Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
Lecture 14
Polar Plot of Frequency Response
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada
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Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
1. Review of Frequency Response
To specify the behavior of a system to a sinusoidal input at a particular angular frequency ω
i.e. the frequency response G(jω), we need both the magnitude | G(jω)| and phase φ of
G(jω).
Recall that both the magnitude and phase are functions of the input frequency and
therefore vary with ω. As we have seen when calculating the magnitude and phase during
the section of our course on Bode Diagrams, G(jω) is a complex number and can be
represented in the form x+jy or the rectangular form, where x and y are the real and the
imaginary parts, respectively. An alternative to represent G(jω) (calculated at frequency ω)
is the use of a polar plot, as shown below.
Im
|G(jω)|
φ
input
Re
In the above diagram, the output, for a unit amplitude sinusoidal input at frequency ω, has
magnitude |G(jω)| (i.e., the length of the vector ) and a phase angle φ.
Example
Find the expression of frequency response for the system with a transfer function of
1
G( s) =
, and then evaluate the magnitude and phase angle of frequency response at
1 + 2s
ω = 0.5 rad/s and represent the result in the complex plane.
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada
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Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
2. Sketch the Polar plot of Frequency Response
To sketch the polar plot of G(jω) for the entire range of frequency ω, i.e., from 0 to infinity,
there are four key points that usually need to be known:
(1)
(2)
(3)
(4)
the start of plot where ω = 0,
the end of plot where ω = ∞,
where the plot crosses the real axis, i.e., Im(G(jω)) = 0, and
where the plot crosses the imaginary axis, i.e., Re(G(jω)) = 0.
Consider a first order system
G (s ) =
1
1 + τs
where τ is the time constant.
Representing G (s ) in the frequency response form G ( jω ) :
G ( jω ) =
1
1 + jωτ
Multiplying both numerator and denominator by the conjugate of denominator, i.e.,
1 − jωτ , one has
G ( jω ) =
− ωτ
1 − jωτ
1
=
+j
2 2
2 2
ω4
τ3 11
+2
+2
1+ ω τ
11
ω 24
τ32
4
4
Re( G ( jω ))
Im(G ( jω ))
The magnitude of G ( jω ) , i.e., G ( jω ) , is
2
2
1
1

  − ωτ 
G ( jω ) = 
+
=
2 2 
2 2 
1+ ω τ  1+ ω τ 
1 + ω 2τ 2
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada
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Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
And the phase of G ( jω ) , denoted by φ , is
 − ωτ

1 + ω 2τ 2
φ = tan −1 
 1 1 + ω 2τ 2

(
(
)

)  = tan
−1


 − ωτ 


 1 
Now that we have expressions for the magnitude and phase of the frequency response, we
can sketch the polar plot using the 4 key points.
Point 1
The start of plot where ω = 0 .
G ( jω ) =
Point 2
1+ 0
0
1
φ = tan −1   = 0
= 1,
The end of plot where ω = ∞
G ( jω ) =
Point 3
1
1
1+ ∞
−∞
o
 = −90
 1 
φ = tan −1 
= 0,
Where the plot crosses the real axis, i.e., Im(G(jω)) = 0.
− ωτ
= 0 ⇒ ω = 0 and ω = ∞ . These are the same as Points 1 and 2.
1 + ω 2τ 2
Point 4
Where the plot crosses the imaginary axis, Re(G(jω)) = 0.
1
1 + ω 2τ 2
= 0 ⇒ ω = ∞ . This is the same as Point 2.
As Point 3 coincides with Points 1 and 2 and Point 4 coincides with Point 2, we need more
1
values of G ( jω ) evaluated at different frequencies. Taking ω = , then one has
τ
G ( jω ) =
1
1+1
=
1
2
,
 −1 
o
 = −45
 1 
φ = tan −1 
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada
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Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
Sketching the Polar plot for 1
(1 + τs ) , one has
Im
ω =0
ω=∞
(1, 0o)
o
(0, -90 )
Re
ω = 1/τ
(1 / 2 , - 45o )
Example
Sketch the polar plot for a third order system with a transfer function of
G ( s) =
1
2s + 3s 2 + 3s + 1
3
Solution
Replacing s in the above transfer function with jω, one has the frequency response
G ( jω ) =
1
2( jω ) 3 + 3( jω ) 2 + 3( jω ) + 1
=
1
(1 − 3ω 2 ) − j (3ω − 2ω 3 )
=
(1 − 3ω 2 ) + j (3ω − 2ω 3 ) (1 − 3ω 2 ) 2 + (3ω − 2ω 3 ) 2
The magnitude and phase of G(jω) are then obtained as:
G ( jω ) =
(Re(G ( jω )))2 + (Im(G ( jω )))2
φ = tan −1
(
=
 − 3ω − 2ω 3
Im(G ( jω ))
= tan −1 
2
Re(G ( jω ))
 1 − 3ω
1
(1 − 3ω ) + (3ω − 2ω )
2 2
3 2
) 


Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada
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Lecture Notes of Control Systems I - ME 431/Analysis and Synthesis of Linear Control System - ME862
Evaluation of the 4 key points of interest and sketch of polar plot will be illustrated in class.
Department of Mechanical Engineering, University Of Saskatchewan, 57 Campus Drive, Saskatoon, SK S7N 5A9, Canada
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