let's consider a generic wave that propagates in space and time, expressed in frequency domain: $$V(x)=A\cdot e^ <-kx>+ B\cdot e^$$ It may represent the voltage along a transmission line, or any physical quantity that appears in a second order differential equation that satisfies the oscillation conditions. In time - domain it becomes something like that: $$V(x)=A\cdot cos(-kx+\omega t) + B\cdot cos(kx+\omega t)$$ The first term is usually called direct or incident wave, while the second one is called reverse or reflected wave. They are two waves that moves in space and time, and precisely the first one moves towards x axis, while the second one moves in the opposite direction to the x axis. My question is about this last statement: why are the term with "-kx" the direct wave and the term with "kx" the reverse wave? I tried to answer my question by using the concept of phase velocity. The phase velocity of a wave is defined as the velocity an observer should have in order not to see any difference of phase, i.e. such that: $$d\cdot (-kx+\omega t) = 0$$ for the first term; $$d\cdot (kx+\omega t) = 0$$ for the second term; So we get: $$-kdx + \omega dt = 0$$ for the first term; $$kdx + \omega dt = 0$$ for the second term; and finally: $$v_ = dx/dt = \omega/k$$ for the first term; $$v_ = dx/dt = - \omega/k$$ for the second term; So the phase velocity of the first term is positive (so, towards x axis), while that of the second term is negative (so, opposite to x axis). What does not convince me of this explanation is the fact that as far as I know the phase velocity does not represent the propagation velocity of the wavefront, but is simply the velocity corresponding to a zero phase difference. And thanks to the fact that it is not a propagation speed, it follows that it can be even greater than the speed of light (it is a known result of the analysis of waveguides, for instance).
I believe you should plot these two waves starting with $t=0$ and then slowly increasing the time.
The red is the incident wave, which travels into the positive $x$ -direction. It is reflected at $x=1.2 \lambda$ . The green wave is the reflected wave. Thus it travels into the negative $x$ -direction. The blue wave is the superposition of these two waves.
answered Apr 9, 2020 at 12:41 9,026 1 1 gold badge 15 15 silver badges 35 35 bronze badges $\begingroup$ Can you please tell me what program did you use yo make this animation? $\endgroup$ Commented Apr 9, 2020 at 13:09$\begingroup$ I used R in combination with the packages ggplot2 and gganimate. However, this is a standard feature in almost all math programs, e.g. matlab, mathematica, python etc. Alternatively, you could probably generate "animated GIFs" using an external program as well. $\endgroup$