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/decadetill 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 = tan1 0.1  tan1 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 
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 = tan1  tan1 0.5  tan1 0.05 = 900 tan1 0.5  tan1 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 – tan1 (0.5×gc) – tan1 (0.05 × gc) = 164.50 PM = 1800  164.50 = 15.50 For phase cross over frequency (pc) = 900 – tan1 (0.5 ) – tan1 (0.05 ) 1800 = 900 – tan1 (0.5 pc) – tan1 (0.05 pc) 900 – tan1 (0.5 pc) – tan1 (0.05 pc) Taking than on both sides Tan 900 = tan1 Let tan1 0.5 pc = A, tan1 0.05 pc = B = 00 = 0 1 =0.5 pc 0.05pc pc = 6.32 rad/sec The plot is shown in figure 2. 
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 = tan1 0.5  tan1 0.2  tan1 0.1  tan1 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 
G(S) = Converting above transfer function to standard from G(S) = =
T1 = , 11= 5 (zero) T2 = 1 , 2 = 1 (pole) 4. 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 


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 = tan1[ 2ξu/(1u2)]   (3) For sinusoidal input the output response for the system is given by C(t) = 1/√(1u2)2 + (2ξu)2Sin[wt  tan1 2ξu/1u2]   (4) The frequency where M has the peak value is known as Resonant frequency Wn. This frequency is given as (from eqn (2)). dM/duu=ur = Wr = Wn√(12ξ2)   (5) from equation(2) the maximum value of magnitude is known as Resonant peak. Mr = 1/2ξ√1ξ2   (6) The phase angle at resonant frequency is given as Φr =  tan1 [√12ξ2/ ξ]   (7)
As we already know for step response of second order system the value of damped frequency and peak overshoot are given as Wd = Wn√1ξ2   (8) Mp = e πξ2√1ξ2   (9) 
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 = tan1(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 ejϴ M = 0 ejϴ = 0 Path P3 : W = ∞ to 0 M = 1/√1+W2 , φ = tan1(W/I) W M φ ∞ 0 900 1 1/√2 450 0 1 00 The Nyquist Plot is shown in fig below 
Path P1 W(0 to ∞) M = 1/√42 + w2 √52 + w2 Φ = tan1(W/4) – tan1(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.ej2ϴ = 0 The plot is shown in fig below. From plot N=0, Z=0, system stable.

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 – tan1(w/10) W M Φ ∞ 0 3 π/2 Ɛ ∞ 1800 Path P3 will be mirror image of P1 about 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) = ∞. ej2ϴ [ 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 right half of S plane so, P = 0 N = Z – 0 
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 – tan1(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) = ∞. ej2ϴϴ(π/2 to π to 3π/2) 2ϴ = (π to 2π to 3π) P4 = 0 