Part of Lecture Series on Automatic Control Systems delivered by me to Final year Diploma in Engg. Students. Easy language and Equally useful for higher level.

1. BEE-502
Automatic Control Systems
Unit-5, PID Controller
Diploma in Engg.
(Electrical/ Instr. & Control)
5th Semester
Instructor: Mohd. Umar Rehman
21. 11. 2017
1
PID Controller: Transfer Function Derivation
Fig. An Electronic PID Controller
The above figure shows an electronic proportional-plus-integral-plus-derivative controller (a
PID controller) using operational amplifiers. The transfer function
i
)
(
(
)E s
E s
is given by:
i
(
)
)
(s
ZE s
E Z
= − 2
1
where, , &||
sR C
Z Z R
sR sC sC
R
R sC
C
=
+
+
= = =+1 2 2
1 1 1 2 2
1 1 2 2
1
1
1
.
Assuming R3 = R4 = R, then we have o
(
(
)
)
s
E s
E
= −1
Hence, the overall transfer function becomes (contd. on next page)

2. BEE-502
Automatic Control Systems
Unit-5, PID Controller
Diploma in Engg.
(Electrical/ Instr. & Control)
5th Semester
Instructor: Mohd. Umar Rehman
21. 11. 2017
2
( )
o
i
( )
[ ing & i
( )
( )( )
(
ng by
)
]
s
sRC sRC
sC R
R sRC sRC
R sR
s R C s RC RCR
R sR
R RC R
E s Z
E Z
R
C
R C
C
C
C
C
sR
R R sRC
  + +           
 + + = × 
  
 + + 
 
 
  

 +
=
=
÷
+
=
+
+

=
2
1
2 2 1 1
2 1
2 2 2 1 1
2
1 2 2
2
1 2 1 2 1 1 2 22
1 2 2
2 1 1 2 2
1 1
1 2 2 2 2
1 1
1 1
1
1
where,
KR sR RC RC
R
C
K
CsKRC K R


 
 
 
 = + + 
 
+

=2 1 1 1 1 2 2
1 2 2 2 2
1
1
Now, compare this expression with the following expression of TF:
PID P D
I
( ) K sG s K
K s
  = + +   
1
1
Then, we get the values of the various gains as follows:
P I D
, ,
KR
K
R K
RC
K
KR
K
C
= = =2 1 1
1 2 2
1
►►►▬◄◄◄