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It is documented that 'one' of the units of the Fourier Transform [of $x(t)$ volt] is volt per Hz. That is $X(\omega)$ components will have units of volt per Hz, where $\omega$ is the angular frequency (along the $x$-axis).

For example, the Fourier Transform of $\cos(\omega_0 t)$ volt has two components - one at $\omega = \omega_0$ and one at $\omega = -\omega_0$, each assigned a 'value' of $\pi$. So it is assumed that the values are '$\pi$ volt per Hz'.

Now, when the plot is adjusted to become a function of cyclic frequency $f$ (in Hz), which means plotting $X(f)$ versus $f$, then the Fourier Transform components of $\cos( 2\pi f_0 t)$ become '1/2 or 0.5' instead of '$\pi$'.

My question is --- why isn't the '$\pi$ volt per hz' preserved when translating between $X(\omega)$ and $X(f)$?

According to tables showing Fourier Transform pairs, the Fourier Transform of $\cos( 2\pi f_0 t)$ volt has two components -- each with a value of '1/2 or 0.5'. For these values, does anyone know what are their units? I had assumed that the values for each component would have been 'preserved' (to be $\pi$ volt per Hz), regardless of whether we do the plot as a function of '$\omega_0$' rad/s or a function of '$f_0$' Hz. Thanks all.

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  • $\begingroup$ remember, both radian and cycle (or "turn") are dimensionless. and remember that the $\mathrm{d}t$ has dimension of time and $\mathrm{d}f$ or $\mathrm{d}\omega$ have dimension of frequency (or 1/time). $\endgroup$ – robert bristow-johnson Mar 19 '18 at 20:40
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For a given signal $g(t)$ with Fourier transform $G(\omega)$:

$$g(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} G(\omega)e^{i\omega t}\textrm{d}\omega$$

If you change variables such that $\omega=2\pi f$

$$g(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} G(2\pi f)e^{i2\pi ft}\textrm{d}(2\pi f)=\int_{-\infty}^{\infty} G(2\pi f)e^{i2\pi ft}\textrm{d}f$$

So if you present your Fourier transform as a function of $f$ instead of $\omega$, your Fourier pair of $g(t)$ will now be $\hat{G}(f)=\frac{G(2\pi f)}{2\pi}$ and:

$$g(t) =\int_{-\infty}^{\infty} \hat{G}(f)e^{i2\pi ft}\textrm{d}f$$

Note that with this definition, there is a scaling of $2\pi$ in comparison with the first equation I wrote. So now, for example, the Fourier transform of a cosine will consist of two deltas not with amplitude $\pi$, but with amplitude $\frac{\pi}{2\pi}=\frac12$, as correctly stated in your question.

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  • $\begingroup$ you have too many "$f$'s" appearing in your integrals (except the top one). there is a reason why electrical engineering textbooks stay away from labeling functions "$f(\cdot)$". $\endgroup$ – robert bristow-johnson Mar 19 '18 at 20:38
  • $\begingroup$ Thanks very much Tendero and robert! Tendero ---- some sources use the units of volt per hz (for the Fourier Transform X(w) if the time function x(t) has units of volts. But if we plot X(f), then what units would that become in relation to the X(w) units? Also, Tendero.... for your second equation, if you have the 1/(2.pi) out the front of the integral, then doesn't this mean that f(t) transforms to 2.pi.F(2.pi.f)? This also means f(t) transforms to 2.pi.F(w), right? I can see that you wrote f(t) transforms to F(2.pi.f)/(2.pi), but don't quite understand how that link was made. Thanks again! $\endgroup$ – Kenny Mar 19 '18 at 21:54
  • $\begingroup$ @robertbristow-johnson You are right, I've changed the names. Thanks! $\endgroup$ – Tendero Mar 20 '18 at 0:21
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    $\begingroup$ @Kenny The units of $X(f)$ and $X(\omega)$ stay the same. It's just the magnitude that changes. Regarding the second question, note the following: $$g(t) = \int_{-\infty}^{\infty} \frac{G(\omega)}{2\pi}e^{i\omega t}\textrm{d}\omega = \int_{-\infty}^{\infty} \hat{G}(f)e^{i2\pi ft}\textrm{d}f$$ $\endgroup$ – Tendero Mar 20 '18 at 0:24
  • $\begingroup$ @Kenny Glad to help. If the answer solved your question, please accept it by clicking on the check mark next to it, so it can help future readers as well. $\endgroup$ – Tendero Mar 20 '18 at 0:32
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Hz is in units of cycles per second.

ω is in units of radians per second.

And there are 2 π radians per cycle.

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  • $\begingroup$ remember, both radian and cycle (or "turn") are dimensionless. $\endgroup$ – robert bristow-johnson Mar 19 '18 at 20:42
  • $\begingroup$ They might be "dimensionless", but they have a ratio between them. $\endgroup$ – hotpaw2 Mar 19 '18 at 21:07
  • $\begingroup$ yup, just like the length of radii of a circle has with the length of the circumference. $\endgroup$ – robert bristow-johnson Mar 19 '18 at 22:30

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