University of Mosul

College of Engineering
Electrical Engineering Department
Master

PID Controller
Younis Abdulelah, Zaid Ali, Omar Haitham and Mohammed Thaer

Supervise by Dr. Mohammed Obead

❖ Introduction to PID
❖ Modes of Control
❑ Proportional Control
❑ Proportional + Integral Control
❑ Proportional + Derivative Control
❑ Proportional + Integral + Derivative Control
❖ PID Tuning Rules
❑ Zeigler-Nichol’s Tuning Rules
• 1st Method
• 2nd Method
 Introduction to PID Control

• The PID controller is the most widely used. Over 85% of all dynamic
controllers are of the PID types.

• There is actually a great variety of types and design methods for the PID
controller.
 What is a PID controller?
• The acronym PID stands for Proportional-Integral-Differential control.

• Each of these, the P , the I and the D are terms in a control algorithm, and
each has a special purpose.

• Sometimes certain of the terms are left out because they are not needed in
the control design.

• This is possible to have a PI, PD or just a P control.It is very rare to have a

ID control.
The standard PID control configuration is as shown below. It is
also sometimes called the “PID parameter form”
 PID CONTROLLER
Proportional: the error is multiplied by a gain Kp. A
very high gain may cause instability, and a very low gain
may cause the system to be very slow.

Integral: the integral of the error is found and

multiplied by a gain. The gain can be adjusted to drive
the error to zero in the required time.

Derivative: The derivative of the error is multiplied by

a gain. The derivative control is used to improve the
transient response by reducing overshoot
• The input–output relationship of a PID controller can be expressed as
1 𝑡 𝑑𝑒(𝑡)
𝑢(𝑡) = 𝐾𝑝 𝑒(𝑡) + ∫0 𝑒(𝑡)𝑑𝑡 + 𝑇𝑑
𝑇𝑖 𝑑𝑡
• where u(t) is the output from the controller and e(t) is the input for PID. Ti
and Td are known as the integral and derivative action time, respectively.
 1 𝑡 𝑑𝑒(𝑡)
𝑢(𝑡) = 𝐾𝑝 𝑒(𝑡) + ∫0 𝑒(𝑡)𝑑𝑡 + 𝑇𝑑
𝑇𝑖 𝑑𝑡

• By taking the Laplace transform of this equation, we can write the

transfer function of a continuous-time PID as

𝑈(𝑠) 𝐾𝑝
= 𝐾𝑝 + + 𝐾𝑝 𝑇𝑑 𝑠
𝐸(𝑠) 𝑇𝑖 𝑠
 PROPORTIONAL CONTROL (P)
• In proportional mode, there is a continuous linear relation between
value of the controlled variable and position of the final control
element.

𝑟(𝑡) 𝑒(𝑡) 𝑐𝑝(𝑡) = 𝐾𝑝 𝑒(𝑡) 𝑐(𝑡)

𝐾𝑝 𝑃𝑙𝑎𝑛𝑡
𝑏(𝑡) - 𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙
𝐶𝑜𝑛𝑡𝑟𝑜𝑙

𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘

• Output of proportional controller is

𝑐𝑝(𝑡) = 𝐾𝑝 𝑒(𝑡)
• The transfer function can be written as

𝐶𝑝(𝑠)
= 𝐾𝑝
𝐸(𝑠) 9
 PROPORTIONAL CONTROL (P)
• As the gain is increased the system responds faster to changes in set-
point but becomes progressively underdamped and eventually
unstable.

10
 PROPORTIONAL PLUS INTEGRAL
CONTROLLERS (PI)
• Integral control describes a controller in which the output rate of
change is dependent on the magnitude of the input.

• Specifically, a smaller amplitude input causes a slower rate of

change of the output.

11
 PROPORTIONAL PLUS INTEGRAL
CONTROLLERS (PI)
• The major advantage of integral controllers is that they have the
unique ability to return the controlled variable back to the exact set
point following a disturbance.

• Disadvantages of the integral control mode are that it responds

relatively slowly to an error signal and that it can initially allow a
large deviation at the instant the error is produced.

• This can lead to system instability and cyclic operation. For this
reason, the integral control mode is not normally used alone, but is
combined with another control mode.

