can someone please explain to me how this works.

y[n] = 2 cos(w0)y[n − 1] − y[n − 2],with no input but with initial conditions y[−1] = 0 and y[−2] = −A sin(w0).

(a)Show that the solution of the above LCCDE is given by the sequence，y[n] = A sin[(n + 1)w0]u[n]. This system is known as a digital oscillator.

(b) For A = 2 and w0 = 0.1π, verify the operation of the above digital oscillator using MATLAB.

Answer 1

[Altaïr]

Hey @Ching-Wen Yu,

For the analytical proof portion (question a), it would be great to see your current approach before discussing further! Feel free to share what you've explored so far.

Regarding the digital oscillator implementation, here's a sample MATLAB code that might help:

% Digital oscillator simulation
A = 2;
w0 = 0.1 * pi;
N = 100;       % Generates y[0] to y[99]

% Initialize with given conditions
y = zeros(1, N);
y_minus1 = 0;                       % y[-1]
y_minus2 = -A * sin(w0);            % y[-2]

% Compute sequence iteratively
for n = 0:(N-1)
    current = 2 * cos(w0) * y_minus1 - y_minus2;
    y(n+1) = current;  
    % Update history buffers
    y_minus2 = y_minus1;
    y_minus1 = current;
end

% Analytical solution comparison
n = 0:(N-1);
y_analytical = A * sin((n + 1) * w0);

% Visualization
figure;
plot(n, y, '-');      % Numerical solution
hold on;
plot(n, y_analytical, 'o'); % Analytical solution
xlabel('n');
ylabel('y[n]');
legend('Numerical','Analytical: A sin((n+1)w0)','Location','best');
title('Digital Oscillator: A=2, w0=0.1\pi');
grid on;

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This code generates both numerical (via recurrence) and analytical solutions, then plots them together for comparison. For additional plotting options, the documentation can be accessed with:

doc plot