# Backward finite difference differentiation filter frequency response

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?

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 $$N \omega / 2$$ .