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I am studying about Fourier Series.

\begin{align} x(t)&=\sum_{n=-\infty}^{\infty}X_ne^{j2\pi nf_0t}\\ X_n&=\frac1{T_0}\int_{T_0}x(t)e^{-j2\pi nf_0t}dt \end{align}

I understand the process eliciting the equations above.

Then, my book says

$$ \mbox{Assuming } x(t) \mbox{ is real,}\\ X_n^*=X_{-n}\mbox{ , where * is conjugate symbol.} $$

Using above equation, my book elicit the following equations. $$ x(t)=X_0+\sum_{n=1}^{\infty}2|X_n|cos(2\pi nf_0t+\angle X_n) $$

I do not know from where $X_n^*=X_{-n}$ is derived.

Can someone help me understand this?

Danny_Kim
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1 Answers1

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The basic fact here is that$\overline{\int f(x)dx}=\int\overline{f(x)}dx$. Take the expression for $X_n$ and take it's complex conjugate. Now we need to figure out how to simplify the expression.

1) $x(t)$ is real-valued, so it's complex conjugate is itself. Same with $\frac{1}{T_0}$

2) $\overline{e^{i\alpha}}=e^{-i\alpha}$

By applying these two rules, we find out that taking the complex conjugate only introduces a negative sign in the exponential. We can group this negative sign with the $-n$ to get $+n$, that is, $$\overline{X_n}=\frac{1}{T_0} \int x(t) e^{+ 2 \pi j n f_0 t} \, dt=\frac{1}{T_0} \int x(t) e^{-2 \pi j (-n) f_0 t} \, dt$$

Substituting $-n$ for $n$ gives the desired result.

Stella Biderman
  • 31,155