I don't understand why the Fourier transoform of $$\delta(t-t_0)$$ is equal to $$e^{-jwt_0}$$ Starting from the beginning, the FT of dirac delta is defined as: $$ \int_{- \infty }^{ \infty }\delta (t - t_0)e^{-j\omega t}dt$$ I have no idea how to calculate integral of dirac delta, for the second term thei ntegral is: $$\int_{- \infty }^{ \infty }e^{-j\omega t}dt = \frac{e^{jwt}}{jw}$$ but this far from expected result.

So could somebody make a step by step Fourier Transofrm for diract delta?

Thanks in advance :)

  • 4
    $\begingroup$ Some care must be taken with distributions in general, but in this case it is straightforward. The general rule with impulse distributions and nice functions is that $\int \delta(t-t_0) f(t) dt = f(t_0)$. Just apply this to $f(t) = e^{-i \omega t}$. $\endgroup$
    – copper.hat
    Jun 24 '18 at 16:26
  • $\begingroup$ Why is it $$f(t_0)$$? $\endgroup$
    – DannyS
    Jun 24 '18 at 20:39
  • $\begingroup$ $\int \delta(t-t_0) f(t) dt = \delta(t) f(t-t_0) dt = f(0-t_0) = f(t_0)$. $\endgroup$
    – copper.hat
    Jun 24 '18 at 20:47
  • $\begingroup$ @copper.hat, okay but I don't get why you moved $t_0$ from delta function to $f(t)$ then $f(t-t_0)$ mysteriously turned out to be $f(0-f_t)$. Could you please explain why and what bases you perform such operations? $\endgroup$
    – DannyS
    Jun 24 '18 at 21:34
  • $\begingroup$ Try the substitution $s=t_0-t$. I have a minor mistake above. In any event, you need to do a little work yourself. $\endgroup$
    – copper.hat
    Jun 24 '18 at 21:50

The Dirac Delta "function" represents a distribution of functions that have an area equal to 1 with infinitely tall height and a correspondingly scaled width. Since we don't care about the derivatives of the delta function in this problem, we can represent the shifted delta function with this sequence of rectangular pulse functions, centered at $t_0$, as the pulse width $\tau \rightarrow 0$:

\begin{equation} p_{\tau}(t-t_0) = \begin{cases} \dfrac{1}{\tau}, & t_0 - \dfrac{\tau}{2}< t < t_0 + \dfrac{\tau}{2}.\\ \\ 0, & \text{otherwise}. \end{cases} \end{equation}

The rest is straightforward:

$$\begin{align*} \mathscr{F}\left\{\delta(t-t_0)\right\} &=\int_{- \infty }^{ \infty }\delta (t - t_0)e^{-j\omega t}dt \\ \\ &= \lim_{\tau \rightarrow 0}\space\int_{-\infty }^{\infty}p_{\tau}(t-t_0)e^{-j\omega t}dt \\ \\ &= \lim_{\tau \rightarrow 0}\space\int_{t_0-\frac{\tau}{2} }^{t_0+\frac{\tau}{2}}\dfrac{1}{\tau}e^{-j\omega t}dt \\ \\ &= \lim_{\tau \rightarrow 0}\space\dfrac{-1}{j\omega\tau}e^{-j\omega t}\biggr\rvert_{t_0-\frac{\tau}{2}}^{t_0+\frac{\tau}{2}} \\ \\ &= \lim_{\tau \rightarrow 0}\space\dfrac{-1}{j\omega\tau}\left[e^{-j\omega \left(t_0 +\frac{\tau}{2}\right)}-e^{-j\omega \left(t_0 -\frac{\tau}{2}\right)}\right] \\ \\ &= \lim_{\tau \rightarrow 0}\space\dfrac{-e^{-j\omega t_0}}{j\omega\tau}\left[e^{-j\omega\frac{\tau}{2}}-e^{j\omega\frac{\tau}{2}}\right] \\ \\ &= \lim_{\tau \rightarrow 0}\space\dfrac{e^{-j\omega t_0}}{\omega\frac{\tau}{2}}\left[\dfrac{e^{j\omega\frac{\tau}{2}}-e^{-j\omega\frac{\tau}{2}}}{2j}\right] \\ \\ &= \lim_{\tau \rightarrow 0}\space e^{-j\omega t_0}\dfrac{\sin\left(\omega\frac{\tau}{2}\right)}{\omega\frac{\tau}{2}} \\ \\ &= e^{-j\omega t_0} \end{align*}$$


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