Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

MODULE-II

COMPENSATOR DESIGN
Every control system which has been designed for a specific application should
meet certain performance specification
...
The
choice of a plant is not only dependent on the performance specification but also on the
size , weight & cost
...
Under this
circumstances it is possible to introduce some kind of corrective sub-systems in order to
force the chosen plant to meet the given specification
...

REALIZATION OF BASIC COMPENSATORS

Compensation can be accomplished in several ways
...
The transfer
function of compensator is denoted as Gc(s), whereas that of the original process of the
plant is denoted by G(s)
...
Mostly
electrical networks are used as compensator in most of the control system
...

Lead Compensator
Lead compensator are used to improve the transient response of a system
...


Fig: Electric Lag Network
Taking i2=0 & applying Laplace Transform, we get
𝑉2(𝑠)
𝑅2𝐶𝑠 + 1
=
𝑉1(𝑠) (𝑅2 + 𝑅1)𝐶𝑠 + 1
Let 𝜏 = 𝑅2 𝐶 ,
𝑉2(𝑠)
𝑉1 (𝑠)

𝜏𝑠+1

= 1+𝜏𝛽𝑠

𝛽 = 𝑅1+𝑅2 > 1
𝑅2

Transfer function of Lag Compensator

33

Fig: S-Plane representation of Lag Compensator
Bode plot for Lag Compensator
Maximum phase lag occurs at 𝑤𝑚 =

1
𝜏 𝛽

Let 𝜙𝑚 = maximum phase lag

sin 𝜙𝑚 =

𝛽=

1−𝛽
1+𝛽

1 − sin 𝜙𝑚
1 + sin 𝜙𝑚

Fig: Bode plot of Phase Lag network

Cascade compensation in Time domain
Cascade compensation in time domain is conveniently carried out by the root locus
technique
...

A compensator is now designed so that the least damped complex pole of the
resulting transfer function correspond to the desired dominant pole & all other closed loop
poles are located very close to the open loop zeros or relatively far away from the jw axis
...

Lead Compensation
• Consider a unity feedback system with a forward path unalterable Transfer function
𝐺𝑓 (𝑠), then let the dynamic response specifications are translated into desired
location Sd for the dominant complex closed loop poles
...
e ∠𝐺𝑓 (𝑠) ≠ ±180° the uncompensated Root
Locus with variable open loop gain will not pass through the desired root location,
indicating the need for the compensation
...
In terms of angle criteria this requires that
∠𝐺𝑐 𝑠𝑑 𝐺𝑓 𝑠𝑑 = ∠𝐺𝑐 𝑠𝑑 + ∠𝐺𝑓 𝑠𝑑 ± 180°
∠𝐺𝑐 𝑠𝑑 = 𝜙 = ±180° − ∠𝐺𝑓 𝑠𝑑
• Thus for the root locus for the compensated system to pass through the desired root
location the lead compensator pole-zero pair must contribute an angle 𝜙
...
The best compensator pole-zero location is the one which gives the
largest value of
...

• The compensator pole then located by drawing a further requisite angle 𝜙 to be
contribute at Sd by the pole zero pair
...
The dominance condition must be checked before completing the
design
...
If the value of the error constant so obtained is unsatisfactory
the above procedure is repeated after readjusting the compensator pole-zero location while
keeping the angle contribution fixed as 𝜙
...
e its root locus plot
passes through(closed to) the desired closed loop poles location Sd
...
This requires that after compensation the root locus should
continue to pass through Sd while the error constant at Sd is raised to 𝐾𝑒𝑐
...
If this

36

pole-zero pair is located closed to each other it will contribute a negligible angle at Sd such
that Sd continues to lie on the root locus of the compensated system
...
that apart from being close to each other the pole-zero pair close to
origin, the reason which will become obvious from discussion below
...
e 𝑎 ≈ 𝑏
i
...

𝐾𝑒𝑐 → is error constant at Sd for compensated system
...
(1)
𝑝𝑐 𝐾𝑒

The 𝛽 parameter of lag compensator is nearly equal to the ratio of specified error constant
to the error constant of the uncompensated system
...

Since the Lag compensator does contribute a small negative angle 𝜆 at Sd , the actual error
constant will some what fall short of the specified value if 𝛽 obtained from equation(1) is
used
...

For the effect of the small lag angle 𝜆 is to give the closed loop pole Sd with specified 𝜁
but slightly lower 𝑤𝑛
...


Cascade compensation in Frequency domain
Lead Compensation
Procedure of Lead Compensation
Step1: Determine the value of loop gain K to satisfy the specified error constant
...

Step2: For this value of K draw the bode plot & determine the phase margin 𝜙 for the
system
...
A guess is made on the value of 𝜖 depending on the slope in this
region of the dB-log w plot of the uncompensated system
...
The guess value may have
to be as high as 15° 𝑡𝑜 20° for a slope of -60dB/decade
...

38

Step5: Find the frequency 𝜔𝑚 at which the uncompensated system will have a gain equals
to −10 log 1 from the bode plot drawn
...

