Consider the function f_{T}(t) represented by the exponential Fourier series

Show that

Hint: try multiplying f_{T}(t) by
then integrating.

Use hint. Multiply

then integrate over one period

Switch order of integration and summation (and pull c_{n} out of
summation)

The function
is periodic and has an integer number of complete oscillations in one
period, T, unless m=n in which case the function is constant with a value of
1. Therefore the integral is 0 unless m=n in which case the integral
is T (integrating 1 over an interval of T). As we perform the
summation we multiply each c_{n} by zero if m≠n, and by T if m=n.
Therefore the only term that makes it out of the summation is when m=n, so

or, by changing the variable and rearranging

For the function, f_{T}, below:

a) Find the coefficients of a Fourier Sine Series (b_{n}).

b) Find the coefficients of an Exponential Fourier Series (c_{n}).

c) Show that the two answers are equivalent (i.e., b_{n}=-2*imag(c_{n})).

a) Exploit symmetries with Fourier Sine Series (function is odd, so we can use only sines).

since f_{T}(t) and sin(nω0t) are odd, their product is even, so we can
take twice the integral from 0 to T/2. Also use the fact that f_{T}(t)=2t
in that interval, and T=2.

Use integration by parts or integration table (or MATLAB/Mathematica)

>> syms t n >> assume(n,'integer') >> b=4*int(t*sin(n*pi*t),0,1) b = (4*(sin(pi*n) - pi*n*cos(pi*n)))/(pi^2*n^2) >> pretty(simple(b)) n 4 (-1) - ------- pi n

b) Exponential series

>> syms t n >> assume(n,'integer') >> c=int(t*exp((0-1j)*n*pi*t),-1,1) c = (exp(-pi*n*i)*(pi*n*i + 1))/(pi^2*n^2) + (exp(pi*n*i)*(pi*n*i - 1))/(pi^2*n^2) >> pretty(simple(c)) n 2 (-1) i --------- pi n

c) b_{n}=-2*imag(c_{n}), so they are
equivalent.

Show that the exponential series coefficients, c_{n}, for the function shown
below (a 60 Hz, half-wave rectified sine wave) are given by

Euler's identity for sin may be useful:
. Also think about the value of e^{jπ} and e^{-jπ}. Be prepared for lots of algebra in the solution.

but ω_{0}T=2π

Make use of the facts

So

Note: in the last line we had to go back to complex exponential form because we set jsin(-πn) to zero earlier (eq. 1), but now we are taking the limit as n→1, so the whole expression in (eq. 1) is going to zero.

Plot the half wave rectified function from the previous problem on the same plot as the sum of the first four non-zero terms of the Fourier Series. Plot for t=0 to 1/30 second. Note that as we add terms, the approximation converges to the function.

Below is some code you can use as a starting point. If you run it as is it plots Fourier sum for square wave.

% skeleton code (use at your own peril) - as is, it plots Fourier Sum for square wave. clear all % Define constants c_order='rgbcm'; %Order of colors used in plot. N=8; %number of Fourier terms to plot T=1/60; wo=2*pi/T; %Period and fundamental frequency t=linspace(0,2*T,1000); %Time vector % % Plot a half-rectified sine wave (for E12 homework problem). % y=sin(2*pi/T*t).*(sin(2*pi/T*t)>0); % plot(t,y,'k:','LineWidth',2); % Plot a square wave (as an example). y = sin(2*pi/T*t)>0; plot(t,y,'k:','LineWidth',2); leg{1}='Function'; %Create legend entry axis([0 2*T -1.1 1.1]); hold on c0=0.5; %c0=DC component of Fourier Series ysum=c0*ones(size(y)); %Set summation to average (or DC) value. plot(t,ysum,'y'); %Plot ysum (the average value) in yellow leg{2}='DC component'; %Create legend entry axis([0 2*T -1.2 1.2]); for n=1:N, %Form Fourier sum cn=j/n/2/pi*(cos(n*pi)-1); %Calculat cn. an=2*real(cn); %calculate an and bn from cn. bn=-2*imag(cn); ysum=ysum + an*cos(n*wo*t) + bn*sin(n*wo*t); %Add to Fourier sum plot(t,ysum,c_order(mod(n,5)+1)); %Plot - using color from c_order leg{end+1} =['Sum of ' num2str(n)]; %Create legend entry end hold off legend(leg,'location','se'); title('Function (black) and Fourier Approximations'); xlabel('t');

