UNIT-5

Frequency Domain Analysis

Q1) Plot polar plot for T(S) = 1/S + 1

Sol: For polar plot substitute S=jw.

TF = 1/1 + jw

(2). Magnitude M = 1 + 0j / 1 + jw = 1/√1 + w2

(3). Phase φ = tan-1(0)/ tan-1w = - tan-1w

W M φ

0 1 00

1 0.707 -450

∞ 0 -900

The plot is shown in fig.

Q2) Plot polar plot for T(S) = 1/(S+1)(S+2)

(1). S = jw

TF = 1/(1+jw)(2+jw)

(2). M = 1/(1+jw)(2+jw) = 1/-w2 + 3jw + 2

M = 1/√1 + w2√4 + w2

(3). Φ = - tan-1 w - tan-1(w/2)

W M Φ

0 0.5 00

1 0.316 -71.560

2 0.158 -108.430

∞ 0 -1800

The plot is shown in fig

Intersection of polar plot with imaginary axis will be when real part of Transfer function = 0

M = 1/(jw + 1)(jw + 2)

= 1/-w2 + j3w + 2

Real part

Re(M) = 1/(2-w2)+j3w x (2-w2)-3jw/(2-w2)-3jw

Re(M) = (2-w2)/(2-w2+9w2) - 3jw/(2-w2) +9w2

Equating Real part = 0

(2-w2)/(2-w2)+ 9w2 = 0

W = +-√2

For w=√2 real part on the polar plot becomes zero.

So, the polar plot intersects the imaginary axis at w=√2 at φ = -900

Q3) Plot polar plot for T(S) = 1/(S+1)(S+2)(S+3)

(1). Substitute S =jw

(2). M = 1/√1+w2 √4+w2 √9+w2

(3). Φ = -tan-1w – tan-1 w/2 – tan-1w/3

W M φ

0 0.16 0

1 0.1 -900

2 0.04 -142.10

∞ 0 -270

(4). The intersection of Polar plot with Real axis

M = 1/(S+1)(S+2)(S+3)

= 1/S3+6S2+11S+6

=1/(jw)2+6(jw)2+11jw+6

= 1/(6-6w2)+j(11w-w3)

Re(M) = 0 [Intersection with imaginary axis ]

M = 6-6w2/(6-6w2)2+(11w-w2)2 - j(11w-w3)/(6-6w2)2+(11w-w2)2

Re(M) = 0

6-6w2 = 0

w=1

Im(M) = 0[Intersection with Real axis]

11w = w3

The plot is shown in fig 3(c)

Q4) For T(S) = 1/S(S+1) plot polar plot?

Sol:

(1). M = 1/W√1+w2

(2). Φ = -900 - tan -1(W/T)

W M φ

0 ∞ -900

1 0.707 -1350

2 0.45 -153.40

∞ 0 -1800

The plot is shown in fig.

Q5) For T(s) = 1/S2(S+1) plot polar plot?

Sol:

(1). M = 1/w2√1+jw

(2). Φ = -1800 – tan-1W/T

The plot is shown in fig.5(b)

Q6) Sketch the bode plot for the transfer function

G(S) =

Sol: 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

= tan-1 0.1 - tan-1 0.001

For phase plot

100 -900

200 -9.450

300 -104.80

400 -110.360

500 -115.420

600 -120.00

700 -124.170

800 -127.940

900 -131.350

1000 -134.420

The plot is shown in figure 1

Q7) For the given transfer function determines

G(S) =

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

Q8. For the given transfer function

G(S) =

Plot the rode plot find PM and GM

T1 = 0.5 1 = = 2 rad/sec

Zero so, slope (20 dB/decade)

T2 = 0.2 2 = = 5 rad/sec

Pole , so slope (-20 dB/decade)

T3 = 0.1 = T4 = 0.1

3 = 4 = 10 (2 pole ) (-40 db/decade)

Phase plot

= tan-1 0.5 - tan-1 0.2 - tan-1 0.1 - tan-1 0.1

500 -177.30

1000 -178.60

1500 -179.10

2000 -179.40

2500 -179.50

3000 -179.530

3500 -179.60

GM = 00

PM = 61.460

The plot is shown in figure 3

Q 9) For the given transfer function plot the bode plot (magnitude plot)

G(S) =

Given the transfer function

G(S) =

Converting above transfer function to standard form

G(S) =

=

T1 = , 11= 5 (zero)

T2 = 1 , 2 = 1 (pole)

4. The initial slope will cut zero dB axis at

(K)1/N = 10

i.e = 10

5. finding n and

T(S) =

T(S)=

Comparing with standard second order system equation

S2+2ns +n2

n = 11 rad/sec

n = 5

11 = 5

= = 0.27

5. Maximum error

M = -20 log 2

= +6.5 dB

6. As K = 10, so whole plot will shift by 20 log 10 10 = 20 dB

The plot is shown in figure 4

Q 10. For the given plot determine the transfer function

From figure 5 we can conclude that

K1/N N = 1

(K)1/N = 10.

3. corner frequency

1 = = 0.2 rad/sec

2 = = 0.125 rad/sec

3. 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

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

5. is a zero at = 8 (-40+(+20)=20)

6. Hence transfer function is

T(S) =

Q11) For T(S) = 1/S plot polar plot?

Sol:

(1). S = jw

(2). M = 1/W

(3). Φ = -tan-1(W/O) = -900

W M φ

0 ∞ -900

1 1 -900

2 0.5 -900

∞ 0 -900

The plot is shown in fig.

