Module-9:Frequency Domain Analysis of Linear Systems
Polar Plot
Akshya Swain
Department of Electrical, Computer & Software Engineering, The University of Auckland,
Auckland, New Zealand.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 1 / 36
Learning Outcome
After completion of this module the students would have learned following:
1. How to draw the polar plot ?
2. How to compute gain and phase margins from the polar plot.
3. How to investigate the relative stability of a system from the polar plot.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 2 / 36
Polar Plot
1. The polar plot of a sinusoidal transfer function G(jω) is a plot of the magnitude of G(jω) versus the phase angle of G(jω) on polar coordinates as ω is varied from zero to infinity.
2. Thus,the polar plot is the locus of vectors as ω is varied from zero to infinity.
3. Note that in polar plots a positive (negative) phase angle is measured counterclockwise (clockwise) from the positive real axis.
4. .Each point on the polar plot of G(jω) represents the terminal point of a vector at a particular value of ω.
5. The projections of G(jω) on the real and imaginary axes are its real and imaginary components.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 3 / 36
Polar Plot-2
I 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. The start of the plot where ω = 0.
2. Theendofplotwhereω=∞.
3. Where the plot crosses the real axis, i.e., Im [G(jω)] = 0 and 4. Where the plot crosses the imaginary axis, i.e., Re [G(jω)] = 0.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 4 / 36
Polar Plot of Pure Integral
I For a pure integrator
G(s)= s, or, G(jω)= jω =−jω = ω\−90
11110 Point-1: The polar plot starts at ω = 0. This gives
|G(jω)|= 1 =∞, φ=−900 ω
Point-2: The polar plot ends at ω = ∞. This gives |G(jω)|= 1 =0, φ=−900
ω
Point-3: The real axis cross-over point of polar plot i.e Im[G(jω)] = 0
1=0, =⇒ ω=∞ ω
Point-4: The imaginary axis cross-over point of polar plot i.e Re[G(jω)] = 0. This does not exist; plot does not cross imaginary axis
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 5 / 36
Polar Plot of Pure Integral
The polar plot of pure integral factor is the negative imaginary axis
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 6 / 36
Polar Plot of Pure Derivative
I For a pure differentiator
G(s) = s, or, G(jω) = jω = ω\+900
Point-1: The polar plot starts at ω = 0. This gives |G(jω)|=ω=0, φ=+900
Point-2: The polar plot ends at ω = ∞. This gives |G(jω)|=ω=∞, φ=+900
Point-3: The real axis cross-over point of polar plot i.e Im[G(jω)] = 0 ω=0
Point-4: The imaginary axis cross-over point of polar plot i.e Re[G(jω)] = 0. This does not exist; plot does not cross imaginary axis
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 7 / 36
Polar Plot of Pure Differentiator
The polar plot of pure derivative factor is the positive imaginary axis
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 8 / 36
Polar Plot of Pure Integral & Derivative Factors
Pure Integral Factor
In this case G(s) = 1
s
1110 or, G(jω)= jω =−jω = ω\−90
Thus the polar plot of pure integral factor is the negative imaginary axis as shown in the figure below.
Pure Derivative Factor
In this case
G(s) = s
or, G(jω) = jω = ω\+900
Thus the polar plot of pure derivative factor is the positive imaginary axis as shown in the figure below.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 9 / 36
Polar Plot of First Order Factor-1
Consider the first order system with transfer function
G(s) = 1 , T = time constant 1+Ts
I Representing this in frequency domain gives the sinusoidal transfer function 1 1 −1
G(jω)= 1+jωT = √1+ω2T2\−tan ωT
I Multiplying both numerator and denominator by the conjugate of denominator,
i.e 1 − jωT gives
G(jω)= 1 = (1−jωT) = (1−jωT)
1+jωT (1+jωT)(1−jωT 1+ω2T2 = 1 −j ωT =X+jY,where
X= 1 ,andY=− ωT 1+ω2T2 1+ω2T2
1+ω2T2 1+ω2T2
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 10 / 36
Polar Plot of First Order Factor-2
I The magnitude of G(jω) i.e. |G(jω)| is given by
1 2 −ωT 2
|G(jω)| = + = 1+ω2T2 1+ω2T2
I The phase of G(jω), denoted as φ is
−1 −ωT/(1 + ω2T2 −1 −ωT
X2 + Y 2
−1
(−ωT)
I Now that we have expressions for the magnitude and phase of the frequency
φ = tan 1/(1 + ω2T2) = tan 1 response, we can sketch the polar plot using the 4 key points.
