I need to compute the Fourier Transform of $x(t) = \cos(t)\sin(t)$.

We know that $\delta(w) = \mathcal{F}[1](w) = \int_{-\infty}^{\infty}e^{-jwt}dt \\$.

And $\mathcal{F}[x(t)](w) = \mathcal{F}[\cos(t)\sin(t)](w) = \frac{1}{2\pi}\mathcal{F}[\cos(t)](w) \circledast \mathcal{F}[\sin(t)](w)$.

So I computed : \begin{split} \mathcal{F}[\cos(t)](w) &= \int_{-\infty}^{\infty}\cos(t)e^{-jwt}dt \\ &= \int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{-jwt}dt \\ &= \frac{1}{2} \Big ( \int_{-\infty}^{\infty}e^{jt}e^{-jwt}dt + \int_{-\infty}^{\infty}e^{-jt}e^{-jwt}dt \Big ) \\ &= \frac{1}{2} \Big ( \int_{-\infty}^{\infty}e^{-jt(w-1)}dt + \int_{-\infty}^{\infty}e^{-jt(w+1)}dt \Big ) \\ &= \frac{1}{2} \big ( \delta(w-1)+\delta(w+1) \big ) \end{split}

and \begin{split} \mathcal{F}[\sin(t)](w) &= \int_{-\infty}^{\infty}\sin(t)e^{-jwt}dt \\ &= \int_{-\infty}^{\infty}\frac{1}{2}j(e^{-jt}-e^{jt})e^{-jwt}dt \\ &= \frac{1}{2}j \Big ( \int_{-\infty}^{\infty}e^{-jt}e^{-jwt}dt - \int_{-\infty}^{\infty}e^{jt}e^{-jwt}dt \Big ) \\ &= \frac{1}{2}j \Big ( \int_{-\infty}^{\infty}e^{-jt(w+1)}dt - \int_{-\infty}^{\infty}e^{-jt(w-1)}dt \Big ) \\ &= \frac{1}{2}j \big ( \delta(w+1)-\delta(w-1) \big ) \end{split}

I finally computed the convolution :

$\begin{split} \mathcal{F}[x(t)](w) = \mathcal{F}[\cos(t)\sin(t)](w) &= \frac{1}{2\pi}\mathcal{F}[\cos(t)](w) \circledast \mathcal{F}[\sin(t)](w) \\ &= \frac{1}{2\pi}\frac{1}{2}\frac{1}{2}j \big ( \delta(w-1)+\delta(w+1) \big ) \circledast \big ( \delta(w+1)-\delta(w-1) \big ) \\ &= \frac{1}{8\pi}j \big ( \delta(w-1)+\delta(w+1) \big ) \circledast \big ( \delta(w+1)-\delta(w-1) \big ) \\ &= \frac{1}{8\pi}j \big ( \delta(w+2)-\delta(w-2) \big ) \end{split}$

Using Matlab and wolframalpha to verify this, I ended up on :

Fourier cos sin

The result of my fourier transform which is the result of the convolution gave me $$\frac{1}{8\pi}j \big ( \delta(w+2)-\delta(w-2) \big )$$

But I don't have any real part anymore.. and my imaginary part is not like on the plot. Using wolframalpha, I should have something like : $$ aj\big (-\delta(w+2)+\delta(w-2) \big )$$ which is correct according to the imaginary plot.

Where is my mistake ? And how can I compute analytically the phase using $\arctan$ if I don't have any real part anymore ? How can I compute analytically the magnitude too ? Wolframalpha's answer doesn't have a real part too, how can I have one in my plots ?

Thanks for the clarifications !

  • $\begingroup$ I am not completely sure where you went wrong, if you did so, but using $\sin(t)\cos(t)=\frac12\sin(2t)$ we can directly conclude that, following your notation, that $$\mathcal F[\sin(t)\cos(t)](\omega)=\mathcal F\left[\frac12\sin(2t)\right](\omega)=\frac1{4j}(\delta(\omega-2)-\delta(\omega+2))$$ Which agrees with WolframAlpha up to a constant factor. $\endgroup$ – mrtaurho Mar 13 at 19:04
  • $\begingroup$ @mrtaurho $\mathscr{F}\{1\}=2\pi \delta(\omega)$. $\endgroup$ – Mark Viola Mar 13 at 19:09
  • $\begingroup$ @MarkViola I was not sure which factor to use here. I am getting always kind of confused regarding all these different normalisations of the Fourier Transform which are used out there. $\endgroup$ – mrtaurho Mar 13 at 19:11

First, $\mathscr{F}\{1\}=2\pi \delta(\omega)$.

Next, note that $f(t)=\cos(t)\sin(t)=\frac12\sin(2t)=\frac{e^{i2t}-e^{-i2t}}{i4}$.

Hence, we see that

$$\mathscr{F}\{f\}=\frac1{i4}\int_{-\infty}^\infty \left(e^{-i(\omega-2)t}-e^{-i(\omega+2)t}\right)\,dt=\frac{\pi}{i2}\left(\delta(\omega-2)-\delta(\omega+2)\right)\tag1$$

One can take the inverse transform, $\frac1{2\pi}\int_{-\infty}^\infty \frac{\pi}{i2}\left(\delta(\omega-2)-\delta(\omega+2)\right)\,e^{i\omega t}\,d\omega$, of the right-hand side of $(1)$ to confirm that $(1)$ is indeed the Fourier transform of $f$.

  • $\begingroup$ Thanks ! It was pretty easy finally .. And I don't know about $\mathscr{F}\{1\}=2\pi \delta(\omega)$ because I have my $\frac{1}{2\pi}$ afterwards, but not really important I guess. How can I compute the magnitude and phase from here ? Where is the real part ? $\endgroup$ – Romain B. Mar 13 at 19:13
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Mar 13 at 19:14
  • $\begingroup$ (I edited my comment, I have few more questions) $\endgroup$ – Romain B. Mar 13 at 19:15
  • $\begingroup$ @RomainB. The Dirac Delta is not a function, it is a distribution (aka, a Generalized Function). As such, it cannot be plotted as a function. You could draw a "pseudo-plot" of the magnitude by drawing (infinite) vertical spikes at $\omega=\pm2$ with Dirac Delta "amplitudes/weights" of $\pi/2$. But what is the point of that? $\endgroup$ – Mark Viola Mar 13 at 19:18
  • $\begingroup$ Thanks a lot ! Last question, what is the real part of it, if any ? Because I want to compute analytically the phase and magnitude. Thanks again for the clarifications :) $\endgroup$ – Romain B. Mar 13 at 21:32

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