I have always been told that for a sinusoidal variable (for instance a voltage signal), the fourier transform coincides with the phasor definition, and this is the reason why the analysis of sinusoidal circuits is done through the phasor method.

But I do not find this correspondence from a mathematical point of view.

Consider for instance this signal and this phasor definition:

enter image description here

If now we calculate its Fourier transform by using its properties, we get:

enter image description here

If we evaluate this one at the pulsation 5, it seems to me different from the phasor (especially because of the presence of the dirac pulse).

  • \$\begingroup\$ Your phasor definition seems wrong. It's lacking any frequency information, and the exponent is not purely imaginary. \$\endgroup\$ May 27, 2019 at 21:59

1 Answer 1


Firstly, your phasor definition should have an \$ e^{jwt}\$ that is usually omitted as we generally deal with a single frequency system and can hence divide out that term.

The only remaining difference between the two is then that the Fourier transform one has a delta function at the positive AND negative frequencies which is missing from the phasor.

This is because a phasor is the simplest example of an analytic signal.

The idea with an analytic signal is that for a real valued function its Fourier transform's amplitude spectra will always be even, i.e. symmetric about the origin. Similarly, the phase spectra is odd, i.e. anti-symmetric about the origin. Therefore, it is pointless to involve both the positive and negative frequencies in calculations because one tells us all the information about the other.

An analytic signal is defined as:

\$ s_a(t) = s(t) + js_h(t)\$

Where \$ s_h(t) \$ is the hilbert transform of the original signal \$ s(t) \$

For a cosine \$ cos(\omega t + \phi)\$

\$ s_a(t) = cos(\omega t + \phi) + jsin(\omega t + \phi) \\ s_a(t) = e^{j(\omega t + \phi)} \\ s_a(t) = e^{j\omega t}e^{j\phi} \$

Which is our familiar phasor, without the time term divided out.

It turns out the spectrum of this analytic signal is a scaled version of the original signal's spectrum but WITHOUT any negative frequencies (There's a factor of 2 carried around the place that I will omit for simplicity, see the link for the details).

So the Fourier transform of our analytic signal will have only one positive Delta function, which simplifies calculations.

It is not strictly correct then that the phasor is equal to the Fourier transform but rather they are closely related. It also turns out that to bring back the negative frequencies all we have to do is take the real part of the analytic signal, which is what we do when converting a phasor back to a cosine!

Analytic signals are even more useful in converting a bandpass signal or system to an equivalent baseband form.

  • \$\begingroup\$ So, if for example there is a filter whose transfer function is A(jw) (so Vout = A(jw) Vin), if Vin is sinusoidal with respect to time, can I put its phasor in the previous espression in order to get Vout? Or should I put its Fourier transform? \$\endgroup\$
    – Kinka-Byo
    May 28, 2019 at 13:03
  • \$\begingroup\$ Either is fine. Just technically the phasor result will be missing the negative frequencies. These end up back in when you convert the phasor back to a cosine. One often plots A(jw) only from 0, not including negative frequencies, because of the symmetry. \$\endgroup\$
    – jramsay42
    May 28, 2019 at 20:29
  • \$\begingroup\$ Perfect, thank you very much. \$\endgroup\$
    – Kinka-Byo
    May 29, 2019 at 12:35

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