Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

enter image description here

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:

enter image description here

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.

share|cite|improve this question
up vote 0 down vote accepted

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 $\theta=\omega_ka$.

share|cite|improve this answer
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 $e^{-i\omega_k (t'+1)} = e^{-i\omega_k t'} e^{-i\omega_k a}$
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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