Loading Documentfind f(4) for F(Z)=(Z−1)(Z2−Z+1)Z2+Z+2
=Z3−Z2+Z−Z2+Z−1Z2+Z+2
=Z3−2Z2+2Z−1Z2+Z+2
Z−1 | 3Z−2 | 6Z−3 | 7Z−4 | |
|---|---|---|---|---|
Z3−2Z2+2Z−1 | Z2 | Z | 2 | |
Z2 | −2Z | 2 | −Z−1 | |
3Z | 0 | Z−1 | ||
3Z | −6 | 6Z−1 | ||
6 | −5Z−1 | |||
6 | 11Z−1 | |||
7Z−1 |
f(4)=7
screw it! second method is better 😡
find the final value ofC(Z)=Z3−1.7Z2+0.8Z−0.1Z2(Z+1)
C(Z)=(Z−1)(Z−0.5)(Z−0.2)Z2(Z+1)
final value =Z→1lim(Z−0.5)(Z−0.2)Z2(Z+1)=0.5×0.82=5
plot the following pole on S and Z planes for T=0.2 sec
S=−2+j7
Z=eST=e0.2(−2+j7)=0.67∠1.4=0.67∠80.2∘
solve the following difference equation
e(k+2)−1.5e(k+1)+0.5e(k)=1
given that e(0)=1&e(1)=2.5
Z2e(Z)−Z2−2.5Z−1.5Ze(Z)+1.5Z+0.5e(Z)=Z−1Z
(Z2−1.5Z+0.5)e(Z)=Z−1Z+Z2+Z
(Z2−1.5Z+0.5)e(Z)=Z−1Z+Z3−Z2+Z2−Z=Z−1Z3
e(Z)=(Z−0.5)(Z−1)2Z3
Ze(Z)=(Z−0.5)(Z−1)2Z2
e(Z)=Z−0.5Z+(Z−1)22Z
e(K)=0.5K+2K assuming T=1
find the z-transform for (where a and b are constants)
(S+a)(S+b)b−a
(S+a)(S+b)b−a=S+aA+S+bB
A=b−ab−a=1
B=a−bb−a=−1
S+a1−S+b1=e−at−e−bt=Z−e−aTZ−Z−e−bTZ
Then, find:
Damping ratio
Maximum overshoot
Settling time for 2% error
Peak time
Find the equivalent continuous system and compare the dynamic characteristics between both the discrete and continuous systems
Find time response for unit step input
ZOH=S1−e−ST
G=(1−e−ST)×S2(S+3)5
=(1−e−ST)(S25/3+S+35/9−S5/9)
=(1−e−ST)(35t+95e−3t−95)
=(ZZ−1)(3(Z−1)25Z+9(Z−e−3)5Z−9(Z−1)5Z)
3(Z−1)5+9(Z−e−3)5(Z−1)−95
95(Z−13+Z−e−3Z−1−1)
95((Z−1)(Z−e−3)3Z−3e−3+Z2−2Z+1−Z2−e−3Z−Z+e−3)
G=9−9Z−9e3Z+9e3Z2−10+5e3−5Z
T.F.=1+GHG=9−9Z−9e3Z+9e3Z2−10+5e3−5Z−10+5e3−5Z
=−1+5e3−14Z−9e3Z+9e3Z2−10+5e3−5Z=99.43−194.77Z+180.77Z290.43−5Z
characteristic equation =99.43−14Z−180.77Z+180.77Z2
poles = 0.5387+j0.5097=0.7416∠0.7577
Damping ratio √0.74162+0.75772ln(0.7416)=0.282
Natural frequency 1√0.74162+0.75772=1.06rad/s
Maximum overshoot e−√1−0.28220.282π=0.3972
Settling time for 2% error 0.282×1.064=13.3815
Damped frequency 1.06√1−0.2822=1.017
Peak time 1.017π=3.089
equivalent continuous system
G=S2+3S5
T.F.=S2+3S+55
S2+3S+5=S2+2ζωnS+ωn2
Damping ratio 2√53=0.7608
Natural frequency √5=2.236rad/s
Maximum overshoot 0.05833
Settling time for 2% error 2.6667
Damped frequency 1.6583
Peak time 1.017π=1.8945
time response for unit step input
Discrete
C=Z−1Z×99.43−194.77Z+180.77Z290.43−5Z
=180.77Z3−194.77Z2+99.43Z−(99.43−194.77Z+180.77Z2)−5Z2+90.43Z
=180.77Z3−375.54Z2+294.2Z−99.4390.43Z−5Z2
f(0)=0,f(1)=−0.0276,f(2)=0.4427,f(3)=0.9648,f(4)=1.2686
continuous
C=S(S2+3S+5)5=S1+S2+3S+5−S−3
=S1−(S+1.5)2+2.75S+3
=S1−((S+1.5)2+2.75S+1.5+(S+1.5)2+2.751.5)
1−e−1.5t(cos(√2.275t)+√2.751.5sin(√2.75t))
Consider a system having the characteristic equation:
F(Z)=Z3+3.3Z2+4Z+0.8=0
Discuss the stability using square matrix test?
and prove the answer using Jury's test
x=⎣⎡103.31⎦⎤
y=⎣⎡40.80.80⎦⎤
x+y=⎣⎡50.84.11⎦⎤>0
x−y=⎣⎡−3−0.82.51⎦⎤>0 Nope!
therefore, the system is unstable
An>0
F(1)>0
F(−1)=−(−1+3.3−4+0.8)=0.9>0
0.8 | 4 | 3.3 | 1 |
|---|---|---|---|
1 | 3.3 | 4 | 0.8 |
-0.36 | -1.36 |
∣A0∣<∣An∣
∣B0∣>∣Bn−1∣ Nope!
therefore, the system is unstable