Question

Exercise 1. Let 0<a<7. Let f(t) denote the 27t-periodic function 10, if a<\t]<1. a) Make a representative plot of f(t) showing at least 3 periods for a = 7/8, 7/2, and 7/8. b) Compute explicitly the Fourier series coefficients as of f(t). c) Make a representative plot of the absolute values of the Fourier coefficients, lax for a= /8, 7/2, and 7/8. The plot must at least show (0-20) through a 2013 d) By looking at the plots in parts a) and c) what general conclusions can you draw? Did you observe that as f(t) gets wider the plot for lax| becomes narrower? This phenomenon is generally referred to as Heisenberg's Uncertainty Principle.

Answer 1

1)Matlab code and plots

ns. I (6) given f(t) = [2 if tica lo if a<lti<a - as To = 2K so wo = 27 = 1 The tourier Series Coefficients of fit) ax = 1/ s fets ej kwotast of f(t) is defined as Coefficients Series ax= att ha e jetzt a una set ax = utama [ ejka ei kar ] = ulana [ ei ka eukary K JK Sinkron] By enter's formula ejo - e jo = ja sino * = with a C j2 Sin Ka] = Tak = ut sine (ka) use this expression in matlab to past Coefficient leal fousier

clear;clc;
T0=2*pi; %time period of f(t)
Ts=0.001; %samplig time
t=-1.5*T0:Ts:1.5*T0-Ts; %time index for f(t) plot upto 3 period
k=-20:20; %harmonic index for ak plots
%% for alpha=pi/8
a1=pi/8;
%f(t) over one period
t1=-T0/2:Ts:T0/2-Ts;
f1=zeros(1,length(t1));
f1(abs(t1)<a1)=1/(2*a1);
% f(t) over 3 period
ft1=repmat(f1,1,3);
% f(t) FS coefficient
ak1=(1/T0)*sinc(k*a1./pi);
figure(1)
plot(t,ft1);
title('plot of f(t) for \alpha =\pi/8');
xlabel('t');ylabel('f(t)');grid on;
figure(2)
stem(k,abs(ak1));
title('plot of absolute value of FS coefficients for \alpha =\pi/8');
xlabel('k');ylabel('|a_k|');grid on;
%% for alpha=pi/2
a2=pi/2;
%f(t) over one period
t2=-T0/2:Ts:T0/2-Ts;
f2=zeros(1,length(t2));
f2(abs(t2)<a2)=1/(2*a2);
% f(t) over 3 period
ft2=repmat(f2,1,3);
% f(t) FS coefficient
ak2=(1/T0)*sinc(k*a2./pi);
figure(3)
plot(t,ft2);
title('plot of f(t) for \alpha =\pi/2');
xlabel('t');ylabel('f(t)');grid on;
figure(4)
stem(k,abs(ak2));
title('plot of absolute value of FS coefficients for \alpha =\pi/2');
xlabel('k');ylabel('|a_k|');grid on;
%% for alpha=7pi/8
a3=7*pi/8;
%f(t) over one period
t3=-T0/2:Ts:T0/2-Ts;
f3=zeros(1,length(t3));
f3(abs(t3)<a3)=1/(2*a3);
% f(t) over 3 period
ft3=repmat(f3,1,3);
% f(t) FS coefficient
ak3=(1/T0)*sinc(k*a3./pi);
figure(5)
plot(t,ft3);
title('plot of f(t) for \alpha =7\pi/8');
xlabel('t');ylabel('f(t)');grid on;
figure(6)
stem(k,abs(ak3));
title('plot of absolute value of FS coefficients for \alpha =7\pi/8');
xlabel('k');ylabel('|a_k|');grid on;

plot of f(t) for a = 7/8 1.2 0.8 0.6 0.4 0.2 10 9 - 20 2 10 4

plot of absolute value of FS coefficients for a = /8 0.16 0.14 0.12 0. 0.08 0.06 0.04 0.02 O -10 OY 15

plot of f(t) for a = 1/2 -10 -8 6 4 2 0 2 4 6 8 10

plot of absolute value of FS coefficients for a = /2 0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 -15 - 10 0 5 10 15 20

plot of f(t) for a =77/8 0.18 0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 - 20

plot of absolute value of FS coefficients for a =771/8 0.16 0.14 * 0.08 0.06 0.04 -20 -15 -10 10 15 20
d)yes as signal strech in time domain its bandwidth decrease or vice-versa