## Control Systems Subject Wise

Question 1 |

that is, The forced response of the system is

U(t) – 2e ^{-t}u(t) + e^{-3t}u(t) | |

U(t) | |

2u(t) – 2e ^{-t}u(t) + e^{-3t}u(t) | |

2u(t) |

Question 2 |

_{p}and the number of system zeros N

_{z}in the frequency range 1 Hz ≤ f ≤ 10

^{7}Hz is

N _{p} = 6, N_{z} = 3 | |

N _{p} = 5, N_{z} = 2 | |

N _{p} = 4, N_{z} = 2 | |

N _{p} = 7, N_{z} = 4 |

Number of poles = 6

Number of zeros = 3

at F = 10 Hz we have one pole

At F = 10

^{2}

Hz we can see two more poles are added as slope is decreased by 40 dB/decade

At F = 10

^{3}Hz we have 1 zero

At F = 10

^{4}Hz we have two zero’s

At F = 10

^{5}Hz we have two pole’s

At F = 10

^{6}we have one pole

Total poles N

_{P}= 6

And total zeros N

_{Z}= 3

**Note**:

When we add one pole, slope becomes –20dB/decade

when we add one zero, slope becomes +20dB/decade

Question 3 |

**k/s**and open-loop transfer function as

,where K > 0,

find the positive value of K for which there are exactly two poles of the unity feedback system on the jω axis is equal to ___________ (rounded off to two decimal places).

Fill in the Blank Type Question |

Question 4 |

Question 5 |

4.50 | |

2.81 | |

5.25 | |

3.89 |

Question 6 |

^{T}and d = 0. Which one of the following options for A and C will ensure that the transfer function of this LTI system is

and C = [1 0 0] | |

and C = [0 0 1] | |

and C = [0 0 1] | |

and C = [1 0 0] |

Question 7 |

**I.**There is no bounded input bounded output (BIBO) stable system with a pole in the right half of the complex plane.

**II.**There is non causal and BIBO stable system with a pole in the right half of the complex plane.

Which one among the following is correct?

Both I and II are true | |

Both I and II are false | |

Only I is true | |

Only II is true |

If a system is non-causal then a pole on right half of the s-plane can give BIBO stable system. But for a causal system to be BIBO all poles must lie on left half of the complex plane.

Question 8 |

Fill in the Blank Type Question |

Question 9 |

__________________________________ |

**Hence Answer=1**

Question 10 |

**In phase lag compensator pole is near to jω- axis,**

Question 11 |

165 V | |

55 V | |

220 V | |

110 V |

Given

=> Vin=220V, Duty cycle=25%

Now, for a chopper circuit, we have:

Average output voltage=Duty cycle x Supply voltage

=> (25/100) x (220)=>55.

So, correct option is B

Question 12 |

_{1}= a

_{2}= a = /2 of both transistors?

V | |

V | |

x 220 V | |

x 220 V |

_{1}=a

_{2}=a=ᴨ/2

Now, we have a single phase full wave AC phase controller. We know the formula for Vout(Mean square value) for a AC phase controller . It is given as:

Vout(mean square)= V

_{m}/[(ᴨα)+sin2α/2]

^{(1/2)}

Here, it is already given in question that

V

_{m}=220V=> 220, and α=ᴨ/2

So, putting all these values in Vout formula, we will get answer as option (a)

Question 13 |

The value of gain for which system is marginally stable is

K = 4 | |

K = 6 | |

K = 10 | |

K = 2 |

We have to determine the range of K for which the system is marginally stable.

Now, for stability=> Routh array should be constructed, and marginally stable means some roots on the imaginary axis, and some roots on the left side of the s plane.

Closed loop transfer function is given as:

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

So, putting the values as given in the question, we get:

k(s+4)/s(s+1) / 1+(k(s+4)/s(s+1))(1/s+2)

So, from we can write the characteristic equation as: s

^{3}+3s

^{2}+s(k+2)+4k=0

Now from this we can construct the routh array as follows:

s

^{3}1 (k+2)

s

^{2}3 4k

s

^{1}(6-k)/3 0

s

^{0}4k

Now, for a marginally stable system, elements the first column should be examined and proper element should be equated to 0.

