# Convolution between a channel with real signal vs. channel with complex signal

The real transmitted signal $$x(t)$$ is convolved with channel $$h$$ in passband as :

$$y(t) = x(t)cos(j2πf_c t) * h(t)$$ = $$Re(x(t)e^{j2πf_c t})*h(t)$$

where $$f_c$$ is the carrier frequency and $$t$$ is the time and $$*$$ is convolution operation. Based on my understanding, different from baseband system, if the modulated signal $$x(t)$$ is real, it means that the received signal $$y(t)$$ will be real too. I am not sure if I am correct or no.

What I am looking for is to get the difference when $$x(t)$$ is real and complex. In baseband mode, it's clear that convolving the real or complex signal with the channel results a complex received signal. However, in passband signal, I think the convolution of real signal with a channel will always result of a real signal. Is that correct?

The baseband received signal may have both in-phase and quadrature components; in other words, the complex envelope of the received signal is complex (not real).

This is quite easy to see. Assume the transmitted signal is an in-phase signal $$s(t) = x(t)\cos(2\pi f_c t)$$, where $$x(t)$$ is a real baseband signal.

Now consider the wireless channel. Let's assume a line-of-sight channel with no reflections, only a delay $$\Delta$$, and a perfectly synchronized receiver. The received signal is $$r(t) = x(t)\cos(2\pi f_c + \phi),$$ assuming $$x(t-\Delta) \approx x(t)$$ (this is reasonable in many cases, since $$x(t)$$ varies very slowly relative to the propagation delay).

At the output of the quadrature receiver, we have, in the in-phase and quadrature branches respectively, \begin{align} r_I(t) &= \text{LPF} \{ x(t)\cos(2\pi f_c t + \phi)\cos(2\pi f_c t ) \} \neq 0, \\ r_Q(t) &= \text{LPF}\{ x(t)\cos(2\pi f_c t + \phi)\sin(2\pi f_c t ) \} \neq 0. \end{align} In other words, the delay in the channel produces a phase-shift in the received carrier, which causes it to become "spread out" in both the I and Q branches. As a consequence, the complex envelope of $$r(t)$$ is complex.

Now, it could be argued that the receiver's synchronizer will cancel out the phase shift $$\phi$$ in the carrier, and that is true for this simple channel. But let's consider a slightly more realistic channel, with one single reflection: $$r(t) = x(t)\cos(2\pi f_c t + \phi) + \alpha x(t)\cos(2\pi f_c t + \theta).$$ Since the reflected signal traverses a longer path than the LOS signal, each path introduces a different phase shift. Assume that the phase shift $$\phi$$ in the LOS path is corrected for by the receiver. You still have a phase shift $$\theta$$ in the reflected path, and in consequence you will have part of the received signal in the I branch, and part in the Q branch.

In summary: in general, over the wireless channel, the I and Q branches of the transmitted signal will get "spread out" into the I and Q branches of the received signal, in amounts that depend on the delays in each path. As a corollary, a real baseband signal will be received as a complex baseband signal.

• Thank you so much for your detailed answer, but do you mean by the phase shift what is called CFO ? I got what you mean, but what's about if we correct that CFO effects before performing the equalization, does that mean the equalizer should have real signal too? Jun 30, 2021 at 14:36
• You're welcome! In general, the transmitter and receiver have different carriers. Part of the reason is that their clocks are different (their measurements of "one second" are slightly different). Also, their initial phases are independent. On top of that, you have the phase shift caused by the channel delay. If there is one single received signal (as in a satellite or a wireline link), the receiver can compensate for all of those and produce a purely in-phase (real) signal. In most cases, there will be multiple reflections, each with a different phase shift.
– MBaz
Jun 30, 2021 at 15:05
• In this case, the ouptut of the quadrature receiver is complex. However, you can still of course further equalize that signal to recover the in-phase, real signal. But that needs to happen after the quadrature receiver. As an example, say you transmit $(I, Q) = (1,0)$ and receive $(0.7, 0.3)$. Applying some equalization method you can still produce $(1,0)$ (a purely real signal).
– MBaz
Jun 30, 2021 at 15:08
• I got it .. thanks again Jul 1, 2021 at 1:30
• I don't know if it's allowed for me to ask about details here or I should raise another question., but I wanted to say that according to your answer, in that case, the phase shift is known at the receiver, right? I mean if we are sure that our transmitted signal is real, and then we receive after the equalizer $I$ and $Q$, that part of $Q$ represents the phase offset, right? so we can get such way to compensate it in order to correct the phase shift. Jul 5, 2021 at 9:22

The following is to add a graphical intuitive view to MBaz's correct and good answer:

The motivation to have complex coefficients for the baseband equivalent of the channel comes down to this:

If we require representing the channel as having spectral asymmetry (putting the carrier at f=0 and considering the spectrum relative to the carrier as the positive and negative frequencies), with asymmetry meaning the positive frequency spectrum is independent of the negative frequency spectrum- then we MUST use complex coefficients for such a representation.

More detail:

The coefficients of the channel represent the channel's impulse response (just like a filter, a "channel" can be considered as a filter). The frequency response of the channel is the Fourier Transform of the impulse response. The Fourier Transform of a real signal will have a complex conjugate symmetric spectrum (the positive frequency spectrum is the complex conjugate of the negative frequency spectrum- so they are dependent and represent the same thing redundantly). In order to not have this symmetry, the impulse response MUST be complex. It is that simple.

In practice the distortion we see as we transmit through arbitrary channels would typically not conveniently have a balanced response on our waveform. For a simple example we may get an increasing loss as we extend higher in frequency such that our equivalent baseband spectrum; once we have moved the signal from the carrier frequency to DC; has higher components in the negative half spectrum than the upper half. Even more typical is to have frequency selective fading where there will be arbitrary dips and nulls across the spectrum with no regard to which half was positive or negative, meaning highly unlikely that the distortion itself maintains the symmetry we need to have real coefficients only.

We can move "0" above to be any carrier frequency without changing any of the characteristics of the waveform (which is what we do when we up-convert, down-convert or frequency translate waveforms.) It is when the waveform is centered at 0 as above where this will require an asymmetric spectrum to represent the same spectral shape that was otherwise real at any other carrier frequency, and as such we will need to use complex numbers to represent the waveform at baseband (which is called the complex equivalent baseband) unless the spectrum was complex conjugate symmetric at the carrier.

Euler's formula and the graphic below helps make this more intuitive. Consider how Euler's formula represents how a (real) cosine can be decomposed as a positive and negative frequency which are complex conjugate symmetric (spin at same rate and opposite direction):

$$2\cos(\omega t) = e^{j\omega t} + e^{-j\omega t}$$

As long as the two phasors represented are of equal magnitude and opposite phase (which is what complex conjugate symmetric means), then the imaginary terms will cancel in the summation of the two, and we are left with what will always stay on the real axis. If we break that symmetry then the summation will include imaginary components and the result is complex.