y(k) — y(k — 1) = x(k) + x(k — XXXXXXXXXXFind the characteristic polynomial, a(z), for this system....

Question

y(k) — y(k — 1) = x(k) + x(k — XXXXXXXXXXFind the characteristic polynomial, a(z), for this system. . Find the input polynomial, b(z) for this system. . If x(k) = .2ku(k), calculate yz,(k), the zero state response of the system.
1. Let
2. Let
x(k) = [2 3] and y(k) = [2 4 6 3] Calculate ryx = D(x)y for this system.
4. Let x(k) = 28(k) + 68(k — 1) — 38(k — 2) and h(k) = [3 0 2] a. Write the summation for y(k) = x(k) * h(k). b. Calculate y(k).
5. Let
y(k) = 2-ku(k — 2) a. Is y(k) periodic? If so, what is the period? b. Is y(k) bounded or unbounded? c. Is y(k) finite or infinite? d. Is y(k) causal? e. Calculate the energy of y(k). Is it finite or infinite?

005_yhwf95k-qbjdswoa.png 005_o3af95k-qcfhylbu.png 005_4naf95k-3hvj4ass.png 005_jxef95k-jru2pbwa.png

Robert · Accepted Answer

2[y(k)− y(k − 1)] = x(k) + x(k − 1)
Taking Z transform,
2[Y (Z)− z−1Y (Z)] = X(Z) + z−1X(Z)
2(1− z−1)Y (Z) = (1 + z−1)X(Z)
Y (Z)
X(Z)
= (1+z
−1)
2(1−z−1) =
z+1
2(z−1) =
b(z)
a(z)
a) Characteristic polynomial a(z) = 2(z-1)
b) Input polynomial b(z) = (z+1)
c) x(k) = 2ku(k)
To find zero state response of system, take Z transform
X(Z) = 11−2z−1
hence, from above equation, Y (Z) = z+1
2(z−1)X(Z)
= z+1
2(z−1)
z
z−2
= (z
2−3z+2)+(4z−2)
2(z2−3z+2)
= 12 +
z−1/2
(z−1)(z−2)
To partial fraction the term z−1/2
(z−1)(z−2) for Z transform,
z−1/2
z(z−1)(z−2) =
A
z−1 +
B
z−2 +
C
z
Hence, z − 12 = Az(z − 2) +Bz(z − 1) + C(z − 1)(z − 2)
Put z = 0, we get C = -1/4
Put z = 1, we get A = -1/2
Put z = 2, we get B = 3/4
Hence,

y(k) — y(k — 1) = x(k) + x(k — XXXXXXXXXXFind the characteristic polynomial, a(z), for this system. . Find the input polynomial, b(z) for this system. . If x(k) = .2ku(k), calculate yz,(k), the zero...

Solution

Answer To This Question Is Available To Download

Related Questions & Answers

Submit New Assignment