Here to find K, let us equate s

^{1}element:

(6-k)/3=0=> k= 6

Question 14 |

Closed loop system with I is stable and with II is unstable | |

Closed loop systems using I and II both are unstable | |

Closed loop system with I is unable and II is stable | |

Closed loop system with I and II are stable |

**->**Both gain and phase margin positive is Stable system,

**->**Both gain and phase margin negative is Unstable system.

From diagram (i)

Gain margin=0-(-4)=4dB

And, phase margin= -160-(-180)=20⁰

So, both positive values is Stable system.

From diagram (II)

Gain margin= 4-(-0)=4dB

Phase margin= -180-(-200)=-20⁰

Negative value is Unstable system.

So, (i) is stable and (ii) in unstable system.

Question 15 |

For the closed loop system shown, the root locus for 0 < K < ∞ intersects the imaginary axis for K = 1.5. The closed loop system is stable for

K >1.5 | |

1 < K <1.5 | |

0 < K <1 | |

no positive value of K |

Question 16 |

^{4}+s

^{2}+1 =0 on the complex plane?

Four left half plane (LHP) roots | |

One right half plane (RHP) root, one LHP root and two roots on the imaginary axis | |

Two RHP roots and two LHP roots | |

All four roots are on the imaginary axis |

Question 17 |

stable for K = 10 and stable for K = 100 | |

stable for K = 10 and unstable for K = 100 | |

unstable for K = 10 and stable for K =100 | |

unstable for K = 10 and unstable for K = 100 |

**Given**

G(s) =

C.E =

If system to stable

24>k+4∩K+4>0

k>-4∩k<20

(i) Stable condition -4

Means If k =10 system stable

k =100 system unstable

Or G(jω)=

If ω G(jω)=

ω G(jω)=

So If k =10 touching point= 0.5

If k =100 touching point =5

N= P-Z, Here P=0

N=-Z

If closed loop system to be stable, then z=0,N=0,

So, k=0 is stable system

Question 18 |

Both the criteria provide information relative to the stable gain range of the system. | |

The general shape of the Nyquist plot is readily obtained from the Bode magnitude plot for all minimum-phase systems. | |

The Routh criterion is not applicable in the condition of transport lag, which can be readily handled by the Nyquist criterion. | |

The closed-loop frequency response for a unity feedback system cannot be obtained from the Nyquist plot. |

**(A)**Both the criteria provides information about the stability of system

**so Option (A) is true**

**(B)**is true as in a minimum-phase system, Bode magnitude plot is enough to obtain a general approximation of its Nyquist plot

**so Option (B) is true**

**(C)**Routh criterion can be applied to any system to check the stability of a system but a transport lag controller can only by explained using Nyquist Criterion.

**so Option (C) is true**

**(D)**We can obtain closed-loop frequency response for Unity Feedback system easily by substituting s = jω, and draw the plot for different values of ω. Usually this is not done as it is not necessary as OLTF is enough to comment on the stability.

**Thus, (D) is false.**

Question 19 |

0 | |

1 | |

2 | |

3 |

**By Rolle’s theorem**

“Between any two real roots of f(n) there exist at least one real root of f'(n) ” But in this question if you observe it is given that there exist no real root of P'(s) and p(s) has 3 roots whereas p’(s) has 2 roots and none of which are real. Thus, p(s) has to have

**1 real**and 2 complex roots.

Question 20 |

Fill in the Blank Type Question |

Question 21 |

Fill in the Blank Type Question |

Question 22 |

The transfer function representation of the system is

Question 23 |

Fill in the Blank Type Question |

Question 24 |

Lead compensator is used to reduce the settling time. | |

Lag compensator is used to reduce the steady state error. | |

Lead compensator improves transient responce of a system. | |

Lag compensator always stabilizes an unstable system. |

Reduces the steady state error

Reduces the speed of response (i.e. decreases)

