Recent content by ace_terabyte

Thanks a lot CEL, I appreciate it.

y(t) = \int ^{t}_{-\infty} e^{-(t-\tau)} x(\tau-2)d\tau So going by what you said: y(t) = \int ^{t}_{-\infty} (\frac{1}{e})^{(t-\tau-2)} x(\tau)d\tau = \int ^{t}_{-\infty} (\frac{1}{e})^{(t-\tau)} (\frac{1}{e})^{(2)}x(\tau)d\tau =(\frac{1}{e})^{(2)} \int ^{t}_{-\infty}...

Thanks for that, the question asks me to find out h(t) and does not give me any other information. I assume it is the same in both frequency and time domain.

Homework Statement y(t) = \int ^{t}_{-\infty} e^{-(t-\tau)} x(\tau-2)d\tau Find h(t), given that the continuous time linear time invariant signal has input x(t) and output y(t). 2. The attempt at a solution I was wondering if I can simply manipulate the expression until i get it in the...