Answer to Question #193691 in Electrical Engineering for Muskan

Question

Design a three-mode temperature control system which inputs error

in 0-4 V range. The output to final control element is 0-8 V. Given:

Kp = 2.4 % per %

Ki = 9 % (%/min)/ %

5

Kd = 0.7 % / (%/min)

Answer 1

kp$= 2.4$ % will be $\frac{100}{x}$ % $=x=\frac{100}{2.4}=41.6$ %

$GP=$ $\frac{(100\%)(V_{out(max)}-V_{out(min)}}{(41.6\%)(V_{in(max)-V_{in(min)}}}=\frac{(100\%)(8V)}{(41.6\%)(4V)}=4.8$

$=K_D=0.7$ %

$[\frac{0.7}{\%min}][\frac{60sec}{1min]}=42\frac{\%}{\%/sec}$

GD$=\frac{(42\%)8v}{1\frac{\%}{sec}(4v)}$

Now period of fastest expected change$=8sec$

$\therefore \frac{R_1}{R_1+R_3}=R_3C=\frac{0.1}{2\pi}\times 8sec=0.1273sec$

We have the relation , GP$=[\frac{R_2}{R_1+R_3}]=4.8$

$G_D=R_3C=60S ec$

$\frac{R_1}{R_1+R_3}=1\%4V=0.004$ and $C=10\mu F$

We will have $R_3=6m\Omega \therefore R_3C=60 sec \space and R_3=\frac{60}{10}=6$

$R_1=0.004R_1+0.004R_3$

$R_1=0.004R_1=0.004(6\Omega)$

$0.996R_1=24k$

$R_1=\frac{24}{0.996}=24.09k\Omega$

$\frac{R_2}{R_1+R_3}=4.8$

$R_2=4.8(R_1)+4.8R_3$

$=4.8(24.09)+4.8(6)$

$=144.432m\Omega$