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I have this signal enter image description here 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 $$\mathrm{rect}(\frac{ t- n \frac{T_0}{4} }{ \frac{T_0}{2} }) - \frac{1}{2} $$. enter image description here 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 $k = 1$.

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    $\begingroup$ What exactly is $\prod$? A rectangular pulse? $\endgroup$ – GKH Jan 17 at 22:39
  • $\begingroup$ Yes :) it’s a rectangular pulse , I’m not so good with latex 🙈 $\endgroup$ – Elena Martini Jan 17 at 23:16
  • $\begingroup$ Can you make a sketch of the periodic pulse? What is the relationship between $T$ and the period, $T_0$? $\endgroup$ – GKH Jan 17 at 23:24
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    $\begingroup$ 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 $\endgroup$ – Elena Martini Jan 17 at 23:32
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    $\begingroup$ why is it "$x_p$" for for even and "$x_d$" for odd? why "p" and "d" for "e" and "o"? $\endgroup$ – robert bristow-johnson Jan 18 at 7:14
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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 $$x_{odd,T_0}(t) = \frac{1}{2}(x_{T_0}(t)-x_{T_0}(-t)) = \left\{\begin{array}{ll} -\frac{1}{2}, & -T_0/2 <t < 0 \\ \frac{1}{2}, & 0 \leq t \leq T_0/2 \end{array}\right.$$

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. :)

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  • $\begingroup$ 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 $\endgroup$ – Elena Martini Jan 18 at 0:28
  • $\begingroup$ Now I have included both ways of solving this. $\endgroup$ – GKH Jan 18 at 12:38
  • $\begingroup$ Thank you so so much for the help !!!!!!!!!!!! I made it and know it’s correct 🥳🥳 $\endgroup$ – Elena Martini Jan 18 at 20:20

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