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I want to ask Question about the Fourier series in continuous time domain. I am following signal and systems 2nd Edition by Alan Oppenheim. I have confusion in understanding the statement that Specifically, suppose that $x(t)$ is real and can be represent in the form 3.25. then since $x^*(t) = x(t)$, we obtain

$$x(t) = \sum^{+\infty}_{k\ =\ -\infty} a^*_k e^{-jk\omega_0t}$$

Then it means the equation 3.25 is for both Real and imaginary?
Equation 3.25

$$x(t) = \sum^{+\infty}_{k\ =\ -\infty} a_k e^{jk\omega_0t}$$

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Under certain conditions, a $T$-periodic function can be represented by its Fourier series

$$x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}\tag{1}$$

with $\omega_0=2\pi/T$. The function $x(t)$ can be complex-valued, i.e. have non-zero real and imaginary parts. Note that generally the Fourier coefficients $a_k$ are also complex-valued (even for real-valued $x(t)$).

Now if $x(t)$ is real-valued, i.e., $x(t)=x^*(t)$, you get

$$x(t)=x^*(t)=\sum_{k=-\infty}^{\infty}a^*_ke^{-jk\omega_0t}= \sum_{k=-\infty}^{\infty}a^*_{-k}e^{jk\omega_0t}\tag{2}$$

Comparing $(2)$ with $(1)$ you see that for real-valued $x(t)$ you get the condition $a_k=a^*_{-k}$, i.e. the Fourier coefficients show Hermitian symmetry. This condition is not satisfied for general complex-valued functions $x(t)$. However, the representation $(1)$ is valid no matter if $x(t)$ is real-valued or complex-valued.

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  • $\begingroup$ When we x(t) is real-valued i.e. x(t)=x*(t) . Then how we can get minus (-) sign in exponential (e ^ -jkwot)? as there was no minus in complex valued x(t) $\endgroup$ – Aadnan Farooq A Oct 5 '15 at 11:18
  • $\begingroup$ @AadnanFarooqA: It's because of taking the complex conjugate: $(e^{jx})^*=e^{-jx}$. $\endgroup$ – Matt L. Oct 5 '15 at 11:22
  • $\begingroup$ means from the complex valued x(t) to real value x*(t), we took complex conjugate, of every term (i.e. ak and e ^ -jkwot) is it right? $\endgroup$ – Aadnan Farooq A Oct 5 '15 at 11:25
  • $\begingroup$ @AadnanFarooqA: You take the complex conjugate of Eq. (1) in my answer, so you have to take the conjugate of each term in the sum. Note that $(a\cdot b)^*=a^*\cdot b^*$. $\endgroup$ – Matt L. Oct 5 '15 at 11:32
  • $\begingroup$ Yes I understand that.. but I have confuson that from Eq. (1) to Eq. (2) you took conjugate to make complex valued to real-valued? $\endgroup$ – Aadnan Farooq A Oct 5 '15 at 11:51

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