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Where did the $k$ of $a_k$ disappear from Fourier Reverse Transform if $\omega=\omega_0k$?

We turn $\omega0$ to be $d\omega$, but $\omega=\omega_0k$, so shouldn’t there be a $k$ in the reverse transform?

Attached reference below: enter image description here

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It is because $\omega_o$ approaches 0 such that $k\omega_o$ becomes a continuous function.

So the discrete function $X(jk\omega_o)$ becomes the continuous function $X(j\omega)$.

This is intuitive if you consider the Fourier Series expansion in that each component of the spectrum is spaced in frequency by $1/T$ Hz, (or $\omega_o = 2\pi/T$ radians/sec). Each of these components are given as $k\omega_o$ with $k$ being the set of all integers. So as $T$ gets larger, $1/T$ gets smaller, and hence $\omega_o$ gets smaller and as $T$ approaches infinity, $k \omega_o$ approaches a continuous function.

The concept behind the Fourier Series Expansion is that any continuous time function defined over time duration $T$ can be decomposed into a sum of sinusoids that each have a fundamental period of $T$ and all higher harmonics such that there are an integer number of cycles in $T$, the the waveform can repeat for all time and start and end the same way at all boundaries $t= NT$ for all integers $N$. For this reason no other frequency components can exist and the non-zero frequency spectrum is therefore discrete.

Fourier Series Expansion

The Fourier Series Expansion given above in terms of sines and cosines is much more compact using exponential frequency notation (and ultimately much more intuitive once you understand that $e^{j\Omega t}$ is a phasor rotating counter-clockwise on the complex plane at rate $\Omega$ radians per second) :

$$x(t) = \sum_{n=-\infty}^{\infty}c_n e^{jn\Omega_o t}$$ Where

$e^{jn\Omega_o t} = \cos(n\Omega_o t) + j\sin(n\Omega_o t)$

$c_n = a_n - jb_n$

And $\Omega_o$ was used in place of $\omega_o$ just to match the plots I already had (where I was distinguishing radian frequency between analog and digital domains).

This becomes the Fourier Transform when we all $T$ to extend to infinity: the spacing given by $1/T$ between each of the components goes to zero, and the frequency spectrum therefore becomes continuous.

Fourier Transform

An easy way to observe this effect is by simply zero-padding a time domain waveform prior to taking the FFT: as you add more zeros, meaning $T$ gets larger, more and more samples will be interpolated in frequency as the spectrum approaches a continuous function.

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  • $\begingroup$ Thanks Dan, I’ll need to go over it and think about what you said. (The first 2 lines) $\endgroup$ – Vitali Pom Apr 29 '20 at 13:48
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    $\begingroup$ @VitaliPom I added a picture, hope it helps! $\endgroup$ – Dan Boschen Apr 29 '20 at 13:49
  • $\begingroup$ Thanks you rock! I’ll need some time to go over it. Many thanks :) $\endgroup$ – Vitali Pom Apr 29 '20 at 13:50
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    $\begingroup$ @VitaliPom I added some more details if it was still confusing. $\endgroup$ – Dan Boschen Apr 29 '20 at 13:55
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    $\begingroup$ OK, caught you up in the last question. Will arrive here soon as well. :’D $\endgroup$ – Vitali Pom May 3 '20 at 20:45
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In the first line they're showing that

$$Ta_k=X(jk\omega_0)\tag{1}$$

That's what they use in the second line, where they also use

$$\frac{1}{T}=\frac{\omega_0}{2\pi}\tag{2}$$

Note that the sum

$$\sum_{k=-\infty}^{\infty}X(jk\omega_0)e^{jk\omega_0t}\omega_0\tag{3}$$

is a Riemann sum which converges to the integral of the inverse Fourier transform:

$$\lim_{\omega_0\to 0}\sum_{k=-\infty}^{\infty}X(jk\omega_0)e^{jk\omega_0t}\omega_0=\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}d\omega\tag{4}$$

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  • $\begingroup$ Yes, but when integrating they’re looking at $d\omega$ not at $dT$. How can dw0 turn into dw...? $\endgroup$ – Vitali Pom Apr 29 '20 at 13:47
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    $\begingroup$ @VitaliPom: $\omega_0$ is the "step size" in the sum (note the $k\omega_0$ in the function). As $T$ is increased, $\omega_0$ becomes smaller and the (Riemann) sum becomes an integral with infinitesimal $d\omega$. $\endgroup$ – Matt L. Apr 29 '20 at 13:49
  • $\begingroup$ But isn’t it like doing to Ct dt and getting C? Alas differentiating constant multiplied by some t by t and getting the constant as a result.....? $\endgroup$ – Vitali Pom Apr 29 '20 at 13:52
  • $\begingroup$ @VitaliPom: We're not differentiating, but integrating. It's a Riemann sum, which converges to the integral as the step size goes to zero. $\endgroup$ – Matt L. Apr 29 '20 at 14:18
  • $\begingroup$ You’re right in differentiation we divide by dt, not multiply. (Regarding Reimann I remember). I’ll think a little bit more about it. Thanks Matt! $\endgroup$ – Vitali Pom Apr 29 '20 at 14:22

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