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Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) PDF Download

Q21: For the transfer function Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) The values of the constant gain term and the highest corner frequency of the Bode plot respectively are  (SET-2 (2014))
(a) 3.2, 5.0
(b) 16.0, 4.0
(c) 3.2, 4.0
(d) 16.0, 5.0
Ans:
(a)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Reducing G(s) in time constant form, we have:
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)∴ The value of constant gain term = 3.2
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Corner frequencies are:
ω= 4 rad/sec
ω= 0.25 rad/sec
ω= 5 rad/sec
∴ Highest corner frequency = ω3 =5 rad/sec  

Q22: The Bode magnitude plot of the transfer function Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) is shown below:
Note that -6 dB/octave = - 20 dB/decade. The value of (a/bK) is _____.   (SET-1 (2014))
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) 0.48
(b) 0.76
(c) 0.92
(d) 0.28
Ans:
(b)
Sol: Given, Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
(1+as) is addition of zero to the transfer function whose contribution in slope = +20 dB/decade or -6 dB/octave.
(1+bs) is addition of pole to the transfer function whose contribution in slope = -20 dB/decade or -6 dB/octave.
Observing the change in the slope at different corner frequencies, we conclude that,
a = 1/4 rad/s and b = 1/24 rad/s
From ω = 0.01 rad/s to ω = 8 rad/s,
slope = -20dB/decade
Let the vertical length in dB by y
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Applying, y − mx + C at ω = 0.01 rad/s,  
We have, 58 = −20log 0.01 + C
C = 58 − 40 = 18  
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q23: The Bode plot of a transfer function G(s) is shown in the figure below.Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)The gain (20 log|G(s)| is 32 dB and -8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative for all ω. Then G(s) is  (2013)
(a) 39.8/s
(b) 39.8/s2
(c) 32/s
(d) 32/s2
Ans: 
(b)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)∵ Gain = 32 dB at ω = 1 rad/sec
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q24: A system with transfer function
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)is excited by sin(ωt) . The steady-state output of the system is zero at  (2012)
(a) ω = 1 rad/s  
(b) ω = 2 rad/s
(c) ω = 3 rad/s
(d) ω = 4 rad/s
Ans:
(c)
Sol: Fpr sinusoidal excitation,
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)For zero steady state output
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)For zero steady state output,
⇒ ω2 = 9
⇒ ω = 3rad/sec

Q25: The frequency response of a linear system G(jω) is provided in the tubular form below
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Gain Margin and phase margin are  (2011)
(a) 6 dB and 30°
(b) 6 dB and -30°
(c) -6 dB and 30°
(d) -6 dB and -30°
Ans:
(a)
Sol: At gain crossover frequency (ωgc), magnitude of G(jω) is 1.
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)At phase cross frequency (ωpc), phase of G(jω) is −180°,
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q26: The frequency response of  Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) plotted in the complex G(jω) plane (for 0 < ω < ∞) is  (2010)
(a) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(b) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(c) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(d) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Ans:
(d)

Q27: The open loop transfer function of a unity feed back system is given by G(s) = (e−0.1s)/s. The gain margin of the is system is  (2009)
(a) 11.95 dB
(b) 17.67 dB
(c) 21.33 dB
(d) 23.9 dB
Ans:
(d)
Sol: Open loop transfer function, Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Put, s = jω
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)At phase crossover frequency (ωpc), phase of OLTF is −180°
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Gain at ωpc ( phase-cross frequency)  
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q28: The asymptotic approximation of the log-magnitude v/s frequency plot of a system containing only real poles and zeros is shown. Its transfer function is  (2009)
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)

(b) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
(c) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
(d) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Ans: (b)
Sol: Initial slope = -40 dB/dec.
Hence the system is type-2. So the corresponding term of the transfer function is 1/s2.
At ω = 2 rad/sec, slope changes by -20 dB/dec from -40 dB/dec to -60 dB/dec. Hence the corresponding term of the transfer function is 1/(1+(s/2)).
At ω = 3 rad/sec, slope changes by 20 dB/dec from -60 dB/dec to -40 dB/dec. Hence the corresponding term of the transfer function is (1 + s/5).
At ω = 25 rad/sec, slope changes by -20 dB/dec from -40 dB/dec to -60 dB/dec. Hence the corresponding term of the transfer function is 1/(1 + s/25).
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q29: The polar plot of an open loop stable system is shown below. The closed loop system is  (2009)
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) always stable
(b) marginally stable
(c) unstable with one pole on the RH s-plane
(d) unstable with two poles on the RH s-plane
Ans:
(d)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Two clockwise encirclement of 1 + j0.
N = -2
Open-loop system is stable ⇒ P = 0
N = P - Z
-2 = 0 - Z
Z = Number of closed loop poles in RHS of 8-plane. Hence the system is unstable.

