Samar Qureshi

1009149583

Part I

Find the FS coefficients for a sawtooth wave. Use the period To = 10^(-3). Choose a suitable number of points and experiment with different numbers of points, e.g. N=100, N=200. Now, determine the original signal from the FS coefficients and see how the accuracy of the reconstructed signal depends on the number of points which is a function of the sampling frequency. Plot the magnitude and phase of the coefficients.

Magnitude and phase plots of the sawtooth FS coefficients.

• The sawtooth is periodic but discontinuous, so its Fourier Series has both the sine and cosine terms
  • The coefficient delay is 1/k, because the discontinuity introduces Gibbs phenomenon (slow convergence)
• As $N$ increases, the RMS reconstruction error falls (better sampling)
• Sawtooth has slowly decaying coefficients $\propto 1/|k|$, so you many need many terms for 99% power (Gibbs ripple also shrink with higher $N$)

To = 1e-3; fo = 1/To;

% compute FS coefficients
function [Xc, kc, fc] = fs_coeffs(x, To)
    N = numel(x);
    X  = (1/N)*fft(x);          % FS coefficients over k = 0..N-1
    Xc = fftshift(X);           % center DC at middle
    kc = (-floor(N/2)):(ceil(N/2)-1);   % harmonic indices (centered)
    fc = kc / To;               % frequencies = k/T0
end

% reconstruct time samples from FS 
function xr = fs_recon(X, To)
    % X is ONE period, (1/N)*fft(x) style; inverse is ifft(N*X)
    N = numel(X);
    xr = ifft(N*X,'symmetric');
end

% 99% BW definition helper
function [K99, frac, order] = fs_99percent(Xc)
    % Xc is centered spectrum (fftshifted)
    Pk = abs(Xc).^2;
    [Pk_sorted, order] = sort(Pk,'descend');
    cumfrac = cumsum(Pk_sorted)/sum(Pk_sorted);
    K99 = find(cumfrac>=0.99,1,'first');
    frac = cumfrac(K99);
end

N = 100;  % try 100, then 200
t = (0:N-1)*(To/N);

% sawtooth on [0,To] with range [-1,1]
x = 2*(t/To) - 1;     
x = x - mean(x);     

% FS coefficients
[Xc, kc, fc] = fs_coeffs(x, To);

% Plots
figure; subplot(2,1,1); stem(kc, abs(Xc), 'filled'); grid on;
xlabel('Harmonic k'); ylabel('|X_k|'); title('Sawtooth: magnitude');
subplot(2,1,2); stem(kc, angle(Xc), 'filled'); grid on;
xlabel('Harmonic k'); ylabel('\\angle X_k (rad)'); title('Sawtooth: phase');

X = ifftshift(Xc); xr = fs_recon(X, To);
fprintf('RMS recon error (N=%d): %g\\n', N, rms(x - xr));

% power capture 
[K99, frac, order] = fs_99percent(Xc);
fprintf('Sawtooth: %d coefficients capture %.1f%% power\\n', K99, 100*frac);

Part II

Find the FS coefficients (plot magnitude and phase) for a half-wave rectified sine wave. Choose the origin so that the signal is even, and To = 10^(-3). Again, consider different choices for the number of points in one period.

• Half-wave rectified sine wave is even if centered properly → only cosine terms appear
• Even signal → pure cosine FS (phases cluster at 0 or $\pi$)
• Strong DC component; harmonics are mostly even or odd depending on symmetry choice
• Large DC term, then even-indexed harmonics dominate afterwards

To = 1e-3; fo = 1/To;

% compute FS coefficients
function [Xc, kc, fc] = fs_coeffs(x, To)
    N = numel(x);
    X  = (1/N)*fft(x);          % FS coefficients over k = 0..N-1
    Xc = fftshift(X);           
    kc = (-floor(N/2)):(ceil(N/2)-1);  
    fc = kc / To;              
end

% reconstruct time samples from FS 
function xr = fs_recon(X, To)
    N = numel(X);
    xr = ifft(N*X,'symmetric');
end

% 99% BW definition helper
function [K99, frac, order] = fs_99percent(Xc)
    Pk = abs(Xc).^2;
    [Pk_sorted, order] = sort(Pk,'descend');
    cumfrac = cumsum(Pk_sorted)/sum(Pk_sorted);
    K99 = find(cumfrac>=0.99,1,'first');
    frac = cumfrac(K99);
end

N = 200; t = (0:N-1)*(To/N); w0 = 2*pi*fo;

% [0, To/2], mirror to [To/2, To]
x_half = sin(w0*t); x_half(x_half<0) = 0;  % half-wave rectified over full grid
x_even = x_half;
x_even( (N/2+1):N ) = x_even( N/2:-1:1 );  n

x = x_even;

[Xc, kc, fc] = fs_coeffs(x, To);

figure; subplot(2,1,1); stem(fc/1e3, abs(Xc), 'filled'); grid on;
xlabel('f (kHz)'); ylabel('|X(f)|'); title('Half-wave rectified sine: |X|');
subplot(2,1,2); stem(fc/1e3, angle(Xc), 'filled'); grid on;
xlabel('f (kHz)'); ylabel('\\angle X'); title('Phase');

[K99, frac] = fs_99percent(Xc);
fprintf('Half-wave: %d coefficients capture %.1f%% power\\n', K99, 100*frac);

Part III:

*Define the time signal x(t) on the interval [0,To], where To = 10^(-3) as x(t) = cos(2pifo(t + (1/(2*To))t^2)) where fo=1/To. This is known as a chirp signal. Find the Fourier series coefficients using a suitable number of points in the sampled signal, i.e a suitable sampling frequency.