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QUESTION: 1

Phase crossover frequency is defined at

Solution:

QUESTION: 2

The Bode plot for a transfer function is shown below:

The steady state error corresponding to a parabolic input is

Solution:

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,

QUESTION: 3

The frequency at which the Nyquist plot crosses the negative real axis is known as

Solution:

At ω = ω_{gc, }ϕ = -180°

i.e. Nyquist plot crosses the negative real axis.

QUESTION: 4

A unity feedback system has a forward path gain of

The phase angle of the Nyquist plot for ω → ∞ is

Solution:

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

QUESTION: 5

The open loop transfer function of a system is given as

The Bode plot of this system is represented as:

Solution:

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.

QUESTION: 6

A second order system has

It’s M_{p} (peak magnitude) will be approximately

Solution:

∴ ξ = 1/√2

So, peak magnitude = 1 + e^{-π }≈ 1

QUESTION: 7

The closed loop transfer function of a control system is given by

C(s)/R(s) = 1/(1+s)

For the input r(t) = sint, the steady state value of c(t) is equal to

Solution:

Given,

C(s)/R(s) = 1/(1+s)

Thus,

Here,

ω = 1 rad/s

and ∠H(jω) = -tan^{-1}ω = -tan^{-1}(1)

∴

QUESTION: 8

A system has transfer function equal to (1-s)/(1+s). Its gain at ω = 1 rad/s is

Solution:

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

QUESTION: 9

The gain margin of a unity negative feedback system having forward path transfer function

Solution:

∠G(jω) H(jω) = -90° - tan^{-1}ωT = ϕ

At ω = ω_{pc},

ϕ = -180º

∴ -90° - tan^{-1}ω_{pc} = -180° or, ω_{pc} = ∞ rad/s

∴

QUESTION: 10

The characteristic equation of a closed loop control system is given by s^{2} + 4s + 16 = 0. The resonant frequency (in radian/sec) of the system is

Solution:

Given, s^{2} + 4s+ 16 = 0

or

∴ ω_{n} = 4 rad/s

and

∴ Resonant frequency,

QUESTION: 11

The gain margin (in dB) of a system having the open-loop transfer function

is

Solution:

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 }= ∞

∴

QUESTION: 12

Nichol’s chart is useful for detailed study and analysis of

Solution:

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.

QUESTION: 13

The forward path transfer function of an unity feedback system is given by

What is the phase margin for this system?

Solution:

Given,

Now,

∠G(jω) = -2tan^{-1}ω = ϕ

At

ω = ω_{gc} = 0 rad/s

ϕ = -2 x 0 = 0º

∴ P.M. = 180° + ϕ

= 180° + 0° = 180° = π

QUESTION: 14

The closed loop system having the open loop transfer function,

is

Solution:

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.

QUESTION: 15

The nyquist plot for a unity feedback control system having open loop transfer function G(s) = K(1-s)/(s+1) is shown in figure below :

The system is

Solution:

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..

QUESTION: 16

A unity feedback control system has a forward path transfer function of Its phase value will be zero at a frequency of ω_{1}.

Which one of the following equation should be satisfied for it?

Solution:

Given,

or, ∠G(jω) H(jω) = -ωT - 90° - tan^{-1} ω

At ω = ω_{1}, ∠G(jω) H(jω) = 0°

∴ -ω_{1} T-90° -tan^{-1}ω_{1} = 0

or, tan^{-1}ω_{1 }= - (ω_{1}T + 90°)

or, ω_{1}, = -tan (90° + ω_{1}T) = cot ω_{1}T

∴ ω_{1} = cot(ω_{1}T)

QUESTION: 17

What is the approximate value of gain margin in the Nyquist diagram shown below?

Solution:

At ω = ω_{gc},

(Using Nyquist diagram)

∴

QUESTION: 18

The magnitude-frequency response of a control system is given in figure below:

The values of ω_{1} and ω_{2} are respectively

Solution:

Number of decade change from first break frequency

Now,

Now, no. of decade change from break frequency

∴

or,

ω_{2} = 400 rad/s

QUESTION: 19

Match List-I with List-Il and select the correct answer using the codes given below the lists:

List-I

A. Resonant peak

B. Damped natural frequency (rad/s)

C. Resonant frequency (rad/s)

D. Peak overshoot

List - II

Codes:

A B C D

(a) 3 4 1 2

(b) 3 1 4 2

(c) 2 4 1 3

(d) 2 1 4 3

Solution:

QUESTION: 20

A second order underdamped system has a damping ratio of 0.8. It is subjected to a sinusoidal input of unit amplitude. It has resonant peak of

Solution:

Resonant frequency is given by

For resonant peak

or,

Since ξ > 0.707, therefore there is no M_{r }(resonant peak).

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