(Exersize in Wiki and LaTex and Hw discussion) |
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[[Category:ECE438Spring2009mboutin]] | [[Category:ECE438Spring2009mboutin]] | ||
| + | == Part 4a == | ||
| − | + | Given | |
| + | |||
| + | <math>y(n) = \frac{x(n) + x(n-1) + x(n-2)}{3} \quad (1)</math> | ||
| + | |||
| + | to find H(w), we let | ||
| + | |||
| + | <math> x(n)\,\! = \exp(jwn) \quad (2)</math> | ||
| + | |||
| + | such that | ||
| + | |||
| + | <math>y(n)\,\! = H(w)\exp(jwn)</math> | ||
| + | |||
| + | substituting equation (1) into (2) | ||
| + | |||
| + | <math>y(n)\,\! = \frac{\exp(jwn) + \exp(jw(n-1)) + \exp(jw(n-2))}{3} \quad (3)</math> | ||
| + | |||
| + | factoring out the <math>\,\! \exp(jwn)</math> from equation (3), we obtain H(w) | ||
| + | |||
| + | <math>\,\! H(w) = \frac{1 + \exp(-jw) + \exp(-jw2)}{3}</math> | ||
| + | |||
| + | == Part 4b == | ||
| + | <math>\left| H(w)\right|</math> | ||
| + | |||
| + | == Part 4c == | ||
| + | To find the the overall frquency response F(w) for this system, I assumed the up/down samplers canceled each other out and so we were left with the LPF and original H(w). | ||
| + | |||
| + | Combining terms | ||
| + | |||
| + | <math>\,\! F(w) = G(w)H(w)G(w)</math> | ||
Revision as of 21:46, 9 February 2009
Part 4a
Given
$ y(n) = \frac{x(n) + x(n-1) + x(n-2)}{3} \quad (1) $
to find H(w), we let
$ x(n)\,\! = \exp(jwn) \quad (2) $
such that
$ y(n)\,\! = H(w)\exp(jwn) $
substituting equation (1) into (2)
$ y(n)\,\! = \frac{\exp(jwn) + \exp(jw(n-1)) + \exp(jw(n-2))}{3} \quad (3) $
factoring out the $ \,\! \exp(jwn) $ from equation (3), we obtain H(w)
$ \,\! H(w) = \frac{1 + \exp(-jw) + \exp(-jw2)}{3} $
Part 4b
$ \left| H(w)\right| $
Part 4c
To find the the overall frquency response F(w) for this system, I assumed the up/down samplers canceled each other out and so we were left with the LPF and original H(w).
Combining terms
$ \,\! F(w) = G(w)H(w)G(w) $
