I'm trying to detect a known signal under heavy interference. I'm working with 2 smartphones and an interfering speaker. I'm sending a signal from one smartphone speaker to the other smartphone microphone while the speaker emits an interfering signal. The bandwidth of the useful signal broadcasted between the 2 smartphones is 2.5 kHz (from 18 to 20.5 kHz), it is a "broadband signal" (imagine something like a spread spectrum) not a tone and the signal emitted by the interfering speaker is a tone at 19 kHz. The distance between the smartphones is 10 cm and the speaker is located 10 cm apart from both of them. I have noticed that as I increase the volume of the interfering tone the signal sent between the smartphones is less accurately detected. I have the following doubts:

  1. Why would this happen if the signal bandwidth is much broader than the interfering tone?

This process also happens when instead of in band, the interfering signal is off band, lets say a tone at 16 kHz. So more questions:

  1. Why does a loud input signal at a given frequency interfere with the performance of another signal (which uses other bandwidth)?

  2. Is microphone saturation a process affecting the whole frequency band in which the microphone is supposed to operate? So that any signal saturating the microphone even if it is at an specific narrow frequency impedes the microphone from working at other frequency ranges?

  3. Is there a technical parameter for a microphone saturation at ultrasound frequencies? Does it have a name?

  4. How does saturation at an stimulus at 1kHz relates to saturation at other frequencies? Which physical process at the microphone relates them?

Note: I would appreciate if every of the 5 questions was answered

  • \$\begingroup\$ Hmm sounds like positive feedback but saturation will attenuate anything else. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 31 '17 at 4:12
  • 1
    \$\begingroup\$ Is it easier or harder to understand a conversation in the presense of a crowd of random people talking to each other? \$\endgroup\$ – Andy aka Mar 31 '17 at 9:03

We may need to separate the "microphone" from any amplifier/filter/quantizer that follows the microphone (where sound becomes electrons).

Sum a 1 volt interferer sinwave with 0.1 volt sinwave desired signal. Run combined signal into an amplifier with 0.5 volt linear range. Whenever the summed input is above 0.5 volt, the amplifier output is driven into saturation/clipping/rail/flatoutput/overload.

This happens regardless of the two frequencies, as long as there is NO FILTERING between microphone and amplifier. This effect is called "blocking", and is a requirement for cellphones to operate in high-energy RF situations, such as when the person sitting next to you is also using their cellphone.

Thus I pick option (3).


Illustration of clipping/overloading/blocking


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ I see this all the time in detector designs used in daylight situations, looking for some optical transmission source they also control. I'd also say #3. \$\endgroup\$ – jonk Mar 31 '17 at 5:38
  • \$\begingroup\$ Blocking may be the cause. Plot quality versus interferer strength, and see if there's a cliff at some power level, or a gradual decline. \$\endgroup\$ – Neil_UK Mar 31 '17 at 9:28
  • \$\begingroup\$ @analogsystemsrf So you think that more than a microphone's saturation is the AD of the smartphone the saturated one? For sure there's no filtering between microphone and AD in an smartphone. This "blocking" effect you refer to, you mention it has to do with cellphone operating high-energy RF situations. What does it has to do with audio recording? \$\endgroup\$ – VMMF Apr 1 '17 at 0:36
  • \$\begingroup\$ @Neil_UK If I build the plot you suggest I would be able to determine over which interference volume the system stops working, I already have that, but I want to know why. What is exactly "blocking"? \$\endgroup\$ – VMMF Apr 1 '17 at 0:39
  • \$\begingroup\$ @VMMF added reply to answer your question. \$\endgroup\$ – Neil_UK Apr 1 '17 at 5:45

It sounds like the system is suffering from 'blocking'. All five of your questions will get answered somewhere along this answer.

Blocking is said to occur when the reception of a wanted signal is disrupted by a strong unwanted signal at a different frequency (far enough away to be removed by the channel filter) although the weak signal would be quite strong enough to be received without error by itself. (If the interferer is close enough in frequency to go through the channel filter, we call it adjacent-channel or co-channel interference).

As in all things communication, there are two ways to look at it, the time domain and the frequency domain. Both treatments will give the same answer if done consistently, but often one or the other will be easier to hand-wave with. Use whichever approach suits your style best.

Consider a receiver system comprising a wide filter, an amplifier, followed by a narrow filter. For the purposes of this illustration, 'wide' means passes the interferer, and 'narrow' means removes the interferer. This is typical of a part of most systems. Depending on which parts are analogue and which digital, the 'amplifier' could comprise one or more amplifiers and an ADC. In a radio system, it will also comprise several stages of frequency conversion through mixers. The important point is that it carries both wanted and interferer signals. There is a difference in that the digital parts will clip or saturate 'harder' than analogue parts, but the general behaviour is the same.

We will look at the signal at the input to the narrow filter.

In the time domain, consider the reception of the small signal only, in the presence of a small out-of-band signal. At all points in time, both signals are present in the system, both signals are amplified all the time. The signal-to-noise, or signal-to-interferer ratio stays the same as at the input. When you eventually pass through the narrow filter, your signal is still there.

