Dados: X(z) = z(z – 6) z2 – 5z + 6 . z2 – 5z + 6 = 0 a = 1; b = – 5; c = 6: Δ = b2 – 4  a  c = (– 5)2 – 4  1  6 Δ = 25 – 24 = 1 z...

Dados: X(z) =
z(z – 6)
z2 – 5z + 6

.

z2 – 5z + 6 = 0

a = 1; b = – 5; c = 6:

Δ = b2 – 4  a  c = (– 5)2 – 4  1  6

Δ = 25 – 24 = 1

z = – b √Δ
2 a

= – (– 5) √1
2 1

= 5 1
2

z1 =
5 – 1
2 =

4
2 = 2

z2 =
5 + 1
2 =

6
2 = 3

z2 – 5z + 6 = a(z – z1)(z – z2) = 1(z – 2)(z – 3)


X(z)
z =

1
z 

z(z – 6)
z2 – 5z + 6

=
z – 6
(z – 2)(z – 3) =

= Az – 3A + Bz – 2B
(z – 2)(z – 3) = (A + B)z – 3A – 2B

(z – 2)(z – 3)

1 = A + B

A + B = 1

B = 1 – A

– 6 = – 3A – 2B

– 6 = – 3A – 2  (1 – A)

– 6 = – 3A – 2 + 2A

3A – 2A = – 2 + 6

A = 4

B = 1 – A = 1 – 4 = – 3

X(z)
z = 4

z – 2 – 3
z – 3

X(z) = z  ( 4
z – 2

– 3
z – 3 ) =

4z
z – 2 –

3z
z – 3

x[n] = Z– 1{X(z)} = Z– 1{ 4z
z – 2

– 4z
z – 3 } = 4Z– 1{ z
z – 2 } – 3Z– 1{ z
z – 3 }

x[n] = 4(2n)u[n] – 3(3n)u[n]

x[n] = [4(2n) – 3(3n)]u[n]

Resp.: Alternativa E.


a) x[n] = 4(2n)u[n] + 3(3n)u[n]
b) x[n] = 4(2n)u[n] + 3(3n)u[n-1]
c) x[n] = 4(2n)u[n-1] + 3(3n)u[n-1]
d) x[n] = 4(2n)u[n-1] - 3(3n)u[n-1]
e) x[n] = 4(2n)u[n] - 3(3n)u[n]