Difference between revisions of "ECE301:SanSummer08:Exam II OldKiwi" - Rhea

Revision as of 16:46, 21 July 2008

Read the discussion page using the discussion tab above. (Aung - 11:09 pm 07/18/08)

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).

Is the output signal y(t) periodic? Answer YES or NO only.

2)b)If your answer to part (a) is YES, then let $ b_k $ be the fourier series coefficients of y(t) and find the values of $ b_k $. You need not simplify the complex numbers. If your answer to part (a) is NO, then explain why y(t) is not periodic.

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)

2)a) YES

2)b)First thing: state $ a_k $ clearly:

$ a_{-3}=-4j $

$ a_{-1}=2 $

$ a_{1}=2 $

$ a_{3}=4j $

Periodic signals have the fourier transform of $ X(e^{jw})=\sum_{k=-\infty}^{infty}2\pi a_k\delta(\omega-k\omega_0) $

T=5

$ w_0=2\pi/T=2\pi/5 $

y(t)=x(t)*h(t) (convulution)

$ Y(e^{j\omega})=X(e^{j\omega})H(e^{j\omega}) $

$ Y(e^{j\omega})=\sum_{k=-\infty}^{infty}2\pi a_k\delta(\omega-k\omega_0)(1/(3+j/omega)) $

$ Y(e^{j\omega})=\sum_{k=-\infty}^{infty}2\pi a_k(1/(3+jk/omega_0)\delta(\omega-k\omega_0)) $

$ Y(e^{j\omega})=\sum_{k=-\infty}^{infty}2\pi(a_k/(3+jk2/pi/5))\delta(\omega-k\omega_0)) $

notice $ b_k=a_k/(3+jk2/pi/5) $

$ b_{-3}=-4j/(3-j6/pi/5) $

$ b_{-1}=2/(3-j2/pi/5) $

$ b_{1}=2/(3+j2/pi/5) $

$ b_{3}=4j/(3+j6/pi/5 $

$ b_k=0 $ for all other k

--Krtownse 19:55, 18 July 2008 (EDT)

Problems

3. Find the Fourier Transform of the following signal x(t)

We were given that x(t) was a triangle with height 1 and a base from -1 to 1, plus a delta function shifted by 2.

For the sake of this problem we'll say: $ x(t) = x_1(t) + x_2(t) similarly; x(j\omega) = x_1(j\omega) + x_2(j\omega) $

The delta part of the function is farely simple, so we'll say

$ x_2(t) = \delta(t-2) $

therefore

$ x_2(j\omega) = e^{-2j\omega} $

The tricky part is the triangle, which we'll say is:

$ x_1(t) $ = square wave from -.5 to .5 w/height of 1 convulved with a square wave from -.5 to .5 w/height of 1.

Lecture 15 pg. 18 gives details of the properties for these types of functions.

Take a look at the image below for a visual representation of $ x_1(t) $

The solution to this problem is the answer given below $ x_1(t) $

--Pmavery_OldKiwi 17:34 p.m. 21, July 2008

Problems

4.

Solutions

Part A

Parts B & C

--Pmavery_OldKiwi 17:45 p.m. 21, July 2008