# Finding $A_k$ coefficients

I was able to demonstrate that for a signal $$x(t)$$ real, we can write the truncated Fourier series as:

$$x_N (t) = A_0 +\sum\limits_{k=1}^{N}A_k\cos(kω_0t + \varphi_k)$$, but now I've been given the signal shown below:

And now I have to find the values of $$A_0, A_1, A_2, A_3, A_4, A_5, A_6, A_7, \varphi_1, \varphi_2, \varphi_3, \varphi_4,\varphi_5, \varphi_6, \varphi_7$$

First attempt:

I know $$A_0$$ is equal to the area of the graph during one period divided by the value of the period. I know $$T_0=0.4$$ s, so $$A_0=\frac{0.3*2}{0.4}=1.5$$ and I wrote $$x(t)$$ between 0 and 0.4 as:

$$x(t)= \begin{cases} 0& 0

I also calculated $$\omega_0=\frac{2\pi}{T_0}=\frac{2\pi}{0.4}=5\pi$$, but then I tried calculating the $$a_k$$ coefficients using the integral definition: $$a_k=\frac{1}{2} \int_{0.1}^{0.2} 20t\ e^{-jk\omega_0t} \,dt\ +\ \frac{1}{2} \int_{0.2}^{0.4} (-10t+8)\ e^{-jk\omega_0t} \,dt$$ but my teacher said the goal wasn't to calculate the integrals and that there was a much easier way to find the values of $$A_0, A_1, A_2, A_3, A_4, A_5, A_6, A_7, \varphi_1, \varphi_2, \varphi_3, \varphi_4,\varphi_5, \varphi_6, \varphi_7$$, but I just don't know where to go from here.

Second attempt:

I know that:

$$\frac{dx}{dt}(t) \rightarrow\ j\omega X(j\omega)$$

And that:

$$x'(t)= \begin{cases} 0& 0

So, if $$x(t)=\frac{1}{2\pi} \int_{-\infty}^{+\infty} X(j\omega)\ e^{j\omega t} \,d\omega$$, then $$x'(t)=\frac{1}{2\pi} X(j\omega)\ e^{j\omega t}$$

So, I find $$X(j\omega)=\frac{x'(t)2\pi}{e^{j\omega t}}$$

Because $$T\cdot a_k=X(jk\omega_0)$$ I assumed that I could write $$a_k$$ as:

$$a_k=\frac{X(jk\omega_0)}{T}=\frac{x'(t)2\pi}{e^{jk\omega_0 t}}$$

$$a_k= \begin{cases} 0& 0

But now I'm stuck again.

• Hint: What is the derivative of $x(t)$, and is it easier to find the Fourier series for the derivative? If so, what is the relationship between the Fourier series of the derivative and the Fourier series of the derivative? Dec 16, 2022 at 14:19
• You have a more basic problem in that you don't understand that Fourier series coefficients are constants rather than functions of time as you find in $$a_k=\frac{x^\prime(t)2\pi}{e^{j\omega t}}.$$ Dec 17, 2022 at 3:42