Increases the gain of original network without affecting stability

Permits the increases of gain if phase margin is acceptable

System becomes lesser stable

Reduces the effect of noise

Decrease the bandwidth

Question 25 |

Fill in the Blank Type Question |

Question 26 |

Where x

_{1}(t) and x

_{2}(t) are the two state variables and r(t) denotes the input. The output c(t) = x

_{1}(t). The system is

Undamped | |

Under damped | |

Critically damped | |

Over damped |

Question 27 |

0 | |

1 | |

2 | |

3 |

**Given**

Question 28 |

X: The system is stable

Y: The system is unstable

Z: The test breaks down

P: when all elements are positive

Q: when any one element is zero

R: when there is a change in sign of coefficients

. | |

Question 29 |

Encircles the s-plane point (-1 + j0) in the counterclockwise direction as many times as the number of right-half s-plane poles. | |

encircles the s-plane point (0–j1) in the clockwise direction as many times as the number of right-half s-plane poles. | |

Encircles the s-plane point (-1 + j0) in the counterclockwise direction as many times as the number of left-half s-plane poles. | |

Encircles the s-plane point (-1 + j0) in the counterclockwise direction as many times as the number of right-half s-plane zeros. |

Question 30 |

The open-loop transfer function of a unity-feedback control system is

The positive value of K at the breakaway point of the feedback control system's root-locus plot is

Fill in the Blank Type Question |

Question 31 |

Fill in the Blank Type Question |

Peak over shoot 10%

The characteristic equation of above transfer function is

Comparing with standard equation

Question 32 |

The number of zeros in the right half of the s-plane is _____.

Fill in the Blank Type Question |

Question 33 |

Question 34 |

reduce the overall gain | |

reduce bandwidth | |

improve disturbance rejection | |

reduce sensitivity to parameter variation |

Negative feedback in a closed loop

**(i)**Increases bandwidth

**(ii)**Reduces gain

**(iii)**Improve disturbance rejection

**(iv)**Reduce sensitivity to parameter variation.

Hence, the correct option is B.

Question 35 |

Fill in the Blank Type Question |

Now, we obtain the Routh array as

Row of s

^{1}to be zero for oscillatory response or for poles to be on imaginary axis.

**Method 2**

For third order system to be marginally stable, IP = EP

**1 * K = 4 * 3**

K = 12

K = 12

Hence, The correct Answer is 12

Question 36 |

first quadrant | |

second quadrant | |

third quadrant | |

fourth quadrant |

Question 37 |

Fill in the Blank Type Question |

Question 38 |

Fill in the Blank Type Question |

Question 39 |

Fill in the Blank Type Question |

Question 40 |

The closed loop transfer function for unity feedback

Using Routh's tabular form:

For system to be stable, the first row should not have any sign change.

To get this, there are two conditions:

And Or

Or

Question 41 |

^{+ }is

Fill in the Blank Type Question |

Since system transfer function to a unit step input G(s) .

T(s) = Y(s)/X(s) = (s-2)/(s+1)(s+3)

where X(s) = 1/s because x(t) = u(t)

Y(s) can be written as in the form of partial fraction

on Solving above equation with residue method to obtain the value for A, B & C

Taking Inverse Laplace of Y(s) -

y(t) = -(2/3) U(t) + (3/2)e^{-t} -(5/6) e^{-3t}

Considering its first derivative

y'(t) = -(2/3) δ(t) - (3/2)e^{-t} + (5/2) e^{-3t}

**Since in the question the value of y'(t) is asked at t = 0 ^{+}. We have already know that the impulse function [δ(t)=1] exist only at t = 0, otherwise its value will be zero.**

