6th Assignment Control Systems Ahmedawad

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Transcript of 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 r 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?

  • =

    =

  • b- Inner loop:

    =

    1 + =

    1 ()

    1 +1

    ()

    =

    1

    (1 + )

    1 +1

    (1 + )

    (1 + )

    Type 1

    c-

  • 4.27 Consider the second-order plant with transfer function:

    () =1

    ( + 1)(5 + 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:

    =

    1 + =

    1 ()

    1 +1

    (), = lim

    0

    1

    1 + ()

    1

    P controller:

    ()

    ()=

    ()

    1 + ()

    () =1

    ( + 1)(5 + 1)=

    1

    52 + 6 + 1

    (0) =19

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

    = lim0

    (1

    1 + 19)

    1

    =

    1

    20

    Type: 0 PD controller:

    ()

    ()=

    ( + )()

    1 + ( + )()

    (0) =4(0) + 19

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

    = lim0

    (1

    1 + 19)

    1

    =

    1

    20

    Type: 0

  • PID controller:

    ()

    ()=

    ( + + )()

    1 + ( + + )()

    (4 + 19 +9.5

    )

    1

    52 + 6 + 1=

    42 + 19 + 9.5

    (52 + 6 + 1)=

    1

    ()

    (0) =42 + 19 + 9.5

    52 + 6 + 1= 9.5

    = lim0

    () = lim0

    1

    1 9.5

    1

    =

    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:

    =

    1 + = (0), = lim

    0 (0)

    1

    P controller:

    () =1

    ( + 1)(5 + 1)=

    1

    52 + 6 + 1

    (0) =1

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

    (0) =(0)

    1 + (0)=

    1

    1 + 19=

    1

    20

    = lim0

    (1

    1 + 19)

    1

    =

    1

    20

    Type: 0

  • PD controller:

    (0) =4(0) + 19

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

    (0) =(0)

    1 + (0)=

    1

    1 + 19=

    1

    20

    = lim0

    (1

    20)

    1

    =

    1

    20

    Type: 0 PID controller:

    (4 + 19 +9.5

    )

    1

    52 + 6 + 1=

    42 + 19 + 9.5

    (52 + 6 + 1)

    (0) =1

    1 + (0)=

    1

    1 + 4 + 19 +9.5

    =

    42 + 20 + 9.5= (0)

    = lim0

    () = lim0

    ( 1

    9.5)

    1

    =

    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-