How do I interpret the result of a Fourier Transform?

For example, I entered the following "equation" into Wolfram|Alpha:

FourierTransform[Piecewise[{{sin[t],t > 0 and t < 2*pi}}, 0], t, \[Omega]]


So as to take the Fourier Transform of a signal like this:

And turn it into some graphs that look like this:

My understanding from constantly — but casually! — hearing about Fourier Transforms (and DFT/FFT) in the context of audio processing and SDR I/Q data processing, is that I now have some information about a portion of the spectrum of the input signal. I'm trying to deepen my understanding, however, and refresh my memory of all I started learning in a Differential Equations class years ago.

Given my more recent practical background I'm left with the following questions understanding what the Fourier transform just did for me:

• Why did I get a complex (real/imaginary) output from a real-only input? (How would I have passed some sort of "I/Q" equivalent in to the "equation" I provided?)
• What units are omega, is it essentially Hertz if t is in seconds? What does negative Hertz mean?
• If the input function were actually a voltage on a wire, would the real and imaginary parts be amplitude and phase (respectively) of the component frequencies?

From https://dsp.stackexchange.com/a/24758/18819, I gather that if I primarily care about power at a given frequency I would take the magnitude of the complex vector. From a Wikipedia article on negative frequency I'm not sure what its "wheel spinning backwards" analogy would mean in the context of e.g. electromagnetic or acoustic waves, but I'll take it to mean maybe it's just "some more of the signals" and combine the values across the y-axis.

So am I correct to interpret this as follows:

• there is no power [correct term?] at 0Hz
• at +0.5Hz and at +1Hz there is 1 unit (?) of "power" but
• …at -0.5Hz there is also 1 unit of power
• …while at -1Hz there is -1 unit of power
• so at "real world" 0.5Hz there might be 2 units of power
• but at "real world" 1Hz there might be 0 units of actual power?

Or what would be the best way to understand the first Fourier plot above, and derive relevant results from it?

The Fourier transform of a function $x(t)$ is defined by

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$

Since $e^{-j\omega t}$ is complex-valued, the Fourier transform $X(\omega)$ is generally a complex-valued function, even if $x(t)$ is real-valued.

$\omega$ is the frequency in radians per second (rad/s), and it is related to the frequency in Hertz $f$ by $\omega=2\pi f$.

Since $X(\omega)$ is generally complex-valued, it can be represented by its real and imaginary parts

$$X(\omega)=X_R(\omega)+jX_I(\omega)\tag{2}$$

or by its magnitude and phase

$$X(\omega)=|X(\omega)|e^{j\phi(\omega)}\tag{3}$$

Clearly, the real and imaginary parts are not the same as magnitude and phase.

Note that for real-valued $x(t)$, the real part $X_R(\omega)$ and the imaginary part $X_I(\omega)$ of the Fourier transform $X(\omega)$ are even and odd functions, respectively. Consequently, the negative frequency components are redundant, and knowledge of the spectrum at positive frequencies is sufficient. If you are interested in power, you should rather plot $|X(\omega)|^2=X_R^2(\omega)+X_I^2(\omega)$ (which, by definition, is non-negative, as it should be the case for energy or power). The function $|X(\omega)|^2$ shows the distribution of the frequency components of $x(t)$.

As you've noted, the DC component of the signal is zero, because its average value over time is zero:

$$X(0)=\int_{-\infty}^{\infty}x(t)dt=0\tag{4}$$

• Great answer!!! But....what do the real part and the imaginary part mean after all? – Daniel Möller Jul 27 '16 at 20:33
• @Daniel: The real part of the Fourier transform corresponds to the even part of the signal, and the imaginary part corresponds to the odd part of the signal. – Matt L. Jul 27 '16 at 21:06
• Thanks @Matt, I was able to interpret both components as components of a phase vector. The real part is the "cosine" component (even, as you said :) ) and the imaginary part is the "sine" component (odd). (Thanks to e^jx = cosx + j.sinx). Sine and cosine are "perpendicular" to one another, and the magnitude of a vector (cos,sen) is 1. Meaning a result without an imag part means the signal is in phase with the cosine function. A result with only the imag part is in phase with the sine function. A combination of both components mean the signal is phased somewhere in between. – Daniel Möller Jul 28 '16 at 12:20
• For a chosen frequency, the magnitude of this vector means: how much of a frequency do I have in the original function? --- Sine and cosine components of this vector mean: how shifted is this frequency? – Daniel Möller Jul 28 '16 at 12:23
• @Daniel: yes, makes sense. – Matt L. Jul 28 '16 at 20:37