**so y'(t) = -(3/2)e ^{-t} + (5/2) e^{-3t}**

**y'(t) at t = 0 ^{+}; y'(0^{+}) = -3/2 + 5/2 = 1**

Question 42 |

Zero | |

one, anti- clockwise | |

One, clockwise | |

two, clockwise |

Question 43 |

The positive value of k for which the gain margin of the loop is exactly 0 dB and the phase margin of the loop is exactly zero degree is _______

Fill in the Blank Type Question |

For marginal stability odd order row of S should be zero. i.e.,

**K = 60 For Marginal Stable**

Question 44 |

_{1}both positive, is shown below. The value of p

_{1}is______

Fill in the Blank Type Question |

_{1}

Phase of transfer function

From the plot at ω=1, Φ=-135°

Solving for p

_{1}, we get p

_{1}= 1.

Question 45 |

Fill in the Blank Type Question |

Question 46 |

*G*(

*s*) is given as

The steady state error

*e*due to a unit step input is

_{ss}0 | |

0.5 | |

1.0 | |

10 |

Question 47 |

The range of

*K*for which the system is stable is

−2.0 < K < 0.5 | |

0< K < 0.5 | |

So, the conditions are and k>-2

and combining k > 0.5

Question 48 |

and H(s)=1

Respectively, If the variable parameter

*K*is real positive, then the location of the breakaway point on the root locus diagram of the system is_____.

Fill in the Blank Type Question |

Characteristic equation is given by

• To find the valid break point we need to find that lies on root locus

• 3.414 lies on root locus

• So break point – 3.414.

Question 49 |

Where K, a and b are positive real numbers. The condition for this controller to act as a phase lead compensator is

a < b | |

a > b | |

K < ab | |

K > ab |

Question 50 |

The value of f

_{H}– f

_{L}(in H’z) is____________.

Fill in the Blank Type Question |

Question 51 |

Where ‘*’ denotes the convolution operation and t is in seconds. The Nyquist sampling rate In samples/sec) for x (t) is_________.

Fill in the Blank Type Question |

Time domain convolution = frequency domain multiplication so, we obtain

Thus, the multiplication will result in maximum frequency of 0.2. Hence,

Nyquist rate = 2

*f*= 2(0.2) = 0.4 sample/sec

_{m}Question 52 |

Fill in the Blank Type Question |

Question 53 |

300 | |

250 | |

400 | |

200 |

Question 54 |

_{1}(s) and G

_{2}(s), one

**CANNOT**realize a transfer function of the form.

G _{1}(s) G_{2} (s) | |

Question 55 |

32 | |

65 | |

80 | |

33 |

**SQNR =31.8dB**

For sinusoidal signal,

signal to quantization noise ratio is given by,

SNR

_{q}= (1.76 + 6n) dB

where n = number of bits

Given SNR = 31.8 dB

So, 1.76 + 6n = 31.8 dB

or 6n + 30

or n = 5

Hence, Levels = 2

^{n}= 2

^{5}= 32

= 32 levels

Question 56 |

The response y(t) is

Sin(t) | |

1 - e ^{t} | |

1 – cos(t) | |

0 |

Since,

where, is state transition matrix given by

Hence,

Question 57 |

The value of K which will place both the poles of the closed-loop system at the same location, is ______.

2.25 | |

2.50 | |

2.75 | |

3.15 |

Question 58 |

Which one of the following conclusions is correct?

G(s) is an all-pass filter | |

G(s) is a strictly proper transfer function | |

G(s) is a stable and minimum-phase transfer function | |

The closed-loop system is unstable for sufficiently large and positive k |

**For larger values of K, the radius of circle increases and it will encircle the critical point (-1+j0), which makes closed-loop system unstable.**

Question 59 |

controllable and observable | |

uncontrollable and observable | |

uncontrollable and unobservable | |

controllable and unobservable |

The system is uncontrollable and observable

Question 60 |

55 | |

40 | |

50 | |

45 |

Question 61 |

The ratio of the power in the 7

^{th}harmonic to the power in the 5

^{th}harmonic for this waveform is closest in value to _______.

