The represents of in static diagram will be:
From options we can easily solve the problem
Forward paths are:
Individual loop,
∴
the angle of departure of the root locus at s = —1 + j is
∴
∴
The openloop transfer function of a system is . The root locus of the system is
The openloop transfer function of a system is G(s) = Indicate the correct root locus diagram is
Breakaway point:
s = —3.4 is a valid breakaway point.
A negative feedback control system has a transfer function G(s)
compensator G_{c}(s) = in order to achieve zero steady state error for a step input. Select 'a' and 'k' so that the overshoot to a step is approximately 5% and the settling time (with a 2% criterion) is approximately 1 second.
The characteristics equation is 1 + G_{c}(s) G(s) I1(s) = 0
compare equation (i) to standard
Now,
The transfer function of a lead compensator is G_{c} s = . The maximum phase shift that can be obtained from this compensator is
The standard transfer function of lead compensator is
......(i)
and given transfer function
put s = j ω then,
......(ii)
Compare equation (i) and (ii) we gt,
∴
∴ Maximum phase shift =
∴
Calculate the sensitivity of the closedloop system shown in figure below with respect to the forward path transfer function at w = 1.3 rad/sec.
Here,
and H(s) = 0.50
∴
The magnitude plot of a transfer function is shown in figure below. The transfer function is
Initially slope is —6 dB/octave i.e. —20 dB/dec. so there must be a pole at origin. At ω = 2 rad/sec slope change to ω db/dec. so there is a zero at ω = 2
and At 0 = 10 rad/sec, slope change to —20 dB/sec. so there is a pole at ω = 10.
∴ Transfer function =
∴
∴ K = 0
∴ Transfer function =
The openloop transfer function of a feedback system is
The gain margin of the system is___
∴
∴
The openloop transfer function of a feedback system is G (s)H (s) =
The value of gain k for obtaining a gain margin of 3 dB for stable system is_________
∴
∴
In the figure given below the phase margin and the gain margin are:
A LTI system is characterized by the homogeneous state equation The initial state is
The state transition matrix 4(t) is
Given
The state equation in the phase canonical form can be obtained from the transfer function by:
The state equation from transfer function by parallel decomposition in the phase canonical form.
Consider the system shown in figure below and investigate whether it is observable or not.
and
and
For observability:
So the System is Observable
For a ()ve feedback second order control system in its step response, the maximum value of the output response in time domain form is given by
Q. The damping factor (damping coefficient) of this system is
From the conventional approach we have,
Give that C(t)I_{max} = 1.75
∴ M_{p} = Maximum overshoot = 1.75  1.00 = 0.75
For a ()ve feedback second order control system in its step response, the maximum value of the output response in time domain form is given by
Q. The openloop zero frequency gain of this given unity feedback control system equals to
Closed loop transfer function
(put, H(s) = 1)
= Open loop transfer function
∴ Open loop Gain at zero frequency i.e. s
The block diogram of a feedback system is shown in figure (a).
Q. Find the minimum value of G for which the step response of the system would
exhibit an overshoot as shown in figure (b).
Closed loop transfer function:
Characteristic equation s^{2} + 3s + G = 0
For minimum valueof G 'ξ’ should be 0.6.
∴
G = 6.25
The block diogram of a feedback system is shown in figure (a).
Q. For G equal to twice this minimum value, find the time period 't' indicated in figure (b).
∴
The system shown in figure below has the oscillation of 2.5 rad/sec, and there are no poles in right half of splane.
Q. Determine the values of K_{mar}.
Since the system oscillates, it is marginally stable. The characteristic equation of the system becomes.
Now the Routh's Array is
At marginal value of k,
Again, at this value of p,
∴
Given, ω = 2.5 rad /sec, therefore
k + 3 = 6.25
∴ k = 3.25
The system shown in figure below has the oscillation of 2.5 rad/sec, and there are no poles in right half of splane.
Q. Also find the value of P.
From the above solution:
at k = 3.25 then,
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