}$

Concept:

To determine Pole-zero plot, we have to first find the transfer function of LTI System, i.e.

$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}$

Calculation:

Input-output difference equation is given as:

$y\left(n\right)=\sum _{K=0}^3{\left(-1\right)}^{k}x\left(n-K\right)$

y[n] can be represented as:

y(n) = x(n) – x(n - 1) + x(n - 2) – x(n - 3)

Taking the Z-transform, we get:

Y(z) = X(z) – z^-1 X(z) + z^-2 X(z) – z^-3 X(z)

$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}=1-z^{-1}+z^{-2}-z^{-3}$

$H\left(z\right)=\frac{z^3-z^2+z-1}{z^3}=\frac{\left(z-1\right)\left(z^2+1\right)}{z^3}$

Hence, Pole-zero plot is:

Concept:

Poles and Zeros of a transfer function are the frequencies for which the value of the denominator and numerator of transfer function becomes zero respectively.

The values of the poles and the zeros of a system determine whether the system is stable, and how well the system performs.

Calculation:

$\begin{array}{l}\frac{\mathrm{Y}\left(\mathrm{z}\right)}{\mathrm{X}\left(\mathrm{z}\right)}=\mathrm{H}\left(\mathrm{z}\right)=\frac{1}{1-\frac{5}{6}{\mathrm{z}}^{-1}+\frac{{\mathrm{z}}^{-2}}{6}}=\frac{{\mathrm{z}}^2}{{\mathrm{z}}^2-\frac{5}{6}\mathrm{z}+\frac{1}{6}}\\ =\frac{{\mathrm{z}}^2}{\left(\mathrm{z}-\frac{1}{2}\right)\left(\mathrm{z}-\frac{1}{3}\right)}\end{array}$

Concept:

Causality and Stability:

For a system with rational transfer function H(z) to be causal, the ROC should lie outside the outermost pole, and for BIBO stability, the ROC should include the unit circle |z| = 1. 

∴ An LTI discrete-time causal system with the rational system function H(z) is said to be stable if all the poles of H(z) lie inside the unit circle.

Calculation:

The transfer function can be written as:

$\mathrm{H}\left(\mathrm{z}\right)=\frac{\mathrm{k}{\mathrm{z}}^4}{\left(\mathrm{z}-0.5-\mathrm{j}0.5\right)\left(\mathrm{z}-0.5+\mathrm{j}0.5\right)\left(\mathrm{z}+0.5+\mathrm{j}0.5\right)\left(\mathrm{z}+0.5-\mathrm{j}0.5\right)}$

$\mathrm{H}\left(\mathrm{z}\right)=\frac{\mathrm{k}{\mathrm{z}}^4}{\left({\mathrm{z}}^2-\mathrm{z}+\frac{1}{2}\right)\left({\mathrm{z}}^2+\mathrm{z}+\frac{1}{2}\right)}$

Now, since the system is causal and stable we have,

$\mathrm{h}\left[0\right]=\underset{\mathrm{z}\to \mathrm{\infty }}{lim}\mathrm{H}\left(\mathrm{z}\right)=\mathrm{k}=1$

Thus, the transfer function will be:

$\mathrm{H}\left(\mathrm{z}\right)=\frac{{\mathrm{z}}^4}{\left({\mathrm{z}}^2-\mathrm{z}+\frac{1}{2}\right)\left({\mathrm{z}}^2+\mathrm{z}+\frac{1}{2}\right)}$

H(z) = 1 - 0.25z^-4.......

h[n] = [1,0,0,0,-0.25,.....]

Now,

h[0] = 1

so,h[n] is real for all n

Note:

Causality:

A linear time-invariant discrete-time system is said to be causal if the impulse response h[n] = 0, for n < 0 and it is therefore right-sided.

The ROC of such a system H(z) is the exterior of a circle. If H(z) is rational, then the system is said to be causal if:

1) The ROC is the exterior of a circle outside the outermost pole; and

2) The degree of the numerator polynomial of H(z) should be less than or equal to the degree of the denominator polynomial.

Stability:

A discrete-time LTI system is said to be BIBO stable if the impulse response h[n] is summable, i.e.

${\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left|h\left[n\right]\right|<\mathrm{\infty }$

z-transform of h[h] is given as:

$H\left(z\right)={\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}h\left[h\right]z^{-n}$

Let z = ejΩ (which describes a unit circle in the z-plane), then

$\left|H\left[e^{j\mathrm{\Omega }}\right]\right|=\left|{\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}h\left[n\right]e^{-j\mathrm{\Omega }n}\right|$

The above equality can be written as:

$\le {\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left|h\left[n\right]e^{-j\mathrm{\Omega }n}\right|$

$={\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left|h\left[n\right]\right|<\mathrm{\infty }$

This is the condition for stability. Thus we can conclude that an LTI system is stable if the ROC of the system function H(z) contains the unit circle |z| = 1

Concept:

The transfer function is given by:

$H\left(z\right)=\frac{b_0+b_1{z}^{-1}+\dots +b_m{z}^{-m}}{a_0+a_1{z}^{-1}+\dots +a_N{z}^{-N}}$

a₀ = b₀ = 1

a₁ is to be determined and all other coefficients are zero.

