Test: Frequency Domain Analysis of Control Systems- 2 - Electrical Engineering (EE) MCQ

We know that, 

Thus, if BW is increased, rise time will decrease and the system response will become fast. Due to faster response of system, damping ratio will decrease.

Given, G(s) = 1/(1+Ts)
It is a type-0 and order-1 system, therefore its polar-plot will be as shown below.

Hence, the polar plot will be semicircular in shape.

The bandwidth is defined as the band of frequencies tying between-3 dB points. Hence, statement-2 is false.
Higher the value of resonant frequency of the system, faster is the time response. Hence, statement-3 is false.

Given, G(s) = (1+sT)/(sT)
The inverse polar plot of G(jω) is the polar plot of 1/G(jω)

Thus, 

and

Hence, inverse polar plot will be as show below,

1/(jω)² represents type-2 factor which will have a slope of -40 db/decade.
We know that,
-20 db/decase = -6 db/octave
∴ -40 db/decase = -12 db/octave

Since, the system is of minimum phase type, it has no poles or zeros in the right hand side of s-plane.
⇒ There will be three corner frequencies at: ω₁ = 5 rad/s
∴ T₁ = 1/ω₁ = 0.2 sec
(one pole added at ω₁ = 5 rad/s)
ω₂ = 40 rad/s
∴ T₂ = 1/ω₂ = 0.025 sec
(two pole added at ω₁ = 40 rad/s)
and ω₃ = 100 rad/s
∴ T₃ = 1/ω₃ = 0.01 sec
(two poles added at ω₃ = 100 rad/s)
So, open loop T.F. is

Now, 20 log K = 100
∴ K = 10⁵
Thus, 

For marginally stable system,
P.M = G.M = 0 and ω_pc = w_gc
Now, phase margin = (180° + ϕ)
Where,
 
Given, P.M = 0° (for marginally stable)
∴ 0º = 180° + ∠G(jω) H(jω)
or, ∠G(jω) H(jω) = -180°
Given, 


Thus,

The OLTF has no-poles in RH s-plane. Thus, for stability the -1 + j0 (critical point) point should not be encircled by Nyquist plot.
For finding the point of intersection of Nyquist piot with the negative reai axis.
Img [G(jω) H(jω)] = 0
or,  -ω² + 15 = 0
or, ω = √15 rad /sec 
Putting, ω = √15 rad/s in equation (1), we have

For system to be stable,

So, 
0 < K < 120 (for stability)

BW ∝ ω_n. Hence, option (d) is not correct. For a prototype second order system,
Thus,

Here,
|X| = 0.25 (Given)

Also, ϕ = -180°
∴  P.M. = 180 + ϕ
= 180°-180° = 0°

Resonant peak is given by,

• For ξ = 0, M_r = ∞
• 
• 
• 

In frequency domainanalysis,

Thus, for ω_r to be real Hence for there is no M_r or M_r is not defined.

Gain margin alone is inadequate to indicate relative stability when system parameters other than the loop gain are subject to variation. Due to variation in some parameters, phase shift of the system can vary which may results instability. Hence, we must also consider the concept of phase margin for determining the relative stability of the system.

Given,


∴ ϕ = ∠G(jω) H(jω)
= -90° - tan^-1ω - tan^-1 (0.1 ω)
At  ω = ω_Pc, ϕ = -180°, therefore
-180° = -90 ° - tan^-1ω - tan^-1 (0.1 ω)
or,

= Phase crossover frequency

∴ When open-loop gain becomes 4-times, GM will become one-fourth.