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‫بسم ا‪ ...‬الرحمن الرحيم‬
‫سیستمهای کنترل خطی‬
‫پاییز ‪1389‬‬
‫دکتر حسین بلندي‪ -‬دکتر سید مجید اسما عیل زاده‬
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Historical note
The first application of PID controller was in 1922 by Minorsky on
ship steering.
Minorsky (1922) “Directional stability of automatically steered
bodies”, J. Am. Soc. Naval Eng., 34, p.284.
This was the first mathematical treatment of the type of controller that
is now used to control almost all industrial processes.
The current situation
Despite the abundance of sophisticated tools, including advanced
controller design techniques, PID controllers are still the most
widely used controller structure in modern industry, controlling
more that 95% of closed-loop industrial processes.
Different PID controllers differ in the way how their parameters be
tuned, manually, or automatically.
Most of the DCS systems have built-in routines to perform autotuning of PID controllers based on the loop characteristics. They
are often called: auto-tuners.
The PID Algorithm
• The PID algorithm is the most popular feedback controller
algorithm used. It is a robust easily understood algorithm
that can provide excellent control performance despite the
varied dynamic characteristics of processes.
• As the name suggests, the PID algorithm consists of three
basic modes:
the Proportional mode,
the Integral mode
& the Derivative mode.
PID controllers
P, PI or PID Controller
• When utilizing the PID algorithm, it is necessary to decide
which modes are to be used (P, I or D) and then specify the
parameters (or settings) for each mode used.
• Generally, three basic algorithms are used: P, PI or PID.
• Controllers are designed to eliminate the need for
continuous operator attention.
 Cruise control in a car and a house thermostat
are common examples of how controllers are used to
automatically adjust some variable to hold a measurement
(or process variable) to a desired variable (or set-point)
Controller Output
• The variable being controlled is the output of the controller
(and the input of the plant):
provides excitation to the plant
system to be controlled
• The output of the controller will change in response to a change
in measurement or set-point (that said a change in the tracking
error)
PID Controller
• In the s-domain, the PID controller may be represented as:
Ki


U ( s)   K p 
 K d s  E ( s)
s


• In the time domain:
de(t )
u (t )  K p e(t )  K i  e(t )dt  K d
0
dt
t
proportional gain
integral gain
derivative gain
PID Controller
• In the time domain:
de(t )
u (t )  K p e(t )  K i  e(t )dt  K d
0
dt
t
• The signal u(t) will be sent to the plant, and a new output y(t)
will be obtained. This new output y(t) will be sent back to
the sensor again to find the new error signal e(t). The
controllers takes this new error signal and computes its
derivative and its integral gain. This process goes on and on.
Definitions
• In the time domain:
de(t )
u (t )  K p e(t )  K i  e(t )dt  K d
0
dt

1 t
de(t ) 

 K p  e(t )   e(t )dt  Td
0
T
dt 
i

t
integral time constant
where Ti 
proportional gain
derivative time constant
Kp
Ki
,
Kd
Td 
Kp
integral gain
derivative gain
PID structures
Standard PID controllers have the following structures:
Proportional only:
Proportional plus Integral:
Proportional plus derivative:
Proportional, integral and derivative:
What is an Op-Amp?
• An Operational Amplifier is an electronic device used to
perform mathematical operations in a circuit – they are
generally abbreviated as “Op-Amps”
• Op-Amps are high gain devices that amplify a signal using
an external power supply
• They are composed of multiple transistors, resistors, and
capacitors
• Common types of op-amps:
•
•
•
•
•
Inverting
Non-Inverting
Integrating
Differential
Summing
What is an Op-Amp?
• All op-amps use a voltage supply (Vcc) to amplify the signal
• The supply voltages can either have equal value but opposite signs, or
the low side is grounded and the high side has a value of twice the
voltage input
• Some common applications of op-amps:
• Low Pass Filters
• Strain Gauges
• PID Controllers
+Vcc
V-
VV
-
V+
V
+
Inverting Input
Vou
tVout
Vou
t
V+
Non-Inverting Input
-Vcc
Typical 8 Pin Op-Amp
Amplifier Gain
• All op-amps can be represented by the
formula:
Vout = K (V+ - V-)
V
Vout
-
• Where K is the gain, and is a property of
the individual op-amp
• This gain should be distinguished from
the gain of the op-amp circuit which is
generally denoted by Av
V
+
Op-Amp
Av = Vout / Vin
• A potential source of confusion comes
from failing to properly distinguish
between the op-amp and the op-amp
circuit
Op-Amp Circuit
Comparator
•
A comparator is an example of an open-loop op-amp
+Vcc
Vout
i- = 0A V- Vin
-
i+ = 0A
+
V+
Vin
+
If V+ > VIf V+ < V-
Vou
+ Vsat
t
Vcc
Vout = Vsat ≈ Vcc
Vout = -Vsat ≈ - Vcc
Vin+ Vin-
- Vsat
Summing Op-Amp
• Application of a non-inverting opamp and Millman‘s theorem
(1) Millman’s theorem:
VinA VinB VinC


