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

• 1(a) .... The period is not 12. Check again.
• 1(b) .... Even with the correct period, the answer is still wrong. Check again.

## 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 4. Therefore N=4

Choose values of $a_k$ that match the forier series repersentation of x[n]

$x[n]=\sum_{k=<n>}^{}a_ke^{jk(2\pi/4)n}$

$a_0=1$

$a_3=\sqrt{2}e^{5} - j\sqrt{2}e^5$

1)b) Some energy therom states that: $\sum_{n=<N>}|x[n]|^2=N*sum_{k=<N>}|a_k|^2$

Therefore:

$(1/N)*\sum_{n=<N>}|x[n]|^2=\sum_{k=<N>}|a_k|^2$

$(1/N)*\sum_{n=<N>}|x[n]|^2=1^2+|(-2je^{\pi/4}e^{5})|^2=1 + 4e^{10}$

$(1/N)*\sum_{n=<N>}|x[n]|^2=1 + 4e^{10}$

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

4.

## Solutions

Part A

Parts B & C

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

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