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# Answer to Question #193691 in Electrical Engineering for Muskan

Question #193691

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)

1
2021-05-31T06:40:12-0400

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"

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