# Fourier coefficients of odd and even part of a signal

I have this signal and I have to find the Fourier coefficients of the odd and even part. First I found that $$x_p(t) = \frac{1}{2} ( x(t) + x(-t) )$$ and I made the graphic of this part and I obtained a straight line in 1/2. I made the same thing for $$x_d (t) = \frac{1}{2} ( x(t) - x(-t) )$$ and making the graphic I obtained the signal $$\mathrm{rect}(\frac{ t- n \frac{T_0}{4} }{ \frac{T_0}{2} }) - \frac{1}{2}$$. Now $$X_{pk} = \frac{1}{T_0} \int_{0}^{T_0} \frac{1}{2} e^{-j 2 \pi k f_0 t } dt$$ and I obtained that if $$k=1$$ $$X_{pk} = 0$$ and if $$k=0$$ $$X_{pk} = \frac{1}{2}$$. This result is the same of my book. For the odd part I should do $$X_{dk} = \int_{0}^{T_0}( \mathrm{rect}(\frac{ t- n \frac{T_0}{4} }{ \frac{T_0}{2} }) - \frac{1}{2})e^{-j 2 \pi k f_0 t }$$ and I obtain $$0$$ if $$k=1$$ and $$1$$ if $$k=0$$ but the result should be $$0$$ if $$k=0$$ and $$\frac{-j}{k \pi }$$ if $$k = 1$$.

• What exactly is $\prod$? A rectangular pulse? – GKH Jan 17 at 22:39
• Yes :) it’s a rectangular pulse , I’m not so good with latex 🙈 – Elena Martini Jan 17 at 23:16
• Can you make a sketch of the periodic pulse? What is the relationship between $T$ and the period, $T_0$? – GKH Jan 17 at 23:24
• I uploaded a sketch of the original periodic signal. Looking at the graph the period shouldn’t be the part of the graph that repeats over time ? In this case the total period should be T0 – Elena Martini Jan 17 at 23:32
• why is it "$x_p$" for for even and "$x_d$" for odd? why "p" and "d" for "e" and "o"? – robert bristow-johnson Jan 18 at 7:14

OK, so your signal is described in one period, $$T_0$$, as $$x_{T_0}(t) = \left\{\begin{array}{ll} 0, & -T_0/2 < t \leq 0 \\ 1, & 0 < t \leq T_0/2 \end{array}\right.$$ Let's keep it that way (without any other shortcuts like $$\prod$$ or $$\mathrm{rect(\cdot)}$$ etc). One way to find the Fourier coefficients of the even and the odd part of a signal is to actually find the even and odd part of the signal, and do a Fourier Series analysis on each.

But, another way is to check Fourier Series properties - again :-) .

If $$x(t)$$ is real - which it is - with Fourier coefficients $$X_k$$, then the odd part of $$x(t)$$, let it be $$x_{odd}(t)$$, has Fourier coefficients $$X^{odd}_k = j\Im\{X_k\}$$, where $$\Im\{X_k\}$$ is the imaginary part of the Fourier coefficients of the original signal $$x(t)$$.

Similarly, the even part of the signal $$x(t)$$, let it be $$x_{ev}(t)$$, has Fourier coefficients $$X^{ev}_k = \Re\{X_k\}$$, where $$\Re\{X_k\}$$ is the real part of the original Fourier coefficients $$X_k$$.

In short: $$x(t) \longleftrightarrow X_k$$ $$x_{odd}(t) \longleftrightarrow j\Im\{X_k\}$$ $$x_{even}(t) \longleftrightarrow \Re\{X_k\}$$

So, you have two options:

1. Find $$x_{odd}(t)$$, $$x_{ev}(t)$$ and find their Fourier coefficients analytically (takes a lot of time and paper)
2. Find the Fourier coefficients, $$X_k$$, of $$x(t)$$ and then take their real and imaginary part (faster and easier)

I would go for the second option, since $$x(t)$$ is very simple and its Fourier coefficients are given by $$X_k = \frac{1}{T_0}\int_0^{T_0}x_{T_0}(t)e^{-j2\pi kf_0t}dt = \frac{1}{T_0}\int_0^{T_0/2}1\cdot e^{-j2\pi kf_0t}dt$$

P.S: In case you wish to go with the first option, the even part of the signal can be computed by first finding $$x_{T_0}(-t) = \left\{\begin{array}{ll} 1, & -T_0/2 \leq t \leq 0 \\ 0, & -T_0 < t < -T_0/2 \end{array}\right.$$ and then $$x_{even, T_0}(t) = \frac{1}{2}(x_{T_0}(t) + x_{T_0}(-t)) = \frac{1}{2}$$

Similarly, the odd part can be obtained as $$x_{odd,T_0}(t) = \frac{1}{2}(x_{T_0}(t)-x_{T_0}(-t)) = \left\{\begin{array}{ll} -\frac{1}{2}, & -T_0/2

So the even part is just the mean value of the signal, $$X_0$$. The odd part of the signal has Fourier coefficients $$X_k = -\frac{1}{T_0}\int_{-T_0/2}^0 \frac{1}{2}e^{-j2\pi kf_0t}dt + \frac{1}{T_0}\int_0^{T_0/2}\frac{1}{2}e^{-j2\pi kf_0t}dt$$

I think you can continue from here. Let me know if you make it till the end. :)

• My exercise explicitly mentions to find Fourier coefficients as the sum of the coefficients of the odd part + the coefficients of the even part. For exercise I already found Fourier coefficients with the second method and the results are right. But I didn’t obtain the right result with the first method. I have the result of the Fourier coefficients and also the results of Fourier coefficients of odd and even part. With the even part I obtained the same result of my book but mine result of Fourier coefficients of the odd part is incorrect and I don’t know why – Elena Martini Jan 18 at 0:28
• Now I have included both ways of solving this. – GKH Jan 18 at 12:38
• Thank you so so much for the help !!!!!!!!!!!! I made it and know it’s correct 🥳🥳 – Elena Martini Jan 18 at 20:20