Re: [問題] 關于噪聲uniformly分布時候比特錯誤概눠…
※ 引述《shiyoung (睽違已久的ID)》之銘言：
: ※ 引述《supsymmetry (supsymmetry)》之銘言：
: : A binary communication system transmits signals s_i(t)(i=1,2).The receiver t
: : est statistic z(T)=a_i+n_0, where the signalcomponent a_i is either a_1=1 or
: : a_2=-1 and the noise component n_0 is uniformly distributed, yielding the c
: : onditional density functions p(z|s_i) given by
: : p(z|s_1)=1/2 for -0.2<=z<=1.8
: : 0 otherwise
: : and
: : p(z|s_2)=1/2 for -1.8<=z<=0.2
: : 0 otherwise
: : Find the probability of a bit error, P_B, for the case of equally likely sig
: : naling and the use of an optimum decision threshold.
: : 我才判決水平應該是\gamma=0,然后bit error probability 應該是
: : \int_{-0.2}^{0}1/2dz=0.1的.
: : 但是怎么都不能從原來的題目的已知中推導出來.
: : 求求解思路.謝謝.
.... Should be Bayesian.
choose S1 if
P(s1 | Z ) > P(s2 | Z) , where Z is the observation.
Assume S1, S2 are equally likely.. etc.
Prob. of Error = 0.1 is a correct answer,
However, you can choose arbitary decision boundary between [-0.2 , 0.2]
and achieve the same prob of error.
Or,in other words, what happens if the likelihood ration is equal to
the threshold ?
: From MAP criterion
: 1
: >
: p(z|s_1) p(z|s_2)
: <
: -1
: therefore P_B = Pr(s_1-s_2 < 0)
what's your symbol Pr(s_1-s_2 < 0) here ?
: s_1
: s_2 ┌─────┐
: ┌──┼──┐ │
: ─┴──┴──┴──┴──
: -1.8 -0.2 0.2 1.8
: P_B = int_{-0.2}^{0.2} int_{x}^{0.2} 0.5 dydx
: = int_{-0.2}^{0.2} 0.5(0.2-x) dx
: = 0.1x - 0.25x^2 |_{-0.2}^{0.2}
: = 0.01 - (-0.03)
: = 0.04
No idea what's talk about.
: 囧rz....跟你算的不一樣
: 我不保證我的就對
: 參考看看吧
: 有錯也指正一下
: 很久沒算這玩意兒了
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