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

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


1 Answer 1


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


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.

  • $\begingroup$ 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) $\endgroup$ Oct 5, 2015 at 11:18
  • $\begingroup$ @AadnanFarooqA: It's because of taking the complex conjugate: $(e^{jx})^*=e^{-jx}$. $\endgroup$
    – Matt L.
    Oct 5, 2015 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? $\endgroup$ Oct 5, 2015 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, 2015 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? $\endgroup$ Oct 5, 2015 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.