# A question about Fourier transformation

Hello, this is my first time actually asking in stackexchange. I am a computer engineering student and currently i am doing a linear system course (i don't really know how this is equivalent in engineering course labeling as my major is apparently in faculty of computer science but it's related a bit toward signal processing), i hope i wasn't in the wrong category

So I was to find the fourier transformation of the graph in the picture, and i need to plot the graph in matlab as well. I was allowed to look into the table but I couldn't find it hence I used brute force(integrals) to find the $$X(\omega)$$ of the signal, which i ended up finding but still there's $$e^{-\omega.j}$$ stuck which confuses me as i believe the plot should be 2 dimensional only (toward x and $$\omega$$, not toward j). can somebody help finding any faults/things i missed here?

• Hello and welcome to StackExchange! Your question fits here well enough though asking on Math.Stackexchange would have been an alternative. I'd highly recommend you to check out the MathJax tutorial that will show you how you can typeset equations in your question. This will make it much easier for us to read your question. – Florian Jul 4 '19 at 10:57
• thanks @Florian, I'm slowly editing my questions for now – Nobel Edgar Jul 4 '19 at 11:04

Regarding your integral, it is not quite correct I think. You used partial integration to solve it, which works. Let's consider $$\int t {\rm e}^{-\jmath \omega t}{\rm d}t$$ and call $$f(t) = t$$ and $$g'(t) = {\rm e}^{-\jmath \omega t}{\rm d}t$$ so that $$g(t) = -\frac{1}{\jmath \omega}{\rm e}^{-\jmath \omega t}$$. Then using $$\int f(t) g'(t){\rm d}t = f(t)g(t)-\int f'(t) g(t) {\rm d}t,$$ we obtain $$\int_0^1 t {\rm e}^{-\jmath \omega t}{\rm d}t = \left[-\frac{t}{\jmath \omega}{\rm e}^{-\jmath \omega t}\right]_0^1 + \frac{1}{\jmath \omega}\int_0^1{\rm e}^{-\jmath \omega t} {\rm d}t.$$ It's close to what you did, but you used $$g'(t)$$ in the first term instead of $$g(t)$$.
Note that $$j$$ is not a variable! It's the imaginary unit $$\sqrt{-1}$$. Hence, your expression only depends on one variable, which is $$\omega$$. Matlab will recognize $$j$$ as the imaginary unit by default, so you should be all set for plotting it.
• oh yeah i did a calculation mistake, thanks for pointing it out. i do understand that j is $\sqrt{-1}$ but i thought that j is outside the graphs as there's a specific plane which maps $Re$ and $j$ implicitly. Thanks for your answer – Nobel Edgar Jul 4 '19 at 11:11
• Well, your spectrum will be complex, i.e, for every frequency $\omega$, you will obtain a complex number. You could plot this on two axis for real and imaginary part. It is much more common though to plot the magnitude $|X(w)|$ which is then $\mathbb{R} \rightarrow \mathbb{R}$ so it can be displayed as an ordinary function graph. This does then not show the phase information, which can be plotted separately. See Bode plot for reference. – Florian Jul 4 '19 at 11:35