I have searched both dsp and electronics stack exchange but I remain confused about the reasons why information is sent in I/Q channels.

For a while, I thought that the main reason for sending I/Q channels was so that you can send 2 independent channels on the same bandwidth and since they were orthogonal to each other you could demodulate them independently as well. Effectively you could double B/W, if one channel was multiplied by cos (for example) and the other by sine. On the receive side you you again mix with carriers that 90 degrees offset and retrieve the information you want. I understood this as QAM. Am I correct? I.e. is QAM two independent channels that mixed with carriers of different phase or are the channels related?

From what I read, I/Q are also used for image rejection. Basically as explained here: Can somebody explain what IQ (quadrature) means in terms of SDR?. Furthermore, from what I have read I/Q modulation is important in SSB/SC but not exactly sure why.

So is I/Q done for both or specifically for one case? Or are there multiple uses for it and you pick it based on application?

  • \$\begingroup\$ There seems to be some confusion here between IQ modulation vs IQ representation. For an IQ representation, the net bandwidth is twice that of either I or Q individually, because having both lets you differentiate positive and negative frequencies. Remove either I or Q and the positive vs. negative frequency components present on the remaining signal become indistinguishable. \$\endgroup\$ Jun 6, 2013 at 17:42
  • \$\begingroup\$ IQ allow you to do more with a lower frequency LO. Updated answer here: electronics.stackexchange.com/questions/39796/… \$\endgroup\$
    – hotpaw2
    Feb 12, 2014 at 21:24

1 Answer 1


I and Q modulation does not double the bandwidth but rather allows you to send twice as much data on a channel (fixed bandwidth) as you can achieve with pulse-amplitude modulation (PAM) on a single carrier signal (which gives a double-sideband (DSB) signal). A quadrature amplitude modulation (QAM) signal is just two different PAM (baseband) signals modulated onto phase-orthogonal carrier signals (the cos and sin as you noted). Both signals are DSB signals. The two carrier signals must be of the same frequency and must differ in phase by 90 degrees. The sum of the two DSB signals is the QAM signal, and it occupies the same bandwidth as the two separate PAM signals. Because the required phase orthogonality is difficult to achieve (and maintain!) with two separate oscillators in the same location, let alone different locations, QAM is not a multiple-access method: the sum of the signals from two different transmitters each creating its own PAM signal and modulating it onto a carrier does not give you QAM unless the two carriers are at exactly the same frequency and differ in phase by 90 degree. Thus, the answer to the question

is QAM two independent channels that mixed with carriers of different phase or are the channels related?

is that QAM is not two independent channels and the carrier phases in the two signals must be carefully controlled and maintained at 90 degrees.

The information in a PAM/DSB signal is carried in both sidebands but can be recovered from only one sideband if need be. So, the bandwidth required to transmit a PAM signal can be reduced by a factor of 2 by filtering the PAM/DSB signal to get a single-sideband (SSB) signal. (Of course, one could create a PAM/SSB signal right from the start instead of filtering a PAM/DSB signal). A separate PAM/SSB signal could be transmitted in the other (unused) sideband and each can be demodulated completely independently of the other. But the sum signal is not a QAM/DSB signal and the demodulation technique is different. Both receivers will use I and Q demodulation and each must filter the incoming signal to eliminate the unwanted sideband which means that filters with narrower bandwidths (and sharper cut-offs) must be used. Two PAM/SSB signals, transmitted separately in the upper and lower sidebands give essentially the same spectral efficiency (bits per Hertz) as QAM but at a higher price. The advantage gained is, of course, that PAM/SSB signals can be used in multiple-access situations.


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.