# Differentiation of discrete sinusoidal signal

What I understand from differentiation of discrete sinusoidal signal is $y(n) = \alpha(x(n)-y(n-1))$ where $\alpha$ may be a factor depending on sampling frequency and signal frequency. Now y(n) represents signal which is $90^\circ$ phase shifted sinusoid.

My question is why to go to Hilbert transform and other complex methods to shift a signal $90 ^\circ$. My concern is with $50\,\text{Hz}$ signal only till $40$-th harmonic.

## 1 Answer

Hilbert transform has the property of harmonic correspondence (Vakman 1996), meaning that the Hilbert transform of a cosine wave is a sine wave etc. So the Hilbert transform is a 90 degree phase shift. In fact the Hilbert transform in the Fourier domain looks quite similar to the differential operator:

$$\mathcal{F}\left[\tfrac{d}{dt} f(t)\right](u) = -iw \mathcal{F}[f(t)](u)$$ $$\mathcal{F}\left[\mathcal{H}[f](t)\right](u) = -i\frac{w}{|w|} \mathcal{F}[f(t)](u)$$

We can see that for a sinusoid of frequency $w_0$, the derivative is the Hilbert transform scaled by $w_0$. e.g. Hilbert transform of $\cos(n t) = -\sin(n t)$, derivative of $\cos(n t) = -n \sin(nt)$. So careful choice of alpha in your formula will give you the phase shift, while compensating for the scaling by frequency, which is the same as the Hilbert transform.

If you have a non-sinuosidal signal, then the 90 degree phase shift must be applied to every sinusoidal component. This is not the same as differentiation, and you couldn't use your formula. e.g. Hilbert transform of square wave (from wikipedia): 