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!

  • @Downvoter, care to comment? – Tanatos Daniel Aug 17 '14 at 21:04
  • 1
    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)?" – Dilip Sarwate Aug 17 '14 at 22:19
  • 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)? – Tanatos Daniel Aug 17 '14 at 23:07
up vote 1 down vote accepted

\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*}

  • 1
    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. – Tanatos Daniel Aug 20 '14 at 1:26

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.

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

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.