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.

  • $\begingroup$ A pair of $\delta()$ functions multiplied by a function is a pair of weighted $\delta()$ functions, not a line. $\endgroup$ – Andy Walls Apr 17 '19 at 23:24
  • $\begingroup$ Oh wow, pretty dumb mistake on my part. $\endgroup$ – Taha Apr 17 '19 at 23:59

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

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

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