As title says what would be frequency response of backward finite difference differential filter, or what would be error of this differential filter, analyzed upon frequency of a signal?


1 Answer 1


In discrete-time signal processing terminology a backward (one sample) difference filter is

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

which is a causal, LTI system with an FIR impulse response of

$$h[n] = \delta[n] -\delta[n-1]$$

The associated frequency response is then the DTFT of the impulse response as

$$\begin{align} H(e^{j\omega}) = \mathscr{F}\{ h[n] \} &= 1 - e^{-j\omega} \\ &= e^{-j\omega/2}(e^{-j\omega/2} - e^{-j\omega/2}) \\ &= 2j e^{-j\omega/2} \ \tfrac1{2j}(e^{-j\omega/2} - e^{-j\omega/2}) \\ &= e^{-j\omega/2} 2j \ \sin(\omega/2) \\ &\approx e^{-j\omega/2} j \ \omega \qquad \text{for } |\omega| \ll \pi \\ \end{align}$$

So you have a delay of $\frac12$ sample, an additional phase shift of +90° (due to the $j$ factor), and a gain proportional to frequency (at least for low frequencies, relative to Nyquist). Other than the half-sample delay, this is what you expect from a differentiator.

If you happen to use an $N$-sample backward difference such as $$y[n] = x[n] - x[n-N]$$

then the associated frequency response will be $$H_N(e^{j\omega}) = 1 - e^{-jN\omega} $$

This is similar to the first orde case, but the delay is $\frac{N}2$ samples, and the gain is $N \omega / 2$ .


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