Digital Signal Processing/Impulse Response

Let's say that we have the following block diagram: center|400px h[n] is known as the 'Impulse Response of the digital system. We will talk about impulse responses here in this chapter. Impulse Function In digital circuits, we use a variant of the continuous-time delta function. This delta function (δ[n]) has a value of 1 at n = 0, and a value of 0 everywhere else. δ[n] = [... 0 0 0 0 1& 0 0 0 0 ...] If we set x[n] = δ[n], then the output y[n] will be the response of the system to an impulse, or simply, the "Impulse Response". We can time-shift the impulse function as such: δ[n-1] = [... 0 0 0 0& 1 0 0 0 0 ...] δ[n+1] = [... 0 0 0 0 1 0& 0 0 0 ...] We can add time-shifted values of the impulse function to create an Impulse Train: y[n] = δ[n] + δ[n-1] + δ[n-2] + [n-4] y[n] = [1& 1 1 0 1] Impulse Trains Impulse Response If we have a difference equation relating y[n] to x[n], we can find the impulse response difference equation by replacing every y with an h, and every x with a δ: y[n] = 2x[n] + 3x[n-1] + 4x[n-2] h[n] = 2δ[n] + 3δ[n-1] + 4δ[n-2] And by plugging in successive values for n, we can calculate the impulse response to be: h[n] = [2& 3 4] Output Now, let's say that we have a given impulse response, h[n], and we have a given input x[n] as such: x[n] = [1& 0 1 2] h[n] = [2& 2 2 1] We can calculate the output, y[n], as the convolution of those 2 quantities: x[n] -> 1& 0 1 2 h[-n] -> 1 2 2 2& h[-n-3] -> 1 2 2 2& -> y[m-3] = 2& h[-n-2] -> 1 2 2 2& -> y[m-2] = 2 h[-n-1] -> 1 2 2 2& -> y[m-1] = 4 h[-n] -> 1 2 2 2& -> y[m] = 7 h[-n+1] -> 1 2 2 2& -> y[m+1] = 6 h[-n+2] -> 1 2 2 2& -> y[m+2] = 5 h[-n+3] -> 1 2 2 2& -> y[m+3] = 2 y[n] = [2& 2 4 7 6 5 2]