Q12. For the transfer function below plot the Nyquist plot and also comment on stability?

G(S) = 1/S+1

Sol:- N = Z – P ( No pole of the right half of S plane P = 0 )

P = 0, N = Z

NYQUIST PATH:-

P1 = W – (0 to - ∞)

P2 = ϴ( - π/2 to 0 to π/2 )

P3 = W(+∞ to 0)

Substituting S = jw

G(jw) = 1/jw + 1

M = 1/√1+W2

Φ = -tan-1(W/I)

for P1 :- W(0 to -∞)

W M φ

0 1 0

-1 1/√2 +450

-∞ 0 +900

Path P2:-

W = Rejϴ R ∞ϴ -π/2 to 0 to π/2

G(jw) = 1/1+jw

= 1/1+j(Rejϴ) (neglecting 1 as R ∞)

M = 1/Rejϴ = 1/R e-jϴ

M = 0 e-jϴ = 0

Path P3:-

W = -∞ to 0

M = 1/√1+W2 , φ = -tan-1(W/I)

W M φ

∞ 0 -900

1 1/√2 -450

0 1 00

The Nyquist Plot is shown in fig 6

From the plot we can see that -1 is not encircled so, N = 0

But N = Z, Z = 0

So, the system is stable.

Q.13. For the transfer function below plot the Nyquist Plot and comment on stability G(S) = 1/(S + 4)(S + 5)

Soln :- N = Z – P , P = 0, No pole on right half of S-plane

N = Z

NYQUIST PATH

P1 = W(0 to -∞)

P2 = ϴ(-π/2 to 0 to +π/2)

P3 = W(∞ to 0)

Path P1 W(0 to -∞)

M = 1/√42 + w2 √52 + w2

Φ = -tan-1(W/4) – tan-1(W/5)

W M Φ

0 1/20 00

-1 0.047 25.350

-∞ 0 +1800

Path P3 will be the mirror image across the real axis.

Path P2: ϴ(-π/2 to 0 to +π/2)

S = Rejϴ

G(S) = 1/(Rejϴ + 4)( Rejϴ + 5)

R∞

= 1/ R2e2jϴ = 0.e-j2ϴ = 0

The plot is shown in fig . From plot N=0, Z=0, system stable.

Q.14. For the given transfer function, plot the Nyquist plot and comment on stability G(S) = k/S2(S + 10)?

Soln: As the poles exist at the origin. So, the first time we do not include poles in the Nyquist plot. Then check the stability for the second case we include the poles at the origin in the Nyquist path. Then again check the stability.

PART – 1: Not including poles at the origin in the Nyquist Path.

P1 W(∞ Ɛ) where Ɛ 0

P2 S = Ɛejϴ ϴ(+π/2 to 0 to -π/2)

P3 W = -Ɛ to -∞

P4 S = Rejϴ, R ∞, ϴ = -π/2 to 0 to +π/2

For P1

M = 1/w.w√102 + w2 = 1/w2√102 + w2

Φ = -1800 – tan-1(w/10)

W M Φ

∞ 0 -3 π/2

Ɛ ∞ -1800

Path P3 will be a mirror image of P1 about a Real axis.

G(Ɛ ejϴ) = 1/( Ɛ ejϴ)2(Ɛ ejϴ + 10)

Ɛ 0, ϴ = π/2 to 0 to -π/2

= 1/ Ɛ2 e2jϴ(Ɛ ejϴ + 10)

= ∞. e-j2ϴ [ -2ϴ = -π to 0 to +π ]

Path P2 will be formed by rotating through -π to 0 to +π

Path P4 S = Rejϴ R ∞ ϴ = -π/2 to 0 to +π/2

G(Rejϴ) = 1/ (Rejϴ)2(10 + Rejϴ)

= 0

N = Z – P

No poles on the right half of the S plane so, P = 0

N = Z – 0

But from the plot shown in fig 8(a). it is clear that a number of encirclements in the Anticlockwise direction. So,

N = 2

N = Z – P

2 = Z – 0

Z = 2

Hence, the system unstable.

PART 2 Including poles at the origin in the Nyquist Path.

P1 W(∞ to Ɛ) Ɛ 0

P2 S = Ɛejϴ Ɛ 0 ϴ(+π/2 to +π to +3π/2)

P3 W(-Ɛ to -∞) Ɛ 0

P4 S = Rejϴ, R ∞, ϴ(3π/2 to 2π to +5π/2)

M = 1/W2√102 + W2 , φ = - π – tan-1(W/10)

P1 W(∞ to Ɛ)

W M φ

∞ 0 -3 π/2

Ɛ ∞ -1800

P3( mirror image of P1)

P2 S = Ɛejϴ

G(Ɛejϴ) = 1/ Ɛ2e2jϴ(10 + Ɛejϴ)

Ɛ 0

G(Ɛejϴ) = 1/ Ɛ2e2jϴ(10)

= ∞. e-j2ϴϴ(π/2 to π to 3π/2)

-2ϴ = (-π to -2π to -3π)

P4 = 0

The plot is shown in fig. from the plot it is clear that there is no encirclement of -1 in the Nyquist path. (N = 0). But the two poles at origin lies to the right half of the S-plane in the Nyquist path.(P = 2)[see path P2]

N = Z – P

0 = Z – 2

Z = 2

Hence, the system is unstable.

Path P2 will be formed by rotating through -π to -2π to -3π