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If the bandwidth of the system is increased, then
We know that,
Thus, if BW is increased, rise time will decrease and the system response will become fast. Due to faster response of system, damping ratio will decrease.
Polar plot of G(s) = 1/(1+Ts) is a
Given, G(s) = 1/(1+Ts)
It is a type0 and order1 system, therefore its polarplot will be as shown below.
Hence, the polar plot will be semicircular in shape.
Consider the following statements related to frequency domain analysis:
1. The cutoff rate is the slope of the log magnitude curve near the cutoff frequency.
2. The bandwidth is defined as the band of frequencies lying between 3 dB points.
3. Higher the value of resonant frequency of the system, slower is the time response.
4. The magnitude of resonant peak gives the information about the relative stability of the system.
Which of these statements is/are not correct?
The bandwidth is defined as the band of frequencies tying between3 dB points. Hence, statement2 is false.
Higher the value of resonant frequency of the system, faster is the time response. Hence, statement3 is false.
The inverse polar plot of the open loop transfer function, G(s) = (1+sT)/(sT) will be re presented by
Given, G(s) = (1+sT)/(sT)
The inverse polar plot of G(jω) is the polar plot of 1/G(jω)
Thus,
and
Hence, inverse polar plot will be as show below,
The constant Mcircles corresponding to the magnitude (M) of the closed loop transfer function of a linear system for value of M less than one lie in the Gplane and to the
The slope of the line due to 1/(jω)^{2 }factor in magnitude part of Bode plot is
1/(jω)^{2} represents type2 factor which will have a slope of 40 db/decade.
We know that,
20 db/decase = 6 db/octave
∴ 40 db/decase = 12 db/octave
The magnitude plot of the open loop transfer function G(s) of a certain system is shown in figure below.
If the system is of minimum phase type, then the openloop transfer function G(s) will be given by:
Since, the system is of minimum phase type, it has no poles or zeros in the right hand side of splane.
⇒ There will be three corner frequencies at: ω_{1} = 5 rad/s
∴ T_{1} = 1/ω_{1} = 0.2 sec
(one pole added at ω_{1} = 5 rad/s)
ω_{2} = 40 rad/s
∴ T_{2} = 1/ω_{2} = 0.025 sec
(two pole added at ω_{1} = 40 rad/s)
and ω_{3} = 100 rad/s
∴ T_{3} = 1/ω_{3} = 0.01 sec
(two poles added at ω_{3} = 100 rad/s)
So, open loop T.F. is
Now, 20 log K = 100
∴ K = 10^{5}
Thus,
A unity feedback system has the open loop transfer function,
For the given system to be marginally stable, the value of gain crossover frequency would be
For marginally stable system,
P.M = G.M = 0 and ω_{pc} = w_{gc}
Now, phase margin = (180° + ϕ)
Where,
Given, P.M = 0° (for marginally stable)
∴ 0º = 180° + ∠G(jω) H(jω)
or, ∠G(jω) H(jω) = 180°
Given,
Thus,
The transfer function of a laglead network is given by
The phase of G_{c} (jω) will become zero at
For the system shown below, what is the range of value of K for stability?
The OLTF has nopoles in RH splane. Thus, for stability the 1 + j0 (critical point) point should not be encircled by Nyquist plot.
For finding the point of intersection of Nyquist piot with the negative reai axis.
Img [G(jω) H(jω)] = 0
or, ω^{2} + 15 = 0
or, ω = √15 rad /sec
Putting, ω = √15 rad/s in equation (1), we have
For system to be stable,
So,
0 < K < 120 (for stability)
Match ListI (Polar plot of systems) with Listll (Transfer functions) and select the correct answer using the codes given below the lists:
ListI
ListII
Codes:
A B C D
(a) 3 1 2 4
(b) 2 4 1 3
(c) 2 1 4 3
(d) 1 2 3 4
Assertion (A): The Nyquist criterion represents a method of determining the location of the characteristic equation roots with respect to the left half and the right half of the splane.
Reason (R): The Nyquist criterion does not give the exact location of the characteristic equation roots.
Which of the following is not true?
BW ∝ ω_{n}. Hence, option (d) is not correct. For a prototype second order system,
Thus,
If the Nyquist plot cuts the negative real axis at a distance of 0.25, then the gain margin and phase margin of the system will be respectively
Here,
X = 0.25 (Given)
Also, ϕ = 180°
∴ P.M. = 180 + ϕ
= 180°180° = 0°
Match ListI (Different values of damping ratio) with ListII (Values of resonant peaks) and select the correct answer using the codes given below the lists:
ListI
A. ξ = 0
B. ξ = 0.707
C. ξ = 0.85
D. ξ = 0.5
ListII
1. More than unity
2. Not defined
3. Unity
4. infinite
Codes:
A B C D
(a) 1 3 2 4
(b) 4 1 2 3
(c) 4 3 2 1
(d) 1 2 3 4
Resonant peak is given by,
In frequency domainanalysis,
Thus, for ω_{r} to be real Hence for there is no M_{r} or M_{r} is not defined.
Assertion (A) : Gain margin alone is adequate to indicate relative stability when system parameters other than the loop gain are subject to variation.
Reason (R) : Gain margin is the amount of gain in dB that can be added to the loop before the closedloop system becomes unstable.
Gain margin alone is inadequate to indicate relative stability when system parameters other than the loop gain are subject to variation. Due to variation in some parameters, phase shift of the system can vary which may results instability. Hence, we must also consider the concept of phase margin for determining the relative stability of the system.
Assertion (A): Phase margin is measured at the gain crossover frequency.
Reason (R): Gain crossover frequency can be easily determined from Bode plot than from the Nyquist plot.
If the magnitude of closed loop transfer function of a system is given by M, then the centre and radius of Mcircle will be respectively given by:
The open loop transfer function of a unity feedback system is given by
The value of ω_{pc}, is
Given,
∴ ϕ = ∠G(jω) H(jω)
= 90°  tan^{1}ω  tan^{1} (0.1 ω)
At ω = ω_{Pc}, ϕ = 180°, therefore
180° = 90 °  tan^{1}ω  tan^{1} (0.1 ω)
or,
= Phase crossover frequency
If the gain of the openloop system is increased to four times, the gain margin of the system
∴ When openloop gain becomes 4times, GM will become onefourth.
22 docs274 tests

22 docs274 tests