12
 PROPORTIONAL PLUS INTEGRAL
CONTROL (PI)

𝐾𝑖 න 𝑒(𝑡) 𝑑𝑡
𝐾𝑖 ∫
𝑟(𝑡) 𝑒(𝑡) 𝐾𝑝 𝑒(𝑡)+
+ 𝑐𝑝𝑖 𝑡 𝑐(𝑡)
𝐾𝑝 𝑃𝑙𝑎𝑛𝑡
𝑏(𝑡) -

𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘

𝑐𝑝𝑖 𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑖 න 𝑒 𝑡 𝑑𝑡
13
 PROPORTIONAL PLUS INTEGRAL CONTROL
(PI)

𝑐𝑝𝑖 𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑖 න 𝑒 𝑡 𝑑𝑡

• The transfer function can be written as

𝐶𝑝𝑖 (𝑠) 1
= 𝐾𝑝 + 𝐾𝑖
𝐸(𝑠) 𝑠

14
 PROPORTIONAL PLUS DERIVATIVE
CONTROL (PD)
• The higher error signal rate of change, soonest the final control element
is positioned to the desired value.

• The added derivative action reduces initial overshoot of the measured

variable, and therefore aids in stabilizing the process sooner.

• This control mode is called Proportional plus Derivative (PD) control

because the derivative section responds to the rate of change of the
error signal

15
 PROPORTIONAL PLUS DERIVATIVE
CONTROL (PD)

𝑑𝑒(𝑡)
𝑑 𝐾𝑑
𝑑𝑡
𝐾𝑑
𝑑𝑡
𝑟(𝑡) 𝑒(𝑡) 𝐾𝑝 𝑒(𝑡)+
+ 𝑐𝑝𝑑 𝑡 𝑐(𝑡)
𝐾𝑝 𝑃𝑙𝑎𝑛𝑡
𝑏(𝑡) -

𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘

𝑑𝑒(𝑡)
𝑐𝑝𝑑 𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑑 𝑑𝑡
16
 PROPORTIONAL PLUS DERIVATIVE
CONTROL (PD)

𝑑𝑒(𝑡)
𝑐𝑝𝑑 𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑑 𝑑𝑡
• The transfer function can be written as

𝐶𝑝𝑑 (𝑠)
= 𝐾𝑝 + 𝐾𝑑 𝑠
𝐸(𝑠)

17
 PROPORTIONAL PLUS DERIVATIVE
CONTROL (PD)
• The stability and overshoot problems that arise when a proportional
controller is used at high gain can be mitigated by adding a term
proportional to the time-derivative of the error signal. The value of
the damping can be adjusted to achieve a critically damped response.

18
 PROPORTIONAL PLUS INTEGRAL PLUS
DERIVATIVE CONTROL (PID)
• Although PD control deals neatly with the overshoot and ringing
problems associated with proportional control it does not cure the
problem with the steady-state error.
• Fortunately it is possible to eliminate this while using relatively low
gain by adding an integral term to the control function which becomes

19
 Tips for Designing a PID Controller
1. Obtain an open-loop response and determine what needs to be
improved
2. Add a proportional control to improve the rise time
3. Add a derivative control to improve the overshoot
4. Add an integral control to eliminate the steady-state error
5. Adjust each of Kp, Ki, and Kd until you obtain a desired overall
response.
• Lastly, please keep in mind that you do not need to implement all
three controllers (proportional, derivative, and integral) into a single
system, if not necessary. For example, if a PI controller gives a good
enough response, then you don't need to implement derivative
controller to the system. Keep the controller as simple as possible.
 PROPORTIONAL PLUS INTEGRAL PLUS
DERIVATIVE CONTROL (PID)

𝐾𝑝 𝑒(𝑡)
𝐾𝑝

𝑟(𝑡) 𝑒(𝑡) 𝐾𝑖 න 𝑒(𝑡) 𝑑𝑡 + 𝑐

𝐾𝑖 ∫
𝑝𝑖𝑑 𝑡 𝑐(𝑡)
+ 𝑃𝑙𝑎𝑛𝑡
𝑏(𝑡) - +
𝑑𝑒(𝑡)
𝑑 𝐾𝑑
𝑑𝑡
𝐾𝑑
𝑑𝑡
𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘

