【正文】 重復(fù)式 ()，接受信號 功率為 PR ： () RTTxR GGPdgP )(10 102?? Alternatively（做為選擇 ,二者擇一地） one could calculate the probability that the ratio of actual signal power received at a mobile terminal to the interference power, the measured SIR, rather than the average SIR, was greater than a specified threshold. Repeating () here, we have the received signal power PR given by： () RTTxR GGPdgP )(10 102?? α是由于多徑造成的小尺度瑞利衰落 和萊斯衰落的因子，這里忽略多徑效應(yīng)。 Consider first a rural（農(nóng)村） region. Say the density of mobile terminals is two terminals per km2. Then the cell radius should be set at about R = km. Say now a suburban region has a mobile terminal density of 100 mobiles/km2. The corresponding cell radius drops to km. In an urban region with a mobile density of 1000 mobiles/km2, the cell radius would reduce further to km. Probabilistic signal calculations 3． 5 信號的概率計算 在一個小區(qū)內(nèi)，更合理的性能測量應(yīng)該 為在這個點所處移動臺（下行鏈路）或基站 （上行鏈路）接收到的信號功率大于任何干 擾信號功率的概率。 郊區(qū)移動終端 的 密度為 100個／ km2移動臺 ，相應(yīng)的半徑為 R= km （ 100（個） =173／ R2 （個 /單位面 積）） 。 To keep the blocking probability PB to less than 1%, the Erlang load allowed turns out to be, using () or tables or plots based on that equation, A = 101 Erlangs. Say a typical user makes 200secondlong calls once every 15 minutes on the average, then the total call arrival rate is λ = call/sec, and 15 x 60/2 = 450 users may be acmodated in each cell. 現(xiàn)在假設(shè)這些用戶在小區(qū)內(nèi)為均 勻分布，半徑為 R的六邊形小區(qū)面積為： 可接受的用戶密度為 : 450/=173／ R2個／單位面積。 Now consider the impact of these calculations on the sizing of a cell. we focus first on an AMPS example. Recall that with a total of 832 frequency channels and C = 7 reuse, 118 channels are available per cell. 要保持阻塞概率 PB小于 1％，通過 使用式 ()和基于此等式的表格和圖 ，愛爾蘭厄蘭負荷為 A=101愛爾蘭厄蘭 。 If the typical call lasts 200 seconds , on the average, but a mobile user attempts a call every ten minutes, on the average, the number of users allowed in the cell is now 250. So cell phone or terminal usage clearly plays a role in determining the number of mobiles that may be acmodated. 現(xiàn)在來考慮小區(qū)尺寸的計算所帶來 的影響。 Consider now the significance of these numbers in relation to mobile usage of a given channel. Say a call lasts 200 seconds on the average. Say a typical user makes a call every 15 minutes, on the average. If 100 channels are available and a blocking probability of 1% is desired, 378 users can be acmodated. (How is this number obtained?) If the average call duration（持續(xù)，持久） increases to 400 seconds, only 189 users can be acmodated. 如果每個呼叫平均持續(xù)了 200 s， 但是平均每隔 10 min用戶進行一次呼 叫，那么小區(qū)中的用戶數(shù)量只能有 250 個。平 均每隔 15 min用戶進行一次呼叫，有 100個有效信道，呼叫阻塞概率為 1％，則可以容納 378個用戶同時 使用 (這個數(shù)字是怎么得到的 ?)。 現(xiàn)在來考慮上面討論的幾個量與給 定信道靈活使用之間的關(guān)系。如果 A增加 13％，到 95愛爾蘭厄蘭，則阻塞概率將變?yōu)? 5％，增加了 5倍。 Consider the converse now. Say the number of channels is fixed. How does the blocking probability vary with Erlang load? 反過來，信道數(shù)量是固定的，阻塞 概率會隨著愛爾蘭厄蘭負荷發(fā)生什么樣 的改變呢 ? Let N = 100 channels, to be concrete. To keep PB at 1% or less, a load of no more than 84 Erlangs can be allowed. If A were to increase by 13%, to 95 Erlangs, the blocking probability would jump to 5%, a fivefold increase! At low blocking probabilit