Control Systems Gate Questions | Control Systems

From the given Bode plot and it’s slopes
Number of poles = 6
Number of zeros = 3
at F = 10 Hz we have one pole
At F = 10²
Hz we can see two more poles are added as slope is decreased by 40 dB/decade
At F = 10³ Hz we have 1 zero
At F = 10⁴ Hz we have two zero’s
At F = 10⁵ Hz we have two pole’s
At F = 10⁶ 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

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.

To get steady state error zero for unit step input and 6 for unit ramp input, the type of the system is one.
Hence Answer=1

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

Here we have to find out the value of DC voltage that is being applied to the motor.
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

Here we have to find the RMS output voltage at delay angles a₁=a₂=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)

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: s³+3s²+s(k+2)+4k=0
Now from this we can construct the routh array as follows:
s³ 1 (k+2)
s² 3 4k
s¹ (6-k)/3 0
s⁰ 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 s¹ element:
(6-k)/3=0=> k= 6

->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.

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

(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.

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.

In terms of State Space form, the Transfer function is given as,

The phase-lead controller adds zero and a pole, with the zero to the right of the pole, to the forward-path transfer function. The general effect is to add more damping to the closed-loop system. The rise time and settling time are reduced in general.
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

Given

K = 2.86
Peak over shoot 10%

The characteristic equation of above transfer function is
Comparing with standard equation

Negative Feedback reduces gain but Bandwidth is increased because gain bandwidth product remains same. So, Negative feedback in a closed-loop control system DOES NOT reduce bandwidth.
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.

Now, we obtain the Routh array as

Row of s¹ 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
Hence, The correct Answer is 12

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

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

The given condition implied marginal stability. One alternative way without going for gain margin, phase margin concepts is find k value for marginal stability using reflection.


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

K = 60 For Marginal Stable

Since it is the phase plot given we can't use the slope concept as these are nonlinear curves. So we can take any phase angle of a given frequency as reference and can obtain P₁
Phase of transfer function

From the plot at ω=1, Φ=-135° 
 
Solving for p₁, we get p₁ = 1.

So, the conditions are and k>-2
and combining k > 0.5

To find break point, from characteristic equation we need to arrange k as function of s, then the root of gives break point
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.

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_m = 2(0.2) = 0.4 sample/sec

Given : 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 = 2ⁿ = 2⁵ = 32
= 32 levels

Since,
where, is state transition matrix given by

Hence,

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.

From the given state model,




The system is uncontrollable and observable

For a periodic sequence wave, nth harmonic component is
power in nth harmonic component is
Ratio of the power in 7^th harmonic to power in 5^th harmonic for given waveform is

Given the natural frequency of an undamped second order system

Answer Range: 38.13 to 39.19

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

From the given signal flow graph, the state model is



it is always controllable

S₁: System is stable (TRUE)
Because h(t) absolutely integrable
is independent of time (TRUE)
(independent of time)
S₃: A non-causal system with same transfer function is stable
(a non-causal system) but this is not absolutely integrable thus unstable.
Only S₁ and S₂ are TRUE

For transfer function

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.

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

The maximum value of p until whichremains stable is 2.

We know that the co-ordinate of point A of the given root locus i.e., magnitude condition

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

For a system to be casual, the R.O.C. of the transfer function of the H (s) system which is rational must be in the right half-plane and to the right of the rightmost pole. For the stability of the LTI system. All the poles of the system must be in the left half of the S-plane and no repeated pole must be on the imaginary axis. Consequently, options A, B and D satisfy both the stability of the LTI system and the causality. However, option C is not true for the stable system because, | S | = 1 also has a pole(1 pole) in the right hand plane.

From Signal Flow Graph, we have two forward paths
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 L₁, L₂, L₃ and L₄ are touching to each other so,

By using Mason’s Gain Formula

With a raised cosine pulse shape, the bandwidth is larger than the minimum required pulse shape and is related as

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/

Transfer function is

Input

So output

Now

this will be zero if s² + ω² will cancel (s² + 9) term as only then final value theorem will be applicable.
So at ω² = 9 => ω = 3 rad/sec, steady state output will be zero.

Probability of error in coherent BPSK is given by

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

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

_S³ + as² + (2+K)s + 1(1+K) = 0

For oscillation

Now
as² + (1+K) = 0
- aω² +(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

for system to be controllable
|Q_c| ≠ 0
⇒ (0 = a₁a₂²) ≠ 0
⇒ a₁≠ 0
a₂ ≠ 0

Phase
for lead comparators phase must be +ve.
For this
So, a = 1, b = 2

The Correct Answer Among All the Options is B
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: s³+3s²+s(k+2)+4k=0
Now from this we can construct the routh array as follows:
s³ 1 (k+2)
s² 3 4k
s¹ (6-k)/3 0
s⁰ 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 s¹ element:
(6-k)/3=0=> k= 6
Refer the Topic Wise Question for Routh-Hurwitz Control Systems

The Correct Answer Among All the Options is A
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

Given
r(t) = 5u(t)
R(s) =
Now

• 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)

Steady state error table: -

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

Given open loop transfer function
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

• 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

• 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.

At resonant frequency where magnitude m has a peak value in frequency response in known as resonant frequency.

G_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

The Correct Answer Among All the Options is A

Converting this into SFG, we get

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

The Correct Answer Among All the Options is B
Damped means gradually reduces to oscillation , but here the oscillations are sustained.
Refer the Topic Wise Question for Basics of Control System Control Systems

The Correct Answer Among All the Options is B

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

The Correct Answer Among All the Options is C
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

The Correct Answer Among All the Options is A
T=(A₁+kA₂)/ (A₃+kA₄)
Sensitivity , =
=
=
On putting given values,

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

The Correct Answer Among All the Options is C
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