𝑑𝑒(𝑡)
𝑐𝑝𝑖𝑑 𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑖 න 𝑒(𝑡) 𝑑𝑡 + 𝐾𝑑 𝑑𝑡 21
PROPORTIONAL PLUS INTEGRAL PLUS
DERIVATIVE CONTROL (PID)

𝑑𝑒(𝑡)
𝑐𝑝𝑖𝑑 𝑡 = 𝐾𝑝 𝑒 𝑡 + 𝐾𝑖 න 𝑒(𝑡) 𝑑𝑡 + 𝐾𝑑
𝑑𝑡

𝐶𝑝𝑖𝑑 (𝑠) 1
= 𝐾𝑝 + 𝐾𝑖 +𝐾𝑑 𝑠
𝐸(𝑠) 𝑠

22
 THE CHARACTERISTICS OF P, I, AND D
CONTROLLERS

CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S-S ERROR

Kp Decrease Increase Small Change Decrease

Ki Decrease Increase Increase Eliminate

Kd Small Change Decrease Decrease Small Change

PID TUNING
RULES
 PID TUNING
• The transfer function of PID controller is given as
𝐶𝑝𝑖𝑑 (𝑠) 1
= 𝐾𝑝 + 𝐾𝑖 +𝐾𝑑 𝑠
𝐸(𝑠) 𝑠
• It can be simplified as
𝐶𝑝𝑖𝑑 𝑠 1
= 𝐾𝑝 (1 + +𝑇𝑑 𝑠)
𝐸 𝑠 𝑇𝑖 𝑠
• Where 𝐾𝑝 𝐾𝑑
𝑇𝑖 = 𝑇𝑑 =
𝐾𝑖 𝐾𝑝
 PID TUNING
• The process of selecting the controller parameters (𝑲𝒑 , 𝑻𝒊 and 𝑻𝒅) to
meet given performance specifications is known as controller tuning.

• Ziegler and Nichols suggested rules for tuning PID controllers

experimentally.

• Which are useful when mathematical models of plants are not

known.

• These rules can, of course, be applied to the design of systems with

known mathematical models.
 PID TUNING
• Such rules suggest a set of values of 𝑲𝒑 , 𝑻𝒊 and 𝑻𝒅 that will give a
stable operation of the system.

• The resulting system may exhibit a large maximum overshoot in the

step response, which is unacceptable, in such a case we need series
of fine tunings until an acceptable result is obtained.

• In fact, the Ziegler–Nichols tuning rules give an educated guess for

the parameter values and provide a starting point for fine tuning,
rather than giving the final settings for 𝑲𝒑 , 𝑻𝒊 and 𝑻𝒅 in a single
shot.
 ZEIGLER-NICHOL’S
PID TUNING METHODS
• Ziegler and Nichols proposed rules for determining values of the
𝑲𝒑 , 𝑻𝒊 and 𝑻𝒅 based on the transient response characteristics of a
given plant.

• Such determination of the parameters of PID controllers or tuning of

PID controllers can be according to the plant needed.

• There are two methods called Ziegler–Nichols tuning rules:

• First method (open loop Method)
• Second method (Closed Loop Method)
 ZEIGLER-NICHOL’S
FIRST METHOD
• In the first method, we
obtain experimentally the
response of the plant to a
unit-step input.

• If the plant involves neither

integrator(s) nor dominant
complex-conjugate poles,
then such a unit-step
response curve may look
S-shaped
 ZEIGLER-NICHOL’S FIRST
METHOD
• This method applies if the response to a step input exhibits an S-
shaped curve.

• Such step-response curves may be generated experimentally or from a

dynamic simulation of the plant.
The response is characterised by two parameters, L the delay time and T the
time constant. These are found by drawing a tangent to the step response at
its point of inflection and noting its intersections with the time axis and the
steady state value. The plant model is therefore
 ZEIGLER-NICHOL’S
SECOND METHOD
• In the second method, we first set 𝑻𝒊 = ∞ and 𝑻𝒅 = 𝟎.

