Review last lectures
procedures for drawing block diagram
1.
2.
3.
4.
Write the equations that describe the dynamic
behavior for each component.
Take Laplace transform of these equations,
assuming zero initial conditions.
Represent each Laplace-transformed equation
individually in block form.
Assembly the elements into a complete block
diagram.
block diagram: example
R
Let consider the
RC circuit:
The equations for
this circuit are:
ei
C
eo
i
ei  eo
i
R
eo
idt


C
block diagram: example
take Laplace transform:
 ei (t )  eo (t )  Ei ( s)  Eo ( s)
I ( s)  L 

R
R




idt
   I s 
Eo s   L 

C  Cs



block diagram: example
block representations for Laplace transforms:
I s 
E o s  
Cs
Ei ( s)  Eo ( s )
I (s) 
R
Ei (s )
+_
Eo s 
1
R
I (s)
I (s)
1
Cs
Eo s 
block diagram: example
Assembly the elements into a complete block diagram.
Ei (s )
+_
Eo s 
1
R
I (s)
I (s)
1
Cs
Eo s 
block diagram reduction
Rules for reduction of the block diagram:

Any number of cascaded blocks can be reduced by a
single block representing transfer function being a
product of transfer functions of all cascaded blocks.

The product of the transfer functions in the
feedforward direction must remain the same.

The product of the transfer functions around the loop
mast remain the same.
block diagram: reduction example
H2
C
_
R
+_
+
+
G1
+
H1
G2
G3
block diagram: reduction example
H2
G1
_
R
+_
+
+
+
C
G1
H1
G2
G3
block diagram: reduction example
H2
G1
_
R
+_
+
+
C
+
G1G2
H1
G3
block diagram: reduction example
H2
G1
C
_
R
+_
+
+
G1G2
+
H1
G3
block diagram: reduction example
H2
G1
_
R
+_
+
G1G2
1  G1G2 H1
C
G3
block diagram: reduction example
H2
G1
_
R
+_
+
G1G2G3
1  G1G2 H1
C
block diagram: reduction example
R
+_
G1G2G3
1  G1G2 H1  G2G3 H 2
C
block diagram: reduction example
R
G1G2G3
1  G1G2 H1  G2G3 H 2  G1G2G3
C
block diagram: reduction example
NOTICE:

Numerator of the closed-loop transfer function
C(s)/R(s) is the product of the transfer functions of
the feedforward path.
1.
The denominator of the closed-loop transfer
function C(s)/R(s) is equal to:
1-Σ( product of the transfer functions around each loop)
2.
The positive feedback loop yields a negative term
in the denominator.
signal flow graph
input node (source)
transmittance
branch
x4
mixed node
node
x1
d
a
b
forward
x2 pathloop
c
input node (source)
mixed node
path
1
x3
x3
output node (sink)
flow graphs of control systems
block diagram:
signal flow graph:
G (s )
C (s )
R(s)
G (s )
R(s)
C (s )
flow graphs of control systems
R(s)
C (s )
E (s )
+_
G (s )
block diagram:
H (s )
signal flow graph:
G (s )
1
R(s)
E (s )
 H (s )
C (s )
flow graphs of control systems
R(s)
E (s )
G (s )
+_
+
N (s )
C (s )
+
block diagram:
H (s )
signal flow graph:
R(s)
N (s )
1
G (s )
E (s )
 H (s )
1
1
C (s ) C (s )
signal flow graph algebra
a
x1
x2
b
a
x1
x2  ax1
x2
ab
x3
x1
x2
signal flow graph algebra
a
x1
x1
x2
x2
b
a
b
x1
x1
c
x3
ab
x4
x2
ac
x4
x2
bc
signal flow graph algebra
x1 a x2 b x3
c
x3  abx1  bcx3
x1
ab
1  bc
x3  bx2
x2  ax1  cx3
x1 ab x3
bc
x3
x3
ab

x1
1  bc
flow graphs of linear systems
x1  a11 x1  a12 x2  a13 x3  b1u1
x2  a21 x1  a22 x2  a23 x3  b2u2
x3  a31 x1  a32 x2  a33 x3
a11
u1
x1
1
b1
x1
a21
a12
a31
a22 x2
a33
x2
a23a32 x3
u2 b2
a13
1
x3
Transient and steady state
response analyses


Input signal not known ahead of time but is
random in nature!
In analyzing and designing we need basic
comparison of performance so we need input
signal by specifying particular test input
signals.
Typical Test Signals
Step functions
Ramp functions
Acceleration (Parabolic) functions



Impulse functions
1 /  , -  t 
f  t   
2
2

Sinusoidal functions
0, otherwise
As   0, f t   Unit impulse function t 
Polynomial functions
1.
2.
3.
4.
5.
6.


