I'm learning how VCOs work, and I'm feeding it a sawtooth waveform to see the impact on my WiFi. This VCO outputs in a 2.4ghz spectrum, and to understand how it sweeps, I'm passing in a sawtooth waveform and watching how it impacts my laptop wireless (ping drops).

But I'm only seeing my WiFi drop out when I send garbage to the VCO. What am I doing wrong? I would expect that the sawtooth would make the VCO sweep across the frequency, knocking out my WiFi.

I've tuned the sawtooth voltage to match the VCO tuning need ( 0 - 2.4v). I have a 555 creating the sawtooth, and a MAXIM 2750 VCO to create the frequency.

Scenario 1: Feeding a sawtooth waveform, WiFi millisecond pings don't change. Blue = waveform OUT to VCO. Yellow = 555 digital OUT

555 Sawtooth. Yellow is 555 digital OUT, Blue is the sawtooth OUT

Scenario 2: I pull the capacitor from the 555 wavetooth circuit, passing noise, and the WiFi drops out. Blue = waveform OUT to VCO. Yellow = 555 digital OUT

Sawtooth circuit without capacitor sink. Yellow is 555 OUT, Blue is waveform OUT

Sawtooth Oscillator

Play with sawtooth circuit here

  • \$\begingroup\$ You'll always cause more interference by transmitting noise than by sweeping across frequencies. The sweep rate on the VCO may also be too fast, as WiFi only uses a 20 MHz channel for uplink/downlink, and the datasheet for the VCO shows a sweep rate of 140 MHz/V applied. \$\endgroup\$ Jul 1, 2016 at 17:18
  • \$\begingroup\$ @captainj2001 that make sense, thanks. And here I thought I was doing the "right" thing by using a sweep, where noise is perfectly fine! \$\endgroup\$ Jul 2, 2016 at 21:00

2 Answers 2


Your WiFi radio is essentially a packet radio, and transmits and receives packets (called "frames" in the 802.11 protocol) of data when there is data traffic.

When you drive the VCO with a sawtooth, if your linearly-sweeping VCO output happens to hit your WiFi channel frequency band during a ping packet transmission, it may cause an error in that packet, and the WiFi protocol will then request a retransmission, by which time the VCO frequency will have "moved on" to a different WiFi channel. Because the VCO sweeps relatively slowly (compared to the ping packet transmission rate), your chances of hitting a packet are quite low, unless you are almost filling the channel with packets (like downloading a large file from your local network drive).

However, when you drive the VCO with noise, it's frequency hops around really rapidly, and the chances of it hitting your WiFi channel frequency during a ping packet transmission are greatly increased, hence the consequent drop of your WiFi channel.

  • \$\begingroup\$ Ahh, I see - thanks. So I'd need to be in lockstep somehow. The VCO manual says it does 140MHZ/V, so I'd have to take the 20MHZ WiFi bandwidth, slow down the VCO sweep to match. \$\endgroup\$ Jul 1, 2016 at 23:12

If you look at the frequency spectrum for your 3 kHz sawtooth, you will only see energy at 3 kHz, 6 kHz, 9 kHz, 12 kHz.... These would show up as offsets from the carrier frequency (e.g., 2.400003 GHz, 2.399997 GHz, 2.400006 GHz, ..., assuming your DC tuning voltage gives exactly 2.4 GHz) The energy is highest at 3 kHz offset and decreases at each harmonic (by 3 dB each, I think).

Depending on what wi-fi channel you're using, this means that you would actually only be generating interference over a quite small range of the spectrum, just a small part of a single channel. Your interferer will have a 3dB bandwidth of just 3 kHz, and a 100 dB bandwidth of just +/- 100 kHz.

  • \$\begingroup\$ This would be true if the sawtoooth were amplitude-modulating the VCO. But the sawtooth is frequency-modulating the VCO, so there is energy at all the intermediate frequencies as well. \$\endgroup\$
    – AndyW
    Jul 1, 2016 at 18:08
  • \$\begingroup\$ This answer is not correct. \$\endgroup\$ Jul 2, 2016 at 22:19

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