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1. TIME RESPONSE
Engr. Joselito O. Daroy

2. POLES, ZEROS, and SYSTEM RESPONSE
TWO RESPONSES OF SYSTEM OUTPUT
1. FORCED RESPONSE
- it is also called the steady-state response or particular
solution
2. NATURAL RESPONSE
- it is also called the homogeneous solution

3. POLES of a TRANSFER FUNCTION
The POLES of a transfer function are:
1. The values of the Laplace transform variable, s, that
cause the transfer function to become infinite or
2. Any roots of the denominator of the transfer function that
are common to roots of the numerator

4. ZEROS of a TRANSFER FUNCTION
The ZEROS of a transfer function are:
1. The values of the Laplace transform variable, s, that
cause the transfer function to become zero or
2. Any roots of the numerator of the transfer function that
are common to roots of the denominator

5. Example: Given the transfer function G(s) in figure 1a
Multiplying the transfer function of figure 1a by a step
function yields:
( )
( )
( )
( )
2
5
5
3
2
5 5
5
+
=
+
= +
+
= +
+
s
C s
s s
A B
s s
C s
s s

6. Taking the inverse Laplace transform
( )
 
( )
1 1
5
3
2
5 5
5
2 3
5 5
 
− −
−
 
 
= +
 
+
 
 
= + t
C s
s s
c t e

7. Figure 1
a. System showing
input and output;
b. pole-zero plot
of the system;
c. evolution of a
system response.
Follow blue arrows
to see the evolution
of the response
component generated
by the pole or zero.

8. 1. A pole of the input function generates the form of the
forced response (that is, the pole at the origin generated
a step function at the output
2. A pole of the transfer function generates the form of the
natural response
3. A pole on the real axis generates an exponential
response of the form e-t, where - is the pole location
on the real axis. Thus, the farther to the left of a pole is on
the negative real axis, the faster the exponential transient
response will decay to zero (see figure 2 for the general
case)
4. The zeros and poles generate the amplitudes for both
the forced and natural responses.

9. Example 1. Given the system in figure 2, write the output,
c(t), in general terms and specify the forced and natural
parts of the solution

10. Solution:
By inspection, each system pole generates an
exponential as part of the natural response.
The input’s pole generates the forced response.
Thus
( )
K
K K K
C s
s s s s
= + + +
+ + +
3
1 2 4
2 4 5

11. Solution:
Taking the inverse Laplace.
( ) t t t
c t K K e K e K e
− − −
= + + +
2 4 5
1 2 3 4
Where:
K1 is the forced response
the exponential parts are the Natural response

12. FIRST-ORDER SYSTEMS
A First-order system without zeros can be described by the
transfer function shown in figure 3a.
If the input is a unit step, where R(s) = 1/s
( ) ( ) ( )
( )
( )
( )
1
C s R s G s
a
C s
s s a
=
=
+

13. FIRST-ORDER SYSTEMS
The step response is given by
( ) ( ) ( )
( ) ( )
1 2
f n
at
c t c t c t
c t e−
= +
= −

14. When t = 1/a
( )
1
1
1
0.37 3
at
t
a
at
t
a
e e
e
− −
=
−
=
=
=
( )
( )
( ) ( )
1 1
1
1
1
1 0.37
0.63 4
at
t t
a a
t
a
t
a
c t e
c t
c t
−
= =
=
=
= −
= −
=

15. Figure 3
a. First-order
system;
b. pole plot

16. Figure 4
First-order
system
response to a
unit
step

17. TIME CONSTANT, Tc
The TIME CONSTANT can be described as the time for e-at
to decay to 37% of its initial value.
alternately, the time constant is the time it takes for the step
response to rise to 63% of its final value
The reciprocal of the time constant is known as frequency
and the parameter a is called exponential frequency
1
c
T
a
=

18. Rise Time is defined as the time for the
waveform to go from 0.1 to 0.9 of its final
value.
Rise Time is found by solving the difference
in time at c(t) = 0.9 and c(t) = 0.1
Hence:
Rise Time, Tr
2.31 0.11
2.2
r
r
T
a a
T
a
= −
=

19. Settling Time is defined as the time for the
response to reach and stay within 2% of its
final value.
Letting c(t) = 0.98 and solving for the time,
the settling time will be:
Settling Time, Ts
4
s
T
a
=