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Question 2

2.1 Find the (causal) inverse z-transform of


F[z] = $\displaystyle \frac{2z(3z+17)}{(z-1)(z^2-6z+25)}$ (12)

using partial fractions.

2.2 Write numerical values for the coefficients of the first two tersm.

Solution.

2.1

The Eq.(13) can be expressed as a factored form

F[z] = $\displaystyle \frac{2z(3z+17)}{(z-1)(z-(3+j4))(z-(3-j4))}$ (13)

First break F(z) into partial fractions


$\displaystyle \frac{F[z]}{z}$ = $\displaystyle \frac{A}{z-1} + \frac{B}{z-(3+j4)} + \frac{C}{z-(3-j4)}$ (14)

where
A = 2 (15)
B = $\displaystyle -1-j\frac{5}{4}$ (16)
C = $\displaystyle -1+j\frac{5}{4}$ (17)

Now F(z) is
F[z] = $\displaystyle \frac{Az}{z-1} + \frac{Bz}{z-(3+j4)} + \frac{Cz}{z-(3-j4)}$ (18)

So applying the z-transform pair
$\displaystyle \frac{cz}{z-p} \leftrightarrow cp^ku[k]$     (19)

We take the inverse z-transform to obtain f[n]

f[n] = Au[n] + B(3+j4)nu[n] + C(3-j4)nu[n] (20)
  = $\displaystyle 2u[n] + (-1-j\frac{5}{4})(3+j4)^nu[n] + (-1+j\frac{5}{4})(3-j4)^nu[n]$ (21)

2.2

The numerical values for the first two terms are

\begin{eqnarray*}n = 0 & : & f[0]=0 \\
n = 1 & : & f[1]=6
\end{eqnarray*}



next up previous
Next: Question 3 Up: Solution to Assignment 4 Previous: Question 1
Hyun Ho Jeon
2001-02-12