Test: Frequency Domain Analysis of Control Systems- 3 - Electrical Engineering (EE)

Initial slope of the system = 0 dB/decade
Therefore, given system is a type-0 system.
For type-0 system, for parabolic input, steady state error is infinite since,

At ω = ω_gc, ϕ = -180°
i.e. Nyquist plot crosses the negative real axis.

At ω → ∞, ϕ = -90° - 90° - 90° = -270° = +90°

Initial slope = -20 dB/decade
= -6 dB/octave (Since Type = 1)
Corner frequency is,
ω_c = (1/a) rad/sec
Initial slope = -20 dB/decade
= -6 dB/octave (Since Type = 1)
Corner frequency is,
ω_c = 1/a rad/sec
It will cross 0 dB line at ω = K and K < 1/a.
So, the Bode plot will be as shown below.

∴ ξ = 1/√2
So, peak magnitude = 1 + e^-π ≈ 1

Given,
C(s)/R(s) = 1/(1+s)
Thus,

Here,
ω = 1 rad/s

and  ∠H(jω) = -tan^-1ω = -tan^-1(1)

∴ 

Given transfer function represents an all pass filter which has gain of 1 at all frequencies.

∠G(jω) H(jω) = -90° - tan^-1ωT = ϕ
At ω = ω_pc,
ϕ = -180º
∴ -90° - tan^-1ω_pc = -180° or, ω_pc = ∞ rad/s

∴ 

Given, s² + 4s+ 16 = 0
or 
∴ ω_n = 4 rad/s
and 
∴ Resonant frequency,

Given,

∴ 
and ∠G{jω)H(jω) = -90° - tan^-1 ω = ϕ
At ω = ω_pc, ϕ = -180°
∴ -90° - tan^-1 ω_pc = -180° or ω_pc = ∞
∴ 

Nichol’s chart consists of magnitude and phase angle of a closed loop system represented as a family of circle known as M and N-circle.

Given,



Now,
∠G(jω) = -2tan^-1ω = ϕ
At
ω = ω_gc = 0 rad/s
ϕ = -2 x 0 = 0º
∴   P.M. = 180° + ϕ
= 180° + 0° = 180° = π

Given,


For finding ω_gc, Img[G(jω)H(jω)] = 0
or, 
Now,


Thus, Nyquist plot will cut real axis at -0.66.
The given system:

of type-1 and order-3.
Therefore, Nyquist plot will be as shown below:

Now, from above Nyquist plot,
N = 0 = Number of encirclements.
Also, P = number of open loop poles in RH s-plane = 0
Now, N = P-Z
or, 0 = 0 - Z or Z = 0
Thus, no. of closed loop poles in RH s-plane = 0.
Hence, system is stable.

Given,
G(s)H(s) = K(1-s)/(s+1)
or,



Thus, the radius of Nyquist plot will be K.
Now, no.of open loop poles in RH s-piane = 0.
When K > 1, no. of encirclement, N = -1
When K< 1, no. of encirclement, N = 0
Thus, for K > 1, N = P - Z
or,    -1 = 0 - Z or Z = 1
∴ System is unstable.
For K < 1, N = P - Z or 0 = 0 - Z or Z = 0
∴ System is stable..

Given,


or, ∠G(jω) H(jω) = -ωT - 90° - tan^-1 ω
At ω = ω₁, ∠G(jω) H(jω) = 0°
∴ -ω₁ T-90° -tan^-1ω₁ = 0
or, tan^-1ω₁ = - (ω₁T + 90°)
or, ω₁, = -tan (90° + ω₁T) = cot ω₁T
∴ ω₁ = cot(ω₁T)

At ω = ω_gc,

(Using Nyquist diagram)
∴ 

Number of decade change from first break frequency

Now,

Now, no. of decade change from break frequency

∴ 
or, 
ω₂ = 400 rad/s

Resonant frequency is given by

For resonant peak

or,

Since ξ > 0.707, therefore there is no M_r (resonant peak).