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Systems and signals square wave

Case I: Square-wave x1(t), 50% duty cycle

f0=500 Hz

To=1/500=0.002s=2 ms

Mathematically, the square-wave can be expressed for a single period as

systems and signals square wave image 1

Therefore the trigonometric Fourier series coefficients of x1(t) are

systems and signals square wave image 2

Case II: Square-wave x2 (t), 25% duty cycle

Mathematically, the square-wave x2 (t) can be expressed for a single period as

systems and signals square wave image 3

Sinusoids from -8 to -1;

systems and signals square wave image 4
        clear
    clc
    f_0= 500; % fundamental frequency
    t= -0.001:0.00001:0.006; %time axis
    Z_50= zeros(1,length(t)); % container for 50% duty cycle waveform
    Z_25= zeros(1,length(t)); % container for 25% duty cycle waveform
    n=1:16;
    n1=sin(pi.*n./2);
    d1=(pi.*n./2);
    n2=sin(pi.*n./4);
    d2=((pi).*n./4);
    a_50=[n1./d1]; %fourier coefficients for 50% duty cycle
    a_25=[0.5*(n2./d2)]; %fourier coefficients for % duty cycle
    for k=1:length(n)
        wa=1*cos(2*pi*(k)*f_0*t);
        Z_50=Z_50+ a_50(k)*wa;
        Z_25=Z_25+ a_25(k)*wa;
    end
    soundsc(Z_50)
    figure (1)
    plot(t,Z_50)
    hold on
    grid on
    axis tight
    xlabel('time')
    ylabel('Amplitude')
    title('Square-wave with 50% duty cycle');

    figure(2)
    plot(t,Z_25)
    grid on
    axis tight
    xlabel('time')
    ylabel('Amplitude');
    title('Square-wave with 25% duty cycle');
    
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