We can describe a BPSK signal, x, as:

$$x=cos(2\pi f_c t+n\pi),\;n\in \{ 0,1 \}$$

The BPSK signal is then transmitted through a narrow band frequency flat Rayleigh channel, described by a single complex exponential $h=Ae^{i\theta}$. $A$ has Rayleigh distribution and $\theta\sim\mathrm{Unif}(0,2\pi)$. The received signal at the destination is given as:

$$y=hx=Ae^{i\theta}[cos(2\pi f_c t+n\pi)].$$

My question is, what does the exponential term of the channel $e^{i\theta}$ represent? Clearly:

$$y\neq A[cos(2\pi f_c t+n\pi + \theta)].$$ This can be verified by expanding the sinusoids terms into their complex exponential forms.

  • 2
    $\begingroup$ The BPSK signal is $\text{Re}(e^{j(2\pi f_ct + n\pi)})$ and the channel changes it to $$\text{Re}(Ae^{j\theta}\cdot e^{j(2\pi f_ct + n\pi)}) = A\cos(2\pi f_ct + n\pi + \theta).$$ Search for "complex baseband representation" to understand the model. $\endgroup$ – Dilip Sarwate Sep 17 '13 at 1:14

The equations are a bit wrong. For a signal $x=cos(2\pi f_{c}t+n\pi), n\in \left\{0,1\right\}$ transmitted through a complex exponential channel $h=Ae^{i\theta}$, the received signal $y$ is not $hx$. $h$ is in the complex baseband notation and your signal $x$ is in the bandpass notation.

To answer your actual question first, $\theta$ is the phase change that is incurred by the signal when it passes through the channel i.e. the delay. And the actual received signal is $y=Re\left\{x_{l}Ae^{i\theta}e^{i2\pi f_ct}\right\}$, where $x_{l}$ is the baseband representation of our signal $x$ which in our case would be something like $x_{l} = e^{i2\pi\frac{(n-1)}{2}}$, where $n\in \left\{0,1\right\}$.

| improve this answer | |
  • $\begingroup$ Thanks for the that. Presumably $\ast$ represents convolution here? $\endgroup$ – Dave S Sep 17 '13 at 10:08
  • 1
    $\begingroup$ I edited that, it was meant to be multiplication actually. $Ae^{i\theta}$ is what every sample of the transmitted signal will get multiplied by(in the baseband representation), where $A$ and $\theta$ are Random Variables as you have said before. There is no convolution here because of the Narrowband model. The bandwidth of the signal is well within the Coherence Bandwidth so that there is no Intersymbol Interference. $\endgroup$ – Sudarsan Sep 17 '13 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.