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1) Discrete
P.T.F.=1+GHG
G=GH=(1−e−ST)×S2(S+1)3
S2(S+1)3=S−3+S23+S+13
=−3+3t+3e−t
=−Z−13Z+(Z−1)23Z+Z−e−13Z
1−e−ST=1−Z−1=1−Z1=ZZ−1
G=(Z−1)(−Z−13+(Z−1)23+Z−e−13)
=−3+Z−13+Z−0.36783(Z−1)
=Z2−1.3678Z+0.3678−3(Z2−1.3678Z+0.3678)+3(Z−0.3678)+3(Z−1)2
=Z2−1.3678Z+0.36781.1034Z+0.7932
P.T.F.=Z2−1.3678Z+0.3678+1.1034Z+0.79321.1034Z+0.7932
P.T.F.=Z2−0.2644Z+1.1611.1034Z+0.7932
Z2−0.2644Z+1.161=0
Z=0.1322±j1.0694=1.0775∠1.4478
ζ=√(lnr)2+θ2lnr=0.0515
ωn=T√(lnr)2+θ2=1.4497
Mp=e−ζπ/√1−ζ2=0.8504
ts=ζωn4=53.57
tp=ωdπ=ωn√1−ζ2π=2.1699
time response for unit step input
C=Z−1Z×Z2−0.2644Z+1.1611.1034Z+0.7932
=Z3−1.2644Z2+1.4254Z−1.1611.1034Z2+0.7932Z
1.1034Z−1 | 2.1883Z−2 | 1.1941Z−3 | |
|---|---|---|---|
Z3−1.2644Z2+1.4254Z−1.161 | 1.1034Z2 | 0.7932Z | |
1.1034Z2 | −1.3951Z | 1.5727 | |
0 | 2.1883Z | −1.5727 | |
2.1883Z | −2.7668 | ||
0 | 1.1941 |
f(0)=0,f(1)=1.1034,f(2)=2.1883,f(3)=1.1941
2) continuous
T.F.=S2+S+33
ωn=√3=1.7320
ζ=2√31=0.2886
Mp=0.3879
ts=8.0023
tp=1.8944
time response for unit step input
C=S(S2+S+3)3=S1+S2+S+3−S−1
=S1−(S+0.5)2+2.75S+1
=S1−(S+0.5)2+2.75S+0.5−(S+0.5)2+2.750.5
1−e−0.5t(cos√2.275t+√2.750.5sin√2.75t)