# Fourier Transform of the Hilbert Transform of cos(t) (using Fourier time-shifting property)

If $$x(t)=cos(t)=\frac{1}{2}e^{jt}+\frac{1}{2}e^{-jt}$$, then $$X(\omega)=\pi \delta(\omega-1)+\pi \delta(\omega+1)$$. If $$y(t)=cos(t-\frac{\pi}{2})=\frac{1}{2}e^{j(t-\frac{\pi}{2})}+\frac{1}{2}e^{-j(t-\frac{\pi}{2})}=\frac{1}{2}e^{jt-j\frac{\pi}{2}}+\frac{1}{2}e^{-jt+j\frac{\pi}{2}}$$, then $$Y(\omega)=\frac{e^{-j\frac{\pi}{2}}}{2}F\{e^{jt}\}+\frac{e^{j\frac{\pi}{2}}}{2}F\{e^{-jt}\}=-j\pi \delta(\omega-1)+j\pi \delta(\omega+1)$$. $$y(t)$$ is the Hilbert transform of $$x(t)$$. This makes sense so far.

The trouble is when I try to evaluate $$Y(\omega)$$ using the time-shift property of the Fourier transform. $$Y(\omega)=F\{x(t-\frac{\pi}{2})\}=X(\omega)e^{-j\omega\frac{\pi}{2}}=(\pi \delta(\omega-1)+\pi \delta(\omega+1))e^{-j\omega\frac{\pi}{2}}$$. If I were to plot the phase of $$Y(\omega)$$, then we would see a line with slope $$-\pi/2$$. This does not seem to match the Hilbert Transform.

• A pair of $\delta()$ functions multiplied by a function is a pair of weighted $\delta()$ functions, not a line. Commented Apr 17, 2019 at 23:24
• Oh wow, pretty dumb mistake on my part.
– Taha
Commented Apr 17, 2019 at 23:59

## 1 Answer

I forgot about the property of a dirac delta function: $$f(t)\delta (t)=f(t)|_{t=0} \delta (t)=f(0)\delta (t)$$

$$Y(\omega)=F\{x(t-\frac{\pi}{2})\}=X(\omega)e^{-j\omega\frac{\pi}{2}}=(\pi \delta(\omega-1)+\pi \delta(\omega+1))e^{-j\omega\frac{\pi}{2}}=\pi \delta(\omega-1)e^{-j(1)\frac{\pi}{2}}+\pi \delta(\omega+1)e^{-j(-1)\frac{\pi}{2}}=-j\pi \delta(\omega-1)+j\pi \delta(\omega+1)$$