Q30: The asymptotic Bode magnitude plot of a minimum phase transfer function is shown in the figure :
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)This transfer function has  (2008)
(a) Three poles and one zero
(b) Two poles and one zero
(c) Two poles and two zero
(d) One pole and two zeros
Ans: 
(c)
Sol: Initial slope is -40 dB/decade, it means there are double pole at origin.
Slope changes from -40 dB/decade to -20 dB/decade. It means there is a zero.
Slope changes from -20 dB/decade to -0 dB/decade at some other frequency.eans ther is one more zero.
Therefore transfer function has two poles and two zeros.

Q31: If x = Re G(jω), and y = lm G(jω) then for ω → 0+, the Nyquist plot for Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) becomes asymptotic to the line  (2007)
(a) x = 0
(b) x = -3/4
(c) x = y - 1/6
(d) x = y/√3
Ans:
(b)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q32: Consider the following Nyquist plots of loop transfer functions over ω = 0 to ω = ∞. Which of these plots represent a stable closed loop system ?  (2006)
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) (1) only
(b) All, except (1)
(c) All, except (3)
(d) (1) and (2) only
Ans:
(d)

Q33: The Bode magnitude plot of
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) is  (2006)
(a) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(b) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(c) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(d) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Ans:
(a)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Corner frequencys
ω1 = 1 rad/sec; ω= 10 rad/sec and ω= 100 rad/sec
For frequency less than ωi.e. ω < ω1
Gain of the system is constant aas there is no pole at origin.
Gain = 20 logk = 20 log 0.1 = −20dB
At ω = ω= 1 rad/sec or ω1 = log1 = 0
There is zero, so system gain increases with slope +20dB/decades and system gain become 0 dB at ω = 10ω = 10 rad/sec or logω∣ω=10 = log10 = 1
At ω2 = 10 or logω∣ω2=10 = log10 = 1
there is pole, so slope is -20 dB/decade.
Overall slope ω2 < ω < ω3
= 20 dB/decade - 20 dB/decade
= 0 dB/decade
So, gain remain constant between  
ω< ω < ω3
or 1 < logω < 2
At ω = ω3 = 100 rad/sec or
logω∣ω3=100 = log100 =2
the double pole are present.
So, system gain decrease with -40 dB/decade.

Q34: If the compensated system shown in the figure has a phase margin of 60° at the crossover frequency of 1 rad/sec, then value of the gain K is  (2005)
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) 0.366
(b) 0.732
(c) 1.366
(d) 2.738
Ans: 
(c)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q35: In the G(s)H(s)-plane, the Nyquist plot of the loop transfer function G(s)H(s) = Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) passes through the negative real axis at the point  (2005)
(a) (-0.25, j0)
(b) (-0.5, j0)
(c) (-1, j0)
(d) (-2, j0)
Ans:
(b)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Putting s = jω, for frequency response
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)at ωpc nyquist plot cuts negative real axis and this frequency ∠G(jωpc) H(jωpc) is −180°
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q36: The gain margin of a unity feed back control system with the open loop transfer function Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) is  (2005)
(a) 0
(b) 1/√2
(c) √2
(d) ∞
Ans:
(d)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)−1 + j0 is not enclosed , system is always stable.
∴ G.M. = ∞

Q37: A system with zero initial conditions has the closed loop transfer function Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) The system output is zero at the frequency  (2005)
(a) 0.5 rad/sec
(b) 1 rad/sec
(c) 2 rad/sec
(d) 4 rad/sec
Ans: 
(c)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)The system output is zero at 2 rad/sec.

