Need more help on z transform(pls check if this is correct)

**TechnicianEngineer** · February 11th 2011, 07:24 PM

I. Find the z transform and ROC of each of the ff sequence
1. $x(n) = 2\delta{n} + 3(\frac{1}{2})^{n}u[n] - (1/4)^{n}u(n)$
$X(z) = 2(1) + 3\frac{1}{ 1-\frac{1}{2}z^{-1} } - \frac{1}{1-\frac{1}{4}z^{-1} }$
ROC: All z, 1/2 <z and 1/4 < z
the total ROC is z > 1/2

II. Use the Z transform to perform the convolution of the following sequence
$x[n] = 3^{n}u(-n)$
since $x(-n)$ 's z transform is $X(z^{-1})$

$X_1(Z) = \frac{1}{1-3z}$

$h[n] = (0.5)^{n}u(n)$
$X_2(Z) = \frac{1}{1-\frac{1}{2}z^-1}$
$x[n]*h[n]$ is equivalent to $X_1(Z)X_2(Z)$

$X_1(Z)X_2(Z) =\frac{1}{(1-3z)(1-\frac{1}{2}z^-1)}$

using wolfram alpha to solve partial fractions
$\frac{2z}{(1-3z)(2z-1)}$

$\frac{2}{3z-1} - \frac{2}{2z-1}$
then simplify
$\frac{ 2z^{-1}\frac{1}{3} }{ 1-\frac{1}{3}z^{-1} } - \frac{ 2z^{-1}\frac{1}{2} }{1-\frac{1}{2}z^{-1} }$
the region of convergence is 1/3 < z and 1/2 < z
then the total ROC is 1/2

the inverse z transform is : using the time shifting property z^-1X(z) = u(n-1)
$\frac{2}{3}(\frac{1}{3})^{n}u(n-1) -(\frac{1}{2})^{n}u(n-1)$

part III. Find the causal signal x(n) fo the following z transforms
$X(Z) = \frac{ 1-z^{-1} }{(1+z^{-1})(1+\frac{1}{2}z^{-1})}$
using wolfram alpha
$\frac{2(z-1)z}{z+1}{2z+1}$
$\frac{3}{2z+1}-\frac{4}{z+1}+1$

Making it into a z transform
$\frac{\frac{1}{2}3z^{-1}}{1-(-\frac{1}{2})z^{-1}} -\frac{4z^{-1}}{(1-(-z^{-1})} +1$
the region of convergence is
-1/2 < z, -1 < z, and the entire plane of z
the total region of convergence is the entire plane of z

the inverse z transform is
$x(n) = \frac{3}{2}(-\frac{1}{2})^{n}u(n-1) - 4(-1)^{n}u(n-1)$

**Ackbeet** · February 14th 2011, 02:22 AM

1.

1/2 <z and 1/4 < z

I think you meant with magnitudes, since complex numbers are not ordered as is, though their magnitudes, being real, are ordered. Otherwise, everything looks correct.

2. Looks correct.

3. You forgot one term, I think: the delta function as the inverse z transform of 1. Otherwise, it looks correct.

You do know that WolframAlpha can do inverse Z transforms? Format would look like this:

InverseZTransform[f(z),z,n]

Here's a simple example.

So you can check your work that way.

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