EET 310
Transient Response of a Second Order System
-Learning Outcomes
Students will demonstrate the ability to:
a- derive a system’s transfer function
b- analyze a system’s transfer function to predict its behavior
c- use LaPlace methods to obtain and analyze response characteristics.
-Problem Statement
For the circuit shown, let vin(t) = 10u(t)
1- Derive the expression for the transfer function G(s)
2- For each of the cases, determine:
a- the natural frequency
b- the damping ratio
c- the damping frequency
d- the poles and zeros of the transfer function
e- plot the poles in the complex plane
f- determine vo(t) due to a step input
g- verify the correctness of your answer in (f) by comparing the response to that of Matlab.
h- from the plot, determine the rise time, settling time, and % overshoot
Let L = 0.2H, C = 0.05F and R = { 8, 4 0, 0.8 Ω}
- First draw the s-domain circuit: XL = Ls, Xc = 1/Cs, Vin(s) and Vo(s)
The transfer function in standard form is:
G(s) =
- Transfer function for this circuit is: G(s) =
by comparison:
and,
, thus:
(1)
√
, thus:
Case -1 ( R =8Ω)
G(s) =
(2)
= 10 rad/sec,
The system is over damped. No oscillation
N/A
From equation (2), the system has no zeros. the poles are located at:
= 0  s1,2 = {-2.68, -37.3 }
The dominant pole is located at -2.68. The transients due to the farther
to the left pole decay faster.
Pole-Zero Map
5
4
3
2
Imaginary Axis
1
0
-1
-2
-3
-4
-5
-40
-35
-30
-25
-20
-15
-10
Real Axis
To determine vo(t):
Vo(s) = Vin(s)G(s) =
By using methods of PFE:
|
 A =10
|
 B = -10.78
|
 C = 0.78
Therefore:
Vo(s) =
From LaPlace transform tables:
{
Check for
}
at t = 0.2 sec.
-5
0
5
10
= 3.693V
Let us verify with Matlab®
num=[100];
den=[1 40 100];
G=tf(num,den);
step(10*G)
Step Response
10
9
8
7
Amplitude
6
5
System: untitled1
Time (sec): 0.2
Amplitude: 3.69
4
3
2
1
0
0
0.5
1
1.5
2
2.5
3
Time (sec)
As can be seen, the system response is in agreement with the derivation.
Step Response
10
System: untitled1
Settling Time (sec): 1.49
9
System: untitled1
Rise Time (sec): 0.823
8
7
Amplitude
6
settling time
5
4
3
rise time
2
1
0
0
0.5
1
1.5
2
Time (sec)
Table 1. Summary of Results for Case -1
Parameter
Response
zeros
poles
tr
ts
%O.S
Value
10 rad/sec.
2
N/A
over-damped,
stable
none
-2.68, -37.3
0.823 s.
1.49 s.
0
2.5
3
Case -2 ( R =4Ω)
G(s) =
(2)
= 10 rad/sec,
The system is critically damped. No oscillation
N/A
From equation (2), the system has no zeros. the system has double
poles located at -10:
= 0  s1,2 = {-10, -10 }
Pole-Zero Map
5
4
3
2
Imaginary Axis
1
0
-1
-2
-3
-4
-5
-20
-15
-10
-5
Real Axis
To determine vo(t):
Vo(s) = Vin(s)G(s) =
By using methods of PFE:
|
|
 A =10
 B = -100
0
5
10
{
}|
|
 C = -10
Therefore:
Vo(s) =
From LaPlace transform tables:
{
}
Check for
at t = 0.2 sec.
= 5.94 Volts
Let us verify with Matlab®
num=[100];
den=[1 20 100];
G=tf(num,den);
step(10*G)
Step Response
10
9
8
7
System: untitled1
Time (sec): 0.2
Amplitude: 5.94
Amplitude
6
5
4
3
2
1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time (sec)
As can be seen, the system response is in agreement with the derivation.