Step6: Corner frequency of the network are calculated as

𝜔1 = 1 = 𝜔𝑚 𝛼 , 𝜔2 =
𝜏

1

𝜏𝛼

=

𝜔𝑚
𝛼
1

𝑠+

Transfer function for compensated system in Lead network 𝐺𝑐 𝑠 =

𝜏

𝑠+

1
𝜏𝛼

Step7: Draw the magnitude & Phase plot for the compensated system & check the resulting
phase margin
...

Lag Compensation
Procedure of Lead Compensation
Step1: Determine the value of loop gain K to satisfy the specified error constant
...

Step3: If ϕs =specified phase margin &
𝜙 = phase margin of uncompensated system (found out from the bode plot drawn)
𝜖 =margin of safety (5° − 10°)
• For a suitable 𝜖 find 𝜙2 = 𝜙𝑠 + 𝜖 ,where 𝜙2 is measured above −180° line
...

Step5: Measure the gain of uncompensated system at 𝜔𝑐2
...
𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛

𝜔𝑐2
2

𝑡𝑜

𝜔𝑐2
10

𝜏

)

Step7: Thus 𝛽 & 𝜏 are determined which can be used to find the transfer function of Lag
compensator
...


39

MATLab Code
Plotting rootlocus with MATLAB(rlocus)
Consider a unity-feedback control system with the following feedforward transfer function:
𝐾
𝐺 𝑠 =
𝑠 𝑠 + 1 (𝑠 + 2)
Using MATLAB, plot the rootlocus
...
If the sampler is used
to represent S/H (Sample and Hold) and A/D (Analog to Digital) operations, it may involve
delays, finite sampling duration and quantization errors
...
Following are two
popular sampling operations:
1
...
Multi-rate sampling
We would limit our discussions to periodic sampling only
...
1 Finite pluse width sampler

In general, a sampler is the one which converts a continuous time signal into a pulse
modulated or discrete signal
...

The symbolic representation, block diagram and operation of a sampler are shown in
Figure 1
...
Uniform rate
sampler is a linear device which satisfies the principle of superposition
...


where 𝑢𝑠(𝑡) represents unit step function
...
Thus 𝑓𝑝∗(𝑡) can be written as

42

Figure : Finite pulse with sampler : (a) Symbolic representation (b) Block diagram (c) Operation

According to Shannon's sampling theorem, "if a signal contains no frequency higher
than wc rad/sec, it is completely characterized by the values of the signal measured at instants
of time separated by T = π/wc sec
...

If the sampling rate is less than twice the input frequency, the output frequency will be
different from the input which is known as aliasing
...

The overlapping of the high frequency components with the fundamental component in the
frequency spectrum is sometimes referred to as folding and the frequency ws/2 is often
known as folding frequency
...

A low sampling rate normally has an adverse effect on the closed loop stability
...

Ideal Sampler : In case of an ideal sampler, the carrier signal is replaced by a train of unit
impulses as shown in Figure 2
...
e
...


43

the output of an ideal sampler can be expressed as

One should remember that practically the output of a sampler is always followed by a hold
device which is the reason behind the name sample and hold device
...
Thus the sampling process can be be always
approximated by an ideal sampler or impulse modulator
...
Then the inverse transform
is not necessarily equal to f(t), rather it is equal to f(kT) which is equal to f(t) only at the
sampling instants
...


The transform can be obtained by using
1
...
Power series
3
...

The Inverse Z-transform formula is given as:

47

MATLab Code to Obtain the inverse Z transform (filter)
Example

Obtain the inverse z transform of 𝑋 𝑧 =

𝑧(𝑧+2)

(𝑧−1) 2

X(z) can be written as
𝑋 𝑧 =

𝑧2 + 2𝑧
𝑧2 − 2𝑧 + 1

num=[1 2 0];
den=[1 -2 1];
u=[1 zeros(1,30)];%If the values of x(k) for k=0,1,2,
...
Let us consider that a discrete time system is described by the following difference
equation
...

We have to find the solution y(k) for k > 0
...

Relationship between s-plane and z-plane

In the analysis and design of continuous time control systems, the pole-zero configuration of
the transfer function in s-plane is often referred
...
Left half of s-plane

...

Unstable region
...

Similarly the poles and zeros of a transfer function in z-domain govern the performance
characteristics of a digital system
...
, in the s-plane where 𝑤𝑠 is the sampling frequency in
rad/sec
...


Figure : Primary and complementary strips in s-plane
The mapping is shown in Figure below
...

Mapping guidelines

1
...

2
...

3
...


Pluse Transfer Function

Pulse transfer function relates Z-transform of the output at the sampling instants to the Ztransform of the sampled input
...


Figure 1: Block diagram of a system subject to a sampled input
The output of the system is C(s) = G(s)R*(s)
...

Thus, for ease of manipulation, it is desirable to express the system characteristics by a
transfer function that relates r*(t) to c*(t), a fictitious sampler output, as shown in Figure 1
...
Similarly,

Since R*(s) is periodic R*( s + jnws ) = R*(s)
...
Sometimes it is also referred to as the starred
transfer function
...
Cascaded elements are separated by a sampler
The block diagram is shown in Figure below
...

2
...


