Chapter 3, Problem 3.13E

Interpretation Introduction

(a)

Interpretation:

The response of the given system is to be estimated to be oscillatory or not for an arbitrary change in u.

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 that 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)

For the general form of the transfer function;

$\frac{Y\left(s\right)}{X\left(s\right)}=\frac{1}{{\tau }^2{s}^2+2\tau \zeta s+1}$

The response of the transfer will be underdamped or oscillatory if $\zeta <1$, critically damped if $\zeta =1$, and overdamped or nonoscillatory if $\zeta >1$.

Interpretation Introduction

(b)

Interpretation:

The steady-state value of yis to be calculated when

$u\left(t-2\right)=1$.

Concept introduction:

The value of $K_p$, steady-state value is calculated by $s=0$ in the transfer function for the unit step disturbance, that is,

$\underset{s\to 0}{\lim }\left[sY\left(s\right)\right]=K_p$

Interpretation Introduction

(c)

Interpretation:

The value of $y\left(t\right)$ for a step-change in u 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 that 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)