The previous solutions are posted on screenshot. please refer to them for gain value. solve : An integrator is added to the PD controller, leading to a PID controller C(s) = kp +ki/s + kds. Keep the PD controllers gains kp and kd from above, and plot the closed-loop poles as the gain ki varies from 0 ! 1 (Hint: The closed-loop poles satisfy 1 + P (s)C(s) = 0, translate this to Evan’s form 1 + kiG(s) = 0).Determine gains such that a load disturbance is rejected in about 10 time units, while keeping the amount of overshoot (from r to y) under 20%. Why can ki not be too large? PlantController1ekp+kass(s+2)a) What type of controller is used?PD controller with Transfer function of (Kp+ K,S) is usedb) Determine the closed-loop transfer function from r to yClosed loop transfer functionKp+K ss(s+2)G(s)1+ G(s)with T(s)G(s)Kp+K s(2+k) .5+KpT(s)=c) s 2/2= 2/2= 0.7072.828о,s225ns,2 = 024s8 0= s3+ (2+ k^)*s + kp = 0kp 8, k4 = 2d) Plot the step response y due to a unit step input in r. Is there a steady stateerror? verify your observation with a hand calculationKp+K ss(s+2)G(s)H(s)for step response,Kp+K ss s+2)Ка(s+2)Крs(s+2)R(s)1lim(s0) G(s)H(s)where the lim part is lim (s ->0)e sse ss 0 20Te) Next, we consider the effect of a load (input) disturbance d, see the figure belowDetermine the (closed-loop) transfer function from d to yControllerPlant1kpkasS(8+2)Closed loop trasnfer functionY(s)1+s+kDs)s 2sf)Does the controller, with the gains computed above, reject a constant loaddisturbance?*D (s), where D(s) is a step funce ss=lim (s ->0)+2sks klim (s 0)1,s+ke sss22s+It is shown that the controlled completely rejects the constant load disturbance.

Draw the block diagram employing the PID controller as shown below:-

Answered: Plant Controller 1 e kp+kas s(s+2) a)…

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The previous solutions are posted on screenshot. please refer to them for gain value.

solve :

An integrator is added to the PD controller, leading to a PID controller C(s) = kp +ki/s + kds. Keep the PD controllers gains kp and kd from above, and plot the closed-loop poles as the gain ki varies from 0 ! 1 (Hint: The closed-loop poles satisfy 1 + P (s)C(s) = 0, translate this to Evan’s form 1 + kiG(s) = 0).

Determine gains such that a load disturbance is rejected in about 10 time units, while keeping the amount of overshoot (from r to y) under 20%. Why can ki not be too large?

Plant
Controller
1
e
kp+kas
s(s+2)
a) What type of controller is used?
PD controller with Transfer function of (Kp+ K,S) is used
b) Determine the closed-loop transfer function from r to y
Closed loop transfer function
Kp+K s
s(s+2)
G(s)
1+ G(s)
with T(s)
G(s)
Kp+K s
(2+k) .5+Kp
T(s)=
c) s 2/2
= 2/2
= 0.707
2.828
о,
s225ns,2 = 0
24s8 0= s3+ (2+ k^)*s + kp = 0
kp 8, k4 = 2
d) Plot the step response y due to a unit step input in r. Is there a steady state
error? verify your observation with a hand calculation
Kp+K s
s(s+2)
G(s)H(s)
for step response,
Kp+K s
s s+2)
Ка
(s+2)
Кр
s(s+2)
R(s)
1lim(s0) G(s)H(s)
where the lim part is lim (s ->0)
e ss
e ss 0

20
T
e) Next, we consider the effect of a load (input) disturbance d, see the figure below
Determine the (closed-loop) transfer function from d to y
Controller
Plant
1
kpkas
S(8+2)
Closed loop trasnfer function
Y(s)
1+
s+k
Ds)
s 2s
f)
Does the controller, with the gains computed above, reject a constant load
disturbance?
*D (s), where D(s) is a step func
e ss=lim (s ->0)
+2sks k
lim (s 0)
1,
s+k
e ss
s22s+
It is shown that the controlled completely rejects the constant load disturbance.

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