for H[z] = z/(z + 0.5)(z - 0.5) and input x[n] = en+1u[n] write the input-output equation in time domain the image shows the zero state response

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Publisher:Robert L. Boylestad
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for H[z] = z/(z + 0.5)(z - 0.5) and input x[n] = en+1u[n]

write the input-output equation in time domain

the image shows the zero state response

z-transform of xin) = x(²)
3
Y(2)
H(2)=
80
inverse
Y(2)
Z
Y(2)
Respose
=
A
B
чего
Z
By partial fraction.
Y(Z)
A
Z
Z-O'S
z +0.5
ze
=
(2²-0₁5) (2-0)/2=-0.5)
ze
5|2=0.5
(z +0.5) (2-e) |2=0.5
0.5e
ze
(0.5+0.5) (0.5-e)
e2
(Z+05)(2-05)/2=6 =
(e +0.5) (e-o.s)
o.se
+
0.se
(oste) (Z+0'S)
(05-e)
-OS)
0.5e
1
+
o.se
(oste) 1+0.52-1
(05-0) 1-0-521
)=[-0.50 (-0·5) " + ose (0.5)
(os-e)
05te
=
X(2)
1
(2+0.5) (2-0
yen) [o.se
e
1-ez-1
z
(2+0.5) (2-0.5)
-X(²) =
+
ze
·5)
+
(z-e)
ze
(Z+0.5) (2-0.5)(z-e)
z-e
-0.5e
- 1x (-0.5-e)
o.se
(0.5+ e)
0.se
(05-e)
e2
(e²-0.25)
e2
e²-0.25 (2-0)
e2
tzte
e2-0.25 1-2-le
e2
22-025 (@jn Juan) B
A
+
Transcribed Image Text:z-transform of xin) = x(²) 3 Y(2) H(2)= 80 inverse Y(2) Z Y(2) Respose = A B чего Z By partial fraction. Y(Z) A Z Z-O'S z +0.5 ze = (2²-0₁5) (2-0)/2=-0.5) ze 5|2=0.5 (z +0.5) (2-e) |2=0.5 0.5e ze (0.5+0.5) (0.5-e) e2 (Z+05)(2-05)/2=6 = (e +0.5) (e-o.s) o.se + 0.se (oste) (Z+0'S) (05-e) -OS) 0.5e 1 + o.se (oste) 1+0.52-1 (05-0) 1-0-521 )=[-0.50 (-0·5) " + ose (0.5) (os-e) 05te = X(2) 1 (2+0.5) (2-0 yen) [o.se e 1-ez-1 z (2+0.5) (2-0.5) -X(²) = + ze ·5) + (z-e) ze (Z+0.5) (2-0.5)(z-e) z-e -0.5e - 1x (-0.5-e) o.se (0.5+ e) 0.se (05-e) e2 (e²-0.25) e2 e²-0.25 (2-0) e2 tzte e2-0.25 1-2-le e2 22-025 (@jn Juan) B A +
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