"Find the Z-transform for the sequence x[n]=2ν[−n]+(14)ν[n−1]" was answered by students like you

Breanna Mcclure

Answered question

2022-04-05

Find the Z-transform for the sequence

x [n] = 2^{ν} [- n] + {(\frac{1}{4})}^{ν} [n - 1]

cutimnm135imsa

Beginner2022-04-06Added 21 answers

Given:

x [n] = 2^{ν} [- n] + {(\frac{1}{4})}^{ν} [n - 1]

Now,

x [n] = {(2^{- 1})}^{- n} u [- n] + {(\frac{1}{4})}^{n + 1 - 1} u [n - 1]

x [n] = {(\frac{1}{2})}^{- n} u [- n] + {(\frac{1}{4})}^{n - 1} (\frac{1}{4}) u [n - 1]

x [n] = {(\frac{1}{2})}^{- n} u [- n] + \frac{1}{4} [{(\frac{1}{4})}^{n - 1} u [n - 1]]

Apply z-transform

x [z] = [\frac{z}{z - \frac{1}{2}}] + \frac{1}{4} (\frac{z}{z - \frac{1}{4}} \times z^{- 1})

x [z] = \frac{\frac{1}{z}}{\frac{1}{z} - \frac{1}{4}} + \frac{1}{4} (\frac{1}{z - \frac{1}{4}})

= \frac{1}{z} \times \frac{2 z}{2 - z} + \frac{1}{4} (\frac{1}{z - \frac{1}{4}})

x [z] = \frac{- 2}{z - 2} + \frac{1}{4} (\frac{1}{z - \frac{1}{4}})

x [z] = \frac{1}{4} (\frac{1}{z - \frac{1}{4}}) - (\frac{2}{z - 2})

= [\frac{\frac{1}{4} (z - 2) - 2 (z - \frac{1}{4})}{(z - 2) (z - \frac{1}{4})}]

x [z] = \frac{\frac{z}{4} - \frac{1}{2} - 2 z + \frac{2}{4}}{}} {(z - 2) (z - \frac{1}{4})}

x [z] = \frac{- \frac{7}{4} z + 0}{(z - 2) (z - \frac{1}{4})}

x [z] = \frac{(- \frac{7}{4}) z}{(z - 2) (z - \frac{1}{4})}

The zero of

x [z]

is the numerator value

- \frac{7}{4} z = 0

z = 0

The pole of

x [z]

is the determinator value

(z - 2) (z - \frac{1}{4}) = 0

z = 2, \frac{1}{4}

The Radius of convergence

x [n] = 2^{ν} [- n] + {(\frac{1}{4})}^{ν} [n - 1]

= {(\frac{1}{2})}^{- n} u [- n] + \frac{1}{4} [{(\frac{1}{4})}^{n - 1} u [n - 1]]

x [n] = {(\frac{1}{2})}^{- n} u [- n] + \frac{1}{4} [{(\frac{1}{4})}^{n - 1} u [n - 1]]

x [n] = (1) + (2)

Since

a^{n} x (n) \Rightarrow | z | > a, u [n] \Rightarrow | z | > 1

a^{- n} x (- n) \Rightarrow | z | < \frac{1}{a}, a^{ν} [n] \Rightarrow | z | > a

ROC (1)

\Rightarrow a = \frac{1}{2}

| z | < \frac{1}{\frac{1}{2}}

| z | < 2

ROC (2)

\Rightarrow a = \frac{1}{4}

| z | > \frac{1}{4}

Radius of convergence is the intersection of ROC (1) and ROC (2)

\frac{1}{4} < | z | < 2

0.25 < | z | < 2

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