## Samstag, 22. Januar 2011

### EE5410 Calculating the impulse response h[n]

EE5410 Calculating the impulse response h[n]

excerpted from : Joyce Van de Vegte : Fundamentals of Digital Signal Processing. page. 123-125

eg. 4.11. Find the first six sample of the impulse response for the differece equation :

y[n] - 0.4y[n-1] = x[n]-x[n-1]

1. First replace x[n] with delta[n] and y[n] with h[n] to give:

h[n] - 0.4h[n-1] = delta[n] - delta[n-1]
or
h[n] = 0.4h[n-1] + delta[n] - delta[n-1]

2. Starting with n =0: therefore,

h[0] = 0.4h[-1] + delta[0] - delta[-1]

The values for the impulse function delta[n] are known: at n=0, it has the value 1, and at all other values of n, it has the value 0.

3. The filter can be assumed tobe casual, which means that the impulse response is zero BEFORE n = 0. Therefore,

h[0] = 0.4(0.0) + 1.0 -0.0 = 1.0

Notice that delta[-1] =0 because zero is the value of the function delta[n] when n= -1, not a consequence of causality.

4 The subsequent impulse response samples are :

h[1] = 0.4h[0] + delta[1] -delta[0] = 0.4(1.0)+0.0-1.0 = -0.6

h[2] = 0.4h[1] + delta[2] - delta[1] = 0.4(-0.6) +0.0-0.0 = -0.24

h[3] = 0.4h[2] + delta[3] -delta[2] = 0.4(-0.24) +0.0- 0.0 = -0.096

h[4] = 0.4h[3] + delta[4] - delta[3] = 0.4(-0.096) +0.0- 0.0 = -0.0384

h[5] = 0.4h[4] + delat[5] -delta[4] = 0.4(-0.0384) + 0.0- 0.0 = -0.01536

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