= tan
Point-1: The polar plot starts at ω = 0. This gives
|G(jω)| = √ 1 = 1, φ = tan−1 0 = 0
1+0 1 Point-2: The polar plot ends at ω = ∞. This gives
|G(jω)|=√ 1 =0, φ=tan−1−∞=−900 1+∞ 1
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 11 / 36
Polar Plot of First Order Factor-3
Point-3: The real axis cross over point of polar plot i.e. Im[G(jω)] = 0. This is computed from
−ωT = 0, =⇒ ω = 0. 1+ω2T2
This is same as Point-1.
Point-4: The imaginary axis cross over point of polar plot i.e. Re[G(jω)] = 0. This is computed from
1 =0, =⇒ ω=∞. 1+ω2T2
This is same as Point-2.
I As Point-3 coincides with Points-1 and Point-4 coincides with Point 2, we need
more values of G(jω) evaluated at different frequencies. I Forexampleatω=1/Ti.e.whenωT=1,gives
1 1 −1−1 0 |G(jω)|= √1+1 = √2, φ=tan 1 =−45
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 12 / 36
Polar Plot of First Order Factor-4
I Let us prove that the polar plot of the first-order factor G(jω) = 1
1+jωT
is a semicircle. Now
G(jω)= 1 = (1−jωT) = (1−jωT)
1+jωT (1+jωT)(1−jωT 1+ω2T2 = 1 −j ωT =X+jY
X= 1 ,andY=− ωT 1+ω2T2 1+ω2T2
where
Then we get
12
X − 2
+ Y
2 1 1 − ω2T2 2 −ωT 2 = 2 1 + ω2T2 + 1 + ω2T2
12 = 2
1+ω2T2 1+ω2T2
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 13 / 36
Polar Plot of First Order Factors-5
I The values of G(jω) at ω = 0 and ω = 1 are respectively T
0110 G(j0)=1\0,andGjT =√\−45
2
I If ω approaches infinity,the magnitude of G(jω) approaches zero and the phase angle approaches −900. The polar plot of this transfer function is a semicircle as the frequency ω is varied from zero to infinity,as shown in Figure.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 14 / 36
Polar Plot of First Order Factors
Thus,in the X-Y plane, G(jω) is a circlewithcenteratX=1,Y =0
2
and with radius 1 as shown in Figure. 2
The lower semicircle corresponds to 0 ≤ ω ≤ ∞,and the upper semicircle corresponds to ̆∞ ≤ ω ≤ 0.
1. Polar plot of the transfer function G(s)=1+jωT.
2. It is simply the upper half of the straight line passing through the point (1, 0) in the complex plane and parallel to the imaginary axis, as shown in Figure.
The polar plot of 1 + jωT has an appearance completely different from that of 1/(1 + jωT ).
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 15 / 36
Polar Plot of Transport Lag
I Let the transport lag be represented as:
G(s)=e−Ls, =⇒ G(jω)=e−jωL
I Note that the magnitude of e−jωL equals to 1 and its phase angle equals to −ωL. Thus
|G(jω)| = 1, φ = \e−jωL = −ωL I The polar plot of the transport lag is shown in the Figure.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 16 / 36
Polar Plot : Example: First Order System with Transportation Lag
I The transfer function of a first order system with transportation lag is given by e−Ls e−jωL
G(s)= 1+sT, =⇒ G(jω)= 1+jωT I This can be written as:
G(jω) = e−jωL 1 1+jωT
I The magnitude and phase angles are given by
−jωL1 1 |G(jω)|=|e |.1+jωT= √1+ω2T2
−jωL 1
φ = \G(jω) = \e +\1+jωT = −ωL−tan
−1
ωT
I Note that here, the magnitude decreases from unity monotonically and the phase angle also decreases monotonically.
I The polar plot is shown ( next slide)
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 17 / 36
Polar Plot of First Order Factor System with Transportation Lag
I
The polar plot for first order system, transportation lag and first order system with transportation lag is shown in the Figures.
1. Polar plot of the transfer function
G(jω) = e−jωL . 1+jωT
Figure: Polar plot: G(jω) = 1 1+jωT
Figure: Polar plot of Transportation Lag
G(jω) = e−jωL
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 18 / 36
Example- Polar Plot
I Consider a second order system with transfer function
1 G(s) =
s(1 + Ts)
I The sinusoidal transfer function can be written as
1 jω(1+jωT)
G(jω) =
= −jω(1 − jωT )
(−jω)(jω)(1 + jωT )(1 − jωT ) = − T −j 1
I The G(jω) plot is asymptotic to the vertical line passing through the point (−T,0).
I Since this transfer function involves an integrator (1/s),the general shape of the polar plot differs substantially from those of second-order transfer functions that do not have an integrator.
