Lecture 14: BER of Multiple Antenna Wireless Systems

[@mvenugopal4129]

At the end 23:00 (1-lamda)^2 should be divided by 4.

[@ayah3672]

clear and NICE!! thank you very much 😍😍😍😍😍

[@lh6619]

What would the BER be if the receiver didn't have the CSI and could not perform MRC? So instead of SNR after MRC you had some sub-optimal scheme at the receiver? Great lectures by the way, I am from a maths background going into comms and these lectures have made things SO much clearer than any of the literature I have found!

[@omar778]

thanks

[@lum2578]

12:50 Does anyone have any documents on how to get that solution? I am a bit confusion there

[@elemento8763]

Hey @Lum, were you able to find the steps to this solution?

[@cyk_official]

Shouldn’t the conclusion be restricted to BPSK only?

[@sunshineqi1861]

good question. I have the same consideration.

[@rajbhushan3541]

hi, u got any explanation for this?

[@desdasdass]

Hi Prof, why we can aprox g as Chi-Squared when g is not a sum of squared normal IID Gaussian RVs but a sum of squared IID Rayleight RVs? Thanks!

[@sandeepmukherjee3909]

desdasdass This is because we are considering h_i = x_i + jy_i, where x_i and y_i are independent normal distributed random variables with zero mean and variance of sigma^2/2. So abs(h_i) is Rayleigh distributed and summation of (abs(h_i))^2 summed from i=1 to N is central chi-squared distribution with 2N degees of freedom, for the special case of N=1, we get the exponential distribution.

[@rays14]

@@sandeepmukherjee3909 And what? We sum (abs(h_i))^2 from i=1 to N and abs(h_i) are Rayleigh distributed but not Gaussian. Hence, we can not say that we have chi-squared distribution. The right version i guess is the following: we have Chi-squared distr for (abs(h_i))^2 by definition and also we have the sum of such values has Chi-squared distribution by definition too.