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.


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


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


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


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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