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.
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*}