• Using the proportional control action only (as shown in figure),

increase Kp from 0 to a critical value Kcr at which the output first
exhibits sustained oscillations.

• If the output does not exhibit sustained oscillations for whatever value
Kp may take, then this method does not apply.
ZEIGLER-NICHOL’S SECOND METHOD

• Thus, the critical gain Kcr

and the corresponding
period Pcr are determined.
Example:1
Using Ziegler-Nichols method to design PID controller for a system
With open- loop transfer function. 2𝑒 −0.6𝑠
𝐺(𝑠) =
(3𝑠 + 1)
Solution:

K = 2, L = 0.6 , T=3
Kp = 1.2∗ TΤK ∗ L = 1.2 ∗ 3Τ( 0.6 ∗ 2) = 3

Ti = 2 L = 2∗ 0.6 = 1.2

Td = 0.5 L = 0.5 ∗ 0.6 = 0.3

EXAMPLE 2
The open-loop unit step response of a thermal system is shown. Obtain the
transfer function of this system and use the Ziegler–Nichols tuning algorithm to
design:
(a) a proportional controller,
(b) a proportional plus integral (PI) controller
(c) a PID controller.
Draw the block diagram of the system in each case.
EXAMPLE 2
SOLUTION:

• From the Figure, the system parameters are obtained as:

K = 40◦C,

L=5s

and T = 20 s,

hence, the transfer function of the plant is

 EXAMPLE 2
(a) a proportional controller
• According to the Table of ZN settings for a proportional controller are:

Thus,

• The transfer function of the Proportional controller is then

Thus,

• And the block diagram of the closed-loop system with the controller is shown below.
EXAMPLE 2
(b) PI controller.
According to the Table, the Ziegler-Nichols settings for a PI controller are

Thus,
0.9 × 20 5
𝐾𝑝 = = 0.09 and 𝑇𝑖 = = 16.5.
40 × 5 0.3
The transfer function of the PI controller is then
𝑈(𝑠) 1 0.09(16.5𝑠 + 1)
Thus, = 0.09 1 + =
𝐸(𝑠) 16.5𝑠 16.5𝑠

and the block diagram of the closed-loop system with the controller is shown in Figure.
EXAMPLE 2
(c) PID controller. According to table, the Ziegler-Nichols settings for a PID controller are

Thus,
1.2 × 20
𝐾𝑝 = = 0.12, 𝑇𝑖 = 2 × 5 = 10, 𝑇𝑑 = 0.5 × 5 = 2.5
40 × 5
The transfer function of the required PID controller is
1
= 0.12 1 + + 2.5𝑠
10𝑠
𝑈(𝑠) 3𝑠 2 + 1.2𝑠 + 0.12
=
or 𝐸(𝑠) 10𝑠
The block diagram of the system, together with the controller, is shown in Figure.
EXAMPLE-3

Consider the control system shown in following figure.

Apply a Ziegler–Nichols tuning rule for the determination of the
values of parameters 𝑲𝒑 , 𝑻𝒊 and 𝑻𝒅 .
EXAMPLE-2

• Transfer function of the plant is

1
𝐺 𝑠 =
𝑠(𝑠 + 1)(𝑠 + 5)

• Since plant has an integrator therefore Ziegler-Nichol’s first

method is not applicable.

• According to second method proportional gain is varied till

sustained oscillations are produced.

• That value of Kc is referred as Kcr.

 EXAMPLE-3
• Here, since the transfer function of the plant is known we can find
𝑲𝒄𝒓 using
– Root Locus
– Routh-Herwitz Stability Criterion

• By setting 𝑻𝒊 = ∞ and 𝑻𝒅 = 𝟎 closed loop transfer function is

obtained as follows.

𝐶(𝑠) 𝐾𝑝
=
𝑅(𝑠) 𝑠 𝑠 + 1 𝑠 + 5 + 𝐾𝑝
 EXAMPLE-3
• The value of 𝐾𝑝 that makes the system marginally unstable so that
sustained oscillation occurs can be obtained as

𝑠 3 + 6𝑠 2 + 5𝑠 + 𝐾𝑝 = 0
• The Routh array is obtained as

• Examining the coefficients of first

column of the Routh array we find
that sustained oscillations will occur
if 𝑲𝒑 = 𝟑𝟎.

