(New page: ==Solution of Week12 Quiz Question 4== ---- a. Linearity ---- Back to Quiz Pool Back to Lab wiki) |
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a. Linearity | a. Linearity | ||
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| + | Given <math>v[n]=ax[n]+by[n]</math> | ||
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| + | Then | ||
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| + | <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> | ||
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Revision as of 13:06, 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} $
