User:Virusys/Temp

just testing some math formatting out


 * $$x[n] = -\frac{1}{3}\left(\frac{1}{2}\right)^n u[n] - \frac{4}{3}(2)^n u[-n -1] $$


 * $$X(z) = \sum_{n=-\infty}^{\infty} -\frac{1}{3}\left(\frac{1}{2}\right)^n u[n] - \frac{4}{3}(2)^n u[-n -1] z^{-n} $$
 * $$X(z) = \left [ \sum_{n=0}^{\infty} -\frac{1}{3}\left(\frac{1}{2}\right)^n - \sum_{n=-\infty}^{-1}-\frac{4}{3}(2)^n \right ] z^{-n} $$
 * $$X(z) = \sum_{n=0}^{\infty} -\frac{1}{3}\left(\frac{1}{2}\right)^n z^{-n} - \sum_{n=-\infty}^{-1}-\frac{4}{3}(2)^n z^{-n} $$
 * $$X(z) = \sum_{n=0}^{\infty} -\frac{1}{3}\left(\frac{1}{2}z^{-1}\right)^n - \sum_{n=-\infty}^{-1}-\frac{4}{3}(2z^{-1})^n $$
 * $$X(z) = -\frac{1}{3}\left(\frac{1}{1 - \frac{1}{2}z^{-1}}\right) - \frac{4}{3}\left(1 - \left(\frac{1}{1-2z^{-1}}\right)\right)$$
 * $$X(z) = -\frac{1}{3}\left(\frac{z}{z - \frac{1}{2}}\right) - \frac{4}{3}\left(1 - \left(\frac{z}{z-2}\right)\right) $$
 * $$X(z) = -\frac{1}{3}\left(\frac{2z}{2z - 1}\right) - \frac{4}{3}\left(\frac{-2}{z-2}\right) $$
 * $$X(z) = -\frac{2z}{6z - 3} + \frac{8}{3z-6} $$
 * $$X(z) = \frac{-6z^2 + 60z - 24}{18z^2 - 45z + 18} $$
 * $$X(z) = \frac{-2z^2 + 20z - 8}{6z^2 - 15z + 6} $$
 * $$X(z) = \frac{-2(z^2 + 10z + 4)}{3(2z-1)(z-2)} $$

1) ii) $$\quad |z| > 2$$
 * $$ \quad |z| < \tfrac {1}{2} $$
 * $$ \quad \tfrac {1}{2} < |z| < 2 $$

iii)
 * $$ \quad H(z) = \frac {1 - z^{-2}}{(1 - \frac{1}{2}z^{-1})(1 - 2z^{-1})} \cdot \frac {3(2z - 1)(z - 2)}{2(z^{2} + 10z + 4)} $$
 * $$ \quad H(z) = \frac {-3(2z - 1)(z - 2))z^{-2}} {2(z^{2} + 10z + 4)(1 - \frac{1}{2}z^{-1})(1 - 2z^{-1})} $$

Therefore, there exist poles at $$\quad \tfrac{-10\pm \sqrt{84}} {2}, $$$$ \tfrac{1}{2} $$ and 2.


 * $$ Y(z) = \frac{1 - z^{-2}}{(1 - \frac{1}{2}z^{-1})(1 - 2z^{-1})}$$
 * $$ Y(z) = \frac{1 - z^{-2}}{1 - \frac{3}{2}z^{-1} + z^{-2}}$$


 * Difference equation:


 * $$ \quad x[n] - x[n - 2] = y[n] - \frac{3}{2}y[n-1] + y[n-2] $$
 * $$ \quad y[n] = x[n] - x[n - 2] + \frac{3}{2}y[n-1] - y[n-2] $$


 * QUESTION TWO:


 * $$ \quad y[n] - a^{-1}y[n - 1] = x[n] - ax[n - 1] $$
 * $$ \quad y[n] = x[n] - ax[n - 1] + a^{-1}y[n - 1] $$


 * In order for this system to be stable, the feedback coefficient must be less than one. Therefore $$ |a| > 1 $$.



h[n] = \begin{cases} \delta[n], & \mbox{if }n\mbox{ = 0} \\ a^{-n} - a^{2-n}, & \mbox{if }n \ge 1 \end{cases} $$


 * $$ H(z) = \left | \frac{1 - a^{-1}e^{-jw}}{1 - ae^{-jw}} \right | = 1 $$ hmm