0
\$\begingroup\$

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?

\$\endgroup\$
  • 1
    \$\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
0
\$\begingroup\$

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 | |
\$\endgroup\$
  • \$\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

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.