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Unit-6Frequency domain analysisQ1) Sketch the bode plot for transfer function G(S) =

A1)

Replace S = j

G(j=

This is type 0 system. so initial slope is 0 dB decade. The starting point is given as

20 log10 K = 20 log10 1000= 60 dB

Corner frequency 1 = = 10 rad/sec

2 = = 1000 rad/sec

Slope after 1 will be -20 dB/decade till second corner frequency i.e. 2 after 2 the slope will be -40 dB/decade (-20+(-20)) as there are poles

2. For phase plot

3. = tan-1 0.1 - tan-1 0.001

For phase plot

100 -900200 -9.450300 -104.80400 -110.360500 -115.420600 -120.00700 -124.170800 -127.940900 -131.3501000 -134.420The plot is shown in figure 1:

Fig 1 Magnitude Plot for G(S) =

Q2) For the given transfer function determine G(S) =

A2) Gain cross over frequency phase cross over frequency phase mergence and gain margin

Initial slope = 1

N = 1, (K)1/N = 2

K = 2

Corner frequency

1 = = 2 (slope -20 dB/decade

2 = = 20 (slope -40 dB/decade

2. phase

= tan-1 - tan-1 0.5 - tan-1 0.05

= 900- tan-1 0.5 - tan-1 0.05

1 -119.430

5 -172.230

10 -195.250

15 -209.270

20 -219.30

25 -226.760

30 -232.490

35 -236.980

40 -240.570

45 -243.490

50 -245.910

Finding gc (gain cross over frequency

M =

4 = 2 ( (

6 (6.25104) + 0.2524 +2 = 4

Let 2 = x

X3 (6.25104) + 0.2522 + x = 4

X1 = 2.46

X2 = -399.9

X3 = -6.50

For x1 = 2.46

gc = 3.99 rad/sec (from plot)

for phase margin

PM = 1800 -

= 900 – tan-1 (0.5×gc) – tan-1 (0.05 × gc)

= -164.50

PM = 1800 - 164.50

= 15.50

For phase cross over frequency (pc)

= 900 – tan-1 (0.5 ) – tan-1 (0.05 )

-1800 = -900 – tan-1 (0.5 pc) – tan-1 (0.05 pc)

-900 – tan-1 (0.5 pc) – tan-1 (0.05 pc)

Taking than on both sides

Tan 900 = tan-1

Let tan-1 0.5 pc = A, tan-1 0.05 pc = B

= 00

= 0

1 =0.5 pc 0.05pc

pc = 6.32 rad/sec

The plot is shown in figure 2.

Fig 2 Magnitude Plot G(S) =

Q3) For the given transfer function G(S) =

Plot the rode plot find PM and GMA3) T1 = 0.5

1 =

= 2 rad/secZero so, slope (20 dB/decade)T2 = 0.2

2 =

= 5 rad/secPole, so slope (-20 dB/decade)T3 = 0.1 = T4 = 0.1

3 =

4 = 10 (2 pole ) (-40 dB/decade)

Initial slope 0 dB/decade till

1 = 2 rad/sec

From

1 to

2 (i.e. 2 rad /sec to 5 rad/sec) slope will be 20 dB/decade

From

2 to

3 the slope will be 0 dB/decade (20 + (-20))

From

3 ,

4 the slope will be -40 dB/decade (0-20-20)

Phase plot

= tan-1 0.5

- tan-1 0.2

- tan-1 0.1

500 -177.301000 -178.601500 -179.102000 -179.402500 -179.503000 -179.5303500 -179.60GM = 00PM = 61.460The plot is shown in figure 3

Fig 3 Magnitude and phase Plot for G(S) =

Q4) For the given transfer function plot the bode plot (magnitude plot) G(S)=

A4) Given transfer functionG(S) =

Converting above transfer function to standard fromG(S) =

As type 1 system, so initial slope will be -20 dB/decade

Final slope will be -60 dB/decade as order of system decides the final slope

Corner frequency

T1 =

11= 5 (zero)T2 = 1 ,

2 = 1 (pole)4. Initial slope will cut zero dB axis at (K)1/N = 10i.e.

= 105. finding

n and

T(S) =

T(S)=

Comparing with standard second order system equationS2+2

ns +

n = 11 rad/sec

n = 5

11 = 5

= 0.276. Maximum errorM = -20 log 2

= +6.5 dB7. As K = 10, so whole plot will shift by 20 log 10 10 = 20 dBThe plot is shown in figure 4

1..JPG

Fig 4 Magnitude Plot for G(S) =

Q5) For the given plot determine the transfer function

Fig 5 Magnitude PlotA5) From figure 5 we can conclude that

Initial slope = -20 dB/decade so type -1

Initial slope all 0 dB axis at

= 10 so

K1/N N = 1(K)1/N = 10.3. corner frequency

1 =

= 0.2 rad/sec

2 =

= 0.125 rad/sec4. At

= 5 the slope becomes -40 dB/decade, so there is a pole at

= 5 as

slope changes from -20 dB/decade to -40 dB/decade

= 8 the slope changes from -40 dB/decade to -20 dB/decade hence

is a zero at

= 8 (-40+(+20) = 20)

