Question

The z-transform of the sequence x[n] is given by \(X\left( z \right) = \frac{1}{{{{\left( {1 - 2{z^{ - 1}}} \right)}^2}}}\), with the region of convergence |z| > 2. Then x[2] is ______

Answer 1

Concept:

Properties:

Differentiation in z domain:

If X(z) is a z transform of x(n), then the z transform of n x(n) is,

\(nx\left( n \right) \leftrightarrow - z\frac{d}{{dz}}\left( {X\left( z \right)} \right)\)

Time-shifting:

If X(z) is a z transform of x(n), then the z transform of x(n – n0) is,

\(x\left( {n - {n_0}} \right) \leftrightarrow {z^{ - {n_0}}}X\left( z \right)\)

Calculation:

Given,

\(x\left[ n \right]\mathop \leftrightarrow \limits^{z - Transform} X\left( z \right) = \frac{1}{{{{\left( {1 - 2{z^{ - 1}}} \right)}^2}}}\;;\left| z \right| > 2\)

We know that:

With x₁[n] = aⁿu[n]

\({x_1}\left[ n \right]\mathop \leftrightarrow \limits^{z - Transform} \;\frac{1}{{\left( {1 - a{z^{ - 1}}} \right)}};\left| z \right| > a\)

The multiplication in time domain property of z-transform states that:

\(n{x_1}\left[ n \right] = n{a^n}u\left[ n \right]\mathop \leftrightarrow \limits^{z - Transform} - z\frac{d}{{dz}}X\left( z \right)\)

\(= - z\frac{d}{{dz}}\frac{1}{{\left( {1 - a{z^{ - 1}}} \right)}}\)

\( = - z\;\left[ {\frac{{ - \left( { + a{z^{ - 2}}} \right)}}{{{{\left( {1 - a{z^{ - 1}}} \right)}^2}}}} \right]\)

\(= \frac{{a{z^{ - 1}}}}{{{{\left( {1 - a{z^{ - 1}}} \right)}^2}}}\)

\(n{a^n}u\left[ n \right] \leftrightarrow \;\;\frac{{a{z^{ - 1}}}}{{{{\left( {1 - a{z^{ - 1}}} \right)}^2}}}\)

\(\frac{{n{a^n}u\left[ n \right]}}{a} \leftrightarrow \frac{{{z^{ - 1}}}}{{{{\left( {1 - a{z^{ - 1}}} \right)}^2}}}\)

Now, according to the time-shifting property of z-transform, we can write:

\(\frac{{\left( {n + 1} \right){a^{n + 1}}u\left[ {n + 1} \right]}}{a} \leftrightarrow \frac{{z\;\;{z^{ - 1}}}}{{{{\left( {1 - a{z^{ - 1}}} \right)}^2}}}\) 

\(= \frac{1}{{{{\left( {1 - a{z^{ - 1}}} \right)}^2}}}\)

\(\left( {n + 1} \right){a^n}\;u\left[ {n + 1} \right]\mathop \leftrightarrow \limits^{z - Transform} \frac{1}{{{{\left( {1 - a{z^{ - 1}}} \right)}^2}}}\)

Let a = 2

\(\left( {n + 1} \right){2^n}\;u\left[ {n + 1} \right]\mathop \leftrightarrow \limits^{z - Transform} \frac{1}{{{{\left( {1 - 2{z^{ - 1}}} \right)}^2}}}\)

Hence,

x[n] = (n + 1) 2ⁿ u[n + 1]

at n = 2

x[n] = 3 ⋅ 2²⋅ 1

= 3 ⋅ 4 ⋅ 1

x[2] = 12