1
$\begingroup$

Given is a baseband signal representation.

A : Complex Signal
R : Real Signal

enter image description here

If I multiply baseband signal with FT(cos) then the signal will look like first plot 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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $r(t)$.

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – LenaPark
    Aug 8, 2018 at 11:40
  • $\begingroup$ @LenaPark: The spectrum of a real-valued signal is always symmetric: $X(j\omega)=X^*(-j\omega)$ $\endgroup$
    – Matt L.
    Aug 8, 2018 at 12:49
  • 1
    $\begingroup$ @LenaPark: If this answer was helpful, please accept it by clicking on the green check mark, thanks. $\endgroup$
    – Matt L.
    Aug 8, 2018 at 12:49
  • $\begingroup$ 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 $\endgroup$
    – LenaPark
    Aug 9, 2018 at 6:43
  • $\begingroup$ @LenaPark: I've added an explanation to my answer; I hope it's clear now. $\endgroup$
    – Matt L.
    Aug 9, 2018 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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