Calculations:

$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}=\frac{1}{1+a_1{z}^{-1}}$

Y(z) [1 + a₁z^-1] = X(z)

Y(z) + a₁z^-1Y(z) = X(z)

y[n] + a₁y[n - 1] = x[n]

y[n] = x[n] - a₁y[n - 1]

at n = 3

y[3] = x[3] - a₁y[2]

0 = 1/2 -a₁(1)

Concept:

To determine Pole-zero plot, we have to first find the transfer function of LTI System, i.e.

$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}$

Calculation:

Input-output difference equation is given as:

$y\left(n\right)=\sum _{K=0}^3{\left(-1\right)}^{k}x\left(n-K\right)$

y[n] can be represented as:

y(n) = x(n) – x(n - 1) + x(n - 2) – x(n - 3)

Taking the Z-transform, we get:

Y(z) = X(z) – z^-1 X(z) + z^-2 X(z) – z^-3 X(z)

$H\left(z\right)=\frac{Y\left(z\right)}{X\left(z\right)}=1-z^{-1}+z^{-2}-z^{-3}$

$H\left(z\right)=\frac{z^3-z^2+z-1}{z^3}=\frac{\left(z-1\right)\left(z^2+1\right)}{z^3}$

Hence, Pole-zero plot is:

Concept:

Poles and Zeros of a transfer function are the frequencies for which the value of the denominator and numerator of transfer function becomes zero respectively.

The values of the poles and the zeros of a system determine whether the system is stable, and how well the system performs.

Calculation:

$\begin{array}{l}\frac{\mathrm{Y}\left(\mathrm{z}\right)}{\mathrm{X}\left(\mathrm{z}\right)}=\mathrm{H}\left(\mathrm{z}\right)=\frac{1}{1-\frac{5}{6}{\mathrm{z}}^{-1}+\frac{{\mathrm{z}}^{-2}}{6}}=\frac{{\mathrm{z}}^2}{{\mathrm{z}}^2-\frac{5}{6}\mathrm{z}+\frac{1}{6}}\\ =\frac{{\mathrm{z}}^2}{\left(\mathrm{z}-\frac{1}{2}\right)\left(\mathrm{z}-\frac{1}{3}\right)}\end{array}$

Concept:

Causality and Stability:

For a system with rational transfer function H(z) to be causal, the ROC should lie outside the outermost pole, and for BIBO stability, the ROC should include the unit circle |z| = 1. 

∴ An LTI discrete-time causal system with the rational system function H(z) is said to be stable if all the poles of H(z) lie inside the unit circle.

Calculation:

The transfer function can be written as:

$\mathrm{H}\left(\mathrm{z}\right)=\frac{\mathrm{k}{\mathrm{z}}^4}{\left(\mathrm{z}-0.5-\mathrm{j}0.5\right)\left(\mathrm{z}-0.5+\mathrm{j}0.5\right)\left(\mathrm{z}+0.5+\mathrm{j}0.5\right)\left(\mathrm{z}+0.5-\mathrm{j}0.5\right)}$

$\mathrm{H}\left(\mathrm{z}\right)=\frac{\mathrm{k}{\mathrm{z}}^4}{\left({\mathrm{z}}^2-\mathrm{z}+\frac{1}{2}\right)\left({\mathrm{z}}^2+\mathrm{z}+\frac{1}{2}\right)}$

Now, since the system is causal and stable we have,

$\mathrm{h}\left[0\right]=\underset{\mathrm{z}\to \mathrm{\infty }}{lim}\mathrm{H}\left(\mathrm{z}\right)=\mathrm{k}=1$

Thus, the transfer function will be:

$\mathrm{H}\left(\mathrm{z}\right)=\frac{{\mathrm{z}}^4}{\left({\mathrm{z}}^2-\mathrm{z}+\frac{1}{2}\right)\left({\mathrm{z}}^2+\mathrm{z}+\frac{1}{2}\right)}$

H(z) = 1 - 0.25z^-4.......

h[n] = [1,0,0,0,-0.25,.....]

Now,

h[0] = 1

so,h[n] is real for all n

Note:

Causality:

A linear time-invariant discrete-time system is said to be causal if the impulse response h[n] = 0, for n < 0 and it is therefore right-sided.

The ROC of such a system H(z) is the exterior of a circle. If H(z) is rational, then the system is said to be causal if:

1) The ROC is the exterior of a circle outside the outermost pole; and

2) The degree of the numerator polynomial of H(z) should be less than or equal to the degree of the denominator polynomial.

Stability:

A discrete-time LTI system is said to be BIBO stable if the impulse response h[n] is summable, i.e.

${\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left|h\left[n\right]\right|<\mathrm{\infty }$

z-transform of h[h] is given as:

$H\left(z\right)={\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}h\left[h\right]z^{-n}$

Let z = ejΩ (which describes a unit circle in the z-plane), then

$\left|H\left[e^{j\mathrm{\Omega }}\right]\right|=\left|{\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}h\left[n\right]e^{-j\mathrm{\Omega }n}\right|$

The above equality can be written as:

$\le {\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left|h\left[n\right]e^{-j\mathrm{\Omega }n}\right|$

$={\sum }_{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left|h\left[n\right]\right|<\mathrm{\infty }$

This is the condition for stability. Thus we can conclude that an LTI system is stable if the ROC of the system function H(z) contains the unit circle |z| = 1