# Transmit/receive signal of two-path propagation model

In wireless communication, when the transmitted signal is $$x$$ and the received signal is $$y$$.

Please tell me why the formula below holds true.

\begin{align} y(t) &= \left( e^{-\jmath 2\pi f_c\frac{d_1}{c}} + e^{-\jmath 2\pi f_c\frac{d_2}{c}} \right) x(t)\\ &= \left( e^{-\jmath 2\pi \frac{d_1}{\lambda}} + e^{-\jmath 2\pi \frac{d_2}{\lambda}} \right) x(t) \end{align}

I don't understand why the exponent of $$e$$ is negative. Also, is the real part signal actually the baseband signal?

The above formula holds true only for a single frequency at $$f_c$$, with $$x(t)$$ as the "analytic signal" (positive frequencies only) and is approximately true when the modulation bandwidth is a small fraction of the carrier.

The general form $$e^{-j2\pi \tau}$$ is a phase shift of -$$2\pi\tau$$ radians which is what a single carrier would experience after propagating over a delay $$\tau$$ seconds long. In general the phase induced by a fixed time delay is proportional to the carrier frequency as given by the Fourier Transform property for a time delay where we see the minus sign appear:

$$x(t-\tau) \leftrightarrow X(f)e^{-j2\pi f \tau}$$

Note that phase in the frequency domain for any specific frequency $$f_c$$ as described by the formula above would be the starting phase at $$t=0$$ for the time domain waveform with frequency $$f_c$$. A real sinusoid with frequency $$f_c$$ has positive and negative frequencies given at $$f_c$$ and $$-f_c$$. The analytic signal would be the positive frequency component only (so for $$\cos(2 \pi f_c t)$$ the analytic signal is $$e^{j2\pi f_c t}$$.

This all may be clear by considering the time delay of a real sinusoid:

After the time delay of $$\tau = d_1/c$$ seconds, the waveform given as $$x(t)=\cos(2\pi f_c t)$$ would be shifted to the right to get:

$$x(t-\tau) = \cos(2\pi f_c (t-\tau))$$

The analytic signal is the equivalent complex signal as the positive only frequency components, resulting in:

$$x_a(t-\tau) = e^{j 2\pi f_c (t-\tau)}$$

From which we get:

$$x_a(t-\tau) = e^{j 2\pi f_c t} e^{-j 2\pi f_c \tau} = x_a(t)e^{-j 2\pi f_c \tau}$$

The real part of this would be the real passband signal $$x(t-\tau)$$.