IQ modulation: baseband to bandpass transformation

Question

can someone explain me these doubts?

it is a baseband signal representation. A- complex signal, R- real.

If I multiply baseband signal with FT( cos) then the signal will look like this ( on the plot the first on the left /For this case, convolution changes nothing). [![enter image description here][1]][1]

I shift the signal in frequency but my passband signal is still complex-valued, so it cannot be transmitted over a single channel ( Is it a reason why I can.t use multiplication with cos to modulate complex baseband signal?). otherwise, I will need two channels, one for the real part and one for the imaginary part.

if I multiply baseband only with FT(cos), why I get the same copy A-R in negative part and A-R in positive? In IQ modulation it is R-A in negative part and A-R in positive

Answer 1

If you have a complex baseband signal

$$x(t)=x_R(t)+jx_I(t)=|x(t)|e^{j\phi(t)}\tag{1}$$

and you multiply it with a (real-valued) sinusoid, the resulting signal is obviously still complex. What happens in IQ-modulation is that you generate a real-valued band-pass signal containing the same information as $x(t)$:

$$s(t)=\text{Re}\{x(t)e^{j\omega_ct}\}=x_R(t)\cos(\omega_ct)-x_I(t)\sin(\omega_ct)\tag{2}$$

Using the magnitude and phase of $x(t)$, the bandpass signal $s(t)$ can also be written in the form

$$s(t)=|x(t)|\cos[\omega_ct+\phi(t)]\tag{3}$$

which shows that in general $s(t)$ is amplitude and phase modulated.

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EDIT:

This is what happens in the frequency domain when you use a sinusoid for modulation as opposed to IQ modulation. Let

$$r(t)=x(t)\cos(\omega_ct)=x(t)\frac12\big[e^{j\omega_ct}+e^{-j\omega_ct}\big]\tag{4}$$

In the frequency domain this corresponds to

$$R(j\omega)=\frac12\big[X(j(\omega-\omega_c)+X(j(\omega+\omega_c))\big]\tag{5}$$

For the IQ modulated signal $s(t)$ we get (from $(2)$)

$$s(t)=\text{Re}\{x(t)e^{j\omega_ct}\}=\frac12\big[x(t)e^{j\omega_ct}+x^*(t)e^{-j\omega_ct}\big]\tag{6}$$

The Fourier transform of $(6)$ is

$$S(j\omega)=\frac12\big[X(j(\omega-\omega_c)+X^*(-j(\omega+\omega_c))\big]\tag{7}$$

Comparing $(5)$ and $(7)$ you see that the part of the spectrum that is centered around $-\omega_c$ is inverted for $s(t)$ (i.e., it is a mirror image of $X(j\omega)$; note the negative sign before $j$), whereas it is just shifted without inversion for the cosine modulated signal $r(t)$.