Chapter 5, Problem 5.1E

Interpretation Introduction

Interpretation:

The time needed for output to reach the steady state depend on the magnitude of input needs to be determined.

Concept introduction:

Fourth order plus time delay system is used for describing the dynamics of chemical process.

Just by selecting proper choice of ${\text{τ}}_{\text{DT}}$ and $\text{τ}$ it represents the dynamics of several industrial processes.

Answer to Problem 5.1E

Magnitude of change increase with $\text{A}$ increased with time and vice versa that is $\text{Y}\left(\text{s}\right)$ is depend on the $\text{A}$.

Explanation of Solution

Given:

For given transformation,

$G\left(S\right)=\frac{K}{\left({\tau }_{1}s+1\right)\left({\tau }_{2}s+1\right)\left({\tau }_{3}s+1\right)\left({\tau }_{4}s+1\right)}{e}^{-\theta s}$

Calculation:

• To find the stead state value of $\text{Y}\left(\text{s}\right)$:

Given transformation-

$\begin{array}{l}\text{G}\left(\text{S}\right)\text{=}\frac{\text{K}}{\left({\text{τ}}_{\text{1}}\text{s+1}\right)\left({\text{τ}}_{\text{2}}\text{s+1}\right)\left({\text{τ}}_{\text{3}}\text{s+1}\right)\left({\text{τ}}_{\text{4}}\text{s+1}\right)}{\text{e}}^{\text{-θs}}\\ \text{G}\left(\text{S}\right)\text{=}\frac{\text{Y}\left(\text{s}\right)}{\text{U}\left(\text{s}\right)}\end{array}$

Step input is applied in system input

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

As magnitude is $\text{1}$

So, $\text{Y}\left(\text{S}\right)\text{=}\frac{\text{1}}{\text{S}}\text{×}\frac{\text{K}}{\left({\text{τ}}_{\text{1}}\text{s+1}\right)\left({\text{τ}}_{\text{2}}\text{s+1}\right)\left({\text{τ}}_{\text{3}}\text{s+1}\right)\left({\text{τ}}_{\text{4}}\text{s+1}\right)}{\text{e}}^{\text{-θs}}$

At the steady state of output,

$\underset{\text{x}\to \text{¥}}{\text{lim}}$

$\text{f}\left(\text{t}\right)$ to $\underset{\text{s}\to \text{¥}}{\text{lim}}$

$\text{f}\left(\text{s}\right)$,

• For the steady state value of

$\begin{array}{l}\text{Y}\left(\text{t}\right)\text{=}\underset{\text{s}\to \text{0}}{\text{lim}}\text{×S×}\frac{\text{1}}{\text{S}}\frac{\text{K}}{\left({\text{τ}}_{\text{1}}\text{s+1}\right)\left({\text{τ}}_{\text{2}}\text{s+1}\right)\left({\text{τ}}_{\text{3}}\text{s+1}\right)\left({\text{τ}}_{\text{4}}\text{s+1}\right)}{\text{e}}^{\text{-θs}}\\ \text{Y}{\left(\text{t}\right)}_{\text{t}\to \text{0}}\text{=}\frac{\text{K}}{\left(\text{1}\right)\left(\text{1}\right)\left(\text{1}\right)\left(\text{1}\right)}{\text{e}}^{\left(\text{0}\right)}\\ \text{Y}{\left(\text{t}\right)}_{\text{t}\to \text{0}}\text{=K}\end{array}$

• If step input magnitude $\text{A}$,

$\text{U}\left(\text{S}\right)\text{=}\frac{\text{A}}{\text{S}}$

When magnitude is $\text{A}$,

$\begin{array}{l}\text{Y}\left(\text{s}\right)\text{=}\frac{\text{A}}{\text{S}}\frac{\text{K}}{\left({\text{τ}}_{\text{1}}\text{s+1}\right)\left({\text{τ}}_{\text{2}}\text{s+1}\right)\left({\text{τ}}_{\text{3}}\text{s+1}\right)\left({\text{τ}}_{\text{4}}\text{s+1}\right)}{\text{e}}^{\text{-θs}}\\ \text{Y}{\left(\text{s}\right)}_{\text{t}\to \text{0}}\text{=}\underset{\text{s}\to \text{¥}}{\text{lim}}\text{×S×Y}\left(\text{s}\right)\\ \text{Y}{\left(\text{s}\right)}_{\text{t}\to \text{0}}\text{=}\frac{\text{A×K}}{\left(\text{1}\right)\left(\text{1}\right)\left(\text{1}\right)\left(\text{1}\right)}{\text{e}}^{\left(\text{0}\right)}\\ \text{Y}{\left(\text{s}\right)}_{\text{t}\to \text{0}}\text{=A×K }\end{array}$

Conclusion

Thus, steady state value of $\text{Y}\left(\text{s}\right)$ is mainly depends on the magnitude of step input in $\text{A}$.