If I have a 20Mhz bandwidth provided by my 2.4ghz router then will my internet speed will be 20M symbols per sec which is equal to 20×4= 80Mbps (considering 16 qam modulator with 4 bits per symbol) Is it true?

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    \$\begingroup\$ Payload data rate will be smaller - overall data rate (including formatting bytes) will be 80 Mbps. \$\endgroup\$ – Andy aka May 12 at 7:28

The successful symbol rate depends on the channel settling, the phase flatness, etc

RF channels tend to have very steep filtering (at the post-down-conversion frequencies where multipole Intermediate Frequency active filters ) to allow packing lots of channels and lots of users into a few hundred MHZ of spectrum (such as 2.4GHz WiFi).

Thus the precise successful symbol rate depends on

  • the interference from adjacent channels,

  • reduced phase warping if adjacent channels are UNUSED on purpose to allow moderate IF filtering

  • the silicon area devoted to IF active filters, where phase-compensation requires MORE opamps and other silicon-area-hogs such as ratiod-capacitors to accurate place the Zeros

If you have a one-pole filter with Tau timeconstant, being a 1-pole allows 6.28 settling Tau in the bandwidth. At 8dB per Tau (a Neper), you have settling to 48dB error flow which allows dense constellations. For a cheap filter.

But one-pole filters prevent dense channel allocation, so are not used.

| improve this answer | |
  • \$\begingroup\$ the silicon area devoted to IF active filters In the last century indeed we did things like that. These days nearly all receivers are zero-IF or low-IF and directly feed the downconverted signal into an ADC. Also not sure how this answer will help OP. \$\endgroup\$ – Bimpelrekkie May 12 at 8:04

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