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I have the following signal:

enter image description here

and this as the solution to the problem:

$$\begin{align} X(j\omega) &= \int_{-\infty}^{+\infty}x(t)e^{-j\omega t}dt \\ &= \int_0^\tau \frac{E}{\tau}t\; e^{-j\omega t}dt + \int_\tau^T E \; e^{-j\omega t}dt \\ &= \frac{jE}{\omega}\left[e^{-j\omega T}-e^{\dfrac{-j\omega \tau}{2}} \mathrm{sinc} \frac{\omega \tau}{2} \right] \end{align}$$

I did the following, but it doesn't seem I'm getting near to their result.

$$\begin{align} \frac{E}{\tau}& \left[\left.-jt\omega\,e^{-j\omega t} \; \right|_0^\tau-\int_0^\tau e^{-j\omega t}dt + \int_\tau^T e^{-j\omega t}dt \right]\\ &=\frac{E}{\tau} -j\omega\tau e^{-j\omega t}-\left.\dfrac{e^{-j\omega t}}{-j\omega} \right|_0^\tau + E\left.\dfrac{e^{-j\omega t}}{-j\omega}\right|_\tau^T \\ &=-j\omega E e^{-j\omega t}+ \dfrac{e^{-j\omega t}}{j\omega}-\dfrac{E}{j\omega t}+E\dfrac{e^{-j\omega T}}{-j\omega}-E\dfrac{e^{-j\omega \tau}}{-j\omega} \end{align}$$

What am I doing wrong?


Could've bet my life that integration by parts formula looked like this:

$$\require{cancel}\cancel{\int u(x)v(x)dx = u(x)v'(x) - \int u'(x)v(x)dx} $$

I would've never guessed I would forget something like this. I saw that I'm prone to integrate instead of derive when applying the formula I thought I knew, but I thought it's just the time that passed since I did integrals. I mean I was doing like above or something like this:

$$\int f(x)g(x)dx = F(x)g(x)-\int f(x)g'(x)dx$$

Please pardon my mistake. Greetings!

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  • $\begingroup$ @Downvoter, care to comment? $\endgroup$ – Tanatos Daniel Aug 17 '14 at 21:04
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    $\begingroup$ Perhaps you should read up on integration by parts. How did you get $t(-j\omega)e^{-j\omega t}$? "First function times derivative of the first minus integral of (derivative of first function) times integral of the second)?" $\endgroup$ – Dilip Sarwate Aug 17 '14 at 22:19
  • $\begingroup$ I guess I got that by doing this: Second function times derivative of the first minus integral of (derivative of first function) times integral of the second)? $\endgroup$ – Tanatos Daniel Aug 17 '14 at 23:07
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\begin{align*} X\left(j\omega \right) &= \int_0^{\tau}\frac{E}{\tau}\cdot t \cdot e^{-j\omega t}dt + \int_{\tau}^T E\cdot e^{-j\omega t}dt\\ &=\frac{E}{\tau} \left[ \frac{-t}{j\omega} e^{-j\omega t}\right]_0^{\tau} - \frac{E}{\tau}\int_0^{\tau}\left( \frac{-1}{j\omega}e^{-j\omega t}\right)dt + E \left[ \frac{e^{-j\omega t}}{-j\omega} \right]_{\tau}^T\\ &=\frac{jE}{\omega}e^{-j\omega \tau} + \frac{E}{\omega^2 \tau}\left( e^{-j\omega \tau} - 1\right) + \frac{jE}{\omega}\left( e^{-j\omega T} - e^{-j\omega \tau}\right)\\ &=\frac{jE}{\omega}\left[ e^{-j\omega T} + e^{-j\frac{\omega \tau}{2}}\left(\frac{e^{-j\frac{\omega \tau}{2}} - e^{j\frac{\omega \tau}{2}}}{2j\frac{\omega\tau}{2}}\right)\right]\\ \Rightarrow X\left(j\omega \right) &= \frac{jE}{\omega} \left(e^{-j\omega T} - e^{-j\frac{\omega \tau}{2}} \text{sinc}\left( \frac{\omega \tau}{2}\right) \right) \end{align*}

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    $\begingroup$ Thank you. I guess someone downvoted because you provided the complete solution. That's because of me and I'm sorry. I'll vote up when I'll have enough points. $\endgroup$ – Tanatos Daniel Aug 20 '14 at 1:26
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Hint:

Your integration by parts is wrong. The second integral looks OK. The first integral should be solved like this (leaving out constant factors):

$$\int_0^{\tau}te^{-j\omega t}dt=\frac{te^{-j\omega t}}{-j\omega}\bigg|_0^{\tau}+\frac{1}{j\omega}\int_0^{\tau}e^{-j\omega t}dt$$

I'm sure you can take it from here.

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  • $\begingroup$ Thanks again. Sorry I did not recap before starting to solve the exercises. I thought I remembered them way better. $\endgroup$ – Tanatos Daniel Aug 17 '14 at 23:03

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