# Document ```4. System Response
• This module is concern with the response of LTI
system.
• L.T. is used to investigate the response of first and
second order systems. Higher order systems can
be considered to be the sum of the response of first
and second order system.
• Unit step, ramp, and sinusoidal signal play
important role in control system analysis. It is
therefore we will investigate this signals.
4. System Response
Review of some LTI properties
We will express system as in figure below,
system input is r(t), output is c(t), and impulse
response is h(t)
r(t)
h(t)
c(t)
1. Impulse response
Impulse response, denoted by h(t), is the
output of the system when its input is
impulse (t). h(t) is called the impulse
response of the system or the weighting
function
2. Convolution
Output of LTI system is the convolution
of its input and its impulse response:
where r’(t) denotes the derivative of r(t)
4. Integral
If the input is r(t)dt then the output is c(t)dt
t
c(t )  r (t ) * h(t )   r ()h(t  )d

t
  r (t  )h()d

Taking the L.T. of (1) yields
C(s) = R(s)H(s)
(2)
Where C(s) is the L.T. of c(t)
R(s) is the L.T. of r(t)
H(s) is the L.T. of h(t)
H(s) is called the transfer function (T.F)
3. Derivative
If the input is r’(t) then the output is c’(t)
(1)
5. Poles and Zero
T.F. is usually rational and therefore can be
expressed as N(s)/D(s). Poles is the values
of s resulting T.F to be infinite. Zeroes is
the values of s resulting T.F to be zero
4.1. Time Response of the First Order Systems.
Here we will investigate the time response of the
first order systems.
The transfer function of a general first order
system can be written as:
G(s) 
C ( s)
K

R( s) τs  1
(1)
We can found the differential equation first
we write (1) as
K
 1
 s  C ( s)  R( s)
τ
τ

(3)
(4)
c(0)
( K / τ)R(s)

s  (1 / τ)
s  (1 / τ)
(5)
The eq. can be shown in the block diagram as
shown in the figure bellow.
R(s)
K
τ
s
c(0)
Now we take the L.T of (3) and include the
initial condition term to get
1
K
sC ( s)  c(0)  C ( s)  R( s)
τ
τ
C ( s) 
(2)
The diff. Eq. is the inverse L.T. of (2)
1
K
c(t )  c(t )  r (t )
τ
τ
Solving for C(s) yields
+
1
C(s)
τ
1
s  1τ
+
Note that the initial condition as an input has
a Laplace transform of c(0), which is
constant.
The inverse L.T of a constant is impulse (t).
Hence the initial condition appears as the
impulse function
Here we can see that the impulse function
has a practical meaning, even though it is not
a realizable signal
4.1. Time Response of the First Order Systems.
Since we usually ignore the initial condition
in block diagram, we use the system block
diagram as shown bellow.
R(s)
K
C(s)
τ
s  1τ
c(t)
Suppose that the initial condition is zero then
( K / τ)
C (s) 
R( s)
s  (1 / τ)
(1)
(2)
Taking the inverse L.T of (2) yields
tτ
c(t )  K (1  e )
τ
K (1  e τ )
t
For unit step input R(s)=1/s, then
( K / τ) 1 K
K
 
( s  1 / τ) s s s  1 / τ
K
t
Unit step response
C (s) 
The first term originates in the pole of input R(s) and
is called the forced response or steady state response
The second term originates in the pole of the transfer
function G(s) and is called the natural response
Figure below plot c(t)
(3)
The final value or the steady state value of c(t) is K
that is
lim c(t)= K
t
c(t) is considered to reach final value after reaching
98% of its final value.
The parameter  is called the time constant. The
smaller the time constant the faster the system reaches
the final value.
4.1. Time Response of the First Order Systems.
A general procedure to find the steady state
value is using final value theorem
lim c(t) = lim sC(s) = lim sG(s)R(s)
t
s0
s0
For Unit step input then the final value is
css(t) = lim G(s)
s0
since R(s) = 1/s
therefore
c(t )  Kt  K  Ke τ )
t
(2)
This ramp response is composed of three term
1.
a ramp
2.
a constant
3.
an exponential.
System DC gain is the steady state gain to
a constant input for the case that the output
has a final value.
Ramp Response
c(t)
r(t)=t
c(t)
For the input equal to unit ramp function
r(t) = t and R(s) = 1/s2, C(s) is
t
C ( s) 
( K / τ) 1 K K
K



(1)
2
2
( s  1 / τ) s
s
s s 1/ τ
4.2. Time Response of Second Order System
Case 3: =1 (real equal poles), c(t) is
The standard form second order system is
2n
C ( s)
G( s ) 
 2
R(s) s  2n s  2n
The poles of the TF is
s = n
c(t )  1  k1et /   k2tet / 
(1)
This system is said to be critically damped
jω(12)
Case 4: =0 (imaginary poles), c(t) is
Where  = damping ratio
n = natural frequency, or undamped frequency.
c(t )  1  sin(nt  )
Consider the unit step response of this system
C ( s )  G ( s ) R( s ) 

(s 2  2n  2n )s
2
n
(2)
For this system we have
Time constant =  = 1/n ; frequency = n
=0
c(t)
Case 1: 0&lt;&lt;1 (complex poles), c(t) is
0.2
(3)
0.7
This system is said to be underdamped
Case 2: &gt;1 (real unequal poles), c(t) is
c(t )  1  k1et / 1  k2et / 2
This system is said to be overdamped
(3)
This system is said to be undamped
define  1   and   tan1 (   ).
c(t )  1  1 ent sin(nt  )
(5)
2
(4)
n t
```