• Thus the critical gain 𝑲𝒄𝒓 is

𝑲𝒄𝒓 = 𝟑𝟎
 EXAMPLE-3
• With gain 𝑲𝒑 set equal to 30, the characteristic equation becomes
𝑠 3 + 6𝑠 2 + 5𝑠 + 30 = 0
• To find the frequency of sustained oscillations, we substitute 𝒔 = 𝒋𝝎
into the characteristic equation.

(𝑗𝜔)3 +6(𝑗𝜔)2 +5𝑗𝜔 + 30 = 0

• Further simplification leads to

6(5 − 𝜔2 ) + 𝑗𝜔(5 − 𝜔2 ) = 0

6(5 − 𝜔2 ) = 0

𝜔 = 5 𝑟𝑎𝑑/𝑠𝑒𝑐
 EXAMPLE-3

𝜔 = 5 𝑟𝑎𝑑/𝑠𝑒𝑐
• Hence the period of sustained oscillations 𝑷𝒄𝒓 is
2𝜋
𝑃𝑐𝑟 =
𝜔
2𝜋
𝑃𝑐𝑟 = = 2.8099 𝑠𝑒𝑐
5
• Referring to the Table

𝐾𝑝 = 0.6𝐾𝑐𝑟 = 18
𝑇𝑖 = 0.5𝑃𝑐𝑟 = 1.405
𝑇𝑑 = 0.125𝑃𝑐𝑟 = 0.35124
 EXAMPLE-3

𝐾𝑝 = 18 𝑇𝑖 = 1.405 𝑇𝑑 = 0.35124
• Transfer function of PID controller is thus obtained as

1
𝐺𝑐 (𝑠) = 𝐾𝑝 (1 + +𝑇𝑑 𝑠)
𝑇𝑖 𝑠

1
𝐺𝑐 (𝑠) = 18(1 + + 0.35124𝑠)
1.405𝑠
EXAMPLE-3
ELECTRONIC PID CONTROLLER

𝑅2
𝐾𝑝 = −
𝑅1

𝑇𝑑 = −𝑅𝐷2 𝐶𝐷

𝑇𝑖 = −𝑅𝐼 𝐶𝐼
Digital implementation of analog controller design
• This section introduces an indirect approach to digital controller design.
• The approach is based on designing an analog controller for the analog
subsystem and then obtaining an equivalent digital controller and using it to
digitally implement the desired control.
• The digital controller can be obtained using a number of recipes that are well
known in the field of signal processing, where they are used in the design of
digital filters.
• In fact, a controller can be viewed as a filter that attenuates some dynamics
and accentuates others so as to obtain the desired time response.
Digital implementation of analog controller design

➢ Forward differencing

➢ Backward differencing

➢ Bilinear transformation
 General PID Direct Digital Control Algorithm

• Since this technique has proven so useful for continuous-time

control systems, it is desirable to develop a digital control algorithm
that will be of similar character to the continuous-time scheme.
• We shall approximate the integral with trapezoidal integration and
the derivative with a backward difference equation.
Proportional Control

A discrete implementation of proportional control action is identical to

continuous, thus

Proportional control is always present either by itself, or allied with

derivative and/or integral control.
Derivative Action

In continuous systems, derivative action has the form

In the discrete domain, differentiation can be approximated by the first

deference, that is,

Derivative action is typically used to improve system stability and

transient response.
Integral Action
Continuous integral action is given by

Discrete integration is a cumulative sum, represented by

NOTE: Gains KI and KD have different values in the continuous and

discrete domains!

• Integral action is typically used to improve system steady-state error

behavior. It usually worsens transient response and stability.
PD Control

The combination of Proportional and Derivative action is given by

𝐾𝐷
Where 𝑎= , 𝐾 = 𝐾𝑝 + 𝐾𝐷
𝐾𝐷 + 𝐾𝑝
PI Control
Proportional plus Integral control is

• which has a pole at z = 1 corresponding to the integration, and a zero on the

positive real axis nearer the origin.
• It is this pole at z = 1 that improves steady-state behavior but hurts stability
and transient response.
• The zero is usually placed fairly near the pole at z = 1 to reduce the effect of
the slow closed-loop pole that will be in this region.
• Note that it is the open-loop pole at z = 1 that improves the steady-state error
by raising the Type of the system.
PID Control
Adding integral action to form three-mode proportional plus integral plus
derivative action yields

which has a pole at z = 1 from the integration, a pole at z = 0, and two

zeros which may be real or complex, depending on the relative values of
the three gains.