RA
RB
RC
V' 
1
1
1


R A RB RC
R2
(2)
R1
Vin RA
A
Vin RB
B
V- -
V+
+
V’
Vin RC
C
(1)
Vout = VinA + VinB + VinC
Vout
if RA = RB = RC = R
V '  1/ 3 ( VinA  VinB  VinC )
(2) Non-inverting Op-Amp:

R2 
Vout  1 
 V'
R1 


R 
if 1  2   3
R1 

Vout  3 x 1/ 3 ( VinA  VinB  VinC )
Substraction (Differential) Op-Amp
• Applying Kirchhoff’s Rules and
Op-Amp Calculation Rules yields:
Vout = VinA - VinB
 R4 
 R2  R3  R4 
VinA   VinB

Vout  
 R1  R2  R3 
 R3 
R4
if R1 = R2 = R3 = R4
Vout  VinA  VinB
Note:
if R1 = R3 = R and R2 = R4 = a R
Vout  a VinA  VinB 
Vin
R3
V- -
B
Vin
A
R1
V+
R2
+
Vout
Derivative Op-Amp
R
Vin
Vin
C
R
V- -
V+ +
(RC)
Vout
d 
dt
Vout
 Applying Kirchhoff’s Rules and Op-Amp Calculation
Rules yields:
dVin ( t )
Vout  ( RC)
dt
Integrating Op-Amp
C
Vin
Vin
R
R
V- V+ +

Vout
1

RC
Vout
 Applying Kirchhoff’s Rules and Op-Amp Calculation
Rules yields:
1 t
Vout  
 Vin  d
RC 0
 dt
PID Controller – System Block Diagram
P
VSET
VERROR
I
Output Process
VOUT
D
VSENSOR
Sensor
•Goal is to have VSET = VOUT
•Remember that VERROR = VSET – VSENSOR
•Output Process uses VERROR from the PID controller to adjust Vout such that it is
~VSET
Applications
PID Controller – System Circuit Diagram
Signal conditioning allows you to
introduce a time delay which could
account for things like inertia
System to control
Calculates VERROR = -(VSET + VSENSOR)
-VSENSOR
Applications
PID Controller – PID Controller Circuit Diagram
VERROR
Adjust
Change
Kp
RP1, RP2
Ki
RI, CI
Kd
RD, CD
VERROR PID
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Controller Effects
• A proportional controller (P) reduces error responses to
disturbances, but still allows a steady-state error.
• When the controller includes a term proportional to the
integral of the error (I), then the steady state error to a
constant input is eliminated, although typically at the cost
of deterioration in the dynamic response.
• A derivative control typically makes the system better
damped and more stable.
Closed-loop Response
Rise time
P
Decrease
Maximum
overshoot
Increase
I
Decrease
Increase
Settling
time
Small
change
Increase
D
Small
change
Decrease
Decrease
Steadystate error
Decrease
Eliminate
Small
change
• Note that these correlations may not be exactly accurate,
because P, I and D gains are dependent of each other.
Example problem of PID
• Suppose we have a simple mass, spring, damper problem.
• The dynamic model is such as:
mx  bx  kx  f
• Taking the Laplace Transform, we obtain:
ms2 X (s)  bsX (s)  kX (s)  F (s)
• The Transfer function is then given by:
X ( s)
1
 2
F ( s) ms  bs  k
Example problem (cont’d)
• Let
m  1kg , b  10 N .s / m , k  20 N / m , f  1N
• By plugging these values in the transfer function:
X ( s)
1
 2
F ( s) s  10s  20
• The goal of this problem is to show you how each of
K p , K i and K d contribute to obtain:
fast rise time,
minimum overshoot,
no steady-state error.
Ex (cont’d): No controller
• The (open) loop transfer function is given by:
X ( s)
1
 2
F ( s) s  10s  20
• The steady-state value for the output is:
X ( s) 1
xss  lim x(t )  lim sX ( s)  lim sF ( s)