0.5 | |

1.0 | |

1.5 | |

2.0 |

power in nth harmonic component is

Ratio of the power in 7

^{th}harmonic to power in 5

^{th}harmonic for given waveform is

Question 62 |

20 rad/sec | |

30 rad/sec | |

35.5 rad/sec | |

38.15 rad/sec |

**Answer Range:**38.13 to 39.19

Question 63 |

approximately

1.44 | |

1.08 | |

0.72 | |

0.36 |

Question 64 |

If the system is connected in a unity negative feedback configuration, the steady state error of the closed loop system, to a unit ramp input, is_________.

0.0 | |

0.5 | |

0.8 | |

1.0 |

Due to initial slope, it is a type-1 system, and it has non zero velocity error coefficient

The magnitude plot is giving 0dB at 2r/sec.

Which gives

The steady state error

given unit ramp input; A 1

Question 65 |

The corresponding system is

always controllable | |

always observable | |

always stable | |

always unstable |

it is always controllable

Question 66 |

Question 67 |

**-3e**, where u(t) is the unit step function, is applied to a system with transfer function. If the initial value of the output is -2, then the value of the output at steady state is __________________.

^{2t}u(t)0 | |

1 | |

2 | |

3 |

Question 68 |

S1: The system is stable.

S2: is independent of t for t=0.

S3: A non-causal system with the same transfer function is stable.

For the above system,

Only S1 and S2 are true | |

only S2 and S3 are true | |

Only S1 and S3 are true | |

S1, S2 and S3 are true |

S

_{1}: System is stable (TRUE)

Because h(t) absolutely integrable

is independent of time (TRUE)

(independent of time)

S

_{3}: A non-causal system with same transfer function is stable

(a non-causal system) but this is not absolutely integrable thus unstable.

Only S

_{1}and S

_{2}are TRUE

Question 69 |

0.5 | |

0.7 | |

0.8 | |

0.9 |

Question 70 |

Question 71 |

There are three poles and one zero. Since only two root loci branches tend to infinity. Option (d) is eliminated.

→ First two dots going to left from origin are poles since they collide and go to infinity. There are poles at s = – 1 and s = – 2. Option (a) is eliminated.

→ Centroid of option (c) is (-3,0)

Not the case hence option (c) is also eliminated.

Question 72 |

– 80 dB/decade | |

– 40 dB/decade | |

+40 dB/decade | |

+80 dB/decade |

Question 73 |

16 | |

4 | |

2 | |

1 |

Question 74 |

Question 75 |

2 | |

4 | |

6 | |

8 |

For stability, first column elements must be positive and non-zero

The maximum value of p until whichremains stable is 2.

Question 76 |

The constant damping ratio line, for intersects the root locus at point A. The distance from the origin to point A is given as 0.5. The value of K at point A is ________ .

0.275 | |

0.345 | |

0.375 | |

0.425 |

Here, the damping factor and the length of OA = 5

Then in the right angle triangle

So, the co-ordinate of point A is

Substituting the above value of A in the transfer function and equating to 1 i.e. by magnitude condition,

__Alternative Explanation__

Question 77 |

To achieve a minimum probability of error the threshold should be

Strictly positive | |

Zero | |

Strictly negative | |

Strictly positive, zero, or strictly negative depending on the nonzero value of |

Question 78 |

The Bode plot of a transfer function G(s) is shown in the figure below.

