Consider the following simple feedback control block diagram. The plant is G_p(s

Question

Consider the following simple feedback control block diagram. The plant is G_p(s) = 2/s + 4. a) What is the bandwidth of the plant alone (assuming there is no feedback) b. Assuming a proportional controller, G_c(s) = k_p, determine the closed loop transfer function, G_0(s) c) Assuming a proportional controller, G_c(s) = k_p, determine the value of k_p so the bandwidth of the closed loop system is 24 rad/sec. d) Assuming the proportional controller from problem c, determine the settling time and the steady state error for a unit step.

Explanation / Answer

A) The plant is :

Gp(s) = 2/ (s+4)

here, The pole is given by,

s= -4 , | jw | = 4

taking mod on both sides we get,

so the magnitude of the bandwidth of the plant is 4

B) The proportional controller is,

Gc(s) = kp

Now,

G(s) = Gc(s) x Gp(s) = kp [2/ (s+4)] ..... which is open loop function

Now closed loop function is given by,

Go(s) = [G(s) x H(s)] / [ 1 + G(s) x H(s) ] = 2kp / (s +4 + 2kp) ... here, H(s) = 1

C) From the closed loop Transfer function it is clear that,

w =4 + 2kp

given the bandwidth of closed loop is 24 rad/s

24 = 4 + 2kp

kp = (24-4)/2 = 10

D) The new closed loop transfer function is,

Go(s) = 2kp / (s +4 + 2kp) = 20 / (s + 24)

Output Y(s) for unit step input is given below,

The laplasce transform of u(t) = 1/s ...... LT of unit step function

Y(s) = 20 /s(s + 24)

Y(s) = (20/24s) - (20/24)/(s+24)

now, y(t) = (20/24) - (20/24) exp (-24t) = (20/24) [1 - exp (-24t) ]

for settling time , let t = 1/24 , y(t) = (20/24) [1 - 0.95] = 0.52 for 63 % of final settled value

Now,

The steady state error is given by,

ess  = lims-> 0 s Y(s) = lims-> 0 s x 20 /s(s + 24) = 20/24 = 0.833 = 83.3 %

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