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The polar plot of the transfer functionfor 0 ≤ ω ≤ ∞ will be in the
Given transfer function
(it is Lead compensator)
Magnitude of the transfer function will be
Phase of the transfer function will be
At ω = 0
Magnitude M_{1} = 1 and phase ϕ_{1} = 0°
At ω = ∞
Magnitude M_{2} = 10 and phase ϕ_{2} = 0°
At ω = 2
Magnitude M_{3} = 2.19 and phase ϕ_{3} = 52.12°
The polar plot considering the above magnitude and phase values will be
Zero is nearer to the imaginary axis than the pole, hence the plot will move in a clockwise direction.
Gain Margin:
Mathematically,
∠G(jω)H(jω) = 180º
If G(jω)H(jω) = a, at ω = ω_{pc}
Then Gain margin,
Phase Margin:
At ω_{pc}, G(jω)H(jω = 1
if, ∠G(jω)H(jω) = ϕ at ω = ω_{gc}
Then, P.M. = 180 + ϕ
Analysis:
For stable systems,
G(jω_{pc})H(jω_{pc} < 1, and
∠G(jω_{gc})H(jω_{gc}) > 180
so that, gain margin and phase margin both are positive
which is possible when ω_{gc} < ω_{pc}
For marginally stable systems,
ω_{gc} = ω_{pc}
Hence option (b) is the correct answer.
Note: For stable systems having two or more gain crossover frequencies, the phase margin is measured at the highest gain crossover frequency.
Directions: It consists of two statements, one labelled as the ‘Statement (I)’ and the other as ‘Statement (II)’. Examine these two statements carefully and select the answers to these items using the codes given below:
Statement (I): The polar plot has limitation for portraying the frequency response of a system.
Statement (II): The calculation of frequency response is tedious and does not indicate the effect of the individual poles and zeros.
Polar plot:
Therefore, Statement (I) is false but Statement (II) is true.
Polar plot:
The polar plot is a plot, that is drawn between the magnitude and the phase angle of G(jω) by varying ω from zero 0 to ∞.
Given,
∠G(jω) = 90°  tan^{1}(ωT_{1})
The polar plot of a transfer function with ω as the parameter is known as the
Nyquist plot:
Nyquist plots are an extension of polar plots for finding the stability of the closedloop control systems. This is done by varying ω from −∞ to ∞, i.e. Nyquist plots are used to draw the complete frequency response of the openloop transfer function.
Method of drawing Nyquist plot:
From the given polar plot,
At ω = 0, magnitude = 1, angle = 0°
Therefore, there are no poles or zeros at the origin.
At ω = ∞, magnitude = 0, angle = 270°
Therefore, the number of poles = 3
From the above two conditions, the transfer function will be
The frequency response ofplotted in the complex G(jω) plane (for 0 < ω < ∞) is
The given frequency response is
In Anticlockwise the phase order is 0, 90, 180, 270, 360 or 0.
In clockwise the phase angle order is 0, 90, 180, 270, 360 or 0.
Here, 90^{∘ } = 270^{∘ } and 270^{∘ }= 90^{∘ }
At ω = ω_{c} (Cutoff frequency), ∠ G(jω) = 180^{∘ }
−90 − tan^{−1}ω_{c} − tan^{−1}0.5ω_{c} = −180
tan^{−1}ωc + tan^{−1}0.5ω_{c} = 90
1  0.5ω_{c}^{2} = 0
ω_{c} = 1.414 rad/sec
At ω = ω_{c}
G(jω) = (1/6)
The polar plot for given transfer function is
The polar plot of the transfer function G (s) = 10 (s + 1) / (s + 10) will be in the
Polar plot:
The Polar plot is a plot, which can be drawn between the magnitude and the phase angle of the transfer function by varying frequency from 0 to ∞.
Given,
the magnitude of the transfer function will be:
The phase of the transfer function will be:
ϕ = tan^{1}(ω)  tan^{1}ω(ω/10)
At ω = 0
M_{1} = 1
ϕ_{1} = 0°
At ω = ∞
M_{2} = 10
ϕ_{2} = 0°
The gain margin of the system is 23.9 dB
∴ phase crossover frequency W_{pc} = 15.7 rad
∴ at W_{pc} = 15.7 rad we will get the phase of 180º
180º = K(900) 90º
∴ K = 90/900
= 0.1
Polar plot:
The polar plot is a plot, that is drawn between the magnitude and the phase angle of G(jω) by varying ω from zero 0 to ∞.
∠G(jω) = 90°  tan^{1}(ωT_{1})
The quadrant at which the polar plot of transfer functionlies is ______
It is a lead compensator
(∴ Polar plot lies in 1^{st} quadrant)
A system with a unity gain margin and zero phase margin is _____
Hence option (3) is the correct answer.
Gain margin and phase margin are frequently used for frequency response specifications by designers. Usually a Gain margin of about 6 dB or a Phase margin of 30  35° results in a reasonably good degree of relative stability.
Important Points
Polar plot:
The sinusoidal transfer function G(jω) is a complex function and it is given by
G(jω) = Re [G(jω)] + j Im [G(jω)]
G(jω) = G(jω) ∠ G(jω)
= M ∠ϕ Polar form
From the above equation,
G(jω) may be represented as a phasor of magnitude M and phase angle ϕ
The starting points (ω = 0) of a polar plot for different types of minimum phase systems is given below:
Therefore, the Polar plot of the sinusoidal transfer function is a plot of magnitude and phase angle.
If the constant 'k' is negative, then what would be its contribution to the phase plot:
Polar plot:
Calculation:
Let the transfer function be:
The total phase shift will be:
∠G(jω)H(jω) = 90°  tan^{1}(ωT)  180° (1)
From equation (1) we can say that constant K has no contribution to the phase plot.
Hence K contributes 180° to the phase plot.
So option (c) is the correct answer.
Transfer function of a system is given as
Polar plot of the same system is
Gain margin (in dB) of the system is ______. (Important  Enter only the numerical value in the answer)
Gain margin (GM) and Phase margin (PM) from the polar plot:
Let us consider the polar plot as shown below,
Figure: A polar plot of a stable system
Gain margin (GM):
The frequency at which the polar plot crosses the negative real axis is called a phase crossover frequency (ω_{pc}).
Let 'a' be the point of intersection and it is the magnitude at the phase crossover frequency.
⇒ a =  G(jω_{pc}) H(jω_{pc}) 
And, we know that
Gain margin (GM) = 1 / G(jω_{pc}) H(jω_{pc}) = 1 / a
Gain margin in dB = 20 log (1 / a)
Calculations:
In the Given polar plot;
Phase margin (PM):
The frequency at which the polar plot crosses the unit circle (formed with the radius of the critical point) is called again crossover frequency (ωgc).
From the polar plot, we can observe that
Phase margin (PM) = 180  ∠G(jω_{gc}) H(jω_{gc})
22 docs274 tests

22 docs274 tests