Hence transfer function is T(S) =

Q6) Draw and explain phase lead compensator?A6) The phase of output voltage leads the phase of input voltage for the applied sinusoidal input. The circuit diagram is shown below. The transfer function is given as,

Fig 6 Phase Lead CompensationVo/Vi = α (1 + ST)/ (1 + S α T)Where, α = R2/R1 + R2 and α< 1T = R1Cw = 1/T lower corner frequency due to zero.w = 1/ αT upper corner frequency due to pole.Mid frequency is given as,wm = 1/T√ αThe maximum phase lead angle is φmΦm = tan-1(1- α)/2√α = Sin-11- α/1 + α Q7) Explain correlation between time and frequency response?A7) The transfer function of second order system is shown as

C(S)/R(S) = W2n / S2 + 2ξWnS + W2n - - (1)

ξ = Ramping factor

Wn = Undamped natural frequency for frequency response let S = jw

C(jw) / R(jw) = W2n / (jw)2 + 2 ξWn(jw) + W2n

Let U = W/Wn above equation becomes

T(jw) = W2n / 1 – U2 + j2 ξU

so,

| T(jw) | = M = 1/√ (1 – u2)2 + (2ξU)2 - - (2)

T(jw) = φ = -tan-1[ 2ξu/(1-u2)] - - (3)

For sinusoidal input the output response for the system is given by

C(t) = 1/√(1-u2)2 + (2ξu)2Sin [wt - tan-1 2ξu/1-u2] - - (4)

The frequency where M has the peak value is known as Resonant frequency Wn. This frequency is given as (from eqn (2)).

dM/du|u=ur = Wr = Wn√(1-2ξ2) - - (5)

from equation (2) the maximum value of magnitude is known as Resonant peak.

Mr = 1/2ξ√1-ξ2 - - (6)

The phase angle at resonant frequency is given as

Φr = - tan-1 [√1-2ξ2/ ξ] - - (7)

As we already know for step response of second order system the value of damped frequency and peak overshoot are given asWd = Wn√1-ξ2 - - (8)Mp = e- πξ2|√1-ξ2 - - (9)The comparison of Mr and Mp is shown in figure (1). The two performance indices are correlated as both are functions of the damping factor ξ only. When subjected to step input the system with given value of Mr of its frequency response will exhibit a corresponding value of Mp.

Fig 7 Frequency Domain SpecificationSimilarly, the correlation of Wr and Wd is shown in fig (7) for the given input step response [ from eqn (5) & eqn (8)]Wr/Wd = √ (1- 2ξ2)/(1-ξ2)Mp = Peak overshoot of step responseMr = Resonant Peak of frequency responseWr = Resonant frequency of Frequency responseWd = Damping frequency of oscillation of step response.From fig (7) it is clear that for ξ> 1/2, value of Mr does not exists. Q8) Explain phase lead lag compensator?A8) To overcome the disadvantages of lead and lag compensation, they are used together. As lead compensation does not provide phase margin but has fast response and lag compensation stabilize the system but does not provide enough bandwidth.

Fig 8 Lag-Lead Compensationeo/ei = (1 + α ST1)/(1 + S

T2)Where, αT1 = R1C1

T2= R2C2we can say in the lag-lead compensation pole is more dominating than the zero and because of this lag-lead network may introduces positive phase angle to the system when connected in series. Q9) List difference between phase lead and phase lag compensator?A9)

Phase Lead Network Phase Lag Network

1>. It is used to improve the 1>It is used to improve the

transient response. Steady state response.

2>. It acts as a high pass filter. 2>It acts as a low pass filter.

3>. The system becomes fast as 3>The Bandwidth decreases

Bandwidth increases as rise through rise time the speed

Time decreases. is slow.

4>. As the circuit acts as 4>Signal to noise ratio is higher as it a differentiator, signal to acts as integrator.

noise ratio is poor.

5>. Maximum peak overshoot 5>It reduces steady state error thus

is reduced. improve the steady state accuracy

Q10) Explain phase lag compensation technique?A10) The phase of the output voltage lags the phase of the input voltage for applied sinusoidal input. The circuit diagram is shown below:

Fig 9 Phase Lag Compensation

Vo/Vi = 1 + ST/1 + SβT

Where, β = R1 + R2/R2, β> 1

T = R2C

w = 1/T upper corner frequency due to zero

w= 1/βT lower corner frequency due to pole

The mid frequency wm,

wm = 1/T√β

The maximum phase lead angle Φm

Φm = tan-11- β/2 √β

= sin-11 – β/1 + β

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