The gain (20 log|G(s)| ) is 32 dB and –8 dB at 1 rad/s and 10 rads/s respectively. The phase is negative for all ω. Then G(s) is

Question 79 |

All the poles of the system must lie on the left side of the jω axis. | |

Zeros of the system can lie anywhere in the s- Plane | |

All the poles must lie within |s| = 1 | |

All the roots of the characteristic equation must be located on the left side of the jω axis. |

Question 80 |

u(t) | |

tu(t) | |

Question 81 |

P

_{k1}=(1)(s

^{-1})(s

^{-1})(1)=(s

^{-2})

P

_{k2}=(1)(s

^{-1})(1)(1)=(s

^{-1})

Since, all the loops are touching to the paths P

_{k1}and P

_{k2}so,

Now, we have

(sum of individual loops)

+ (sum of product of nontouching loops)

Here, the loops are

As all the loop

*L*

_{1},

*L*

_{2},

*L*

_{3}and

*L*

_{4}are touching to each other so,

**By using Mason’s Gain Formula**

Question 82 |

The state variable equations of the system in the figure above are:

Question 83 |

Question 84 |

1750 | |

2625 | |

4000 | |

5250 |

The term α is called the excess bandwidth factor which is given here 0.75

W=3500 Hz, the goal is to find the maximum possible symbol rate 1/T

Refer : http://www.dsplog.com/2012/11/01/gate-2012-ece-q3-communication/

Question 85 |

_{1}and C

_{2}are ideal capacitors. C

_{1}has been charged to 12 V before the ideal switch S is closed at t = 0. The current i(t) for all t is

zero | |

a step function | |

an exponentially decaying function | |

an impulse function |

Question 86 |

ω = 1 rad/s | |

ω = 2 rad/s | |

ω = 3 rad/s | |

ω = 4 rad/s |

Input

So output

Now

this will be zero if s

^{2}+ ω

^{2}will cancel (s

^{2}+ 9) term as only then final value theorem will be applicable.

So at ω

^{2}= 9 => ω = 3 rad/sec, steady state output will be zero.

Question 87 |

_{0}/2, uses equi-probable signal

over the symbol interval (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45 ° with respect to the received signal, the probability of error in the resulting system is

If the local oscillator in a coherent receiver is ahead in phase by with respect to the received signal then Probability of error in coherent BPSK is given by

Question 88 |

K = 2 and a = 0.75 | |

K = 3 and a = 0.75 | |

K = 4 and a = 0.5 | |

K= 2 and a = 0.5 |

1+G(s)H(s)=0

_{S}

^{3}+ as

^{2}+ (2+K)s + 1(1+K) = 0

For oscillation

Now

as

^{2}+ (1+K) = 0

- aω

^{2}+(1+K) = 0

Given ω = 2rad/sec

-4a+(1+K) = 0

-4(1+K)+(2+K)(1+K) = 0

(1+K)[(2+K)-4] = 0

K=-1,2

but K = -1 is not possible as system will not oscillate for this as a = 0

So, K = 2

Question 89 |

Where y is the output and u is the input. The system is controllable for

a _{1} ≠ 0, a_{2} = 0, a_{3} ≠ 0 | |

a _{1} = 0, a_{2} ≠ 0, a_{3} ≠ 0 | |

a _{1} = 0, a_{2} ≠ 0, a_{3} = 0 | |

a _{1} ≠ 0, a_{2} ≠ 0, a_{3} = 0 |

for system to be controllable

|Q

_{c}| ≠ 0

⇒ (0 = a

_{1}a

_{2}

^{2}) ≠ 0

⇒ a

_{1}≠ 0

a

_{2}≠ 0

Question 90 |

G

_{c}(s) is a lead compensator if

a = 1, b = 2 | |

a = 3, b = 2 | |

a = –5, b = –1 | |

a = 3, b = 1 |

Phase

for lead comparators phase must be +ve.

For this

So, a = 1, b = 2

Question 91 |

The value of gain for which system is marginally stable is

K = 4 | |

K = 6 | |

K = 10 | |

K = 2 |

Given: G(s)= k(s+4)/s(s+1) and H(s)=1/s+2

We have to determine the range of K for which the system is marginally stable.

Now, for stability=> Routh array should be constructed, and marginally stable means some roots on the imaginary axis, and some roots on the left side of the s plane.

