Lecture 07: Bit Error Rate (BER) of AWGN Channels

Поделиться
HTML-код
  • Опубликовано: 11 ноя 2024

Комментарии • 22

  • @mouctechy
    @mouctechy 3 года назад +4

    YOU'RE A GREAT LECTURE, THIS FIELD IS NOT EASY BUT YOU'RE MAKING IT AWESOME FOR US

  • @neelapuharshavardhanreddy1669
    @neelapuharshavardhanreddy1669 8 лет назад +4

    Thank you sir !! u saved my life

  • @yashiroisana2622
    @yashiroisana2622 7 лет назад +1

    thanks for the lecture

  • @pulkitmathur5942
    @pulkitmathur5942 6 лет назад

    Thank-you so much!

  • @flokkykari9429
    @flokkykari9429 4 года назад

    thank you sir

  • @unnikrishnan6588
    @unnikrishnan6588 7 лет назад

    good explanation sirthank you very much

  • @nuwanatthanayake
    @nuwanatthanayake 5 лет назад

    Sir,Your work is marvellous and magnificent

    • @jambajuice07
      @jambajuice07 Год назад

      where are u working currently ??

  • @blutzp6703
    @blutzp6703 3 года назад +2

    At 17:40 1/2*e^(-5) = 3.4 * 10^(-3) ..... what am I missing?

  • @yusufunlu2772
    @yusufunlu2772 4 года назад +1

    Sir, I love you.

  • @utkarshshukla2340
    @utkarshshukla2340 4 года назад

    sir,when you are talking about average power it is whether rms power because average power of sinusoidal wave is zero

  • @Khanofkhans00
    @Khanofkhans00 Год назад

    what is the mean of P in this video? I try to make my homework but the threshold voltage is 0.13. And if voltage greater than 0.13 the transmitted bit is 0. AWGN channel has 5dB SNR. I must derivate a probability function for the situation that is the transmitted bit is 0 but received bit is 1. Can anybody help me

  • @arpitwashal6852
    @arpitwashal6852 4 года назад +1

    can anyone give the matlab code for implementing the models for calculation of ber of various digital modulation techniques such as ASK, FSK, PSK, M-PSK. Also implement models to predict propagation losses for communication microwave signals

  • @anuraagtummanapally1395
    @anuraagtummanapally1395 5 лет назад +2

    At 17:32
    Q(x) EQ #0
    for 10 dB SNR, SNR = 10
    Pe EQ #1
    At 14:42:
    Pe = 7.82 * 10^(-4) ------------> EQ #2
    The result from EQ #1 is less than EQ #2
    But according to EQ #0:
    Pe

    • @blutzp6703
      @blutzp6703 3 года назад +1

      It's a mistake. the correct answer should be 3.37*10^(-3).
      3.91*10^(-4) = 1/2 * 7.82*10^(-4). Looks like the numbers got mixed up in his notes.

  • @Mk-mj4pp
    @Mk-mj4pp 3 года назад +1

    8:22

  • @GiantDanish
    @GiantDanish 6 лет назад +3

    Are you sure that the 1/2 coefficient in the approximation belongs there? The approximated value in the example seems to be pretty much exactly 1/2 of the exact value.

    • @PTAlisPT
      @PTAlisPT 6 лет назад

      You are right, the 1/2 shouldn't be there

    • @konchady1
      @konchady1 5 лет назад +1

      Check the Wikipedia article. The half is there for q function approx.

  • @zahraazaidan1152
    @zahraazaidan1152 3 года назад

    at 5:40 why does the lower limit of integration becomes( p/standard deviation) instead of p... can any one make it clear please

    • @tvsmohan184
      @tvsmohan184 3 года назад

      because we have to use new limits of integration when we substitute t=n/sigma. The previous limits belong to variable n so by substituting n limits of integration in the new variable t =n/sigma we get new limits as infinity and p/standard deviation.

  • @mohammadalmaazi
    @mohammadalmaazi 5 лет назад

    Indian . you are the worst one who try to explain communication theory. You never concern about the quality of your lectures but view number of your videos.
    Actually, you are disturbing people from finding the highest quality information