(New page: ==Solution of Week12 Quiz Question 4== ---- a. Linearity ---- Back to Quiz Pool Back to Lab wiki) |
|||
| (One intermediate revision by the same user not shown) | |||
| Line 3: | Line 3: | ||
---- | ---- | ||
a. Linearity | a. Linearity | ||
| + | |||
| + | Given <math>v[n]=ax[n]+by[n]</math> ,then | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | V(\omega ,n) &= \sum_k v[k]w[n-k]e^{-j\omega k} \\ | ||
| + | &= \sum_k (ax[k]+by[k])w[n-k]e^{-j\omega k} \\ | ||
| + | &= \sum_k ax[k]w[n-k]e^{-j\omega k}+\sum_k by[k]w[n-k]e^{-j\omega k} \\ | ||
| + | &= aX(\omega ,n)+bY(\omega ,n) | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | b. Modulation | ||
| + | |||
| + | Given <math>v[n]=x[n]e^{j\omega_0n}</math> ,then | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | V(\omega ,n) &= \sum_k v[k]w[n-k]e^{-j\omega k} \\ | ||
| + | &= \sum_k (x[k]e^{j\omega_0 k})w[n-k]e^{-j\omega k} \\ | ||
| + | &= \sum_k x[k]w[n-k]e^{-j(\omega -\omega_0)k} \\ | ||
| + | &= X(\omega -\omega_0 ,n) | ||
| + | \end{align} | ||
| + | </math> | ||
---- | ---- | ||
Latest revision as of 13:09, 10 November 2010
Solution of Week12 Quiz Question 4
a. Linearity
Given $ v[n]=ax[n]+by[n] $ ,then
$ \begin{align} V(\omega ,n) &= \sum_k v[k]w[n-k]e^{-j\omega k} \\ &= \sum_k (ax[k]+by[k])w[n-k]e^{-j\omega k} \\ &= \sum_k ax[k]w[n-k]e^{-j\omega k}+\sum_k by[k]w[n-k]e^{-j\omega k} \\ &= aX(\omega ,n)+bY(\omega ,n) \end{align} $
b. Modulation
Given $ v[n]=x[n]e^{j\omega_0n} $ ,then
$ \begin{align} V(\omega ,n) &= \sum_k v[k]w[n-k]e^{-j\omega k} \\ &= \sum_k (x[k]e^{j\omega_0 k})w[n-k]e^{-j\omega k} \\ &= \sum_k x[k]w[n-k]e^{-j(\omega -\omega_0)k} \\ &= X(\omega -\omega_0 ,n) \end{align} $
