Impulse-response coefficients (that represent a differentiation)

**courteous** · November 27th 2011, 08:42 AM

These weighted-impulse-response $h$ coefficients (should) represent a differentiator:
$h = \frac{1}{3}(-1, -2, 0, 2, 1)$

If $x[n]$ is the input sample, what is then the output $y[n]$ ?

Is it this: $y[n] = \frac{1}{3}\big(-x[n-4] - 2x[n-3] + x[n-1] + 2x[n]\big)$ ?

Thank you.

**chisigma** · November 27th 2011, 09:12 AM

courteous wrote...

... these weighted impulse response h coefficients should represent a differentiator:

$h= \frac{1}{3}\ (-1,-2,0,1,2)$

A differentiator has typically linear phase and that means that its impulse response must be antysimmetrical, so that [probably] the impulse response is...

$h= \frac{1}{3}\ (-1,-2,0,2,1)$

Kind regards

$\chi$ $\sigma$

**courteous** · November 27th 2011, 11:44 AM

chisigma, you're (of course) correct ... that was a typo (fixed it in OP)

Can you help me out?

**chisigma** · November 27th 2011, 11:54 AM

The response to an imput sequence x(n) is...

$y(n)= \sum_{k=0}^{4} h(k)\ x(n-k) = \frac{1}{3}\ \{- x(n) -2\ x(n-1) +2\ x(n-3) + x(n-4)\}$ (1)

Kind regards

$\chi$ $\sigma$

**courteous** · November 27th 2011, 12:08 PM

Thank you! What is the (1) thing at the end? Just a reference mistakenly copied along?

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