# Representing a sinusoidal signal in terms of positive and negative frequency components?

I am reading "signal processing first " by Mcclean

https://www.amazon.com/Signal-Processing-First-James-McClellan/dp/0130909998

In article 2.5.4, author discusses about representation of real sinusoidal signals in terms of positive and negative frequency components

Can someone please kindly explain this idea of representation of real sinusoidal signals in terms of positive and negative frequency components? In simple words ?preferably with the example/case discussed by author in fig 2.13(b) I have also attached a snapshot and encircled confusing lines

Here is my view. As you can see in most books related to DSP, a complex exponential $$e^{j\omega_0 t + \phi}$$ is a fundamental signal representing a point on unit circle on complex plane at the angle $$\omega_0 t + \phi$$ at the instant $$t$$. It is a periodic signal. The whole of signal processing concept rests on the fundamental operation of representing any signal as a linear combination of complex exponential signal (Fourier Transform) which you will encounter as you learn more about DSP. The signal $$A\cos(\omega_0 t + \phi)$$ as per Eulers Formula (https://en.wikipedia.org/wiki/Euler%27s_formula) is represented as sum of complex exponential - one having angular frequency of $$\omega_0$$ and another $$-\omega_0$$. The former represents a complex exponential rotating in anti-clockwise direction (increasing angle), latter representing a complex exponential rotating in clockwise direction (decreasing angle). Your cosine signal is the sum of these 2 vectors at any time if you visualize them rotating on a complex plane.
In the above example, after the first step, you can re-write the sum as $$A\cos(\omega_0 t + \phi)$$ = $$(A/2)e^{j\phi}e^{j\omega_0t}$$ + $$(A/2)e^{-j\phi}e^{-j\omega_0t}$$ . So $$X = Ae^{j\phi}$$, $$X^{*} = Ae^{-j\phi}$$.
$$z(t) = Xe^{j\omega_0 t}$$, $$z^{*}(t) = X^{*}e^{-j\omega_0 t}$$. Hence,
$$A\cos(\omega_0 t + \phi)$$ = $$\frac{1}{2}z(t) + \frac{1}{2}z^{*}(t)$$ = $$Re(z(t))$$. You need to remember that $$Re()$$ operation adds half of signal with its own half conjugate which is the last step.