3. Introduction
In previous chapters we learn to analyze
and design a control system using root
locus method.
Another method that can be used is
frequency response.
4. Introduction
What is frequency response?
The frequency response is a representation of
the system's open loop response to sinusoidal
inputs at varying frequencies.
The output of a linear system to a sinusoidal
input is a sinusoid of the same frequency but
with a different amplitude and phase.
The frequency response is defined as the
amplitude and phase differences between the
input and output sinusoids.
5. Introduction
Why do we use frequency response?
The open-loop frequency response of a system
can be used to predict the behaviour of the
closed-loop system .
we directly model a transfer function using
physical data.
6. Introduction
Frequency response is consists of:
Magnitude frequency response, M(ω)
Phase frequency response, ø(ω)
M o (ω )
M (ω ) =
M i (ω )
φ (ω ) = φo (ω ) − φi (ω )
7. Introduction
A transfer function Laplace form can be
change into frequency response using the
following expression:
G ( jω ) = G ( s ) s → jω
We can plot the frequency response in two
ways:
Function of frequency with separate magnitude
and phase plot.
As a polar plot.
8. Introduction
Magnitude and phase plot
Magnitude curve can be plotted in decibels
(dB) vs. log ω, where dB = 20 log M.
The phase curve is plotted as phase angle vs.
log ω.
Data for the plots can be obtained using
vectors on the s-plane drawn from the poles
and zeros of G(s) to the imaginary axis.
9. Introduction
Magnitude response at a particular frequency
is the product of the vector length from the
zeros of G(s) divided by the product of the
vector lengths from the poles of G(s) drawn to
points on the imaginary axis.
jω
jω1
A
O
O
B
X
C
X
D
A•B
M ( jω1 ) =
C•D
σ
10. Introduction
The phase response is the sum of the
angles from the zeros of G(s) minus the
sum of angles from the poles of G(s)
drawn to points on imaginary axis.
jω
jω1
θ1
O θ1
O
θ2
X
θ3
X
θ3
σ
φ ( jω1 ) = [θ1 + θ 2 ] − [θ3 + θ 4 ]
11. Introduction
Example 10.1
Find the analytical expression for the
magnitude frequency response and the phase
frequency response for a system G(s) = 1/
(s+2). Also, plot both the separate magnitude
and phase diagrams and the polar plot.
12. Introduction
Solution:
First substitute s=jω in the system function and
obtain
G ( jω ) =
1
jω + 2
We always put complex number as numerator so we
will multiply the above transfer function with the
complex conjugate.
1
jω + 2
1
− jω + 2
=
×
jω + 2 − jω + 2
(2 − jω )
= 2
(ω + 4)
G ( jω ) =
13. Introduction
In order for us to plot the magnitude frequency
response we must find the magnitude of the transfer
function.
Magnitude G(jω), M(ω)
G ( jω ) = G ( jω )g ( jω )∗
G
Where G(jω)* is the conjugate of G(jω), so the
magnitude for transfer function in the question is
G ( jω ) =
=
(2 − jω ) (2 + jω )
g 2
2
(ω + 4) (ω + 4)
1
(ω 2 + 4)
14. Introduction
The phase angle of G(jω), ø(ω)
B
φ (ω ) = − tan −1 ÷
A
(2 − jω )
A
B
(ω 2 + 4)
1
= 2
× (2 − jω )
(ω + 4)
G ( jω ) =
ω
φ (ω ) = − tan −1 ÷
2
15. Introduction
We can plot the magnitude frequency response and
phase frequency response
20 log M (ω ) = 1
ω 2 + 4 vs. log ω
φ (ω ) = − tan −1 ω 2 vs. log ω
21. Introduction
Nyquist criterion
Nyquist criterion relates the stability of a
closed-loop system to the open-loop frequency
response and open-loop pole location.
This concept is similar to the root locus.
The most important concept that we need to
understand when learning Nyquist criterion is
mapping contours.
25. Introduction
When checking the stability of a system,
the shape of contour that we will use is a
counter that encircles the entire right halfplane.
26. Introduction
The number of closed-loop poles in the right half
plane (also equals zeros of 1+ G(s)H(s)), Z
The number of open-loop poles in the right half
plane , P
The number of counterclockwise rotations about
(-1,0), N
N=P-Z
The above relationship is called the Nyquist
Criterion; and the mapping through G(s)H(s) is
called the Nyquist Diagram of G(s)H(s)
27. Introduction
Examples to determine the stability of a
system
P = 0, N = 0,
Z = P − N = 0, the system is stable
P = 0, N = −2, (clockwise '− ve ')
Z = 0 − (−2) = 2,system unstable
28. Sketching the Nyquist Diagram
The contour that encloses the right half-plane
can be mapped through the function G(s)H(s)
by substituting points along the contour into
G(s)H(s).