t 
s 0
s 0
F ( s) 20
Ex (cont’d): Open-loop step response
• 1/20=0.05 is the final value
of the output to an unit step
input.
• This corresponds to a
steady-state error of 95%,
quite large!
• The settling time is about
1.5 sec.
Ex (cont’d): Proportional
Controller
• The closed loop transfer function is given by:
Kp
X ( s)

F ( s)
Kp
s 2  10s  20 
Kp
s 2  10s  (20  K p )
1 2
s  10s  20
Ex (cont’d): Proportional control
• Let K p  300
• The above plot shows that
the proportional controller
reduced both the rise time
and the steady-state error,
increased the overshoot, and
decreased the settling time
by small amount.
Ex (cont’d): PD Controller
• The closed loop transfer function is given by:
K p  Kd s
X ( s)

F ( s)
K p  Kd s
s 2  10s  20 
K p  Kd s
s 2  (10  K d ) s  (20  K p )
1 2
s  10s  20
Ex (cont’d): PD control
• Let K p  300, K d  10
• This plot shows that the
proportional derivative
controller reduced both
the overshoot and the
settling time, and had
small effect on the rise
time and the steady-state
error.
Ex (cont’d): PI Controller
• The closed loop transfer function is given by:
K p  Ki / s
X ( s)

F ( s)
K p s  Ki
s 2  10s  20 
K p  Ki / s
s 3  10s 2  (20  K p ) s  K i
1 2
s  10s  20
Ex (cont’d): PI Controller
• Let
K p  30, K i  70
• We have reduced the proportional
gain because the integral controller
also reduces the rise time and
increases the overshoot as the
proportional controller does
(double effect).
• The above response shows that the
integral controller eliminated the
steady-state error.
Ex (cont’d): PID Controller
• The closed loop transfer function is given by:
K p  K d s  Ki / s
X ( s)

F ( s)
K d s 2  K p s  Ki
s  10s  20
 3
K p  K d s  K i / s s  (10  K d ) s 2  (20  K p ) s  K i
1
s 2  10s  20
2
Ex (cont’d): PID Controller
• Let
K p  350, K i  300,
K d  5500
• Now, we have obtained
the system with no
overshoot, fast rise time,
and no steady-state
error.
Ex (cont’d): Summary
P
PD
PI
PID
PID Controller Functions
• Output feedback
 from Proportional action
compare output with set-point
• Eliminate steady-state offset (=error)
 from Integral action
apply constant control even when error is zero
• Anticipation
 From Derivative action
react to rapid rate of change before errors grows too big
Effect of Proportional,
Integral & Derivative Gains on
the
Dynamic Response
Proportional Controller
• Pure gain (or attenuation) since:
the controller input is error
the controller output is a proportional gain
E ( s) K p  U ( s)  u (t )  K p e(t )
Change in gain in P controller
• Increase in gain:
 Upgrade both steadystate and transient
responses
 Reduce steady-state
error
 Reduce stability!
P Controller with high gain
Integral Controller
• Integral of error with a constant gain
 increase the system type by 1
 eliminate steady-state error for a unit step input
 amplify overshoot and oscillations
t
Ki
E ( s)
 U ( s)  u (t )  Ki  e(t )dt
s
0
Change in gain for PI controller
• Increase in gain:
 Do not upgrade steadystate responses
 Increase slightly
settling time
 Increase oscillations
and overshoot!
Derivative Controller
• Differentiation of error with a constant gain
 detect rapid change in output
 reduce overshoot and oscillation
 do not affect the steady-state response
de(t )
E ( s) K d s  U ( s )  u (t )  K d
dt
Effect of change for gain PD
controller
• Increase in gain:
 Upgrade transient
response
 Decrease the peak time
and rise time
 Increase overshoot
and settling time!
Changes in gains for PID Controller