Step Response
10
System: untitled1
Settling Time (sec): 0.583
9
System: untitled1
Rise Time (sec): 0.336
8
7
Amplitude
6
5
4
3
2
1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Time (sec)
Table 2. Summary of Results for Case -2
Parameter
Response
zeros
poles
tr
ts
%O.S
Value
10 rad/sec.
1
N/A
critically damped,
stable
none
-10, -10 (double
poles)
0.336 s.
0.583 s.
0
0.9
1
Case -3 ( R = 0 Ω)
G(s) =
(2)
= 10 rad/sec,
The system has no damping
√
=
= 10 rad/sec.
From equation (2), the system has no zeros. the system has a complex pair
= 0  s1,2 = {+j10, -j10 }
Pole-Zero Map
15
10
Imaginary Axis
5
0
-5
-10
-15
-1
-0.8
-0.6
-0.4
To determine vo(t):
Vo(s) = Vin(s)G(s) =
By using methods of PFE:
-0.2
0
Real Axis
0.2
0.4
0.6
0.8
1
In class work here :
Thus, A = 10, B = -5, B* = -5
Therefore:
Vo(s) =
And, verify that:
= 10 -10cos(10t) u(t)
Check for
at t = 0.2 sec.
= 14.16 Volts
Let us verify with Matlab®
num=[100];
den=[1 0 100];
G=tf(num,den);
>> t=0:0.0001:5;
>> step(10*g,t)
Step Response
20
18
System: untitled1
Time (sec): 0.2
Amplitude: 14.1
16
14
Amplitude
12
10
8
6
4
2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Time (sec)
As can be seen, the system response is in agreement with the derivation.
Table 3. Summary of Results for Case -3
Parameter
Response
zeros
poles
tr
ts
%O.S
Value
10 rad/sec.
0
10 rad/sec. (theoretical)
no damping,
Marginally stable
none
complex pair at +/j10
N/A
N/A
N/A
Case -4 ( R = 0.8 Ω)
G(s) =
(2)
= 10 rad/sec,
The system is underdamped
=
√
9.8 rad/sec.
From equation (2), the system has no zeros. the system has
a complex pair
= 0  s1,2 = {-2 ± j 9.8 }
Pole-Zero Map
15
System: g
Pole : -2 + 9.8i
Damping: 0.2
Overshoot (%): 52.7
Frequency (rad/sec): 10
10
Imaginary Axis
5
0
-5
-10
-15
-10
System: g
Pole : -2 - 9.8i
Damping: 0.2
Overshoot (%): 52.7
Frequency (rad/sec): 10
-8
-6
-4
To determine vo(t):
Vo(s) = Vin(s)G(s) =
By using methods of PFE:
In class work here :
-2
0
Real Axis
2
4
6
8
10
Thus, A = 10, B =
Therefore:
Vo(s) =
To find vo(t)
, B* =
And, verify that:
= 10 + 10.2
Check for
at t = 0.2 sec.
= 11.222 Volts
u(t)
Let us verify with Matlab®
num=[100];
den=[1 4 100];
G=tf(num,den);
>> t=0:0.0001:5;
>> step(10*g,t)
Step Response
16
14
System: untitled1
Time (sec): 0.2
Amplitude: 11.3
12
Amplitude
10
8
6
4
2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Time (sec)
Step Response
16
System: untitled1
Peak amplitude: 15.3
Overshoot (%): 52.7
At time (sec): 0.321
14
12
Amplitude
10
System: untitled1
Settling Time (sec): 1.96
System: untitled1
Rise Time (sec): 0.12
8
6
4
2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Time (sec)
As can be seen, the system response is in agreement with the derivation.
Table 4. Summary of Results for Case -4
Parameter
Response
zeros
poles
tr
ts
%O.S
Theoretical estimates:
In class
Value
10 rad/sec.
0.2
9.8 rad/sec.
(theoretical)
under-damped
none
complex pair at
-2 +/- j9.8
0.12 sec.
1.96 sec.
52.7