1 + ω2T 2 ω(1 + ω2T 2) I The low-frequency portion of the
polar plot becomes
lim G(jω) = −T − j∞
ω→0
I The high-frequency portion of the
polar plot becomes
lim G(jω)=0−j0 ω→∞
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 19 / 36
Example: Polar Plot
I Consider a second order system with transfer function G(s) = K
(1 + T1s)(1 + T2s) I The sinusoidal transfer function can be written as
G(jω) = K
(1 + jωT1)(1 + jωT2)
I This can be expressed by the form u + jv as
G(jω) = K(1 − jωT1)(1 − jωT2)
(1 + jωT1)(1 + jωT2)(1 − jωT1)(1 − jωT2) = K(1−ω2T1T2) −j ω(T1 +T2)
(1 + ω2T12)(1 + ω2T2) (1 + ω2T12)(1 + ω2T2)
I The polar plot intersects the imaginary axis where the real part of G(jω) = 0.
This is computed by solving
K(1 − ω2T1T2) = 0, =⇒ ω = √ 1 (1 + ω2T12)(1 + ω2T2) T1T2
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 20 / 36
Example: Polar Plot
√
I The magnitude of G(jω) at this frequency i.e at ω = 1/ T1T2 equals to √
K T1T2 T1 + T2
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 21 / 36
Example: Polar Plot of Third Order System
I Consider a third order system with transfer function G(s) = K
s(1 + T1s)(1 + T2s) I The sinusoidal transfer function can be written as
G(jω) = K
jω(1 + jωT1)(1 + jωT2)
I This can be expressed by the form u + jv as
G(jω) = K(−jω)(1 − jωT1)(1 − jωT2)
(jω)(−jω)(1 + jωT1)(1 + jωT2)(1 − jωT1)(1 − jωT2) = −Kω2(T1 + T2) − jKω(1 − ω2T1T2)
ω2(1 + ω2T12)(1 + ω2T2)
= −K(T1 + T2) − j (K/ω)(1 − ω2T1T2)
1 + ω2(T12 + T2) + ω4T12T2 1 + ω2(T12 + T2) + ω4T12T2
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 22 / 36
Example: Polar Plot of Third Order System
I The polar plot intersects the real axis where the imaginary part of G(jω) = 0. This is computed by solving
−(K/ω)(1 − ω2T1T2)
1 + ω2(T12 + T2) + ω4T12T2
ω=√1 T1 T2
= 0
This gives
I The magnitude of the real part of G(jω)H(jω) at this frequency is obtained by computing the real part at this frequency. This gives
−K(T1+T2) =− KT1T2 1 + ω2(T12 + T2) + ω4T12T2 (T1 + T2)
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 23 / 36
Polar Plot of Third Order Transfer Function
I The polar plot is shown below for various values of gain K
I From the plot, it is evident that for the system to be stable
KT1T2 <1, =⇒ K
1+2ζj ω +j ω 2 ωn ωn
where ζ > 0. Then
lim G(jω) = 1\00, and
ω→0
lim G(jω) = 0\−1800,
ω→∞
I Thus the polar plot starts at 1\00 and ends at 0\1800 as ω increases from zero to infinity.Thus,the high-frequency portion of G(jω) is tangent to the negative real axis.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 25 / 36
Polar Plots Some Common Transfer Functions
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 26 / 36
Stability from Frequency Response
Review of Stability Studies:
1. We have discussed Routh-Hurwitz criterion which enables us to determine whether a system is stable or unstable ( absolute stability) by examining the roots of characteristic polynomial, corresponding to the denominator of the closed loop transfer function.
2. The root locus method enables us to determine the relative stability in terms of the damping ratio of the dominant poles.
3. Both of these methods require the knowledge of the transfer function of the open-loop system, which must be a rational function of the conmplex variable s – that is, a ratio of two finite-degree polynomials of s.
4. These methods can not be used when the system contains an ideal delay of the form e−Ts, although it is possible to obtain approximate analysis by replacing e−Ts by a truncated power series or by a rational function such as a Pade approximant.
Objective: To utilize the frequency response of the open-loop system a. To determine the absolute stability of the closed-loop system.
b. To evaluate some measures of relative stability
Advantage of Stability Investigation in Frequency Domain:
– There is no need to know the transfer function of the system since the frequency response can be measured experimentally.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 27 / 36
Stability from Frequency Response
I Consider the closed-loop system shown in the Figure.
What is the main reasaon for instability ?
a. Intuitively, the the main reason for possible instability is that what is designed as negative feedback may turn out to be positive feedback if at some frequency, a phase lag of 1800 occurs in the loop.
b. If the feedback at this frequency is sufficient to sustain oscillations, the system will act as an oscillator.
