5
$\begingroup$

I have a simple question.

Suppose I have a signal : $f(t)=\sin(2\pi \, 25t)$

We know that it's CTFT(continuous time fourier transform) will result in 2 delta's at +- $2\pi\,25$ [rad/sec]. Or conversely at +- $25$ [Hz].

And that makes sense because $f(t)$ has a fundamental period of $T=1/25=0.04$ [sec]. But that also means that $f(t)$ is periodic with any integer multiple of $T$. Therefore, wouldn't that mean that we would expect to see all these harmonics as well in the Fourier Transform?

Meaning peaks at 12.5 Hz, 6.25 [Hz], … ?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Harmonics are multiples of the fundamental frequency $f_0$ which is $25$ Hz in this instance, so $2f_0 = 50$ Hz, $3f_0 = 75$ Hz etc. What you have listed are subharmonics $f_0/2$, $f_0/4$ and so on. $\endgroup$
    – Dilip Sarwate
    Sep 19 at 19:03

2 Answers 2

Reset to default
8
$\begingroup$

A $T$-periodic signal doesn't necessarily have a (sinusoidal) fundamental with period $T$. It never has one if the signal is also periodic with period $T/2$ or $T/3$ etc. (as in your example).

But even if $T$ is the smallest period of a signal, there needn't be a fundamental with period $T$. As an example take the signal

$$x(t)=\sin(2\omega_0t)+\sin(3\omega_0t)\tag{1}$$

with arbitrary $\omega_0>0$. Note that the fundamental frequency of $x(t)$ is not $2\omega_0$ but $\omega_0$. Yet, $x(t)$ has no sinusoidal component with frequency $\omega_0$. In other words, the period of $x(t)$ is $T=2\pi/\omega_0$, but there's no sinusoid with that period contained in $x(t)$.

In general, a $T$-periodic function can be represented by its Fourier series:

$$x(t)=\sum_{k=0}^{\infty}c_k\cos(2\pi k/T+\phi_k)$$

The point here is that any of the coefficients $c_k$ can be zero. If $c_1=0$, then $x(t)$ has no fundamental component with period $T$, but it is still $T$-periodic (unless all $c_k$ with odd indices $k$ are zero, in which case the period of $x(t)$ would be $T/2$).

Improve this answer
$\endgroup$
7
  • $\begingroup$ I'll elaborate: if a function $f(t)$ is periodic with a period of $T$ , then $f(t)=f(t+T)$ , which implies that $f(t)$ is also periodic with a period of $2T,3T...$ or any integer multiple of this fundamental period . Therefore if I look at it's spectrum I would've thought that they would show up as well, but they don't. Only the fundamental frequency is present while the rest has 0 energy. $\endgroup$
    – Sammy Apsel
    Sep 19 at 19:52
  • $\begingroup$ @SammyApsel: Yes, but even the fundamental needn't be present, as I've shown in the example in my answer. $\endgroup$
    – Matt L.
    Sep 19 at 20:17
  • $\begingroup$ Hmm, this is actually a real clever observation because one would be inclined to think that the fourier spectrum of a signal represents it's periodic components but that is somewhat wrong I guess then. Your example shows that even if a signal has period $T$, it doesn't necessarily mean that the corresponding frequency $\omega_0=2\pi/T$ will have non-zero energy,meaning it could be zero. But what is then the correct conclusion? That there must be some integer multiple of the fundamental which has non-zero energy? (in your example that is $2\omega_0$ and $3\omega_0$) $\endgroup$
    – Sammy Apsel
    Sep 19 at 21:29
  • $\begingroup$ @SammyApsel You are nearly there. At least two harmonics must have nonzero amplitudes. One odd harmonic and one even harmonic will work, but not two even harmonics, as noted in MattL's answer. Some pairs of odd harmonics will work, such as $3\omega_0$ and $5\omega_0$, but not other such as $3\omega_0$ and $9\omega_0$. Can you see why? $\endgroup$
    – Dilip Sarwate
    Sep 20 at 2:08
  • 1
    $\begingroup$ @SammyApsel: The fundamental frequency is the largest $\omega_0$ for which all harmonics satisfy $\omega_i=k_i\omega_0$ for integer $k_i$. $\endgroup$
    – Matt L.
    Sep 20 at 8:37
0
$\begingroup$

Therefore, wouldn't that mean that we would expect to see all these harmonics as well in the Fourier Transform?

No. It's simple as that – this intuition doesn't hold, and isn't based on the actual math you're doing when you're evaluating the Fourier integral!

Remember that the $s_0(t)=e^{j2\pi f_0t}$ is orthogonal to any other $s_1(t)=e^{j2\pi f_1t}$, under the inner product $\langle f_0, f_1\rangle=\lim\limits_{T\to\infty}\int\limits_{-T/2}^{+T/2} \frac1T f_0(t)\cdot f_1^*(t)\,\mathrm dt$. Compare that to the Fourier transform's integral!

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ This answer is missing the point. $\endgroup$
    – Dilip Sarwate
    Sep 19 at 19:04
  • $\begingroup$ hm, if you say so! What's the point I'm missing? The question is "Therefore, wouldn't that mean that we would expect to see all these harmonics as well in the Fourier Transform?", in verbatim, and I very much addressed that, no, periodity with that does not imply energy at that the inverse of the period. The misuse of the word (sub)harmonic contributes little to the question. $\endgroup$
    – Marcus Müller
    Sep 19 at 19:09
  • 2
    $\begingroup$ See Matt L's answer. $\endgroup$
    – Dilip Sarwate
    Sep 20 at 13:47
  • $\begingroup$ @DilipSarwate yep, thanks $\endgroup$
    – Marcus Müller
    Sep 20 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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