I have a homework problem that I'm not quite sure how to complete. The problem is as follows:


Write the definition of the Fourier coefficients, and show that

$$f(t + \frac{1}{2}T) = f(t) \rightarrow c_{2n+1} = 0$$ $$f(t + \frac{1}{2}T) = -f(t) \rightarrow c_{2n} = 0$$

Also show that this is a consequence of the similarity theorem in the Fourier analysis.


I have been able to solve the first part. The definition of Fourier coefficients is:

$$c_n = \frac{1}{T} \int_{0}^{T} f(t) \exp(-2 \pi i n t/T) dt$$

Now let $f(t) = f(t + \frac{1}{2}T)$. If we plug this in we get:

$$c_n = \frac{1}{T} \int_{0}^{T} f(t + \frac{1}{2}T) \exp(-2 \pi i n (t + \frac{1}{2}T)/T)dt$$

$$ = \frac{1}{T} \int_{0}^{T} f(t + \frac{1}{2}T) \exp(-2 \pi i n t/T) \exp(- \pi i n) dt$$

From this last expression we se that when $n$ is even, $\exp(- \pi i n)$ is $1$, and when $n$ is odd, $\exp(- \pi i n)$ is $-1$. Thus we see that if we want $f(t + \frac{1}{2}T) = f(t)$ we must have all odd Fourier coefficients equal to $0$, and if we want $f(t + \frac{1}{2}T) = -f(t)$ then all even Fourier coefficients must be equal to $0$.

I don't know quite how to approach the last question though - how to show that this is a consequence of the similarity theorem. Of course, the definition for the similarity theorem is that:

$$f(t/ \gamma) \leftrightarrow |\gamma| F(\gamma \omega)$$

and a property is that if we "stretch" the function in the time domain, then we "shrink" it in the time domain, and vice versa. But how this apply to the problem above? My intuition tells me that since every second Fourier coefficient is set equal to $0$ we are "stretching" the function in the frequency domain, but I don't know how to show that we then get a "shrinking" in the time domain.

If anyone can give me some help in how to attack this last part of the problem, then I would be extremely grateful!


1 Answer 1


If you have a $T$-periodic function $g(t)$ with Fourier coefficients $c_n$


and you create a new function $f(t)$ with Fourier coefficients $d_n$ defined by

$$d_n=c_{n/2},\quad n\textrm{ even}\\ d_n=0,\quad n\textrm{ odd}$$

i.e. the spectrum is expanded by a factor two by inserting zeros between the original Fourier coefficients. Then $f(t)$ is given by

$$f(t)=\sum_{n=-\infty}^{\infty}d_ne^{in\omega_0t}= \sum_{n\textrm{ even}}c_{n/2}e^{in\omega_0t}= \sum_{n=-\infty}^{\infty}c_ne^{i2n\omega_0t}=g(2t)$$

So you get a compressed time domain signal with period $T/2$, as directly shown in the first equation of the problem formulation in your question.

  • $\begingroup$ Matt: Thanks a lot! This was really very helpful and easy to understand! Really appreciate it. $\endgroup$
    – Kristian
    Commented Apr 14, 2014 at 14:02

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