IQ modulation: baseband to bandpass transformation

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). 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

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.

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 $r(t)$.

• ok, it makes sense. But I realy can't understand why after IQ modulation in the plot the R-real part of the baseband is A-complex? – LenaPark Aug 8 '18 at 11:40
• @LenaPark: The spectrum of a real-valued signal is always symmetric: $X(j\omega)=X^*(-j\omega)$ – Matt L. Aug 8 '18 at 12:49
• @LenaPark: If this answer was helpful, please accept it by clicking on the green check mark, thanks. – Matt L. Aug 8 '18 at 12:49
• it is lit a bit difficult to understand the part with IQ modulation. Ok, the spectrum of the real-valued signal is symmetric. The output of IQ modulation is the multiplication of the complex baseband with FT( cos) and FT(sin). it means that I shift the baseband in frequency. It makes sense. But 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 – LenaPark Aug 9 '18 at 6:43
• @LenaPark: I've added an explanation to my answer; I hope it's clear now. – Matt L. Aug 9 '18 at 8:39

I have a question according to this one. if i modulate the baseband signal ( not complex), then i shift the signal in Fc and -Fc ( signal* cos(2*piFc)) and in Fc and -Fc ( signal* sin(2*piFc)). But in the case with sin, the signal in -Fc go down on the plot ( Fourier transform of the sin function). My quastion is: What is the sum, output then? How to add two results?

PS Sorry i can't comment becouse i don't have 50 reputation