Note: which input signals we must use?
Depend on the system normal operation inputs
Test Input Signals

The impulse input is useful when we consider the convolution
integral for the output y(t) in terms of an input r(t):
yt  
t
1




Gs Rs 
g
t



t
d


L



This relationship is shown in the block diagram:

If the input is a unit impulse function then
yt   g t 
Transient response and Steady- State
response

Time response parts
Steady state
Output which behaves as t
approach infinity
 Transient
Which Goes from initial
state to the final state
Or output minus steady
state output!

First order systems response
Final value Theorem for
steady state response
Second order systems response
(steady state part)
Second order systems types

Over damped


Critically damped


real poles
Real equal poles
Under damped

Imaginary poles
Over damped response
Critically damped
Under damped
Under damped 2
Performance of a SecondOrder System

Consider the system:

The closed loop output is:
Y s  
G s 
K
Rs   2
Rs 
1  G s 
s  ps  K
Or
n2
Y s   2
Rs 
2
s  2 n s  n
W it ha unit st ep input ,
n2
Y s  
s s 2  2 n s  n2 
T he transientoutputis :
1
y t   1  e  t sin t       n

  n 1   2 ,   cos1  , 0    1
Stable
Unstable
Performance of a SecondOrder System


The transient response of this second-order system is shown below.
As ζ decreases, the closed-loop roots approach the imaginary axis,
and the response becomes oscillatory.
1
Based on : yt   1 
e t sin n 1   2 t  
2
n
1 


Performance of a SecondOrder System

For the unit impulse (R(s)=1) the output is:
n2
Y s   2
s  2 n s  n2

The transient response is:
yt  

n
1  2
ent sin n 1   2 t
The impulse response of the
second order system is shown here
Performance of a SecondOrder System

Standard performances are usually defined in terms of the step
response of a system as shown below:

The swiftness of the
response is measured
by the rise time (Tr),
and the peak time (Tp).
Performance of a SecondOrder System

The percent overshoot for the unit step input is defined as:
P.O.



M pt  fv
fv
100%
Mpt: is the peak value of the time response
fv: is the final value of the response
Normally fv is the magnitude of the input, but many systems have a
final value that is different from the desired input magnitude.
Performance of a SecondOrder System



Settling time (Ts): the time required for the system to settle
within a certain percentage, δ, of the input amplitude.
The settling time is four time constants (τ=1/ζωn) of the
dominant roots of the characteristic equation.
The steady-state error of the system may be measured on the
step response of the system as shown in the previous figure.
Performance of a SecondOrder System

Consider the second order system with closed-loop damping
constant ζωn and a response described by
yt   1 

1
1  2

ent sin n 1   2 t  

We seek to determine the time, Ts, for which the response remains
within 2% of the final value.
e  nTs  0.02
 nTs  4
Ts  4 
4
 n
Performance of a SecondOrder System

The transient response of the system may be
described in terms of two factors:


The swiftness of the response, as represented by
the rise time and the peak time
The closeness of the response to the desired
response, as represented by the overshoot and
settling time
Performance of a SecondOrder System

The peak time relationship for this second-order system is:
Tp 


n 1   2
The peak response is:
M pt  1  e


The percent overshoot is:
P.O.  100e

1 2


1 2
Performance of a SecondOrder System

The swiftness of step response can be measured as the time it takes
to rise from 10% to 90% of the magnitude of the step input. This is the
rise time (Tr).
 
Tr 






  n 1   2 ,   cos 1  , 0    1
The swiftness of a response to a step input is dependent of ζ and ωn.
For a given ζ, the response is faster for larger ωn.
The overshoot is independent of ωn.
For a give ωn, the response is faster for lower ζ.
The swiftness of the response will be limited by the overshoot that
can be accepted
Quiz and Home work


Don’t forget quiz in next time
Home works of chapter 1

No: 1, 2, 3, 4, 5, 6, 7, 8, 13.