Loading DocumentConsider the open loop transfer function
G(s)H(s)=S(S+5)(S2+2S+2)K
a. Draw approximate complete root loci.
b. Find range of K to be critical stable system.
c. Find range of K to be overdamped system.
poles: 0,−5,−1±1j zeros: ∄
asymptotic angles =4−0180+(360×i)∣i=0,1,2,3=±45∘,±135∘
centroid =4−5−1−1+j−1−j=−2
break points dSdK=0K=−GH1=−(S4+7S3+12S2+10S)
10+24S+21S2+4S3=0⟹S=−3.8649 or S≈−0.69253±0.40896i (rejected)
departure angle for (−1+1j):
90+135+tan−141+ϕ=180∘⟹ϕ=270∘=−59∘
departure angle for (−1−1j):ϕ=59∘
to findKcritical:
characteristic equation 1+GH=0
S4+7S3+12S2+10S+K=0
S4 | 1 | 12 | K |
|---|---|---|---|
S3 | 7 | 10 | |
S2 | 77×12−10=774 | K | |
S | 74/7740/7−7K | ||
1 | K |
Find K critical:
K=0
74/7740/7−7K=0⟹Kcritical=49740≈15.102
Find intersection with imaginary axis
774S2+49740=0⟹S≈±1.1952i
Range of K to be critical stable system: K=0 or K=49740≈15.102
Range of K to be overdamped system: system won't be overdamped for any K
since there's always poles with imaginary component resulting in oscillatory response
For the following system:
G(s)H(s)=S(S+4)(1+10S)(1+8S)K(S+2)
a. Draw Bode Diagram for K=40.
b. Find GM and PM.
c. Find the new value of GM and PM for K=300.
Find corner frequencies: ω=[0.1,2,4,8]
ω | 0.01 | 0.1 | 1 | 2 | 4 | 8 | 10 | 100 | 1000 |
|---|---|---|---|---|---|---|---|---|---|
magnitude (dB) | 65.9 | 43 | 6.6 | -4.2 | -15 | -27.7 | -32.5 | -89.9 | -149.8 |
magnitude (mm) | 42.8 | 27.9 | 4.3 | -2.7 | -9.7 | -18 | -21.1 | -58.4 | -97.4 |
phase (deg) | -95.6 | -134.2 | -168.8 | -172.7 | -186.6 | -211.7 | -220.2 | -264.2 | -269.4 |
phase (mm) | -13.8 | -19.3 | -24.3 | -24.9 | -26 | -30.5 | -31.8 | -38.1 | -38.9 |
phase crosses -180 between ω=[2,4]
find ωp with bisection method
ωp=3.1
GM=−20log∣GH(ωp)∣=10.97
magnitude crosses 0 dB between ω=[1,2]
find ωg with bisection method
ωg=1.5
PM=180+∠GH(ωg)=9.5
ωp=3.1 => won't get affected by gain changes
GM=−20log∣GH(ωp)∣=−6.5
find ωg with bisection method
ωg=4.62
PM=180+∠GH(ωg)=−11.28
G(s)H(s)=S(S+4)(1+10S)(1+8S)K(S+2)
GH(ω)=ω∠90×√16+ω2∠tan−1ω/4×√1+100ω2∠tan−110ω×√1+ω2/64∠tan−1ω/8K∠0×√4+ω2∠tan−1ω/2
=ω×√16+ω2×√1+100ω2×√1+ω2/64K√4+ω2∠0+tan−1ω/2−90−tan−1ω/4−tan−110ω−tan−1ω/8
find ωp:∠GH(ωp)=−180
0+tan−1ω/2−90−tan−1ω/4−tan−110ω−tan−1ω/8=−180
ωp=3.1084343
GM=−20log∣GH(ωp)∣
=−20log(ω×√16+ω2×√1+100ω2×√1+ω2/6440√4+ω2)=11.01332515
find ωg:∣GH(ωg)∣=1
ω×√16+ω2×√1+100ω2×√1+ω2/6440√4+ω2=1
ωg=1.516902766
PM=180+∠GH(ωg)
=180+tan−1ω/2−90−tan−1ω/4−tan−110ω−tan−1ω/8=9.445676523
ωp=3.1084343 => won't get affected by gain changes
GM=−20log∣GH(ωp)∣=−6.487900123
find ωg:∣GH(ωg)∣=1
ω×√16+ω2×√1+100ω2×√1+ω2/64300√4+ω2=1
ωg=4.625734406
PM=180+∠GH(ωg)=−11.32981624