Figure : Block diagram of a closed loop system with a sampler in the forward path
The objective is to establish the input-output relationship
...
The input to the sampler is regarded as
another output
...


Figure 2: Block diagram of a closed loop system with a sampler in the feedback path
The corresponding input output relations can be written as:

𝐸 𝑠 = 𝑅 𝑠 − 𝐻 𝑠 𝐶∗(𝑠)
… … … … … …
...


Since the input r(t) is not sampled, the sampled signal 𝑟∗(𝑡) does not exist
...


Stability Analysis of closed loop system in z-plane

Similar to continuous time systems, the stability of the following closed loop system

can also be determined from the location of closed loop poles in z-plane which are the
roots of the characteristic equation

1
...
Otherwise the system would be unstable
...
If a simple pole lies at 𝑧 = 1, the system becomes marginally stable
...

Multiple poles at the same location on unit circle make the system unstable
...

→ Jury Stability test
→ Routh stability coupled with bi-linear transformation
...

Jury Table

Rest of the elements are also calculated in a similar fashion
...
0756
58

All criteria are satisfied
...


59

MODEL QUESTIONS
Module-2
Short Questions each carrying Two marks
...
Determine the maximum phase lead of the compensator
0
...
5s + 1
2
...
Draw a representative
sketch of a lag-lead compensator
...
Derive Z transform of the following
X(1)=2; X(4)=-3; X(7)=8
and all other samples are zero
...


4
...

The figures in the right-hand margin indicate marks
...
The unity feedback system has the open loop plant 𝐺 𝑠 =

+

-

Lag
compensator

G(s)

1

𝑠 𝑠+3 (𝑠+6)

C

Design a lag compensation to meet the following specifications:
(i)
Step response settling time < 5s
...

6
...

(ii)
Find the output y(k) for the input u(k)=1 for ≥ 0
...
Use Jury‟s test to show that the two roots of the digital system F(z)=Z2+Z+0
...

[3]
8
...
[3]
(b) A type-1 unity feedback system has an open-loop transfer function
𝐾
𝐺 𝑠 =
𝑠 𝑠 + 1 (0
...
A discrete-time system is described by the difference equation
y(k+2)+5y(k+1) +6y(k) =u(k)
y(0)=y(1)=0; T=1 sec
(i) Determine a state model in canonical form
(ii) Find the state transition matrix
(iii) For input u(k)=1 for k ≥ 0, find the output y(k)
[5+5+6]
10
...
Write short notes on
(a) Feedback compensation
(b) stability analysis of sampled data control system
(c) R-C notch type a
...
lead network
(d) Hold circuits in sample data control
(e) Network compensation of a
...
(a) Describe the effect of:
(i) Lag and
(ii) Lead compensation on a root locus
[4]
(b) Design a suitable phase lag compensating network for a type-1 unity feedback
system having an open-loop transfer function
𝐾
𝐺 𝑠 =
𝑠 0
...
2𝑠 + 1)
to meet the following specifications:
Velocity error constant Kv = 30sec−1 and phase margin ≥ 40°
[12]
13
...
Find the inverse z-transform of
10𝑧
(i) 𝐹 𝑧 =
𝑧−1 (𝑧−2)

(ii) 𝐹 𝑧 =

−𝑎𝑇

𝑧(1−𝑒

𝑧−1 (𝑧−𝑒

)

−𝑎𝑇

[4+4]

)

61

15
...
Explain the relationship between s-plane & z-plane
...
How do you find out response between sampling instants?
18
...
𝟏𝟓𝒔)

[7]

[3]
[4]
[15]

Add series lag compensation to the servo mechanism to give a gain margin of ≥
15dB and a phase margin ≥ 45◦
...

19
...
(a) Derive the transfer function of zero order hold circuit
[4×3]
(b) State the specification in time domain and frequency domain used for the design of
continuous time linear system
...

21
...

[8]
1
i) G(s)= (𝑠+𝑎)
2
𝑠+𝑏
ii) G(s)=(𝑠+𝑎)
2+
𝑤

2

22
...

[8]

62

23
...

24
...


[4]
[4]

25
...

[7]
(b) Check the stability of the linear discrete system having the characteristics equation:
𝑧4 − 1
...
04𝑧2 − 0
...
024 = 0
26
...

(c) Explain the significance of shanon‟s theorem in sampling process
...
A linear control system is to be compensated by a compensating network having

[8]

[7]
[4]
[4]

𝐺𝐶(𝑆)= 𝐾𝑝+ 𝐾𝐷𝑠+ 𝐾𝑖𝑠
The system is shown in figure below

Find 𝐾𝑝,𝐾𝐷𝑠 and 𝐾𝑖 so that the roots of the characteristics equation are placed at s= -50, -5 +
...

28
...

Design a suitable lag network and compute the value of network components assuming any
suitable impedance level
...
(a) Find the z transform of the following:
[4+4]
−𝑎𝑡
2
i)y(t)=𝑒 𝑡
ii)G(s)= 𝑠+𝑎2 2
𝑠+𝑎

+𝑤

(b) For the system shown in fig below

G(z)=

[8]

𝑘(𝑧+0
...
7)

Determine the range of k for stability

---