One strategy for PID design is as follows:

1. Place one zero and pole as for the PD design.

2. Place the second pole at z = 1 (the integration).
Example : Consider the block diagram of a robot-arm control system shown in the
Figure below.
Let T=0.1 sec ,damping ratio ζ = 0.5 and settling time ts = 2 sec for 2%
tolerance band . Design PD controller.
 𝑎 𝑧ൣ 𝑎𝑇 − 1 + 𝑒 −𝑎𝑇 𝑧 + 1 − 𝑒 −aT − 𝑎𝑇𝑒 −aT ൧
2 =
𝑠 𝑠+𝑎 𝑎(𝑧 − 1)2 𝑧 − 𝑒 −a𝑇

𝑧−1 2 0.01873𝑧 + 0.0175

𝐺 𝑧 = 𝒵 2 =
𝑧 𝑠 𝑠+2 𝑧 − 1 𝑧 − 0.818

Solution:
4 4
𝑇𝑠 = 2= 𝜔𝑛 = 4 𝑟𝑎𝑑/𝑠𝑒𝑐
𝜁 𝜔𝑛 0.5 𝜔𝑛

S1,2 = - ζ 𝜔𝑛 ∓ 𝑗 𝜔𝑛 1 − ζ2

S1,2 = -2 ∓ j 3.46

𝑧 = 𝑒 −𝑇ζ𝜔𝑛 = 𝑒 −0.1∗0.5∗4 = 0.82

 ∠Z = T 𝜔𝑛 1 − 𝜁 2 = 0.1 ∗ 4 1 − 0.52

180
Z =0.346 rad * 𝜋
= 19.85°

0.82 ∠19.85° = 0.77 ∓ 𝑗 0.28

Transfer Function for PD controller

𝑘 𝑧−𝑎
𝐷(𝑧) =
𝑧
 𝑘 𝑧−𝑎
𝐷(𝑧) =
𝑧

෍ 𝜃𝑧 − ෍ 𝜃𝑝 = ±180

0.28
𝜃𝑧1 = tan−1
0.77 + 0.935
𝜃𝑧1 = 9.32∘ P3 P1
Z1 Z2 P2

0 − 28
𝜃𝑝1 = 180 − tan−1 = 129.4∘
1 − 0.77

0.28
𝜃p2 = tan−1 = 20∘
0.77

−1
0.28
𝜃𝑝3 = 180 − tan = 99.72
0.818 − 0.77
𝜃𝑧2 + 9.32 − (129.4 + 20 + 99.72) = −180

𝜃𝑧2 = −180 + 239 ⋅ 8 = 59.8∘

0.28
tan 59.8 = ⇒ 𝑥 = 0.163
𝑥

0.163 − 0.77 = −0.6

𝑘(𝑧 + 0.6)
𝐷(𝑧) =
𝑧

Now we must found the value of K

0.01873 (𝑧 + 0.935) 𝑘(𝑧 − 0.6)
× =1
𝑧 − 1)(𝑧 − 0.8187 𝑧

𝑧(𝑧 − 1)(𝑧 − 0.8187)

𝑘=
0.01873(𝑧 + 0.935)(𝑧 − 0.6) 𝑧=0.77+𝑗0.28

|0.77 + 𝑗0.28||0.77 + 𝑗0.28 − 1||0.77 + 𝑗0.28 − 0.8187|

𝑘=
0.01873 0.77 + 𝑗0.28 + 0.935 |0.77 + 𝑗0.28 − 0.6|

0.819 × 0.362 × 0.284

𝑘= =8
0.032 × 0.327
 𝑘 𝑧−𝑎 8(𝑧 − 0.6)
𝐷(𝑧) = =
𝑧 𝑧

𝑘𝑃 + 𝑘𝐷 = 8 … . 1

𝑘𝐷
= 0.6 … … 2
𝑘𝑃 + 𝑘𝐷

𝑘𝐷 = 4.8 , 𝑘𝑝= 3.2

𝑧−1
𝐷(𝑧) = 3.2 + 4.8
𝑧
Example:

