Test: Controllers & Compensators - 2 - Electronics and Communication Engineering (ECE) MCQ

Here, reason (R) is false because the lag network allows to pass low frequencies and high frequencies are attenuated.

The transfer function of a phase-lag network is:

The pole-zero plot is shown below:

Here, pole is nearer to the imaginary axis and origin compared to zero.
Hence, reason (R) is true.
Introduction of phase lag network in forward path reduces the phase shift. Hence, assertion (A) is a false statement.

A PID controller is shown below. 

Here,
G(s) = O.L.T.F

Due to PD controller, a zero is added to the forward path which results in a lead compensator. Therefore, steady state error remais the same whiie rise time of the system is reduced (i.e. stability is increased).

For the given system, characteristic equation is

or,
s² + 1.4s + 14s T_d + 14 = 0 
or, 
s² + (1.4 + 14T_d) s + 14 = 0
Comparing with s² + 2 ξω_n + ω_n² =0, we have:
ω_n = √14 rad/s and 2ξω_n = (1.4 + 14T_d)
or, 2 x 0.7 x √14 = 1.4 + 14 T_d (∴ ξ = 0.7, given)
or,

or,

= 0.274

For the given system, forward path transfer function is

For parabolic input, steady state error is 

where,


∴     

The forward path gain of the given system is:

∴ Characteristic equation is
1 + G(s)H(s) = 0
or

or, s² + (2 + 10 K_e)s +10 = 0
Here,
ω_n = √10 rad/s
and 2ξω_n = (2 + 10K_e)
Given, ξ = 0.5
So, 2 x 0.5 x √10 = (2 + 10K_e)
or, 

= 1.16/10 = 0.116
Also, 2% settling time

The characteristic equation of the given system is
1+G(s) H(s) = 0
Here,


∴ Characteristic equation is
10

or, s² + (K₀ + 2)s +10 = 0
Comparing with s² + 2ξω_ns + ω_n² = 0 we have 
ω_n = √10 rad/s
and 2ξω_n = K₀ + 2
Given, ξ = 0.6
∴ 2 x 0.6 x √10 = K₀ + 2
or, K₀ = 1.792
For a unit ramp input, e_ss = 1/K_v
where,


∴ 
= 0.3792 ≈ 0.38 radian

Case-1: For integral type

∴ 
∴ e_ss = T_i/2
Case-2: For derivative type

G(s)H(s) = (10_sT_d)/(s+5)
∴ 

∴ e_ss = 1/K_v = ∞
Case-3: For proportional type
G(s)H(s) = 10K_p/(s+5)
∴ 

∴ 
Case-4: For proportional plus derivative type

∴ 

∴ e_ss = 1/K_v = ∞

Redrawing the given circuit in s-domain, we have:

Now,

From virtual ground concept,


Now, 

or,
2V_A(s) = E_i(s) + E₀(s)
E₀(s) = 2V_A(s) - E_i(s)
or,
or,

Since the nature of command is in the form of pulses, therefore stepping motors are compatible with modern digital systems.

For given system,

With proportional controller,


∴ 
(for a unit step input)
Given, e_ss = less than 2% of input
i.e. e_ss < 0.02
Now, characteristic equation is:
1 + G(s)H(s) = 0
or, s³ + 4s² + 5s + 2 + 2K_c = 0
Routh’s array is:

For stability of given system,
2 + 2K_c > 0 or K_c > -1

or K_c < 9
Hence, -1 < K_c < 9 (for stability) ....... (1)
Now,

From equations (1) and (2), we conclude that the minimum value of steady state error = 0.1 when K_c = 9.
Also, for all value of K_c between -1 and 9, value of e_ss will be more than 0.02 i.e. more than 2% of input.
Therefore, there is no such value of K_c for given system i.e. can not be determined.

Given,

The pole-zero plot is shown below.


Since zero is near to origin, therefore it represents a lead network.

Given,

Here, T₁ = 3T and αT₁ = T
∴ α = T/T₁ = T/3T = 1/3
∴
or,  ϕ_m = 30° = maximum phase shift