With amplitude modulation the signal only has a real component and the modulated signal has two sidebands that are symmetric around the carrier frequency. My understanding is that in QAM the sidebands are necessarily not symmetric. The point of confusion for me is that QAM is effectively just two separate signals amplitude modulated and then summed together. So how does summing two signals (each of which has symmetric sidebands on their own) result in a combined signal which has asymmetric sidebands?


1 Answer 1


The problem in your thinking is that you forget about the phase difference between the carriers of the two amplitude modulated signals. A QAM signal has the following form:


where $x_I(t)$ and $x_Q(t)$ are the in-phase and quadrature components, respectively. Note that the carriers $\cos(\omega_ct)$ and $\sin(\omega_ct)$ have a phase difference of $\pi/2$.

Since both signals $x_I(t)$ and $x_Q(t)$ are real-valued, their spectra satisfy

$$X_I(\omega)=X_I^*(-\omega)\quad\textrm{and}\quad X_Q(\omega)=X_Q^*(-\omega)\tag{2}$$

where $^*$ denotes complex conjugation. With the usual assumption that $x_I(t)$ and $x_Q(t)$ are lowpass signals, and that the carrier frequency $\omega_c$ is much greater than their bandwidth, the spectrum of the QAM signal $(1)$ at positive frequencies is given by


Let $\Delta\omega=\omega-\omega_c$ denote the frequency deviation from the carrier. With $(2)$ and $(3)$ we obtain



$$X^*(\omega_c-\Delta\omega)=\frac12\left[X_I^*(-\Delta\omega)-jX_Q^*(-\Delta\omega)\right]\neq X(\omega_c+\Delta\omega)$$

which shows that the spectrum is not (conjugate) symmetric about the carrier frequency. As mentioned above, the reason is the phase difference between the two carriers.

  • $\begingroup$ Could you elaborate why (3) needs "the carrier frequency $\omega_c$ is much greater than their bandwidth" as the premise? You can just cite any source, I can look it up. I remember this is a very basic stuff. Thanks. $\endgroup$
    – Rokai
    Feb 11 at 18:41
  • 2
    $\begingroup$ @Rokai: The modulation moves the baseband spectrum to around $+\omega_c$ and to around $-\omega_c$. You don't want those two contributions to overlap, so you require $\omega_c>W$, where $W$ is the bandwidth of the baseband signal. In practice you choose $\omega_c\gg W$. In other words, after modulation you want a bandpass signal with no content around DC. $\endgroup$
    – Matt L.
    Feb 11 at 18:48

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