% Plots Fourier Sum for half-wave rectified signal. clear all % Define constants c_order='rgbcm'; %Order of colors used in plot. N=7; %number of Fourier terms to plot T=1/60; wo=2*pi/T; %Period and fundamental frequency t=linspace(0,2*T,1000); %Time vector % Plot a half-rectified sine wave y=sin(2*pi/T*t).*(sin(2*pi/T*t)>0); plot(t,y,'k:','LineWidth',2); leg{1}='Original'; hold on c0=1/pi; %c0=DC component of Fourier Series ysum=c0*ones(size(y)); %Set summation to DC value. plot(t,ysum,'y'); %Plot ysum in yellow leg{2}='DC component'; %Create legend entry axis([0 2*T -0.2 1.2]); for n=[1 2 4 6 8], %Form Fourier sum nc=n+0.001; %change n slightly to avoid 0/0 cn=-1/2/pi./(nc.^2-1).*(exp(-j*pi*nc)+1); %Calculat cn. an=2*real(cn); %calculate an and bn from cn. bn=-2*imag(cn); ysum=ysum + an*cos(n*wo*t) + bn*sin(n*wo*t); %Add to Fourier sum plot(t,ysum,c_order(mod(n,5)+1)); %Plot - using color from c_order leg{end+1} =['Sum of ' num2str(n)]; %Create legend entry end hold off legend(leg,'location','ne'); title('Function (black) and Fourier Approximations'); xlabel('t');

Prove that if

then

where the c_{n,x} terms are the Fourier series coefficients of
x_{T}(t), and the c_{n,y} terms are for y_{T}(t)

Note the similarity to the time shift property of the Laplace Transform.

The third line uses the fact that if we are integrating over a period, it doesn't matter if the function is shifted or not.

In class we showed that the Fourier Series coefficients of the function shown
(call it Π_{T}(t/T_{p}) - a rectangular pulse of
period T, and width T_{p}, centered on the origin).

are given by

Use this result, and the time delay property derived
previously, to find the coefficients, c_{n}, of the function shown
below without performing any integration. Once you have an expression for
c_{n} there is no need to simplify.

Time delay is T_{p}/2, so Fourier Series coefficients for bottom function are

Find the coefficients, c_{n}, of the function shown below without
performing any integration by representing the function as the differences of
two pulse functions centered at t=0.

The function is the sum of a pulse of amplitude 2 and width 6 minus a pulse of amplitude 3 and width 4.

Note that c_{0}=0, and that the average value of the function is
zero.

Briefly explain in your own words:

a) ...why the Fourier Series of* x _{T}(t)* (periodic function with period=

*T*) uses only sinusoids with discrete frequencies (

*nω*,

_{0}*ω*)

_{0}=2π/Tb) ... why the Fourier Transform of *x(t) *has sinusoids with a continuum of frequencies (*ω*).

I'm looking for a fairly basic answer here, not a complicated discussion.

a) Since *x _{T}(t)* is periodic (with period=

*T*), all of its composite functions must also be periodic with period

*T*. A sinusoid at frequency

*nω*, is periodic with period

_{0}*T*(and has

*n*oscillations in that time). The change in frequency as

*n*increments by one is

*ω*.

_{0}=2π/T= Δωb) If we consider the function* x(t)* to be the function *x _{T}(t)* as

*T→∞*(so it becomes aperiodic) the the change in frequency as

*n*increments is

*ω*- in other words, the frequency becomes a continuum.

_{0}=2π/T=Δω→0