Question

Problem 2 Periodic Force First Cycle The graph at the right depicts the first period of a non-harmonic periodic force (measured in Newtons). This first cycle is described by the piecewise function F(t) below the graph. Per the definition of a periodic function, the function repeats every T seconds. Note that T = 1 s. 1.8 a. What is the angular frequency wT of the periodic function?2 Include units. b. What is the Fourier Series representation of this function? c. Using MATLAB, plot ONE cycle of the original function. d. On a separate graph, plot TWO cycles of the Fourier Series 0.8 0.6 0.4 02 approximation of this function for n = 2 terms. Then super- impose the original function on this graph in a different color 0.2 0.4 0.6 0.8 t (s) (this will demonstrate the accuracy of the approximation) e. Repeat the previous step for n 5 terms f. Repeat the previous step for n = 20 terms. You should end up with 4 separate graphs. Please include a printout of each one in your homework submission. You should be able to fit all 4 graphs on one page. Please title each graph

Answer 1

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4 (e -j2An 2 汉Kn -J -(2 Kn 2

T212 )끈 T/2 リ2 .? 3丁 (s :+[4 +も-v4 t)1T/L+ 8(止至り13仆. (et-ot. 3t/y

32 & 2
1)matlab code and result

clear;clc;
%% for orignal time signal
T=1;%time period of signal
t=0:0.001:(T-0.001);%time index vector upto one period
t1=0:0.001:(2*T-0.001);%time index vector upto two period
xt=zeros(1,length(t));
%time signal over one period
xt(t>=0&t<T/4)=1;
xt(t>=T/4&t<T/2)=1-4*(t(t>=T/4&t<T/2)-T/4);
xt(t>=T/2&t<3*T/4)=8*(t(t>=T/2&t<3*T/4)-T/2);
xt(t>=3*T/4&t<T)=2-4*(t(t>=3*T/4&t<T)-3*T/4);
xt1=repmat(xt,1,2);%time signal over two periods
%% for approximation ofsignal X_N(t) N=2
N1=2;
K1=-N1:N1;
CK1=-(1-exp(-j*0.5*pi*K1)).*(1+3*exp(-j*pi*K1)).*(1./((pi*K1).^2));
CK1(K1==0)=1;
xN1=CK1*exp(j*pi*2*K1'*t1);
%% for approximation ofsignal X_N(t) N=5
N2=5;
K2=-N2:N2;
CK2=-(1-exp(-j*0.5*pi*K2)).*(1+3*exp(-j*pi*K2)).*(1./((pi*K2).^2));
CK2(K2==0)=1;
xN2=CK2*exp(j*pi*2*K2'*t1);
%% for approximation ofsignal X_N(t) N=20
N3=20;
K3=-N3:N3;
CK3=-(1-exp(-j*0.5*pi*K3)).*(1+3*exp(-j*pi*K3)).*(1./((pi*K3).^2));
CK3(K3==0)=1;
xN3=CK3*exp(j*pi*2*K3'*t1);
%% for signals plots
figure(1)
plot(t,xt)
title('plot of orignal signal X(t) over one period');
xlabel('time(second)');ylabel('X(t)');grid on;
figure(2)
plot(t1,xt1,t1,xN1)
title('plot of orignal signal X(t) and it''s aproximation for N=2');
xlabel('time(second)');ylabel('magnitude');grid on;
legend('orignal X(t)','X_N(t) N=2');
figure(3)
plot(t1,xt1,t1,xN2)
title('plot of orignal signal X(t) and it''s aproximation for N=5');
xlabel('time(second)');ylabel('magnitude');grid on;
legend('orignal X(t)','X_N(t) N=5');
figure(4)
plot(t1,xt1,t1,xN3)
title('plot of orignal signal X(t) and it''s aproximation for N=20');
xlabel('time(second)');ylabel('magnitude');grid on;
legend('orignal X(t)','X_N(t) N=20');

plot of orignal signal X(t) over one period 2 1.8 1.6 1.4F 1.2 0.8 0.6 0.4 0.2 0 0.2 0.6 0.7 0.1 0.3 0.4 0.5 0.8 0.9 0 time(second)

plot of orignal signal(t) and it's aproximation for N-2 orignal X(t) 1.8 1.6 1.2 0.8 0.6F 0.4 0.2 0.2 1.2 1.6 0 0.4 0.6 0.8 1.4 2 1.8 time(second)

plot of orignal signal X(t) and it's aproximation for N-5 2 orignal X(t) 1.8 1.6 1.4 1.2 0.8 0.6 0.4 0.2 0 0.6 1.4 0.4 0.8 1.2 1.6 1.8 2 0.2 time(second)

plot of orignal signal X(t) and it's aproximation for N-20 orignal X(t) X) N-20 1.8F 1.6 1.4F 1.2 0.8F 0.6 0.4 0.2 0 1.6 1.8 0 0.2 0.4 0.6 0.8 1.2 4 time(second)