6th Assignment Control Systems Ahmedawad

4.7 Consider the DC-motor control system with rate (tachometer) feedback shown in Fig. 4.44(a).

Figure 4.44: Control system for Problem 4.17

(a) Find values for K' and k't so that the system of Fig. 4.44(b) has the same transfer function as the system of Fig.4.44 (a).

(b) Determine the system type with respect to tracking θr and compute the system Kv in terms of the new parameters K' and k't.

(c) Does the addition of tachometer feedback with positive kt increase or decrease Kv?

𝐾′ =𝐾𝐾𝑚𝐾𝑝

𝑘

𝑘′ =𝑘𝑘𝑡

𝐾𝑝

b- Inner loop:

𝑌 =𝐺

1 + 𝐺𝑅 =

1𝑠𝑛 𝐺𝑜(𝑠)

1 +1

𝑠𝑛 𝐷𝐺𝑜(𝑠)

𝜃 =

1𝑠

×𝐾′

(1 + 𝜏𝑚𝑠)

1 +1𝑠

× (1 + 𝑘′𝑡𝑠) ×

𝐾′

(1 + 𝜏𝑚𝑠)

𝜃𝑟

Type 1

c-

4.27 Consider the second-order plant with transfer function:

𝐺(𝑠) =1

(𝑠 + 1)(5𝑠 + 1)

And in unity feedback a) Determine the system type and error constant with respect to tracking

polynomial reference inputs of the system for P, PD, and PID controllers Let kp = 19, kI = 9.5, and kD = 4.

b) Determine the system type and error constant of the system with respect to disturbance inputs for each of the three regulators in part with respect to rejecting polynomial disturbances w(t) at the input to the plant.

c) Is this system better at tracking references or rejecting disturbances? Explain your response briefly.

d) Verify your results for parts (a) and (b) using MATLAB by plotting unit step and ramp responses for both tracking and disturbance rejection.

With respect to reference:

𝐸 =𝐺

1 + 𝐺𝑅 =

1𝑠𝑛 𝐺𝑜(𝑠)

1 +1

𝑠𝑛 𝐷𝐺𝑜(𝑠), 𝑒𝑠𝑠 = lim

𝑠→0𝑠 ×

1

1 + 𝐺𝑜(𝑠)×

1

𝑠

P controller:

𝑌(𝑠)

𝑅(𝑠)=

𝐾𝑝𝐺(𝑠)

1 + 𝐾𝑝𝐺(𝑠)

𝐺(𝑠) =1

(𝑠 + 1)(5𝑠 + 1)=

1

5𝑠2 + 6𝑠 + 1

𝐺𝑜(0) =19

5(0)2 + 6(0) + 1= 19

𝑒𝑠𝑠 = lim𝑠→0

𝑠 × (1

1 + 19) ×

1

𝑠=

1

20

Type: 0 PD controller:

𝑌(𝑠)

𝑅(𝑠)=

(𝐾𝐷𝑠 + 𝐾𝑝)𝐺(𝑠)

1 + (𝐾𝐷𝑠 + 𝐾𝑝)𝐺(𝑠)

𝐺𝑜(0) =4(0) + 19

5(0)2 + 6(0) + 1= 19

𝑒𝑠𝑠 = lim𝑠→0

𝑠 × (1

1 + 19) ×

1

𝑠=

1

20

Type: 0

PID controller:

𝑌(𝑠)

𝑅(𝑠)=

(𝐾𝐷𝑠 + 𝐾𝑝 + 𝐾𝐼 𝑠⁄ )𝐺(𝑠)

1 + (𝐾𝐷𝑠 + 𝐾𝑝 + 𝐾𝐼 𝑠⁄ )𝐺(𝑠)

(4𝑠 + 19 +9.5

𝑠)

1

5𝑠2 + 6𝑠 + 1=

4𝑠2 + 19𝑠 + 9.5

𝑠(5𝑠2 + 6𝑠 + 1)=

1

𝑠× 𝐺𝑜(𝑠)

𝐺𝑜(0) =4𝑠2 + 19𝑠 + 9.5

5𝑠2 + 6𝑠 + 1= 9.5

𝑒𝑠𝑠 = lim𝑠→0

𝑠𝐸(𝑠) = lim𝑠→0

𝑠 ×1

1𝑠

× 9.5×

1

𝑠=

𝑠

9.5= 0

Type: 1

Matlab: nom1 = [ 5 6 1]; dom1 = [ 5 6 20 ]; P = tf( nom1,dom1); t = 0:0.1:15; y1 = step (P,t); nom2 = [ 5 6 1]; dom2 = [ 5 10 20]; PI= tf( nom2,dom2); t = 0:0.1:15; y2 = step (PI,t); nom3 = [ 5 6 1 0]; dom3 = [ 5 10 20 9.5]; PID= tf( nom3,dom3); t = 0:0.1:15; y3 = step (PID,t); plot ( t,y1,'r',t,y2,'b',t,y3,'g')

With respect to disturbance:

𝐸 =𝐺

1 + 𝐷𝐺𝑊 = 𝑠𝑛𝑃𝑜(0), 𝑒𝑠𝑠 = lim

𝑠→0𝑠 × 𝑠𝑛𝐸𝑜(0) ×

1

𝑠

P controller:

𝐺(𝑠) =1

(𝑠 + 1)(5𝑠 + 1)=

1

5𝑠2 + 6𝑠 + 1

𝐺𝑜(0) =1

5(0)2 + 6(0) + 1= 1

𝐸𝑜(0) =𝐺𝑜(0)

1 + 𝐷𝐺𝑜(0)=

1

1 + 19=

1

20

𝑒𝑠𝑠 = lim𝑠→0

𝑠 × (1

1 + 19) ×

1

𝑠=

1

20

Type: 0

PD controller:

𝐺𝑜(0) =4(0) + 19

5(0)2 + 6(0) + 1= 19

𝐸𝑜(0) =𝐺𝑜(0)

1 + 𝐷𝐺𝑜(0)=

1

1 + 19=

1

20

𝑒𝑠𝑠 = lim𝑠→0

𝑠 × (1

20) ×

1

𝑠=

1

20

Type: 0 PID controller:

(4𝑠 + 19 +9.5

𝑠)

1

5𝑠2 + 6𝑠 + 1=

4𝑠2 + 19𝑠 + 9.5

𝑠(5𝑠2 + 6𝑠 + 1)

𝐸𝑜(0) =1

1 + 𝐷𝐺𝑜(0)=

1

1 + 4𝑠 + 19 +9.5𝑠

=𝑠

4𝑠2 + 20𝑠 + 9.5= 𝑠𝑃𝑜(0)

𝑒𝑠𝑠 = lim𝑠→0

𝑠𝐸(𝑠) = lim𝑠→0

𝑠 × (𝑠 ×1

9.5) ×

1

𝑠=

𝑠

9.5= 0

Type: 1

Matlab: nom1= [ 1]; dom1= [ 5 6 20 ]; P = tf( nom1,dom1); t = 0:0.1:15; y1 = step (P,t);

nom2 = [ 1]; dom2 = [ 5 10 20]; PI= tf( nom2,dom2); t = 0:0.1:15; y2 = step (PI,t); nom3 = [ 1 0]; dom3 = [ 5 10 20 9.5]; PID= tf( nom3,dom3); t = 0:0.1:15; y3 = step (PID,t); plot ( t,y1,'r',t,y2,'b',t,y3,'g')

c-