| Line 4: | Line 4: | ||
<math>x[n]=1 - 2je^{5+j(3\pi/2*n+\pi/4)}</math> | <math>x[n]=1 - 2je^{5+j(3\pi/2*n+\pi/4)}</math> | ||
| − | 1)b) | + | 1)b)Evaluate the value of <math>(1/N)*\sum_{n=<N>}|x[n]|^2</math> for the signal x[n] given in part (a). |
| + | |||
| + | 2)a)Let's consider a continuous-time periodic signal x(t) with period T = 5 whose non-zero Fourier series coefficients <math>a_k</math> are given by | ||
| + | |||
| + | <math>a_1=a_{-1}=2,a_3=a*_{-3}=4j</math> | ||
| + | |||
| + | If x(t) is the input to a particular LTI system characterized by the frequency response | ||
| + | |||
| + | <math>H(j\omega)=1/(3+j\omega)</math> | ||
| + | |||
| + | then the output is y(t). | ||
| + | |||
==Solutions== | ==Solutions== | ||
| Line 22: | Line 33: | ||
<math>a_9=\sqrt{2}e^{5} - j\sqrt{2}e^5</math> | <math>a_9=\sqrt{2}e^{5} - j\sqrt{2}e^5</math> | ||
| + | 1)b) | ||
| + | Some energy therom states that: | ||
| + | <math>\sum_{n=<N>}|x[n]|^2=N*sum_{k=<N>}|a_k|^2</math> | ||
| + | |||
| + | Therefore: | ||
| + | |||
| + | <math>(1/N)*\sum_{n=<N>}|x[n]|^2=sum_{k=<N>}|a_k|^2</math> | ||
| + | |||
| + | <math>(1/N)*\sum_{n=<N>}|x[n]|^2=1^2+|(-2je^{\pi/4}e^{5})|^2=1 + e^{10}</math> | ||
| + | |||
| + | <math>(1/N)*\sum_{n=<N>}|x[n]|^2=1 + e^{10}</math> | ||
--[[User:Krtownse|Krtownse]] 15:53, 18 July 2008 (EDT) | --[[User:Krtownse|Krtownse]] 15:53, 18 July 2008 (EDT) | ||
[[Category:ECE 301 San Summer 2008]] | [[Category:ECE 301 San Summer 2008]] | ||
Revision as of 17:23, 18 July 2008
Problems
1)a)If a discrete time singal x[n] is periodic with period N, then the Fourier series coefficients $ a_k $ of the signal x[n] is also periodic with period N. For the periodic signal x[n], find the values of $ a_0,a_1,...,a_{N-1}. $ Express your answer in a + jb form.
$ x[n]=1 - 2je^{5+j(3\pi/2*n+\pi/4)} $
1)b)Evaluate the value of $ (1/N)*\sum_{n=<N>}|x[n]|^2 $ for the signal x[n] given in part (a).
2)a)Let's consider a continuous-time periodic signal x(t) with period T = 5 whose non-zero Fourier series coefficients $ a_k $ are given by
$ a_1=a_{-1}=2,a_3=a*_{-3}=4j $
If x(t) is the input to a particular LTI system characterized by the frequency response
$ H(j\omega)=1/(3+j\omega) $
then the output is y(t).
Solutions
1)a) Change the equation to look like a forier series. $ x[n]=1-2je^{5+j(3\pi/2*n+\pi/4)}= $ $ x[n]=1 - 2je^{\pi/4}e^{5}e^{j(3\pi/2)*n}= $ $ x[n]=1 + (\sqrt{2}e^{5} - j\sqrt{2}e^5)e^{j(3\pi/2*n)} $
find the period N. $ \omega_0=3\pi/2 $. $ P=2\pi/\omega_0=4/3 $. The LCM of 1 and 4/3 is 12. Therefore N=12
Choose values of $ a_k $ that match the forier series repersentation of x[n]
$ x[n]=\sum_{k=<n>}^{}a_ke^{jk(2\pi/12)n} $
$ a_0=1 $
$ a_9=\sqrt{2}e^{5} - j\sqrt{2}e^5 $
1)b) Some energy therom states that: $ \sum_{n=<N>}|x[n]|^2=N*sum_{k=<N>}|a_k|^2 $
Therefore:
$ (1/N)*\sum_{n=<N>}|x[n]|^2=sum_{k=<N>}|a_k|^2 $
$ (1/N)*\sum_{n=<N>}|x[n]|^2=1^2+|(-2je^{\pi/4}e^{5})|^2=1 + e^{10} $
$ (1/N)*\sum_{n=<N>}|x[n]|^2=1 + e^{10} $
--Krtownse 15:53, 18 July 2008 (EDT)
