enter image description hereI have to calculate the Fourier coefficients of this signal. I found that signal equation is $$ y = \frac {A(2t-T)}{T} $$

To find Fourier coefficients I wrote

$$ x_k = \frac{2A}{T} \int_{0}^{T/2} \frac{2t-T}{T} e^{-i2 \pi k f_0’ t } $$

In this case $$ f_0’ = f_0 $$ because the period is T_0

I calculate the integration by parts of the first integral and I obtained $$ - \frac{-e^{-i \pi k }( i \pi k ) +1 - e^{-i \pi k }}{4 (\pi k f_0 )^ 2} $$

( i control This with wolfram and should be correct, I hope !)

After solving the integral by parts I obtained that

$$ x_k = \frac{A( e^{-i \pi k} + 1 - \pi i k )} { (\pi k )^2} $$ but the result should be $$ \frac{iA}{k \pi} $$

I can also post all my resolution. Can someone help me ? Thank you

  • 1
    $\begingroup$ why do you integrate from 0 to T/2. Shouldn't you integrate from 0 to T ? $\endgroup$
    – Ben
    Commented Jan 6, 2020 at 16:12
  • 2
    $\begingroup$ no, @Ben is right. $\endgroup$ Commented Jan 6, 2020 at 16:57

4 Answers 4


This is one of the best problems to demonstrate Fourier Series properties, and specifically the time derivative property: $$\frac{d}{dt}x(t) \overset{FS}\longleftrightarrow j2\pi kf_0 X_k$$

Instead of computing the integral $$X_k=\frac{1}{T_0}\int_{T_0}x(t)e^{-j2\pi kf_0t}dt$$ which is time consuming, you can take the first derivative of your signal and compute the Fourier coefficients of the derivative. At discontinuities, you will have Delta functions, and everywhere else you will have a constant equal to the slope of your $y$ signal, which is $2A/T_0$. In other words, inside the time interval $[0,T_0)$, your derivative can be expressed as $$\frac{d}{dt}x_{T_0}(t) = \Bigg\{\begin{array}{ll}\frac{2A}{T_0}, & t \in (0,T_0) \\ -2A, & t=0\end{array} = \frac{2A}{T_0}\mathrm{rect}\Big(\frac{t-\frac{T_0}{2}}{T_0}\Big)-2A\delta(t)$$ I will not use the last equality for integration, but you now see it is much easier to integrate $\frac{d}{dt}x_{T_0}(t)$. Let's say $z(t)=\frac{d}{dt}x(t)$, so its Fourier coefficients would be $$\begin{align} Z_k &= \frac{1}{T_0}\int_0^{T_0}\frac{2A}{T_0}e^{-j2\pi kf_0t}dt -\frac{2A}{T_0}\int_{0}^{T_0}\delta(t)e^{-j2\pi kf_0t}dt \\ &= \frac{2A}{T_0^2}\frac{1}{(-j2\pi kf_0)}e^{-j2\pi kf_0t}\Big|_{0}^{T_0} - \frac{2A}{T_0}e^{-j2\pi kf_0t}\Big|_{t=0} \\ &= \frac{2A}{-j2\pi kf_0T_0^2}(e^{-j2\pi kf_0T_0}-1)-\frac{2A}{T_0} \\ &= 0 - \frac{2A}{T_0} = -\frac{2A}{T_0} \end{align}$$ Can you apply the time derivative property to find your $X_k$ coefficients?

P.S: If you are not aware of the Delta function and its properties, or even the properties of Fourier Series, then just skip my answer. :)


Not a full solution but a few hints:

  • Note that your function has an odd symmetry. Hence the even Fourier coefficients should be zero. They don't seem to be in your case and I think that's a consequence of how you integrate: you integrate from 0 to $T/2$ instead of $-T/2$ to $T/2$. If you did both halves, they would turn out with opposing signs and then cancel.
  • For odd $k$, notice that $e^{-i \pi k} = {-1}^k = -1$. And hence you're already very close, only off by a minus sign. Try integrating from $-T/2$ to $T/2$ instead.

Your problem is at this step. You're doing half of a shortcut, but not the other half:

$$ x_k = \frac{2A}{T} \int_{0}^{T/2} \frac{2t-T}{T} e^{-i2 \pi k f_0’ t } $$

Start with the Fourier series definition (with the notation tidied up): $$ x_k = \frac{A}{T_0} \int_{-T_0/2}^{T_0/2} \frac{2t-T_0}{T_0} e^{-i2 \pi \frac{k}{T_0} t } dt$$

What you did with this was that while you recognized that you only needed to do the integration from $0$ to $T_0/2$, you failed to realize that you needed to modify it at the same time. Specifically, you have to note that the function is odd about $t = 0$, which means that the odd-symmetric part of $e^{i x}$ must be eliminated. You need to first observe that

$$ x_k = \frac{A}{T_0} \int_{-T_0/2}^{T_0/2} \frac{2t-T_0}{T_0} e^{-i2 \pi \frac{k}{T_0} t } dt = \frac{A}{T_0} \int_{-T_0/2}^{T_0/2} \frac{2t-T_0}{T_0} \left(-i \sin \left(2 \pi \frac{k}{T_0} t\right) \right) dt $$

because the function is odd around $t = 0$. Then you can observe that the argument to the integral is even-symmetrical around $t = 0$, which allows you to take the shortcut

$$ x_k = \frac{2A}{T_0} \int_{0}^{T_0/2} \frac{2t-T_0}{T_0} \left(-i \sin \left(2 \pi \frac{k}{T_0} t\right) \right) dt $$

Try that. See if it helps.

  • $\begingroup$ I solve it from 0 to T_0 ( that should be the same of to $$ \frac{-T_0}{2} to \frac{T_0}{2} $$ now I obtained $$ \frac{-2i \pi k e^{-2i \pi k } - e^{-2i \pi k } + 1}{(2i \pi k f)^2} $$ as result of integral by parts. Using this to find xk I obtained $$ \frac{ A ( i \pi k e^{-2i \pi k } + e^{-2i \pi k } - 1 + i \pi k } {(\pi k )^2 } $$ after I wrote e^ as cos + isen and I obtained $$ \frac{2A}{i\pi } $$ $\endgroup$ Commented Jan 8, 2020 at 7:20

For anyone else visiting this question trying to solve for coefficients, I just had something complex like:

$$ \begin{align} a_k &= \frac{1}{T_0} \int^{T_0}_0 x(t) e^{-j2\pi f_0 k t}\,\, dt\\ &= \frac{1}{8} \int^8_0 x(t) e^{-j2\pi f_0 k t}\,\, dt\\ &= -\dfrac{16\left(\left({\pi}k-\mathrm{i}\right)\sin\left(2{\pi}k\right)+\left(\mathrm{i}{\pi}k+1\right)\cos\left(2{\pi}k\right)+\mathrm{i}{\pi}k-1\right)}{{\pi}^2k^2} \end{align} $$

But as $k \in \mathbb{Z}$ and $y(t) = \sum^\infty_{k=-\infty}{a_k e^{j2 \pi f_0 k t}}$, I could simplify $\sin\left(2{\pi}k\right) = 0$ and $\cos\left(2{\pi}k\right) = 1$ which boiled down to:

$$ a_k = -\frac{4i}{\pi k} $$


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