The frequency spectrum of a time domain signal x(t) can either be written as X(f) or $X(j\omega)$. But how is the later correct? I mean, the frequency spectrum is clearly dependant on the frequency, not on a $j\omega$ part, is it?
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1$\begingroup$ I don't understand your question. "Can either be written" is from some book or other source.. and I'm getting tired of repeating|this|repeatedly, but if you have a formula from somewhere that you've got a question about, properly cite and give a bit of background. $\endgroup$– Marcus MüllerCommented Feb 19, 2017 at 12:32
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$\begingroup$ I'd normally not be as persistent about this, but you're dealing with spectra now for quite a while now, and you should by now be very able to write down the Fourier transform of $x(t)$ from the top of your head. Thus, you can probably understand and represent very much more precise where that $j\omega$ is used later on, or where it came from. $\endgroup$– Marcus MüllerCommented Feb 19, 2017 at 12:37
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$\begingroup$ @MarcusMüller Chill out man.... $\endgroup$– Robert L.Commented Feb 19, 2017 at 13:38
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$\begingroup$ @CarlosDanger extremely chill here :) $\endgroup$– Marcus MüllerCommented Feb 19, 2017 at 13:47
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2 Answers
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Basicaly it's just a convention and the Fourier transform is a function of $\omega$ .
The imaginary argument $j\omega$ can be understood in different perspectives. In one convention, for the class of signals $x(t)$ which are absolutely summable (i.e., stable) , the continuous time Fourier Transform (CTFT) $X(j\omega)$ is considered to be derived from the existing Laplace transform $X(s)$, with $s$ being a complex number, by replacing $s$ with $j\omega$, which amounts to evaluating the Laplace transform on the imaginary axis ( where $s =j\omega$ ) of the complex plane. Hence in this interpretation, the argument $j\omega$ retains the conceptual and mathematical link between the Laplace transform $X(s)$ and the CTFT $X(j\omega)$ where we say $X(j\omega) = X(s)|_{s=j\omega}$
A similar argumentation also holds for the discrete time Fourier transform (DTFT) $X(e^{j\omega})$ as such: For the class of signals $x[n]$ which are absolutely summable (i.e., stable), the discrete time Fourier Transform $X(e^{j\omega})$ is considered to be derived from the existing $Z$-transform $X(z)$, where $z$ is a complex number, by replacing $z$ with $e^{j\omega}$, which amounts to evaluating the $Z$-transform on the unit circle ( where $z = e^{j\omega}$) of the complex plane. Hence in this interpretation, the argument $e^{j\omega}$ retains the conceptual and mathematical link between the $Z$-Transform $X(z)$ and the DTFT $X(e^{j\omega})$ where we say $X(e^{j\omega}) = X(z)|_{z=e^{j\omega}}$
However, note that, there exist certain very important class of signals (such as the pure sine wave or the impulse response of an ideal Low-Pass filter) which are not absolutely summable (unstable) therefore whose formal Fourier transforms (as obtained by replacing $s$ or $z$ by the corresponding $j\omega$ or $e^{j\omega}$) do not exist. Those Fourier transforms are obtained from another perspective by utilizing the impulse function $\delta(\omega)$ or discontinuities in the frequency domain. In such a case the argument is misleading and there exist no coresponding $X(s)$ or $X(z)$ which you can obtain by replacing $j\omega$ or $e^{j\omega}$ with $s$ or $z$ in the Fourier transform $X(j\omega)$ or $X(e^{j\omega})$, hence in that case it's just a convention.
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Both. For a strictly real time domain signal, $X(j\omega)$ is conjugate symmetric, and thus X(f) is identical to folding the complex range (with a scale factor of 2 pi thrown in).
