Impulse-response coefficients (that represent a differentiation)

• Nov 27th 2011, 08:42 AM
courteous
Impulse-response coefficients (that represent a differentiation)
Quote:

These weighted-impulse-response $\displaystyle h$ coefficients (should) represent a differentiator:
$\displaystyle h = \frac{1}{3}(-1, -2, 0, 2, 1)$
If $\displaystyle x[n]$ is the input sample, what is then the output $\displaystyle y[n]$?

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

Thank you.
• Nov 27th 2011, 09:12 AM
chisigma
Re: Impulse-response coefficients (that represent a differentiation)
courteous wrote...

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 27th 2011, 11:44 AM
courteous
Re: Impulse-response coefficients (that represent a differentiation)
chisigma, you're (of course) correct ... that was a typo (fixed it in OP)

Can you help me out?
• Nov 27th 2011, 11:54 AM
chisigma
Re: Impulse-response coefficients (that represent a differentiation)
The response to an imput sequence x(n) is...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 27th 2011, 12:08 PM
courteous
Re: Impulse-response coefficients (that represent a differentiation)
Thank you! What is the (1) thing at the end? Just a reference mistakenly copied along?