Fourier series: conjugate symmetry

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 $$x(t)$$.

Thank you.

• 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 ? – P. Quinton Jan 27 at 7:28
• 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 $x(t)=x^{\ast}(t)$. – macy Jan 27 at 11:14
• Yes but $e^{j k \omega_0}=e^{-jk(-\omega_0)$ so you directly get the result without $x$ being real right ? – P. Quinton Jan 27 at 11:17
• 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 $c^\ast_{k}$. – macy Jan 27 at 11:23
• 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 $e^{j k\omega_0}=e^{-jk(-\omega_0)}$ – P. Quinton Jan 27 at 11:32

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*}