# What are real-valued and complex signals and why is the Fourier transform of a real-valued signal Hermitian?

I have confusions in these concepts:

1. What are real-valued signals?
2. What is the difference between it and complex signals?
3. Why is it so that for real-valued signals, the spectrum of negative frequencies is the complex conjugate of the positive frequencies?
• 1. Signals with values that are real numbers, as the name says 2. they are not complex-valued. 3. Math, it's definitely in the book you're currently reading, no matter what that is. But even if not, this question has been answer multiple times on this website. – Marcus Müller Feb 3 '19 at 13:21

When one analyses data, outcomes both depends on the nature of the data, and their handling (eg a transformation, like the Fourier transform). One can consider binary data, categorical data, data in two dimensions, etc. To develop a theory and the related tools, it is useful to define where the d ata dwell.

A real signal is simply a signal that takes values that are real. The signal can be discrete or continuous, mono- or mulitdimensional, the observed values are real. The set of real values $$\mathbb{R}$$ is quite interesting. Since it is a continuous field, it allow a lot of "signal-like" mathematical operations: addition, product, convolution, and a lot of properties related to limits. $$\mathbb{R}$$ is one of the simplest mathematical structure allowing such operations.

But a natural tool to study signals, spectra, is related to Fourier analysis. And Fourier uses complex arguments. Why shall we use complex transforms to study real signals? That is a complicated story, but let us say this is mandatory, to cope with linear and time-invariant systems. See for instance: Why cosine is not an eigen signal?

So one can also use complex-valued data. The complex field $$\mathbb{C}$$ is, maybe, the second simplest mathematical structure allowing the same operations as $$\mathbb{R}$$, without pain. And a real signal is just a subset of complex-valued data: take a complex object, whose imaginary part of zero, and you have a real signal.

For the last question: real signals are just a subset of complex signals. So when you analyze them (the reals) with a fully complex transformation, you can expect that the outcome will be restricted too. One of the restriction is that real signals have certain symmetries in the Fourier domain, that fully complex signals don't have

A signal is mathematically a function, $$x(t)$$, (or a sequence $$x[n]$$ in discrete-time) which is a mapping from its domain set to its range set; i.e., $$x(t): D_x \to R_x$$

While the domain set determines the arguments of the signal, the range set determines the values of it.

Therefore a real valued signal is a function whose range set is the real numbers.

Consequently a complex valued signal is the one whose range set is the complex numbers.

Another way to think about it is that all signals are complex. A real-valued signal is just a complex signal where all the imaginary components of all the complex values are strictly zero.

1st (real valued: only datapoints from the domain of real numbers) and 2nd (contains points from the domain of complex numbers) point was already answered in detail so I focus on the 3rd.

Every function $$f(t)$$ can be written as:

$$f(t) = f_{even}(t)+f_{odd}(t)$$

(Ie every function is decomposable to an even and to an odd part.)

The source of the following screenshot is this.

Proof:

Fourier transform of a real even function is real:

Fourier transform of an even function is real:

Fourier transform of a real odd function is imaginary:

The Fourier transform of an odd function is odd:

If $$f(t)$$ is real-valued, so is $$f_{even}$$ and $$f_{odd}$$. So the Fourier transform of $$f_{even}$$ is real and even, the FT of $$f_{odd}$$ is odd and imaginary. The Fourier transform of $$f(t)$$ is the sum of the Fourier transforms of $$f_{even}$$ and $$f_{odd}$$ (decomposition property).

Putting it together:

$$f(t)\xrightarrow{\mathscr{F}}F_{real\&even}+F_{real\&odd} = F(\omega)$$

So from here we see that F is hermetian, ie in OP's terms, the spectrum of negative frequencies is the complex conjugate of the positive frequencies:

$$F(-\omega)=F^*(-\omega)$$

• sorry there was a typo now i believe it's correct – zabop Feb 5 '19 at 20:56
• zabop, this looks like a good answer. might be nice if the math was done with $\LaTeX$ instead of a bitmap. – robert bristow-johnson Feb 7 '19 at 3:54
• yep, will change it when I have the time, anyone who feels like changing it is encouraged to do so. – zabop Feb 7 '19 at 7:45
• if you do change it, i recommend changing "$F$" to "$X$" and "$f$" to "$x$" and then change "$s$" to "$f$" and leave $s$ for the Laplace Transform ($s=j 2 \pi f$). it will make the notation more consistent with continuous-time Fourier transform (and in the Electrical Engineering context). – robert bristow-johnson Feb 7 '19 at 20:03