# Digital demodulator (SDR) for receiving legacy analog modulated wave

I understand that in SDR, after the ADC the signal processing is fundamentally based on I/Q modulated waves where any discrete value that can be transmitted corresponds to a symbol which is represented as amplitude and phase through I and Q signals.

How can a modern SDR receive legacy AM waves?

The SDR uses I,Q components whereas legacy AM doesn't have any such component applicable.

Do current SDRs receiving AM waves switch to a different technique wherein it doesn't use I/Q demodulation?

Sure, an SDR can transmit and receive conventional, analog AM. I've done it many times as an introductory exercise in my courses.

A quadrature signal $$s_{\text{QAM}}$$ is made up of two signals, $$s_I(t)$$ and $$s_Q(t)$$, that are transmitted over the same bandwidth and modulated in quadrature (i.e. with carriers phase-shifted by 90 degrees):

$$s_{\text{QAM}}(t) = s_I(t)\cos(2\pi f_c t) + s_Q(t)\sin(2\pi f_c t)$$

There is no restriction on the shape of these signals, beyond the physical limits of the hardware or the channel (in bandwidth or in power). In particular, they can be a DSB-LC AM signal!

$$s_{\text{QAM}}(t) = (A_m + m(t))\cos(2\pi f_c t) + 0\sin(2\pi f_c t)$$

The regular downcoversion and filtering process applied to $$s_{\text{QAM}}$$ will produce $$A_m + m(t)$$. You can also apply a traditional envelope detector. That's the beauty of SDR: it does not care about the meaning or structure of the in-phase and quadrature signals; you define that in software.

To conclude, note that not only it is indeed possible to transmit/receive AM with an SDR; you can actually transmit/receive two AM signals on the same bandwidth!

$$s_{\text{QAM}}(t) = (A_{m1} + m_1(t))\cos(2\pi f_c t) + (A_{m2} + m_2(t))\sin(2\pi f_c t)$$

You select one of the IQ branches and ignore the other, and then you can listen to one "station" or the other. Of course, this assumes that you are transmitting your own quadrature AM signals (for example, over one of the ISM bands).

• Thank you very much MBaz for the reply. But, QAM is not the legacy AM modulation; right? The legacy AM I suppose do not have any I/Q components at transmitter side. There is no symbol mapping in legacy AM. So, how does it make sense or what can an SDR do if legacy AM received is sampled and each sample split into I and Q components? – DevBee Apr 19 at 8:14
• The I samples correspond to the AM signal and the Q samples are zero. It's as simple as that. Think about it this way: Each of the I and Q signals is simply a train of pulses upconverted using AM-DSB SC. There is no fundamental difference between BPSK ($s(t) = m_1(t)\cos(2\pi f_c t)$) and AM ($s(t) = m_2(t)\cos(2\pi f_c t)$). Sure, $m_1(t)$ and $m_2(t)$ are quite different, but the upconversion/transmission/downconversion process doesn't care about that -- and that process is exactly what the analog front end of an SDR does. The rest is up to software. – MBaz Apr 19 at 12:33
• Thanks MBaz. plz confirm - If am right, a true SDR using direct sampling needs to know if the received signal is legacy AM or IQ modulated (say, by the user through external interface) so as to process the received samples as containing I component only or both I and Q components. Otherwise splitting a wave which is not in-fact generated from two quadrature components into I and Q components would only end up in receiving junk bit pattern. – DevBee Apr 22 at 11:47
• That is incorrect. The SDR does not need to know anything about the signal except its center frequency, bandwidth, etc. Think of an SDR receiver as a device whose input is a bandpass signal $s_{\text{BP}}(t)$, and its output are the samples of its complex envelope $s_{\text{BB}}(t)$. All bandpass signal, whether a simple AM DSB-LC signal or a complicated 1024-QAM signal, have a complex envelope. The interpretation and further processing of the CE is done in software and is thus completely flexible. – MBaz Apr 22 at 12:54
• Got it. Thanks a lot MBaz! – DevBee Apr 23 at 11:09