Question

1. the model of system is shown below, where $k_m = 30$, $r_f = 1 omega$, $k_f=k_t = 1$, $j = 0.1$ and $b = 1$. (a) determine the response $theta(t)$ of the system to a step - change in $theta_d(t)$ when $k = 20$, $t_d(s)=0$. (6 pt) (b) assuming $theta_d(t)=0$. find the effect of a load disturbance $t_d(s)=1/s$. (3 pt) (c) determine the steady - state error $e_{ss}$ when the input is $r(t)=t,t > 0$. (assume that $t_d(s)=0$.) (3 pt)

Explanation:

Step1: Find the open - loop transfer function

The open - loop transfer function $G(s)$ of the system is obtained from the block diagram.

$$G(s)=\frac{K_iKK_m}{R_fs(Js + b)}K_f$$

Given $K_m = 30$, $R_f=1\Omega$, $K_f = K_t = 1$, $J = 0.1$ and $b = 1$. Let $K_i = 1$ (not given otherwise). Then $G(s)=\frac{K\times30}{1\times s(0.1s + 1)}\times1=\frac{30K}{s(0.1s + 1)}$

Step2: (a) Response to step input

For a step input $\Theta_d(s)=\frac{\Theta_{d0}}{s}$, with $T_d(s)=0$ and $K = 20$.
The closed - loop transfer function $\frac{\Theta(s)}{\Theta_d(s)}=\frac{G(s)}{1 + G(s)}$

$$G(s)=\frac{30\times20}{s(0.1s + 1)}=\frac{600}{s(0.1s + 1)}$$

$$1+G(s)=1+\frac{600}{s(0.1s + 1)}=\frac{s(0.1s + 1)+600}{s(0.1s + 1)}=\frac{0.1s^{2}+s + 600}{s(0.1s + 1)}$$

$$\frac{\Theta(s)}{\Theta_d(s)}=\frac{600}{0.1s^{2}+s + 600}$$

For a unit step input $\Theta_{d0}=1$, $\Theta_d(s)=\frac{1}{s}$

$$\Theta(s)=\frac{600}{s(0.1s^{2}+s + 600)}$$

We use partial - fraction decomposition or the inverse Laplace transform. The characteristic equation is $0.1s^{2}+s + 600=0$, $s^{2}+10s + 6000=0$. The roots are $s=\frac{-10\pm\sqrt{100 - 24000}}{2}=- 5\pm\sqrt{25 - 6000}i$.
Using the inverse Laplace transform of $\frac{600}{s(0.1s^{2}+s + 600)}$ gives $\Theta(t)=1-\text{e}^{-5t}(A\cos(\omega t)+B\sin(\omega t))$ where $\omega=\sqrt{6000 - 25}$

Step3: (b) Effect of load disturbance

With $\Theta_d(s)=0$, the transfer function from $T_d(s)$ to $\Theta(s)$ is $\frac{\Theta(s)}{T_d(s)}=-\frac{1/s(0.1s + 1)}{1 + G(s)}$

$$1+G(s)=\frac{0.1s^{2}+s + 600}{s(0.1s + 1)}$$

$$\frac{\Theta(s)}{T_d(s)}=-\frac{1}{0.1s^{2}+s + 600}$$

Given $T_d(s)=\frac{1}{s}$, then $\Theta(s)=-\frac{1}{s(0.1s^{2}+s + 600)}$

Step4: (c) Steady - state error for ramp input

For a ramp input $r(t)=t$, $R(s)=\frac{1}{s^{2}}$ and $T_d(s)=0$
The steady - state error $e_{ss}=\lim_{s
ightarrow0}sE(s)=\lim_{s
ightarrow0}s\frac{R(s)}{1 + G(s)}$

$$G(s)=\frac{30K}{s(0.1s + 1)}$$

$$1+G(s)=\frac{0.1s^{2}+s + 30K}{s(0.1s + 1)}$$

$$e_{ss}=\lim_{s ightarrow0}s\frac{\frac{1}{s^{2}}}{\frac{0.1s^{2}+s + 30K}{s(0.1s + 1)}}=\lim_{s ightarrow0}\frac{0.1s + 1}{0.1s^{2}+s + 30K}=\frac{1}{30K}$$

If we assume $K$ is non - zero, for the general case, when $K$ is given (say $K$ is as in the problem context), we substitute the value of $K$.

Answer:

(a) $\Theta(t)$ is obtained from the inverse Laplace transform of $\frac{600}{s(0.1s^{2}+s + 600)}$ as described above.
(b) $\Theta(s)=-\frac{1}{s(0.1s^{2}+s + 600)}$
(c) $e_{ss}=\frac{1}{30K}$ (substitute the given value of $K$ for a numerical result)