I am trying to work out what happens when you downconvert an RF signal, such as what happens in an SDR device when tuning to a given frequency. For example if a device using a Zero-IF is tuned to 399MHz, then whatever signal you see at 400MHz will appear at 1MHz where it is then digitised.

Now imagine at 400MHz you see a signal consisting of just a carrier, which switches on and off very rapidly. On for one cycle, off for one cycle, on for another cycle, off again. If you assign a binary 1 to the 'on' cycle and a binary 0 to the 'off' cycle, I believe this would allow you to transmit 400,000,000 bits per second.

Now what happens if this signal is downconverted to 1MHz ready for the SDR to digitise? If the 1MHz carrier switches on and off at a rate of one cycle at a time, there will only be 1,000,000 transitions, although the original signal had 400,000,000 transitions in the same time period.

So what happens in this case? Does the 1MHz carrier cycle on and off at the original 400MHz frequency? Does that allow you to transmit your original 400,000,000 bits per second on a 1MHz carrier frequency? Or are the extra cycles lost somehow? What would the resulting signal at 1MHz look like?


The thing to realize here is that if you take a sinusoidal carrier and switch it on and off, change its amplitude, frequency, or modulate it in any way, then it can be shown mathematically, but somewhat counter-intuitively, that what you are doing is introducing sinusoidal components at other frequencies. In fact, any periodic waveform can be represented as a sum of sine waves. Take for example, the square wave here:

enter image description here

The mathematical tool that allows this transformation is the Fourier transform. Here in the case of the square wave we can see it is made of the fundamental frequency, plus all of its odd harmonics. Even if the signal we care about isn't strictly periodic (they usually aren't), we can pick some segment of the signal that is periodic, or mostly so, and analyze that.

Similarly, your example of switching a carrier on and off also introduces higher frequency components than your carrier. In fact, any rapid departure from a perfect sine wave creates high-frequency components. This explains how information is not lost: these high-frequency components are also down-converted and detectable by your SDR, provided it has sufficient bandwidth to see them all.

It also explains why this modulation scheme is not used in practice: each switch on and off would create a lot of noise far away from the carrier spectrum. In fact, this might be one of the oldest modulation problems in radio: CW (the usual way to modulate Morse code, simply switching a carrier on and off) is exactly what you describe, albeit at a much slower rate. While it would be conceptually simplest to switch the carrier hard on and hard off, this creates what's called "key clicks", undesirable interference on other frequencies, as well as an audible "click" resulting from those high-frequency components being converted down to audio frequencies. Consequently, the carrier is actually slowly tapered on and tapered off to reduce the bandwidth occupied by the signal. The tapering is fast enough it's not perceived by the listener as a taper, but slow enough that the high-frequency components are negligible compared to the carrier.

  • \$\begingroup\$ Nice graphic! and nicely fills in some of what I over-simplified. \$\endgroup\$ – Brian Drummond Mar 12 '13 at 14:16
  • \$\begingroup\$ Ahh, I think this is what I missed - that turning a carrier on and off is no different to modulating it with a lower frequency. Now I understand! Thanks! \$\endgroup\$ – Malvineous Mar 12 '13 at 19:51
  • \$\begingroup\$ Yes, nice graphic. How did you go about creating it, or where did you get it from? \$\endgroup\$ – akohlsmith Mar 12 '13 at 22:22
  • \$\begingroup\$ @AndrewKohlsmith it's from the linked Wikipedia article. \$\endgroup\$ – Phil Frost Mar 12 '13 at 22:59

Practically, information is not lost in a properly designed down-conversion process, but that's because the RF signal is rarely used in the way you suggest (and when it is, you wouldn't down-convert it).

Take the process you described : adding information to a 400MHz carrier - this process is called modulation. Simply turning the carrier on or off for an entire cycle is the same as switching its amplitude to 0% or 100% in "off" or "on" cycles; and this is a scheme called "amplitude modulation".

Now, over-simplifying, this means the signal is no longer just a 400MHz carrier; the carrier alone carries no information. It has a bandwidth ranging from (carrier - max modulation frequency) aka "lower sideband" to (carrier + max modulation frequency) aka "upper sideband" and if you are transmitting 400Mbits across it, that means 0 to 800MHz, or 800MHz bandwidth. And any other signals between 0 and 800MHz will interfere with your transmission.

(There are communication systems called "spread spectrum" that actually work like this, but we'll ignore them. With due care, you can also transmit N bits as on-off signals in N/2 bandwidth, and use more levels to reduce bandwidth further, but let's concentrate on the basic point : information needs bandwidth).

If you did downconvert it, the lower sideband would run from -399 to 1 MHz, the upper from 1 to 401MHz. The "negative frequency" part is also called "spectral folding" and results in a signal up to 399MHz which, being superimposed on the upper sideband is very difficult to decode; information is not actually lost, but probably corrupted beyond repair.

More usually, you would transmit a lower rate over this carrier : say 0.5 million bits/second, on a 400MHz carrier, in 1MHz bandwidth (399.5 to 400.5 MHz).

And in this case when you down-convert to 1 MHz, you will have a signal between 0.5MHz and 1.5 MHz (roughly, the AM band!) with no lost information.

  • \$\begingroup\$ Right, this makes perfect sense now I realise that turning a carrier completely on and off, even only at the zero point, is still the same as modulating it. Thanks for the detailed explanation! \$\endgroup\$ – Malvineous Mar 12 '13 at 19:53

The actual answer depends on what you mean by downconversion.

If you simply mean mixing with a local oscillator, then, at least in the IQ case, information is not necessarily lost, since your quite wideband signal will still be present at the output of a perfect mixer, only nearly half of its bandwidth will now be at negative frequencies.

If you use only a single mixer instead of a dual IQ setup, then those negative frequencies are still present, but unable to be distinguished from positive frequencies of the same magnitude, so that confusion would arguably constitute information loss.

However, most actual conversion schemes also include filters often both before and after the mixing, and filtering is basically the act of intentionally throwing away uninteresting information in order to concentrate on the interesting information. So that is information loss; though usually of a desirable sort.

Finally you have to consider that all of this processing - (analog, conversion to digital, and often truncation within the digital realm as well) - introduces error, which can also easily lead to loss of information.


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