Now consider that the interferer signal is much larger, and some part of the amplifier saturates, or clips, for 50% of the time. While it is clipped, there is zero amplification of the wanted signal, we have lost half of our signal power. But it gets worse. Clipping an amplifier this hard may disturb the amplifier bias conditions, so that even when the output signal returns to the valid range, the amplifier still doesn't immediately start amplifying the small signal again. One way or another, we rapidly lose wanted signal power as the interferer takes the amplifier into saturation.

In the frequency domain, a saturated amplifier becomes a multiplier. Signals that previously passed through the amplifier without disturbing each other now alter each others frequencies, by generating sum and difference terms. Signal power is thrown out to other frequencies. Noise power may be thrown into the wanted signal band. The effect is to reduce the signal-to-noise ratio of the wanted signal.

There are two obvious ways to reduce susceptibility to blocking. One is to make the amplifier tolerate a larger signal before overload. The other is to add a linear filter (that does not distort or clip) at the front of the system to attenuate the blocker. Ultimately though, the design of this filter will be wider than the channel filter, and you will have parts of the system subject to the interferer signal. If the filter can be a passive physical filter before the first transducer, say a Helmholtz arrangement of cavities for audio before the microphone, or an IR filter for optical, then so much the better.

A more subtle way to reduce susceptibility to blocking may be available through the signal design, and channel coding. Reduction of the information bandwidth in a way that can be recovered by signal processing, for instance signal spreading and/or forward error correction, can make the receiver much more tolerant to poor signal-to-noise ratio. However, it's better to prevent the damage being done to the signal in the first place, rather than trying to dig it out of the dirt later.

  • \$\begingroup\$ Excellent explanation, very didactic. I found "the 50% of the time" paragraph very illustrative and thank you for explaining non-linear behavior of a saturated amp so well. However I have a couple of original topic oriented questions. First I'm not dealing with RF, I'm dealing with audio. How can I match your explanation with a microphone's behavior? Will clipping do the same thing there? Will it behave non-linearly as well? Do you consider that audio amps are saturating? Sorry I can't modify or alter the hardware, I have to deal with the smartphone's inherent recording mechanism. \$\endgroup\$ – VMMF Apr 2 '17 at 5:20
  • \$\begingroup\$ Audio amplifiers block, just like RF amplifiers. The RF circuits will have some wideband tuned circuits between antenna and LNA, but that LNA is obliged to tolerate nearby strong channels while amplifying both nearby and the desired channel whether it is weak or strong. The strong nearby channels are the blockers, because the cellphones cannot afford to include 100 or 500 RF filters; the filtering is done after down-conversion to IF (intermediate frequency) where 8-pole or 10-pole filters are low-power and accurate. \$\endgroup\$ – analogsystemsrf Apr 5 '17 at 5:34

If you are using a correlator to improve the channel robustness, the resolution of the digitizer matters. As well as IP2 and IP3 etc of the front-end linearity.

With the correlator performance in mind, I agree with (1) and (2): for full correlator benefit, using e.g. sinusoids, the sinusoids must be of zero distortion. As you are discovering, channel overload degrades the dataeye, and the SNR at the dataslicer/comparator, and the BER. Thus channel-coding rears its head.

With OFDM, you can use the entire available set of sinusoids. Its your task to ensure the OFDM is noise-like and coexists with voice/music.

  • \$\begingroup\$ I'm using a matched filter (equivalent response to correlator at maximum value) for useful signal detection. The resolution of the digitizer is fixed is the AD of a Samsung S6. What do you mean "channel-coding rears its head"? \$\endgroup\$ – VMMF Apr 1 '17 at 0:43
  • \$\begingroup\$ Read my 3rd paragraph: consider using a wider bandwidth, randomly. \$\endgroup\$ – analogsystemsrf Apr 1 '17 at 3:24
  • 1
    \$\begingroup\$ Use the methods of Qualcomm, in their channel-sharing CDMA networks. Each phone gets a private pseudo-random pattern that has very low crosstalk with the other phones. Consider how 1010101010101010, multiplied by 1100110011001100, always comes to ZERO when filtered through a LowPass. You want that kind of channel robustness. By the way, that example does not use pseudo-random patterns. You'll have to study up on them. \$\endgroup\$ – analogsystemsrf Apr 2 '17 at 1:59
  • \$\begingroup\$ I have I doubt about your comment. I'm not performing "base band" transmission so I would have to modulate your zeros and ones sequences into "pass band" symbols. Wouldn't that make me lose any good properties the multiplication of the bit sequences had in the first place? When it came to simultaneous pass band symbols transmission, the multiplication of them wouldn't yield a good differentiation at the receiver side, would it? \$\endgroup\$ – VMMF Apr 2 '17 at 5:43
  • \$\begingroup\$ Using wide bandwidth transmission, and then reducing the bandwidth ---- perhaps by convolving with PRN that are precisely time-aligned with symbols ---- is a standard method to maintain a datalink even in presence of narrow-band undesired tones or in presence of uncorrelated PRN. You have the compute power, so use it. You can handshake occasionally to discuss the BER; if BER rises, then switch to PRN-modulated symbols. Plan a robust, multi-mode link. \$\endgroup\$ – analogsystemsrf Apr 2 '17 at 22:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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