Question

Question 2 (Generalized Controller Design): Consider the open loop system GH (=)=- (=-0.75)(=-1)² a) Design a controlle...


Question 2 (Generalized Controller Design)

Consider the open loop system GH=

Transcribed: Question 2 (Generalized Controller Design): 1 Consider the open loop system GH (2) =- (z-0.75)(z-1)* a) Design a controller that assigns the closed-loop system poles to z,2 = 0.25± j0.25 and the rest poles to z =0.

Answer

Closed loop system:

It is a system in which the output is dependent on the input of the system. The output has a feedback system that ia connected with the input. Examples are AC, voltage stabilizer, electric iron, etc.

Open loop system has GH(S)=1Z-0.75Z-12

The general controller block diagram is shown below.

Electrical Engineering homework question answer, step 2, image 1

the transfer function of the block diagram is

T.F.=YZRZ=CZ GHZ1+CZ GHZCZ=Kp+KiZ+Kd.ZCZ=Kd.Z2+KpZ+KiZ

Put C(Z) in the transfer function equation.

T.F.=Kd.Z2+KpZ+KiZ 1z-0.75z-121+Kd.Z2+KpZ+KiZ 1z-0.75z-12=Kd.Z2+KpZ+KiZz-0.75z-12+Kd.Z2+KpZ+Ki=Kd.Z2+KpZ+KiZ4-2.75 Z3+2.5+KdZ2+Kp-0.75Z+Ki.......(1)

Poles are at

Z1, 2 =0.25±j0.25Pole, P1=0.25+j0.25Pole, P2=0.25-j0.25

The polynomial is

Polynomial =Z-0.25+j0.25Z-0.25-j0.25=Z2-0.5Z+0.125.......(2)

Compare (1) with (2) and get

Ki=0.125.......(3)2.5+Kd=1Kd=-1.50......(4)KP-0.75=-0.50KP=0.25.........(5)

Put the above value in the C(Z) equation.

CZ=Kd.Z2+KpZ+KiZCZ=-1.50Z2+0.25Z+0.125Z

Put the value (3), (4), and (5) in the eq(1).

The transfer function with the controller is

T.F.=-1.50.Z2+0.25Z+0.125Z4-2.75 Z3+2.5-1.50Z2+0.25-0.75Z+0.125T.F.=-1.50.Z2+0.25Z+0.125Z4-2.75 Z3+Z2+-0.5Z+0.125

 

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