Lets say I have a cosine function

$f(t) = Acos(\omega_0t) $.

I find the Fourier transform $\mathfrak{F}[f(t)] = \frac{A}{2}\left [ \delta (\omega - \omega_0 ) + \delta (\omega + \omega_0 )\right ]$ which is two delta spikes symmetric about zero at $\omega_0$ and $-\omega_0$. I think this is correct.

Now, if I phase shift $f(t) $ to get

$f_s(t) = Acos(\omega_0t+\theta)$

my intuition tells me that the $\mathfrak{F}[f_s(t)] = \mathfrak{F}[f(t)]$ because the phase shift should not impact the frequency content of the signal.

However, when I do the work, I get the following:

$f_s(t)=\frac{A}{2}\left [ e^{i(\omega_0t+\theta)}+e^{-i(\omega_0t+\theta)}\right ]$ (Euler's formula)


$\mathfrak{F}[f_s(t)]=\frac{A}{2}\left [ \int_{-\infty}^\infty e^{i(\omega_0t+\theta)}e^{-i\omega t}dt + \int_{-\infty}^\infty e^{-i(\omega_0t+\theta)}e^{-i\omega t}dt\right ]$

$\, \,\,\,\,\,\,\,\,\,\,=\frac{A}{2}e^{i\theta}\int_{-\infty}^\infty e^{i(\omega_0-\omega)t}dt+\frac{A}{2}e^{-i\theta}\int_{-\infty}^\infty e^{-i(\omega_0+\omega)t}dt$

$\, \,\,\,\,\,\,\,\,\,\,=Acos(\theta)\left [\delta(\omega+\omega_0) + \delta(\omega-\omega_0) \right ]$

And thus,

$\mathfrak{F}[f_s(t)] \neq \mathfrak{F}[f(t)] $ which goes against my intuition.

Is the math wrong or is my intuition wrong?

  • $\begingroup$ Unfortunately, both the intuition and the analysis have flaws. Please see the posted solution and please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Oct 1 '15 at 16:39

If we define the Foruier Transform of $f$ to be

$$\mathscr{F}\left(f(t)\right)(\omega)=\frac{1}{2\pi}\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt$$

then the Foruier Transform of $f(t+t_0)$ is

$$\begin{align} \mathscr{F}\left(f(t+t_0)\right)(\omega)&=\frac{1}{2\pi}\int_{-\infty}^\infty f(t+t_0)e^{i\omega t}\,dt\\\\ &=e^{-i\omega t_0}\frac{1}{2\pi}\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt\\\\ &=\bbox[5px,border:2px solid #C0A000]{e^{-i\omega t_0}\mathscr{F}\left(f(t)\right)(\omega)} \tag 1 \end{align}$$

This general result shows that a time-shift transforms to a phase multiplication.

Now, if $f(t)=A\cos \omega_0t$, then its Fourier Transform is given by

$$\begin{align} \mathscr{F}\left(A\cos \omega_0t\right)&=\frac{1}{2\pi}\int_{-\infty}^\infty A\cos (\omega_0t)\,e^{i\omega t}\,dt\\\\ &=\frac{1}{2\pi}\int_{-\infty}^\infty A\left(\frac{e^{i\omega_0t}+e^{-i\omega_0t}}{2}\right)\,e^{i\omega t}\,dt\\\\ &=\frac A2\left(\frac{1}{2\pi}\int_{-\infty}^\infty e^{i(\omega+\omega_0)t}\,dt+\frac{1}{2\pi}\int_{-\infty}^\infty e^{i(\omega-\omega_0)t}\,dt\right)\\\\ &=\frac A2\left(\frac{1}{2\pi}2\pi \delta(\omega+\omega_0)+\frac{1}{2\pi}2\pi \delta(\omega-\omega_0)\right)\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac A2\left(\delta(\omega+\omega_0)+ \delta(\omega-\omega_0)\right)} \tag 2\\\\ \end{align}$$

Finally, if $f(t)=A\cos (\omega_0 t+\theta)=A\cos (\omega_0 (t+t_0))$, where $t_0=\theta /\omega_0$, then we see that this added phase is tantamount to a time shift. Using $(1)$ and $(2)$ reveals that

$$\begin{align} \mathscr{F}\left(A\cos (\omega_0t+\theta)\right)&=\frac A2e^{-i\theta \omega/\omega_0}\left(\delta(\omega+\omega_0)+ \delta(\omega-\omega_0)\right)\\\\ &=\frac A2\left(e^{i\theta}\delta(\omega+\omega_0)+ e^{-i\theta}\delta(\omega-\omega_0)\right)\\\\ \end{align}$$

$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\left(A\cos (\omega_0t+\theta)\right)=\frac A2\left(e^{i\theta}\delta(\omega+\omega_0)+ e^{-i\theta}\delta(\omega-\omega_0)\right)}$$

  • $\begingroup$ Thanks for the detailed response. What is the physical significance of the $e^{i\theta}$ term being multiplied by the infinite delta spike? Also, why does the $\frac{\omega}{\omega_0}$ term disappear in your final line? $\endgroup$ – Darcy Oct 1 '15 at 18:20
  • $\begingroup$ You're welcome. My pleasure. The reason that the $\omega/\omega_0$ terms simplify is due to the "sifting" property of the Dirac Delta. In general, we have $$f(x)\delta(x-a)=f(a)\delta(x-a)$$Here, we have $x=\omega$, $a=\pm \omega_0$, and $f(x)=e^{-i\theta x/a}$. $\endgroup$ – Mark Viola Oct 1 '15 at 18:46
  • $\begingroup$ @MarkViola: I think it should be ${\mathscr{F}\left(A\cos (\omega_0t+\theta)\right)=\frac A2\left(e^{i\theta}\delta(\omega-\omega_0)+ e^{-i\theta}\delta(\omega+\omega_0)\right)}$ $\endgroup$ – Meet Jan 31 at 17:15
  • $\begingroup$ @Meet And why do you believe that to be the case? It is not, in fact, the case. The posted solution is correct given the definition of the Fourier Transform as defined herein. $\endgroup$ – Mark Viola Jan 31 at 17:33
  • $\begingroup$ Using the Euler's Identity \begin{align} Acos(w_0t+\theta)&=\frac{A}{2}\bigg[e^{j(w_0t+\theta)}+e^{-j(w_0t+\theta)}\bigg]\\ &=\frac{A}{2}e^{j\theta}e^{jw_0t}+\frac{A}{2}e^{-j\theta}e^{-jw_0t} \end{align} So on taking Fourier transform \begin{align} \mathscr{F}[cos(w_0t+\theta)]&=\frac{A}{2}e^{j\theta}\mathscr{F}[e^{jw_0t}]+\frac{A}{2}e^{-j\theta}\mathscr{F}[e^{-jw_0t}]\\ &=A\pi e^{j\theta}\delta(w-w_0)+A\pi e^{-j\theta}\delta(w+w_0)\\ &=A\pi\bigg\{e^{j\theta}\delta(w-w_0)+e^{-j\theta}\delta(w+w_0)\bigg\} \end{align} Do correct me if I'm making mistake. $\endgroup$ – Meet Jan 31 at 17:58

That is correct. If you want another way of looking at it you can use the fact that cos(a+b)= cos(a)cos(b)-sin(a)sin(b).

However, I think a more interesting thing to note is that if you look at the average value of $\mathfrak{F}[f_s]$ (and by average value I mean the root mean square), then you end up with $\mathfrak{F}[f].


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