Closed loop transfer function is given as:

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

So, putting the values as given in the question, we get:

k(s+4)/s(s+1) / 1+(k(s+4)/s(s+1))(1/s+2)

So, from we can write the characteristic equation as: s

^{3}+3s

^{2}+s(k+2)+4k=0

Now from this we can construct the routh array as follows:

s

^{3}1 (k+2)

s

^{2}3 4k

s

^{1}(6-k)/3 0

s

^{0}4k

Now, for a marginally stable system, elements the first column should be examined and proper element should be equated to 0.

Here to find K, let us equate s

^{1}element:

(6-k)/3=0=> k= 6

Refer the Topic Wise Question for Routh-Hurwitz Control Systems

Question 92 |

Closed loop system with I is stable and with II is unstable | |

Closed loop systems using I and II both are unstable | |

Closed loop system with I is unable and II is stable | |

Closed loop system with I and II are stable |

Here, the bode plot is given and we have to determine which system is stable and which one is not.

**Important concept:**Both gain and phase margin positive=> Stable system, and both gain and phase margin negative=>Unstable system.

Now, looking at diagram (i) given in the question, we can say

Gain margin=0-(-4)=4dB

And, phase margin= -160-(-180)=20⁰

So, both positive values=> Stable system.

Again for diagram (ii):

Gain margin= 4-(-0)=4dB

Phase margin= -180-(-200)=-20⁰

Negative value=> Unstable system.

So, (i) is stable and (ii) in unstable system.

Refer the Topic Wise Question for Routh-Hurwitz Control Systems

Question 93 |

The response y(t) for a step input r(t) = 5u(t) will be Where u(t) is a unit step input.

r(t) = 5u(t)

R(s) =

Now

Question 94 |

loss of system gain | |

rise of system gain | |

improvement in transient response, delayed response | |

poor transient response |

- Negative feedback reduces the system gain which is one of the disadvantages of feedback
- With feedback, sensitivity improved by factor of (1/1+GH) Hence gain reduces by factor (1/1+GH)

Question 95 |

The asymptotes of the three branches of root locus plot of this system will form the following angles with the real axis

60 ^{o}, 120^{o} and 300^{o} | |

60 ^{o}, 120^{o} and 180^{o} | |

60 ^{o}, 180^{o} and 300^{o} | |

40 ^{o}, 120^{o} and 200^{o} |

Question 96 |

s

^{4}+ 20s

^{3}+ 15s

^{2}+ 2s + K = 0

then the range of values of K for the system to be stable will be

1 < K < 2.49 | |

0 < K < 1.49 | |

1 < K < 4.49 | |

0 < K < 3.49 |

Question 97 |

0 and ∞ | |

∞ and 0 | |

0 and 0 | |

∞ and ∞ |

∴For type – 2 system steady state error for unit step and unit ramp input will be zero

Question 98 |

The characteristic equation of the unity negative feedback will be

(s + 1) (s + 4) + K(s + 2) = 0 | |

(s + 2) (s + 1) + K(s + 4) = 0 | |

(s + 1) (s – 2) + K(s + 4) = 0 | |

(s + 2) (s + 4) + K(s + 1) = 0 |

G(S) =

and unity feedback ∴ H (S) = 1

For a unity negative feedback system, the characteristic equation is given by

∴ q (S) = 1 + G(s) H (S) = 0

∴ Option A

Question 99 |

magnitude response | |

transient response | |

steady-state response | |

frequency response |

- Frequency response of a system defines the magnitude response and phase response at steady state for a sinusoidal input.
- Here we are talking about relationship between amplitude of input & output are well as relationship between phase of input & output which means frequency of the input signal is varied which Is known as frequency response of an LTI system

Question 100 |

a minimum-phase function | |

a complex transfer function | |

an all-pass transfer function | |

a maximum-phase transfer function |

- If all the poles & zeros of a transfer function lie in the left half of the S–plane the transfer function is called minimum phase transfer function except poles and zeros at origin & infinity.
- A maximum phase transfer function has poles & zeros on right half of s-plane.
- An all pass transfer function has poles & zeros symmetrically located w.r.t. jω axis.
- A complex transfer function has complex poles & zeros, not necessarily on a particular half.

