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Given the Fourier series expansion of a signal $x(t)$ with fundamental frequency $\omega_0$ gives $c_k$, I am trying to find the Fourier coefficients $b_k$ for the signal $x(1-t)$. Instead of using the time shifting property directly, I am trying to obtain the result by first principles.

$$b_k=\frac{1}{T} \, \int_T \, x(1-t) \, \text{e}^{-j\, k\, \omega_0\, t} \, \text{d}t$$

Making a substitution $u=1-t$, I obtain

$$b_k= - \frac{1}{T} \, \int_T \, x(u) \, \text{e}^{-j\, k\, \omega_0\, (1-u)} \, \text{d}u$$

$$=- \frac{1}{T} \, \int_T \, x(u) \, \text{e}^{-j\, k\, \omega_0} \, \text{e}^{j\, k\, \omega_0 \, u} \, \text{d}u$$

$$=- \text{e}^{-j\, k\, \omega_0} \, \frac{1}{T} \, \int_T \, x(u) \, \text{e}^{j\, k\, \omega_0 \, u} \, \text{d}u$$

I can see that the inetgral is the conjugate of the synthesis equation.

If $x(t)$ is a real signal, then $c_k=c^{\ast}_{-k}$ or $c^{\ast}_k=c_{-k}$. Thus I get $b_k=-\text{e}^{-j\, k\, \omega_0} \, c^{\ast}_k$ or $b_k=-\text{e}^{-j\, k\, \omega_0} \, c_{-k}$.

I would say that for $x(t-1)$, its Fourier series coefficients are $b_k=-\text{e}^{-j\, k\, \omega_0} \, c_{-k}$ when $c_k$ are the Fourier series coefficients of $x(t)$.

Is the above answer correct?

Thank you.

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  • $\begingroup$ yes that's correct but I don't see why you use conjugation and the fact that $x$ is real. For any $x$ you obtain that $b_k = -e^{-jk\omega_0}c_{-k}$ directly from the last line of your derivation right ? $\endgroup$ – P. Quinton Jan 27 at 7:28
  • $\begingroup$ Thank you. Conjugation because in the last line with the integration we have $\text{e}^{j \, k \, \omega_0 \, u}$ and not $\text{e}^{-j \, k \, \omega_0 \, u}$. If only $x(t)$ is real then we have $x(t)=x^{\ast}(t)$. $\endgroup$ – macy Jan 27 at 11:14
  • $\begingroup$ Yes but $e^{j k \omega_0}=e^{-jk(-\omega_0)$ so you directly get the result without $x$ being real right ? $\endgroup$ – P. Quinton Jan 27 at 11:17
  • $\begingroup$ These two exponential terms in your comment are only equal if $x(t)$ is real. I made the assumption that $x(t)$ is real after my last inegration line so that I could write the coefficient $c^\ast_{k}=c_{-k}$. If $x(t)$ was not real then I would have been stuck with $c^\ast_{k}$. $\endgroup$ – macy Jan 27 at 11:23
  • $\begingroup$ Can you read again my equality ? There are no $x(t)$ term, it is always true. Also I made a tex error here it is $e^{j k\omega_0}=e^{-jk(-\omega_0)}$ $\endgroup$ – P. Quinton Jan 27 at 11:32
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Your derivation is correct but the argument can be generalized for any complex signal $x(t)$. You already had \begin{align*} b_k=-e^{-jk\omega_0}\frac{1}{T}\int_T x(u) e^{jk\omega_0 u}du \end{align*} and by observing that for any signal $x(t)$, $c_{-k}=\frac{1}{T}\int_T x(u) e^{-j(-k)\omega_0 u}du=\frac{1}{T}\int_T x(u) e^{jk\omega_0 u}du$ hence \begin{align*} b_k &=-e^{-jk\omega_0} c_{-k} \end{align*}

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  • $\begingroup$ I understand your point. It's clear now. $\endgroup$ – macy Jan 30 at 3:21

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