Design PI controller with ζ=0.7 , Wn=4 rad/sec and T=0.1 sec for below
system:
Solution:

S1,2 = - ζ 𝜔𝑛 ∓ 𝑗 𝜔𝑛 1 − ζ2

S1,2 = -2.8 ∓ j 2.865

𝑧 = 𝑒 −𝑇ζ𝜔𝑛 = 𝑒 −0.1∗0.7∗4 = 0.755

∠z = T 𝜔𝑛 1 − 𝜁 2 = 0.1 ∗ 4 1 − 0.72

180
z =0.285 rad * = 16.36°
𝜋

0.755 ∠16.36° = 0.724 ∓ 𝑗 0.213

 𝑎 𝑧ൣ 𝑎𝑇 − 1 + 𝑒 −𝑎𝑇 𝑧 + 1 − 𝑒 −aT − 𝑎𝑇𝑒 −aT ൧
2 =
𝑠 𝑠+𝑎 𝑎(𝑧 − 1)2 𝑧 − 𝑒 −a𝑇

0.018(z − 0.935)
𝐺 𝑧 =
𝑧 − 1 𝑧 − 0.818
Transfer Function for PI controller
𝑘 𝑧−𝑎
𝐺𝐷 (𝑧) =
(𝑧 − 1)

𝐾𝑃
𝑎= 𝑘 = 𝐾𝑝 + 𝐾𝐼
𝐾𝑝 + 𝐾𝐼

𝑘 𝑧−𝑎 0.02(z − 0.935)

𝐺 𝑧 = ×
(𝑧 − 1) 𝑧 − 1 𝑧 − 0.818
෍ 𝜃𝑧 − ෍ 𝜃𝑝 = ±180

0.213
𝜃𝑧1 = tan−1
0.724 + 0.935

𝜃𝑧1 = 7.31∘

0.213
𝜃𝑝1,p2 = 180 − tan−1 = 142.34∘
1 − 0.724

−1
0.213
𝜃𝑝3 = 180 − tan = 113.8∘
0.8187 − 0.724
෍ 𝜃𝑧 − ෍ 𝜃𝑝 = ±180

𝜃𝑧a = ±180 + 142.3 + 142.3 + 113.5 − 7.32

𝜃𝑧a = 210.94∘

0.213
tan 210.9 = ⇒ 𝑎 = 1.079
𝑎 − 0.72

𝑘 𝑧 − 1.079
𝐺𝐷 (𝑧) =
(𝑧 − 1)
 (𝑧 − 1)2 𝑧 − 0.8187
|𝐾| = อ
0.02(𝑧 − 1.079)(𝑧 + 0.935)
𝑧=0.724±𝑗0.123

0.724 + 𝑗0.213 − 1)2 (0.724 + 𝑗0.213 − 0.8187

𝐾 =
0.02(0.724 + 𝑗0.213 − 1.079)(0.724 + 𝑗0.213 + 0.935)

(0.123∠ − 75.31) + 0.233. ∠113.96

𝐾 =
0.02 ∗ 0.414∠149.03

0.123 × 0.233
|𝐾| = ⇒ 𝐾 = 2.07
0.02 ∗ 0.414 × 1.672
 𝐾𝑃
𝑎=
𝐾𝑝 + 𝐾𝐼

𝐾𝑃
1.079 = 2.97

𝑘𝑃 = 2.233

𝑘 = 𝑘𝑃 + 𝑘𝐼 = 2.07

𝑘𝐼 = 0.163

2.07 𝑧 − 1.079
𝐺𝐷 (𝑧) =
(𝑧 − 1)

2.07 𝑧 − 1.079 0.02(z − 0.935)

𝐺 𝑧 = ×
(𝑧 − 1) 𝑧 − 1 𝑧 − 0.818