Chapter 4, Problem 4.1E

Interpretation Introduction

(a)

Interpretation:

The steady-state gain of the equation $\frac{\text{Y}\left(\text{s}\right)}{\text{U}\left(\text{s}\right)}\text{=}\frac{\text{d}}{\text{bs+c}}$ needs to be stated.

Concept introduction:

Steady-state gain is the steady-state value of transfer function which has been attended by a given unit step in the input.

Answer to Problem 4.1E

The steady-stategain of equations as: $\frac{\text{d}}{\text{c}}$.

Explanation of Solution

Given:

The transfer equation is $\frac{\text{Y}\left(\text{s}\right)}{\text{U}\left(\text{s}\right)}\text{=}\frac{\text{d}}{\text{bs+c}}$.

Calculation:

The steady-state gain can be expressed as:

$\text{U}\left(\text{s}\right)\text{=}\frac{\text{1}}{\text{s}}$

Substituting in the given equation,

$\begin{array}{l}\text{Y}\left(\text{s}\right)\text{=}\frac{\text{ds}}{\text{bs+c}}\text{×U}\left(\text{s}\right)\\ \text{Y}\left(\text{s}\right)\text{=}\frac{\text{ds}}{\text{bs+c}}\text{×}\frac{\text{1}}{\text{s}}\end{array}$

Steady-state value can be given as:

$\begin{array}{l}\underset{\text{s}\to \text{0}}{\text{lim}}=\text{s×}\frac{\text{d}}{\text{bs+c}}\text{×}\frac{\text{1}}{\text{s}}\\ \underset{\text{s}\to \text{0}}{\text{lim}}=\frac{\text{d}}{\text{c}}\end{array}$

Hence, the steady-state gain of equations is $\frac{\text{d}}{\text{c}}$.

Interpretation Introduction

(b)

Interpretation:

Whether the output will be bounded for all the values of $\text{b,}\text{c}\text{ and }\text{d}$ constants for the step change of magnitude $\text{M}$ in the input or not needs to be determined.

Concept introduction:

Steady-state gain is the steady-state value of transfer function which has attended by given unit step in the input.

Answer to Problem 4.1E

The $\frac{\text{Md}}{\text{c}}$ has bounded response for a finite value of $\text{d}$ and $\text{c}$ or finite ratio $\frac{\text{d}}{\text{c}}$.

If the ratio of $\text{d}$ and $\text{c}$ will infinite then it will tend to unbounded.

Explanation of Solution

Given:

The transfer equation is $\frac{\text{Y}\left(\text{s}\right)}{\text{U}\left(\text{s}\right)}\text{=}\frac{\text{d}}{\text{bs+c}}$.

Calculation:

The steady-state gain can be expressed as:

$\begin{array}{l}\text{U}\left(\text{t}\right)\text{=M}\\ \text{U}\left(\text{s}\right)\text{=}\frac{\text{M}}{\text{s}}\end{array}$

Substituting in the given equation,

$\begin{array}{l}\text{Y}\left(\text{s}\right)\text{=}\frac{\text{d}}{\text{bs+c}}\text{×M}\\ \text{Y}\left(\text{s}\right)\text{=}\frac{\text{d}}{\text{bs+c}}\text{×}\frac{\text{M}}{\text{s}}\end{array}$

Steady-state value can be given as:

$\begin{array}{l}\underset{\text{s}\to \text{0}}{\text{lim}}=\text{s×}\frac{\text{d}}{\text{bs+c}}\text{×}\frac{\text{M}}{\text{s}}\\ \underset{\text{s}\to \text{0}}{\text{lim}}=\frac{\text{Md}}{\text{c}}\end{array}$

Hence, the steady-state gain of equations is $\frac{\text{Md}}{\text{c}}$.

This $\frac{\text{Md}}{\text{c}}$ has bounded response for a finite value of $\text{d}$ and $\text{c}$ or finite ratio $\frac{\text{d}}{\text{c}}$.

If ratio of $\text{d}$ and $\text{c}$ will infinite then it will tend to unbounded.