Chapter 3, Problem 3.19E

Interpretation Introduction

(a)

Interpretation:

The liquid level response, $y\left(t\right)$ for the sudden change in $u\left(t\right)$ from $0{\text{ to 1 m}}^3$ at $t=0$ is to be calculated.

Concept introduction:

For a function $f\left(t\right)$, the Laplace transform is given by,

$F\left(s\right)=\mathcal{L}\left[f\left(t\right)\right]={\displaystyle {\int }_0^{\infty }f\left(f\right)e^{-st}dt}$ …… (1)

Here, $F\left(s\right)$ represents the Laplace transform, $s$ is a variable which is complex and independent, $f\left(t\right)$ is any function of time which is being transformed, and $\mathcal{L}$ is the operator which is defined by an integral.

$f\left(t\right)$ is calculated by taking inverse Laplace transform of the function $F\left(s\right)$.

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

Laplace transform of higher order derivatives is given by:

$\mathcal{L}\left(\frac{d^{n}f}{d{t}^n}\right)=s^{n}F\left(s\right)-s^{n-1}f\left(0\right)-s^{n-2}{f}^{\left(1\right)}\left(0\right)-\cdots -s{f}^{\left(n-2\right)}\left(0\right)-f^{\left(n-1\right)}\left(0\right)$ …… (2)

Interpretation Introduction

(b)

Interpretation:

It is to be determined if the tank will overflow or not when the height of the tank is 2.5 m.

Concept introduction:

For large value of time, the asymptotic value of $y\left(t\right)$ can be calculated using Final Value theorem (FVM) as shown below:

$\underset{t\to \infty }{\lim }y\left(t\right)=\underset{s\to 0}{\lim }\left[sY\left(s\right)\right]$ ........ (3)

This theorem is applicable only if $\underset{s\to 0}{\lim }\left[sY\left(s\right)\right]$ exists for all values of $\mathrm{Re}\left(s\right)\ge 0$.

Interpretation Introduction

(c)

Interpretation:

The maximum flow change, $u_{\max }$ without the tank being overflowing is to be calculated.

Concept introduction:

For a function $f\left(t\right)$, the Laplace transform is given by,

$F\left(s\right)=\mathcal{L}\left[f\left(t\right)\right]={\displaystyle {\int }_0^{\infty }f\left(f\right)e^{-st}dt}$ ........ (1)

Here, $F\left(s\right)$ represents the Laplace transform, $s$ is a variable which is complex and independent, $f\left(t\right)$ is any function of time which is being transformed, and $\mathcal{L}$ is the operator which is defined by an integral.

$f\left(t\right)$ is calculated by taking inverse Laplace transform of the function $F\left(s\right)$.

PFE is the partial fraction expansion is the method of expanding the denominator of a fraction into simpler terms.

For large value of time, the asymptotic value of $y\left(t\right)$ can be calculated using Final Value theorem (FVM) as shown below:

$\underset{t\to \infty }{\lim }y\left(t\right)=\underset{s\to 0}{\lim }\left[sY\left(s\right)\right]$ ........ (3)

This theorem is applicable only if $\underset{s\to 0}{\lim }\left[sY\left(s\right)\right]$ exists for all values of $\mathrm{Re}\left(s\right)\ge 0$.