Friday, 10 May 2013

PI CONTROLLER

It decreases the steady state error without effecting the stability

Its transfer function is KP + KI / S = (KPS + KI)/S

PI controller adds one pole at origin which increases TYPE. As TYPE increases steady state error decreases.

It also adds one finite zero in left of s-plane which avoids the effect on stability
Example
 
Characteristic equation is S2 + 5S + 1=0

Applying controller
TYPE increases. So steady state error decreases

Characteristic equation S3 + 5S2 + KPS+ KI = 0

All powers of S are there in above equation. So no effect o stability


EFFECT ON DAMPING FACTOR

 
Characteristic equation is S2 + 5S + 1 =0
From the above characteristic equation

              ξ= 2.5

After applying controller (S+ 5)/S

              G(S) H(S) = 1/S2

Characteristic equation is S2 + 1 =0

              ξ= 0

We can infer that there is a decrement in the value ξ  

As ξ decreases peak over shoot increases

From above calculations

1) Damping factor decreases.

2) Peak over shoot increases.

3) Rise time decreases.( To verify this substitute two different values of ξ in Tr formula )

4) Bandwidth increases, since Tr is inversely proportion to bandwidth.

5) Settling increases as it is inversely proportional to ξ.( To verify this substitute two different values of ξ in Ts formula )

6) Affect on gain margin and phase margin


GAIN MARGIN

Example

Without controller

 
Solution

For ωpc

-180 = -90 – tan-1(ω)- tan-1(ω/2)

By solving above equation we get

 ωpc = √2 rad/sec

Gain margin = -20 log | G(s) H(s) | at ωpc

| G(s) H(s) | at ωpc = 1/6

Gain margin= 15.5 db

For simplicity for calculation I assume controller (s + 1)/s

 
 For ωpc
-180 = -180 – tan-1(ω) - tan-1(ω/2) + tan-1(ω)

ωpc = 0

| G(s) H(s) | at ωpc = ∞

Gain margin= -∞

So, We Can Infer That There Is A Decrease In Gain Margin



PHASE MARGIN

Without controller


Solution
For ωgc
| G(s) H(s) | at ωgc = 1

 
ω4+2ω2-1 =0
finding roots we get ωgc = 0.6435
PM = 180 + angle (G(s) H(s) | at ωgc
     = 180 -90 - tan-1(ω/2)
     = 90- tan-1(0.6435/2)
     = 72.16

For simplicity for calculation I assume controller (s + 2)/s

 For ωgc
| G(s) H(s) | at ωgc = 1
       1/ω2 = 1
ω2 = 1
finding roots we get ωgc = 1
PM = 180 + angle (G(s) H(s))| at ωgc
     = 180 -180
     = 0
From Above Two PM’s we can conclude that PM decreases




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