Test: Frequency - Domain Analysis - Electronics and Communication Engineering (ECE) MCQ

From the fig. P6.5.1-2, |T(j0)| = 1

The peak value of T(jω) occurs when the denominator of function |T(jω)|² is minimum i.e. when

The Bode plot is as shown in fig.

From Fig.:

If Nyquist plot encloses the point (-1, j0), the system is unstable and gain margin is negative.

Hence, the asymptote of the Nyquist plot tends to an angle of -90° as ω→ 0.

Since system is stable, it will cross at s = -0.1

Hence (B) is correct option.

Due to s there will be a infinite semicircle. Hence (C) is correct option. 

∠GH(jw) = -270°+ 2 tan-1ω 
For ω = 0, GH(jω) = ∞∠ -270°
For ω = 1 , ∠GH(jω) =  -180°
For ω = ∞, GH(jω) = 0 ∠ - 90°
As ω increases from 0 to ∞, phase goes -270° to -90°.
Due to s³ term there will be 3 infinite semicircle. 

∠GH(j(ω) = -90 °- tan^-1 ω - tan^-1 2ω - tan ^-1 3ω,
For ω = 0, GH(jω) = ∞∠ - 90°,
For ω = ∞, GH(jω) = 0∠- 360°,
Hence (A) is correct option. 

The open-loop poles in RHP are P = 0. Nyquist path enclosed 2 times the point (-1 + j0). Taking clockwise encirclements as negative N = -2.
N = P - Z, -2 = 0 -Z , Z = 2 which implies that two poles of closed-loop system are on RHP. 

RHP poles of open-loop system P = 1, Z = P - N .
For closed loop system to be stable, Z = 0, 0 =1 -  N ⇒N = 1
There must be one anticlockwise rotation of point (-1+ j0). It is possible when K > 1.

The intersection with the real axis can be calculated as {GH(jω)} = 0, The condition gives ω(2ω² -1) = 0 

With the above information the plot in option (C) is correct. 

The Nyquist plot crosses the negative real axis

Hence phase crossover frequency is 

The frequency at which magnitude unity is

∠GH(jω) = -90°- tan^-1 ω- tan^-1 2ω ,
At unit gain ω₁ = 0.57 rad/sec,
Phase at this frequency is ∠GH(jω₁) = -90°- tan^-1 0.57 -tan^-1 2(0.57) = -168.42°
Phase margin = -168.420+180° = 11.6°
Note that system is stable. So gain margin and phase margin are positive value. Hence only possible option is (D).