Real Periodic Fourier Transforms

Question

Suppose we have some input data (only real numbers) and the data is periodic. Is it true that the Fourier coefficients will also be only real numbers?

You need to elaborate your question more. Fourier transforms are usually defined for integrable (or square integrable) signals. Are you asking about Fourier series maybe? — copper.hat, Mar 15 '13 at 06:48

score 2 · Answer 1 · answered Nov 16 '13 at 11:02

If your input is an analog signal $t\mapsto f(t)\in{\mathbb R}$ of period $T>0$ then you can expand it into a Fourier series of the form $$f(t)={a_0\over2}+\sum_{k=0}^\infty \left(a_k\cos{2k\pi t\over T}+b_k\sin{2k\pi t\over T}\right)$$ with real coefficients $a_k$, $b_k$.

When the input data represent a discrete signal $k\mapsto y_k\in{\mathbb R}$ $\>(k\in{\mathbb Z})$ of period $N$ then the so-called discrete Fourier transform, which works with the complex exponential function, produces complex Fourier coefficients $c_j$ $\>(0\leq j<N)$. The fact that the $y_k$ are real is then reflected in the "symmetry" $c_{-j}\bigl(=c_{N-j}\bigr)=\overline{c_j}$.

score 0 · Answer 2 · answered Nov 16 '13 at 10:34

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Nope, it is definitely not true! You can try to check the fourier transform of the sequence {1 2}

answered Nov 16 '13 at 10:34

meta_warrior

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score 0 · Answer 3 · edited Sep 29 '15 at 21:46

No. For example if you want to do the Fourier transform of

$$ f : \mathbb R \to [0,1]: t \to \sin(\omega t ) $$

Than we get using the Euler formula

$$ \frac{1}{\omega} \int_{0}^{\omega} \sin(\omega t ) \exp(i \omega k t ) dt= \frac{1}{\omega} \int_{0}^{\omega} \sin(\omega t ) \cos (\omega k t )dt + i\frac{1}{\omega} \int_{0}^{\omega} \sin(\omega t ) \sin (\omega k t ) dt $$

The first integral will clearly be $0$ due to symmetry, and the second coefficient will be one only be different from $0$ if $k=1$. Hope this helps a bit.

Real Periodic Fourier Transforms

3 Answers3