Convolution with a time shifted box function
6 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Hello, I have the input:
Its a box funciton x2(t-0.00225) with L = 0.0005 where it's 1 when 0.002 <= t<= 0.0025 and 0 elsewhere. I want to convolve this with the impulse funciton of a low pass filter:
My hand calculation gives the result for when (-0.00225 + t ) > 0 is y(t) = 1 - e^ (0.00225-t)/RC):
And my code is:
Ts=1e-6; % Set up time and frequency variables
t = 0:Ts:0.01-Ts; N = length(t);
fs=1/Ts; F=fs/N; f=(-fs/2):F:(fs/2)-F;
R = 2000; % Resistance
C = 1.5*10^(-8); % Capacitance (15 nF)
RC = R*C; % Time contant
h = (1/RC) .* exp(-t/RC) .* heaviside(t); % Impulse response of RC LPF
x2 = heaviside(t-0.002) - heaviside(t-0.0025); % box function
y2 = conv(h, x2)*Ts;
figure(3) % Plot result
subplot(2,1,2), plot(t, y2(1:N), 'Linewidth', 2.0)
axis([0 0.004 0 1.2])
xlabel('Time, sec', 'Fontsize', 14), ylabel('y_2(t)', 'Fontsize', 14)
grid on
Which has given the graph:
The graph itself seems correct but when substituting random values of t, the graph is incorrect (it should be 1 from 2.5 and beyond, not reducing to 0 like here). Am I missing something here? Maybe in hand calculation it shouldn't be -tau but I have to take into account the -0.00225 shift for tau too? That would create a big e^(0.00225/RC) constant which I'm skeptical
Thank you
0 comentarios
Respuesta aceptada
Paul
el 7 de Abr. de 2024
Hi Nghi,
Why should the output of system remain at 1 for t > 2.5? For t > 2.5, the input is 0 and so we'd expect the output to decay to zero from that time because the system is stable.
heaviside(0)
which is not really what you want when using a discrete-time convolution sum with conv to approximate the continuous-time convolution integral. That can be changed with sympref, or write your own unit step function
u = @(t) double(t>=0);
0 comentarios
Más respuestas (0)
Ver también
Categorías
Más información sobre Simultaneous and Synchronized Operations en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!