文档预览： A律十三折线 c1 - 符号位；c2c3c4 - 段落码；c5c6c7c8 - 段内码。共8bit。 480k Baud 建立物理模型！ 5Challenge the future 5 2012-5-16 通信原理第一过零点带宽 = ... 点击下载

文档预览： C2C3C4 段内码的权值 C5C6C7C8 B1 B2 B3 B4 B5 B6 B7 B8 B9 B10 B11 B12* 1024 512 256 128 64 32 16 8 4 2 1 0.5 8 111 512 256 128 ... 点击下载

文档预览： Lemma A5: If c1, c2, c3, c4 is a test sequence, then ρ（ci, c1c2c3c4） = 1 /4 for all i = 1, 2, 3, 4. Proof: Let c = c1c2c3c4 and without loss of generality assume ρ... 点击下载

文档预览： ＂CONDL＂,＂04/05/01＂,＂10/05/02＂,＂04/05/11＂,＂C2C3C4＂＂ALTUSPUBLICSCHOOLDISTIRCT#18＂,＂WNC714＂,＂1194＂,,＂ALTUS,OK＂,＂IF＂,＂... 点击下载

文档预览： Lyx, lyxose; Ara, arabinose; Rib, ribose)ConRgurationan(C±O±C) Ring and side group vibrations (C±C), (C±OH), (C±H)Glc C2C3C4 1149 1106 1079 1035 994... 点击下载

文档预览： J16 = d5c01(c2c3c4 s2s4) d5c4s3s01; J17 = 0; J21 = (1/2)rs0 c(ac0 + c01((d3 + d5c4)s2 + d5c2c3s4) d5s3s4s01); J22 = (1/2)rs0 + c(ac0 + c01((d... 点击下载

文档预览： 6, the mast consists of equilateral triangle C2C3C4hinged via a revolute joint to boom CIBo at point C4.The key aspect of this mast concept is that theconfiguration o f ... 点击下载

文档预览： then ρ（ci, c1c2c3c4） = 1/4 for all i =1, 2, 3, 4.Proof: Let c = c1c2c3c4 and without loss of generality assume ρ（ci, c） ≥ ρ（cj, c） wheneveri ≤ j. Hence, by Lemma A4,... 点击下载

文档预览： bugsy@bugsbennett.com＂,＂PAULBENNETT＂,＂C2C3C4i＂＂1616＂,＂BUGSLTD＂,＂POBOX270413＂,＂LASVEGAS＂,＂NV＂,＂89127＂,＂702-648-6030＂,,＂... 点击下载

文档预览： (1-c2c4)(c8 +c3c7) c3c4 -1 cot毸1毵A21 = c2 -c3 c3c4 -1 tan毸1毵+ c3 -c2c3c4 c3c4 -1 cot毸1毵A22 = c3c4 -c2c4 c3c4 -1 tan毸1毵- c2c4 -1 c3c4 -... 点击下载