written 2.5 years ago by |
Solution:
Foster −1 from:
z(s)=3(s+2)(s+4)(s)(s+3)....(1)=3s2+18s+24s2+3s∴z(s)=3+9S+24s2+3S=Z1(s)+Z2(s)...(2)
Where
z1(s)=3
And
z2(s)=9s+24(s)(s+3)=As+Bs+3}....(3)
A=9S+24(s)(S+3)|S=0=0+240+3=8B=9S+24S|S=−3=−27+24−3=1
from (3)
z2(s)=8s+1s+3=Z′2(s)+z′′2(s)...(4)
Where
Z′2(s)=8s=118s
And
z′′2(s)=1s+3=1y′′2(s)∴y′′2(s)=s+3
Foster −1 form is
Foster −2 from:
V(s)=1z(s)=(s)(s+3)3(s+2)(s+4)....(1)=s2+3s3s2+18s+24
∵ polarity of remainder is -ve, divide both side of equation by S.
∴y(s)S=S+33(S+2)(S+4)=AS+2+BS+4}...(2)
A=s+33(s+4)|s=−2=−2+33(−2+4)=16
B=s+33(s+2)|S=−4=−4+33(−4+2)=16
from eq−1
y(s)s=1/6s+2+1/6s+4
y(s)s=16s+12+16S+24
Y(S)=S6S+12+S6S+24
y(s)=y1(s)+y2(s)...(3)
Where y1(s)=s6s+12=1[6s+12s]
=16+125=1Z1(s)
∴Z1(s)=6+129=6+1112s
And
y2(s)=s6S+24=1[6s+24s]=1[6+24s]=1Z2(S)∴Z2(S)=6+24s=6+1124s