2
Solutions to Questions
Question 1
a) System poles occur at the roots of the denominator of
5
5
=
s2 + 11s + 10
(s + 1)(s + 10)
G(s) =
The poles are at s = −1, s = −10. There are no zeros.
The poles are in the left half plane so the system is stable.
b) The steady state gain of the system is 0.5 = −3dB. There are break points at frequencies s = 1rad/s
and s = 10 rad/s, with attenuation of 40 dB per decade at high frequencies. As there are 2 poles
and no zeros the total phase lag is 180o at high frequencies.
One way to draw the Bode plot in Matlab is to use the commands:
>> s=tf(’s’);
>>G=5/(s^2+11*s+10);
>>bode(G);
The result is shown in Figure 2.1.
Bode Diagram
Magnitude (dB)
0
−20
−40
−60
−80
Phase (deg)
−100
0
−45
−90
−135
−180
−2
10
−1
10
0
1
10
10
2
10
Frequency (rad/sec)
Figure 2.1: Bode plot of G(s)
4
3
10
The Nyquist plot starts at the point (0.5+0j), proceeds with negative phase, crosses the imaginary
axis and terminates at the origin. It has a phase of −180o as it approaches the origin.
The Nyquist plot is drawn with the command:
>>nyquist(G);
The result is shown in Figure 2.2.
Nyquist Diagram
0.3
0.2
Imaginary Axis
0.1
0
−0.1
−0.2
−0.3
−0.4
−0.2
−0.1
0
0.1
0.2
Real Axis
0.3
0.4
0.5
Figure 2.2: Nyquist plot of G(s)
c)
i) The closed loop transfer function from r to y is
Gyr (s) =
K(s)G(s)
1 + K(s)G(s)
ii) The closed loop transfer function from r to e is
Gye (s) =
1
1 + K(s)G(s)
Gyr (s) is then
1
50(1+s)
s(s+1)(s+10)
50(1+s)
+ s(s+1)(s+10)
50
s(s + 10) + 50
50
= 2
s + 10s + 50
=
In the transfer function from r to y there are no closed loop zeros; the closed loop poles are at
s = −5 + 5i and s = −5 − 5i.
The steady state gain in closed loop is 1 because of the presence of the integrator in the loop.
5
Bode Diagram
Magnitude (dB)
0
−20
−40
−60
Phase (deg)
−80
0
−45
−90
−135
−180
−1
10
0
1
10
10
2
10
3
10
Frequency (rad/sec)
Figure 2.3: Bode plot of Gyr (s)
One way to calculate the closed loop transfer function and its poles using Matlab is as follows:
>> K=10*(1+s)/s;
>> Gyr=G*K/(1+G*K)
Transfer function:
50 s^4 + 600 s^3 + 1050s^2 + 500 s
--------------------------------------------------s^6 + 22 s^5 + 191 s^4 + 820 s^3 + 1150 s^2 + 500 s
>> Gyr=minreal(G*K/(1+G*K))
Transfer function:
50
--------------s^2 + 10 s + 50
>> pole(Gyr)
ans =
-5.0000 + 5.0000i
6
Bode Diagram
Magnitude (dB)
10
0
−10
−20
−30
Phase (deg)
−40
90
45
0
−1
10
0
1
10
10
2
10
Frequency (rad/sec)
Figure 2.4: Bode plot of Ger (s)
-5.0000 - 5.0000i
Where the command minreal removes the pole zero cancellations in Gyr.
√
10
The natural frequency is ωn = 50 = 7.07 and the damping factor is ζ = 2ω
= 0.707.
n
d) The closed loop step response from r to y is obtained with
>> step(Gyr)
The result is shown in Figure 2.6.
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Bode Diagram
40
Magnitude (dB)
20
0
−20
−40
−60
Phase (deg)
−80
−90
−135
−180
−1
10
0
10
1
10
Frequency (rad/sec)
2
10
3
10
Figure 2.5: Bode plot of K(s)G(s)
e) Matlab code for the Bode plot of
i) Gyr (s) is
>> bode(Gyr)
ii) and for the Bode plot of Ger (s)
>> Ger=1/(1+G*K);
>>bode(Ger);
The results are shown in Figures 2.3 and 2.4.
f) The closed loop response from r to y has 0 dB gain at low frequency, i.e. steady state gain of
1. The response is critically damped with natural frequency ωn = 7 radians/s. The rise time is
approximately
1.7
τr ≈
= 0.24s
ωn
g) The open loop Bode plot of G(s)K(s) can be generated in Matlab with the command
bode(G*K)
The Bode plot is shown on Figure 2.5. At zero frequency the bode plot has infinite gain, which
correlates to a closed loop steady state gain of 1. The magnitude plot crosses 1 (0dB) at frequency
4.5radians/s, this is approximately the closed loop bandwidth.
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Step Response
1.4
1.2
Amplitude
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
1.2
Time (sec)
Figure 2.6: Closed loop step response of Gyr (s)
The phase margin is 65.5o , while the gain margin is infinite. A rule of thumb relating the damping
factor ζ to the phase margin PM is
PM
ζ=
100
Using this rule yields a closed loop damping factor of approximately 0.65. The actual value is 0.707,
from part c.
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Question 2
The unforced solution can be obtained as follows. The homogeneous equation is
ẋ = −3x
Using the fact that
d 3t
(e x) = e3t (ẋ + 3x) = 0
dt
integrating both sides gives
e3t x(t) = c
Here c takes the value x0 , so the homogenous solution is
x(t) = x0 e−3t
The solution with an input u(t) uses
d 3t
(e x) = e3t (ẋ + 3x) = e3t u
dt
Integrating yields
Z
x(t) = x0 e−3t +
t
e−3(t−τ ) u(τ )dτ
0
which for the given input is
Z
−3t
x(t) = x0 e
t
+ 1.5
e−3(t−τ ) dτ
0
or
x(t) = x0 e−3t + 0.5(1 − e−3t )
Question 3
a) Using the formula for 2 × 2 matrices, one obtains
·
−1
A
−5
=
−2
¸
6
2
The characteristic polynomial is
det(sI − A)
For the given matrix we have
(s + 5)(s − 2) + 12 = s2 + 3s + 2
10
The eigenvalues of A are the roots of the characteristic polynomial, s = −1 and s = −2.
An eigenvector of A corresponding to an eigenvalue λ is a vectors q that satisfies
Aq = λq
So with λ1 = −1
Aq1 = −q1
This leads to two equations in the elements of q1 with solution
· ¸
2
q1 = ρ1
1
where ρ1 6= 0 is an arbitrary scaling factor. Similarly,
· ¸
3
q2 = ρ2
2
The eigenvalue-eigenvector decomposition is then
A = QΛQ−1
or
·
−0.8944
A=
−0.4472
¸·
−0.8321
−0.5547
−2
0
¸·
0
−1
¸
−4.4721 6.7082
3.6056 −7.2111
where the factors ρi have been used to scale the eigenvectors to norm 1.
b) The Matlab commands are as follows:
>>A=[-5 6;-2 2];
>> Ainv=inv(A)
Ainv =
1.0000
-3.0000
1.0000
-2.5000
>> [e,ev]=eig(A)
e =
-0.8944
-0.8321
-0.4472
-0.5547
ev =
-2
0
0
-1
>> poly(A)
ans =
1
3
2
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