Lesson 3b — Negative Frequencies

Yes, but why do we care? Real electrical signals are real, aren’t they?

Well, yes, they are. As you can verify from Euler’s identity, \begin{equation}\label{eq:Euler} e^{i \phi} = \cos\phi + i \sin\phi \end{equation} we can express the cosine of \(\phi\) as \begin{equation} \cos \phi = \frac{1}{2} (e^{i\phi} + e^{-i\phi}) \end{equation} for any phase \(\phi\), including \(\phi = 2\pi f t\). That is, a real cosine wave is composed of equal parts of positive and negative frequencies: \begin{equation} \cos(2\pi f t) = \frac{1}{2} ( e^{i\,2\pi f t} + e^{-i \,2\pi f t}) \end{equation}

When a software-defined radio mixes such a wave with a high-frequency carrier, it multiplies the cosine wave by a signal of the form \(2 e^{-i 2\pi f_0 t}\). That produces \begin{equation} 2\cos(2\pi f t) e^{-i 2\pi f_0 t} = [ e^{i\,2\pi (f-f_0) t} + e^{-i \,2\pi (f+f_0) t}] \end{equation} The first term oscillates at the difference frequency \(f - f_0\), and the second oscillates at the sum frequency \(f + f_0\). For frequencies \(f\) close to the carrier frequency \(f_0\), the difference frequency will be much lower than the sum frequency. The second term oscillates at almost twice the carrier frequency. Passing the result through a low-pass filter, which strongly attenuates high frequencies, effectively removes the second term, leaving just \(e^{i\,2\pi (f-f_0) t}\), which is clearly a complex sinusoidal signal. Thus, the frequency shifting done by a software-defined radio necessarily generates complex signals.

Illustrations in GNU Radio

time sinks
frequency sinks
constellation plots

Exercises

The Taylor series for the exponential function is \begin{equation}\label{eq:exposeries} e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \end{equation} where \(3! = 3\times 2\times 1 = 6\) and \(n! = n (n-1) (n-2) \cdots (2) (1)\) is the factorial of \(n\).

a. Using this expression, show that \(\frac{d e^x}{dx} = e^x\), an important property of the exponential function.

b. The Taylor series for sine and cosine are \begin{equation} \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \end{equation} \begin{equation} \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \end{equation} Use these three Taylor series to show Euler’s identity, \begin{equation} e^{i\phi} = \cos\phi + i \sin\phi \end{equation} by substituting \(i\phi\) for \(x\) in the Taylor series for the exponential, Eq. \eqref{eq:exposeries}.