The points along the positive extension of the
imaginary axis yield the polar frequency
response of G(s)H(s).
Approximation can be made to G(s)H(s) for
points around the infinite semicircle by
assuming that the vectors originate at the
origin.
29. Sketching the Nyquist Diagram
Example 10.4
Sketch a nyquist diagram based on the block
diagram below.
30. Sketching the Nyquist Diagram
Solution
The open loop transfer function G(s),
500
G (s) =
( s + 1)( s + 10)( s + 3)
Replacing s with jω yields the frequency response of
G(s)H(s), i.e.
G ( jω ) =
500
( jω + 1)( jω + 10)( jω + 3)
(−14ω 2 + 30) − j (43ω − ω 3 )
= 500
(−14ω 2 + 30) 2 + (43ω − ω 3 ) 2
32. Sketching the Nyquist Diagram
Using the phase frequency response and magnitude
frequency response we can calculate the key points
on the Nyquist diagram. The key points that we will
calculate are:
Frequency when it crosses the imaginary and real
axis.
The magnitude and polar values during the
frequency that crosses the imaginary and real
axis.
The magnitude and polar values when frequency
is 0 and ∞.
33. Sketching the Nyquist Diagram
When a contour crosses the real axis, the imaginary
value is zero. So, the frequency during this is,
(−14ω 2 + 30) − j (43ω − ω 3 )
500
(−14ω 2 + 30) 2 + (43ω − ω 3 ) 2
(−14ω 2 + 30)
(43ω − ω 3 )
= 500
−j
2
2
3 2
(−14ω + 30) + (43ω − ω )
( −14ω 2 + 30) 2 + (43ω − ω 3 ) 2
real
imaginary
(43ω − ω 3 )
=0
2
2
3 2
(−14ω + 30) + (43ω − ω )
34. Sketching the Nyquist Diagram
We need to find the frequency when imaginary is
zero by finding the value of ω that could produce
zero imaginary value.
There are actually two conditions that could produce
zero imaginary.
First
0
=0
2
2
3 2
(−14ω + 30) + (43ω − ω )
Second
(43ω − ω 3 )
=0
∞
35. Sketching the Nyquist Diagram
For the first condition, in order to get the numerator
equals to zero we must find the root value of the
numerator polynomial.
(43ω − ω 3 )
0
=
2
2
3 2
(−14ω + 30) + (43ω − ω )
( −14ω 2 + 30) 2 + (43ω − ω 3 ) 2
(43ω − ω 3 ) = 0
ω1 = 0
ω2 = 6.5574
ω3 = −6.5574
There are three frequencies
where the contour crosses the
real axis.
36. Sketching the Nyquist Diagram
For the second condition, the frequency values in the
denominator that could produce zero imaginary value
is infinity, ∞.
(43ω − ω 3 )
(43ω − ω 3 )
=
2
2
3 2
(−14ω + 30) + (43ω − ω )
∞
(−14ω 2 + 30) 2 + (43ω − ω 3 ) 2 = ∞
ω =∞
38. Sketching the Nyquist Diagram
There are two conditions that could produce zero real
value.
First
0
=0
2
2
3 2
(−14ω + 30) + (43ω − ω )
Second
(−14ω 2 + 30)
=0
∞
39. Sketching the Nyquist Diagram
Calculate the frequency values for the first condition.
(−14ω 2 + 30)
0
=
2
2
3 2
(−14ω + 30) + (43ω − ω )
(−14ω 2 + 30) 2 + (43ω − ω 3 ) 2
(−14ω 2 + 30) = 0
ω = ±1.4639
Calculate the frequency values for the second
condition
(−14ω 2 + 30)
( −14ω 2 + 30)
=
2
2
3 2
(−14ω + 30) + (43ω − ω )
∞
ω=∞
40. Sketching the Nyquist Diagram
Now that we know the frequencies of the key points
in our polar plot we will now calculate the
magnitudes and phase for each key points frequency.
Cross real
Cross imaginary
41. Sketching the Nyquist Diagram
The new contour can be plot based on the key points
in the previous table.
ω = 6.5574
ω =∞
C
ω = 1.4639
A
ω=0
42. Sketching the Nyquist Diagram
Note that the semicircle with a infinite
radius, i.e., C-D, is mapped to the origin if
the order the denominator of G(s) is
greater than the order the numerator of
G(s).
