# Fourier Series: Shifting in time domain

I am reading "Fourier Transformation for Pedestrians" from T. Butz. He speaks about what happens to the Fourier coefficients when the function is shift in time. I have copied the equation I have a problem with:

I don't understand the logic behind going from $f(t-a)$ in the first integral to $f(t')e^{-iw_kt'}e^{-iwk_a}dt'$. It seems like he is using the identity $e^{a+b}=e^ae^b$ but I don't understand the complete logic.

Also he then applies the same thing without the complex notation:

Why does that work? Why does shifting in time correspond to multiplying $A_k$ for example by $cos \omega_k a - B_k sin \omega_k a$. I don't understand where this is the case. If someone could explain it would be great. Thank you.

-

He is doing $t'=t-a$, note that he is also shifting the interval of the integral, and also $t=t'+a$, replaced in the exponent and applying the exponent rule.

For the non-complex, he is shifting from $\mathbb C$ to $\mathbb R^2$, but just by notation.

$$C_k=A_k+iB_k=\{A_k;B_k\}\\ e^{i\theta}=\cos\theta+i\sin\theta=\{\cos\theta;\sin\theta\}\\ C_ke^{i\theta}=(A_k\cos\theta-B_k\sin\theta)+i(A_k\sin\theta+B_k\cos\theta)=\{A_k\cos\theta-B_k\sin\theta;A_k\sin\theta+B_k\cos\theta\}\\$$

Where $\theta=\omega_ka$.

-
Thank you for the 2 answers. –  Marc Ourens Oct 16 '13 at 20:21

In the first step you have a question about, he replaces $t-a$ with $t'$. So, everywhere $t$ appears it becomes $t' + a$. Thus

• $f(t -a)$ becomes $f(t')$
• $e^{-i\omega_k t}$ becomes $e^{-i\omega_k (t'+1)} = e^{-i\omega_k t'} e^{-i\omega_k a}$
-