Lecture 09: Exact BER Expression for Rayleigh Fading Wireless Channel

[@VishalSingh-el7rx]

गुरू गोविन्द दोऊ खड़े, काके लागूं पांय।
बलिहारी गुरू अपने गोविन्द दियो बताय।।
u r great!! sir

[@refatkhan9256]

thanks you very much sir for such a beautiful explanantion.

[@justinle998]

Best explaination... very thorough thanks!

[@sachinramsuran7372]

Very thorough. Thank you

[@Dr-Asim_ihsan]

Sir, Very helpful.

[@prabhavathit4097]

sir , really this video for my project works

[@saurabhchandiiit]

Thank You Sir

[@karthikeyini1215]

Sir please explain Exact BER Expression for Nakagami Fading Wireless Channel

[@sachidanandprajapati9446]

YOu are a king

[@BlueAustralian]

So this video is nothing but the derivation of the formula that had been written directly in the previous video right?

[@prabhavathit4097]

sir, i need BER,SNR and achievable rate for wireless channel from equation y=hx +n how to find?

[@isisisis123]

why does u equals sqrt(2/mu)tan(theta)??

[@basics7930]

sir what about BER in MIMO fading channels

[@konchady1]

This proof is unnecessarily complicated. A much simpler proof can be got by swapping integrals, integrating rayleigh first and then completing the squares. You do have to use the expression for variance of gaussian also.

[@isisisis123]

Could you provide me with the proof you mentioned?? It would be helpful for my research...

[@konchady1]

@@isisisis123 I apologize to the professor for my choice of words. I was probably in a bad mood when I posted this. But it is true that a simpler proof is there. Let me try to explain it in subsequent posts.

[@konchady1]

@@isisisis123 Let us look at 3:40. Swap the integrals. Inner integral will be from 0 to sqrt(t/mu). Integrating this for 2ae^(-a^2) is easy. You get 1 - e^(-(t^2)/mu). The rest should be a simple Gaussian integral manipulation .Write it out and see.

[@suryateja9003]

@@konchady1 can you send me in telegram?