Question 101 |

normalized frequency | |

resonant frequency | |

peak frequency | |

tuned frequency |

Question 102 |

1)

2)

3) τ > 0

4) τ < 0

Which of the above are correct?

1 and 4 | |

1 and 3 | |

2 and 4 | |

2 and 3 |

_{C}(S) =

For lead compensator zero is dominant over pole

|Zc| < |Pc|

∴

α < 1

∴ α =

& is the time constant it can’t be negative

∴ > 1

Question 103 |

Question 104 |

Converting this into SFG, we get

Refer the Topic Wise Question for Basics of Control System Control Systems

Question 105 |

Damped Forced Oscillation | |

Undamped Forced Vibration | |

Damped Vibration | |

None of the above |

Damped means gradually reduces to oscillation , but here the oscillations are sustained.

Refer the Topic Wise Question for Basics of Control System Control Systems

Question 106 |

_{t}= 0 and K

_{a}= 5, then steady state error for unit ramp input is 0.2. What will be the new value of K

_{t}and K

_{a}if damping ratio is increased to 0.5 without affecting steady state error.

K _{t} = 1.5, K_{a} = 1.25 | |

K _{t} = 1.5, K_{a} = 12.5 | |

K _{t} = 15, K_{a} = 12.5 | |

K _{t} = 15, K_{a} = 1.25 |

G(s) =

G(s)= & H(s)=1

Therefore, =

=

On comparing it with standard 2

^{nd}order equation which is

We get , →

Also , given,

………………………………….(1)

Also steady state error,

As, R(t)=tu(t)→R(s)=

Therefore,

=

Given ,

…………………………………………………..(2)

On solving (1) and (2), we get

Refer the Topic Wise Question for Time Domain Analysis Control Systems

Question 107 |

G(s) = 0.25/((s

^{2}+1)(8s+3))

Transfer function of a lead compensator aimed at achieving gain crossover frequency of 0.5rad/sec and phase margin of 30 deg is

Putting ω=0.5rad/sec in all the options, only option (C) gives |G(jω)|=1

G(jω)=

|G(jω)|= 1

Refer the Topic Wise Question for Compensators and Controllers Control Systems

Question 108 |

_{1}+kA

_{2})/ (A

_{3}+kA

_{4}) with respect to parameter k is given by

k(A _{2}A_{3} – A_{1}A_{4})/ ((A_{3} + kA_{4}) (A_{1} + kA_{2})) | |

(A _{2}A_{3} – A_{1}A_{4})/ ((A_{3} + kA_{4})^{2}) | |

k(A _{2}A_{3} – A_{1}A_{4})/ ((A_{3} + kA_{4})^{2}) | |

(A _{2}A_{3} – A_{1}A_{4})/ ((A_{3} + kA_{4}) (A_{1} + kA_{2})) |

T=(A

_{1}+kA

_{2})/ (A

_{3}+kA

_{4})

Sensitivity , =

=

=

On putting given values,

=

Refer the Topic Wise Question for Feedback Principle and Frequency Response Control Systems

Question 109 |

_{1}(t) and x

_{2}(t) plus Additive White Gaussian Noise at the input of matched detector. If the noise power spectral density (N

_{0}) is 10

^{-11}W/Hz, compute E

_{b}/N

_{0}(in dB). Assume system characteristics impedance as 1 .

3 dB | |

4 dB | |

7 dB | |

10 dB |

As the signal and noise are in additive form, we can do the separate analysis of signal and noise.

Binary ‘1’ ,

Binary ‘0’ ,

Energy per bit , for binary 1; =

=

=

= 5

Given,

= 10 log 5

=7dB

Refer the Topic Wise Question for SNR, BER and Bandwidth Control Systems