I Superficially, it appears that the closed-loop system will be stable, provided the open-loop transfer function G(jω)H(jω) does not have gain of 1 or more at the frequency where the phase shift is 1800.However, this is not always true.
I Consider the following polar plot.
I For two values of ω, the phase shift is 1800 and gain more than one ( at points A and B), and yet the closed loop system is stable.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 28 / 36
Relative Stability Analysis using Polar Plot
I Phase margin:The phase margin is that amount of additional phase lag at the gain crossover frequency required to bring the system to the verge of instability. .The phase margin γ is 1800 plus the phase angle φ of the open-loop transfer function at the gain crossover frequency,or
γ = 1800 + φ
a. In the polar plot,a line may be drawn from the origin to the point at which the unit circle crosses the G(jω) locus.If this line lies below (above) the negative real axis,then the angle γ is positive (negative).
b. The angle from the negative real axis to this line is the phase margin.
c. The phase margin is positive for γ > 0 and negative for γ < 0 < 0.
d. For a minimum phase system to be stable,the phase margin must be positive.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 29 / 36
Relative Stability Analysis using Polar Plot: Gain Margin
I Gain margin: The gain margin is the reciprocal of the magnitude |G(jω)| at the frequency at which the phase angle is −1800. Defining the phase crossover frequency ω1 to be the frequency at which the phase angle of the open-loop transfer function equals −1800 gives the gain margin Kg
I In terms of decibels
Kg= 1 |G(j ω1 )|
Kg dB = 20log Kg = −20log|G(jω1)|
a. The gain margin expressed in decibels is positive if Kg is greater than unity and negative if Kg is smaller than unity.Thus,a positive gain margin (in decibels) means that the system is stable, and a negative gain margin (in decibels) means that the system is unstable
b. For a stable minimum-phase system,the gain margin indicates how much the gain can be increased before the system becomes unstable.
c For an unstable system,the gain margin is indicative of how much the gain must be decreased to make the system stable.
d. The gain margin of a first or second-order system is infinite since the polar plots for such systems do not cross the negative real axis. Thus, theoretically, first- or second order systems cannot be unstable.
e. A stable non-minimum-phase system will have negative phase and gain
margins.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 30 / 36
Relative Stability Analysis using Polar Plot & Bode Plot
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 31 / 36
Relative Stability Analysis using Polar Plot & Bode Plot
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 32 / 36
Few Comments on Phase and Gain Margins
I The phase and gain margins of a control system are a measure of the closeness of the polar plot to the −1 + j0 point and may be used as design criteria.
1. Note that either the gain margin alone or the phase margin alone does not give a sufficient indication of the relative stability. Both should be given in the determination of relative stability.
2. For a minimum-phase system to be stable,,both the phase and gain margins must be positive.Negative margins indicate instability.
3. Proper phase and gain margins ensure us against variations in the system components and are specified for definite positive values.The two values bound the behavior of the closed-loop system near the resonant frequency.
4. For satisfactory performance,the phase margin should be between 300 and 600,and the gain margin should be greater than 6dB. With these values,a minimum-phase system has guaranteed stability,even if the open loop gain and time constants of the components vary to a certain extent.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 33 / 36
Few Comments on Phase and Gain Margins
5. The requirement that the phase margin be between 300 and 600 means that in a Bode diagram the slope of the log-magnitude curve at the gain crossover frequency should be more gradual than −40dB/decade.
6. In most practical cases,a slope of −20dB/decade is desirable at the gain crossover frequency for stability.
6. If it is −40dB/decade,the system could be either stable or unstable.(Even if the system is stable,however,the phase margin is small.)
7. If the slope at the gain crossover frequency is −60dB/decade,or steeper,the system is most likely unstable.
8. For non-minimum-phase systems,the correct interpretation of stability margins requires careful study.
9. The best way to determine the stability of non-minimum-phase systems is to use the Nyquist diagram approach rather than Bode diagram approach.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 34 / 36
Few Comments on Gain and Phase Margin
I In many cases,phase margin provides a better measure of relative stability.
I For example, consider a unity feedback system with forward transfer function
G(s)= K s+2
I The polar plot is shown in the figure
I It is evident that for both the cases, the gain margin is infinity since the polar plot never crosses the negative real axis.
I However, for K = 10, the phase margin is 350 and for K = 50, the phase margin is 160.
I Thus it is much better to have K = 10.
I In practice, a phase margin of 450 to 600 is usually considered satisfactory.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 35 / 36
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 36 / 36