Q38: The open loop transfer function of a unity feedback control system is given as
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)The value of 'a' to give a phase margin of 45° is equal to  (2004)
(a) 0.141
(b) 0.441
(c) 0.841
(d) 1.141
Ans:
(c)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)ωg = gain crossover frequency at which open gain is loop
Phase margin = 180 +∠G(jω)
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q39: In the system shown in figure, the input x(t) = sint. In the steady-state, the response y(t) will be  (2004)
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)

(b) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
(c) sin(t − 45°)
(d) sin(t + 45°)
Ans: 
(b)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)For sinusoidal input, s = jω = j
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q40: The Nyquist plot of loop transfer function G(s)H(s) of a closed loop control system passes through the point (-1, j0) in the G(s)H(s) plane. The phase margin of the system is  (2004)
(a) 0°
(b) 45°
(c) 90°
(d) 180°
Ans:
(a)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)ωg is gain cross over frequency at which the gain ∣G(jω) H(jω)∣ becomes unity.
In this case, phase ∠G(jω)H(jω) is −180° at ω = ωg.
ϕ = −180°
So phase margin = 180° + ϕ = 0°

Q41: The asymptotic Bode plot of the transfer function Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) is given in figure. The error in phase angle and dB gain at a frequency of ω = 0.5 are respectively  (2003)
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) 4.9°, 0.97db
(b) 5.7°, 3db
(c) 4.9,3db4.9°, 3db
(d) 5.7°, 0.97db
Ans:
(a)
Sol: Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)From the plot
magnitude = 20log⁡k
∴ error in dB gain = 0.97 dB
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q42: The asymptotic approximation of the log-magnitude versus frequency plot of a minimum phase system with real poles and one zero is shown in figure. Its transfer functions is  (2001)
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)

(b) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
(c) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
(d) Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Ans: (d)
Sol: Type: 2
Poles : 0, 0, 2, 25
Zeroes : 5
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q43: The polar plot of a type-1, 3-pole, open-loop system is shown in figure. The closed loop system is  (2001)
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) always stable
(b) marginally stable
(c) unstable with one pole on the right half s-plane
(d) unstable with two poles on the right half s-plane
Ans:
(d)
Sol: Nyquist plot will be
Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE)∴ Number of encirclement of (-1, j0) = -2 and number of right sided poles in open loop system = 0
 N = P−Z
∴ −2 = 0−2
 ⇒ Z = 2
∴ Closed loop system is unstable with two poles on the right half of s-plane.

The document Previous Year Questions- Frequency Response Analysis - 2 | Control Systems - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Control Systems.
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FAQs on Previous Year Questions- Frequency Response Analysis - 2 - Control Systems - Electrical Engineering (EE)

1. What is frequency response analysis and why is it important in control systems?
Ans.Frequency response analysis examines how a system responds to different frequencies of input signals. It is crucial in control systems as it helps in understanding the stability and performance of the system, allowing engineers to design controllers that ensure desired behavior over a range of operating conditions.
2. How do you calculate the frequency response of a system?
Ans.The frequency response of a system can be calculated by applying a sinusoidal input of varying frequencies and measuring the steady-state output. This is often represented using Bode plots, which illustrate the magnitude and phase shift of the output relative to the input across a range of frequencies.
3. What are Bode plots and how are they used in frequency response analysis?
Ans.Bode plots are graphical representations of a system's frequency response, showing the gain (magnitude) and phase shift across a range of frequencies. They are used to analyze system stability, design controllers, and predict how the system will behave in response to different inputs.
4. What is the significance of the gain margin and phase margin in frequency response analysis?
Ans.Gain margin and phase margin are measures of a system's stability in frequency response analysis. The gain margin indicates how much gain can be increased before the system becomes unstable, while the phase margin indicates how much phase shift can occur before instability. Both margins help in designing robust control systems.
5. How does frequency response analysis differ from time-domain analysis?
Ans.Frequency response analysis focuses on how a system behaves in response to sinusoidal inputs of various frequencies, while time-domain analysis examines the system's response to time-varying inputs directly. Frequency response is useful for analyzing stability and dynamic characteristics, whereas time-domain analysis provides insights into transient responses and system behavior over time.
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