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Phase crossover frequency is defined at
The Bode plot for a transfer function is shown below:
The steady state error corresponding to a parabolic input is
Initial slope of the system = 0 dB/decade
Therefore, given system is a type-0 system.
For type-0 system, for parabolic input, steady state error is infinite since,
The frequency at which the Nyquist plot crosses the negative real axis is known as
At ω = ωgc, ϕ = -180°
i.e. Nyquist plot crosses the negative real axis.
A unity feedback system has a forward path gain of
The phase angle of the Nyquist plot for ω → ∞ is
At ω → ∞, ϕ = -90° - 90° - 90° = -270° = +90°
The open loop transfer function of a system is given as
The Bode plot of this system is represented as:
Initial slope = -20 dB/decade
= -6 dB/octave (Since Type = 1)
Corner frequency is,
ωc = (1/a) rad/sec
Initial slope = -20 dB/decade
= -6 dB/octave (Since Type = 1)
Corner frequency is,
ωc = 1/a rad/sec
It will cross 0 dB line at ω = K and K < 1/a.
So, the Bode plot will be as shown below.
A second order system has
It’s Mp (peak magnitude) will be approximately
∴ ξ = 1/√2
So, peak magnitude = 1 + e-π ≈ 1
The closed loop transfer function of a control system is given by
C(s)/R(s) = 1/(1+s)
For the input r(t) = sint, the steady state value of c(t) is equal to
Given,
C(s)/R(s) = 1/(1+s)
Thus,
Here,
ω = 1 rad/s
and ∠H(jω) = -tan-1ω = -tan-1(1)
∴
A system has transfer function equal to (1-s)/(1+s). Its gain at ω = 1 rad/s is
Given transfer function represents an all pass filter which has gain of 1 at all frequencies.
The gain margin of a unity negative feedback system having forward path transfer function
∠G(jω) H(jω) = -90° - tan-1ωT = ϕ
At ω = ωpc,
ϕ = -180º
∴ -90° - tan-1ωpc = -180° or, ωpc = ∞ rad/s
∴
The characteristic equation of a closed loop control system is given by s2 + 4s + 16 = 0. The resonant frequency (in radian/sec) of the system is
Given, s2 + 4s+ 16 = 0
or
∴ ωn = 4 rad/s
and
∴ Resonant frequency,
The gain margin (in dB) of a system having the open-loop transfer function
is
Given,
∴
and ∠G{jω)H(jω) = -90° - tan-1 ω = ϕ
At ω = ωpc, ϕ = -180°
∴ -90° - tan-1 ωpc = -180° or ωpc = ∞
∴
Nichol’s chart is useful for detailed study and analysis of
Nichol’s chart consists of magnitude and phase angle of a closed loop system represented as a family of circle known as M and N-circle.
The forward path transfer function of an unity feedback system is given by
What is the phase margin for this system?
Given,
Now,
∠G(jω) = -2tan-1ω = ϕ
At
ω = ωgc = 0 rad/s
ϕ = -2 x 0 = 0º
∴ P.M. = 180° + ϕ
= 180° + 0° = 180° = π
The closed loop system having the open loop transfer function,
is
Given,
For finding ωgc, Img[G(jω)H(jω)] = 0
or,
Now,
Thus, Nyquist plot will cut real axis at -0.66.
The given system:
of type-1 and order-3.
Therefore, Nyquist plot will be as shown below:
Now, from above Nyquist plot,
N = 0 = Number of encirclements.
Also, P = number of open loop poles in RH s-plane = 0
Now, N = P-Z
or, 0 = 0 - Z or Z = 0
Thus, no. of closed loop poles in RH s-plane = 0.
Hence, system is stable.
The nyquist plot for a unity feedback control system having open loop transfer function G(s) = K(1-s)/(s+1) is shown in figure below :
The system is
Given,
G(s)H(s) = K(1-s)/(s+1)
or,
Thus, the radius of Nyquist plot will be K.
Now, no.of open loop poles in RH s-piane = 0.
When K > 1, no. of encirclement, N = -1
When K< 1, no. of encirclement, N = 0
Thus, for K > 1, N = P - Z
or, -1 = 0 - Z or Z = 1
∴ System is unstable.
For K < 1, N = P - Z or 0 = 0 - Z or Z = 0
∴ System is stable..
A unity feedback control system has a forward path transfer function of Its phase value will be zero at a frequency of ω1.
Which one of the following equation should be satisfied for it?
Given,
or, ∠G(jω) H(jω) = -ωT - 90° - tan-1 ω
At ω = ω1, ∠G(jω) H(jω) = 0°
∴ -ω1 T-90° -tan-1ω1 = 0
or, tan-1ω1 = - (ω1T + 90°)
or, ω1, = -tan (90° + ω1T) = cot ω1T
∴ ω1 = cot(ω1T)
What is the approximate value of gain margin in the Nyquist diagram shown below?
At ω = ωgc,
(Using Nyquist diagram)
∴
The magnitude-frequency response of a control system is given in figure below:
The values of ω1 and ω2 are respectively
Number of decade change from first break frequency
Now,
Now, no. of decade change from break frequency
∴
or,
ω2 = 400 rad/s
Match List-I with List-Il and select the correct answer using the codes given below the lists:
List-I
A. Resonant peak
B. Damped natural frequency (rad/s)
C. Resonant frequency (rad/s)
D. Peak overshoot
List - II
Codes:
A B C D
(a) 3 4 1 2
(b) 3 1 4 2
(c) 2 4 1 3
(d) 2 1 4 3
A second order underdamped system has a damping ratio of 0.8. It is subjected to a sinusoidal input of unit amplitude. It has resonant peak of
Resonant frequency is given by
For resonant peak
or,
Since ξ > 0.707, therefore there is no Mr (resonant peak).
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21 docs|263 tests
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