Difference between revisions of "W7Q4Sol" - Rhea

Latest revision as of 07:24, 4 October 2010

Solution to the Question 4

a.

$ \begin{align} X(e^{j\omega}) &= \sum^{\infty}_{n=-\infty} x(n)e^{-j\omega n}=\sum^{N-1}_{n=0}e^{j\omega _0 n}e^{-j\omega n} \\ &= \sum^{N-1}_{n=0}e^{j(\omega _0 -\omega)n} \\ &= \left\{\begin{array}{ll} \frac{1-e^{j(\omega _0 -\omega)N}}{1-e^{j(\omega _0 -\omega)}} & \text{if } \omega _0 \text{ is not equal to } \omega , \\ N & \text{if } \omega _0 =\omega. \end{array} \right. \\ &= \left\{\begin{array}{ll} e^{j(\omega _0 -\omega)(N-1)/2}\frac{sin\frac{\omega _0 -\omega}{2}N}{sin\frac{\omega _0 -\omega}{2}} & \text{if } \omega _0 \text{ is not equal to } \omega , \\ N & \text{if } \omega _0 =\omega. \end{array} \right. \end{align} $

b.

$ \begin{align} X[k]&=\sum^{N-1}_{n=0}x[n]e^{-j\frac{2\pi kn}{N}}=\sum^{N-1}_{n=0}e^{j\omega _0 n}e^{-j\frac{2\pi kn}{N}} \\ &=\sum^{N-1}_{n=0}e^{j(\omega _0 -\frac{2\pi k}{N})n} \end{align} $

c.

For $ \omega_0 =\frac{2\pi k_0}{N}, \text{ where } k_0 \text{ is an integer} $

$ X[k]=\sum^{N-1}_{n=0}e^{j(k_0 -k)\frac{2\pi n}{N}} $

i. if $ k_0 -k=mN, \text{ where } m \text{ is an integer} $

$ x[k]=\sum^{N-1}_{n=0}1=N $

ii. otherwise

$ X[k]=\frac{1-e^{j(k_0 -k)2\pi}}{1-e^{j(k_0 -k)\frac{2\pi}{N}}}=\frac{1-1}{1-e^{j(k_0 -k)\frac{2\pi}{N}}}=0 $

Combine the two,

$ X[k]=\left\{\begin{array}{ll} N, & k_0 -k=0,\pm N,\pm 2N, \text{...} \\ 0, & \text{ otherwise} \end{array} \right. $

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