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 type0 system.
For type0 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 = ω_{gc, }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,
ω_{c }= (1/a) rad/sec
Initial slope = 20 dB/decade
= 6 dB/octave (Since Type = 1)
Corner frequency is,
ω_{c} = 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 M_{p} (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^{1}ω = tan^{1}(1)
∴
A system has transfer function equal to (1s)/(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^{1}ωT = ϕ
At ω = ω_{pc},
ϕ = 180º
∴ 90°  tan^{1}ω_{pc} = 180° or, ω_{pc} = ∞ rad/s
∴
The characteristic equation of a closed loop control system is given by s^{2} + 4s + 16 = 0. The resonant frequency (in radian/sec) of the system is
Given, s^{2} + 4s+ 16 = 0
or
∴ ω_{n} = 4 rad/s
and
∴ Resonant frequency,
The gain margin (in dB) of a system having the openloop transfer function
is
Given,
∴
and ∠G{jω)H(jω) = 90°  tan^{1} ω = ϕ
At ω = ω_{pc}, ϕ = 180°
∴ 90°  tan^{1} ω_{pc} = 180° or ω_{pc }= ∞
∴
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 Ncircle.
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^{1}ω = ϕ
At
ω = ω_{gc} = 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 ω_{gc}, Img[G(jω)H(jω)] = 0
or,
Now,
Thus, Nyquist plot will cut real axis at 0.66.
The given system:
of type1 and order3.
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 splane = 0
Now, N = PZ
or, 0 = 0  Z or Z = 0
Thus, no. of closed loop poles in RH splane = 0.
Hence, system is stable.
The nyquist plot for a unity feedback control system having open loop transfer function G(s) = K(1s)/(s+1) is shown in figure below :
The system is
Given,
G(s)H(s) = K(1s)/(s+1)
or,
Thus, the radius of Nyquist plot will be K.
Now, no.of open loop poles in RH spiane = 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 ω_{1}.
Which one of the following equation should be satisfied for it?
Given,
or, ∠G(jω) H(jω) = ωT  90°  tan^{1} ω
At ω = ω_{1}, ∠G(jω) H(jω) = 0°
∴ ω_{1} T90° tan^{1}ω_{1} = 0
or, tan^{1}ω_{1 }=  (ω_{1}T + 90°)
or, ω_{1}, = tan (90° + ω_{1}T) = cot ω_{1}T
∴ ω_{1} = cot(ω_{1}T)
What is the approximate value of gain margin in the Nyquist diagram shown below?
At ω = ω_{gc},
(Using Nyquist diagram)
∴
The magnitudefrequency response of a control system is given in figure below:
The values of ω_{1} and ω_{2} are respectively
Number of decade change from first break frequency
Now,
Now, no. of decade change from break frequency
∴
or,
ω_{2} = 400 rad/s
Match ListI with ListIl and select the correct answer using the codes given below the lists:
ListI
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 M_{r }(resonant peak).
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