《CMOS receiver front-ends....》学习问题讨论，请进
时间：10-02
整理：3721RD
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各位好，我最近才开始学射频方向的书，正在看Johan Janssens的 CMOS Cellular receiver front-ends from specification to realization 一书。
我现在碰到一个问题，怎么也想不通啊。
这本书的page 41, 3.5.2节里， 作者说：
“The required noise figure of the complete receive path is 8.87 dB, of which 3 dB has already been
assigned to the LNA. Taking into account the 18 dB gain of the LNA, the required noise figure
for the quadrature down-converter becomes 25.6 dB. However, in order to get some margin, the
noise figure specification of the receive path has been set to 5 dB, requiring a noise figure less
than 18.6 dB from the quadrature down-converter. For a single down-converter this boils down
to a required noise figure of 21.6 dB, provided that all the output noise between DC and 200 kHz
is taken into account. The detailed proof is given in Subsection A.3.2. ”
我在subsection A.3.2里找的公式，算出来，对于8.87dB的总NF, 来说，down converter的NF 应该为16dB多一点，也不该是25.6dB啊。
如果按照 拉扎维的 射频微电子 一书中，找的计算级联系统的NF公式来算的话，也得不到25.6dB啊。
我非常想知道，是我算错了，还是我有地方没考虑清楚。或者是作者的一点小失误啊。但作者是鲁文大牛，我也不能轻易怀疑。
所以，我想请各位读过这本书的朋友看下。 教下我。在此谢谢了~
我现在碰到一个问题，怎么也想不通啊。
这本书的page 41, 3.5.2节里， 作者说：
“The required noise figure of the complete receive path is 8.87 dB, of which 3 dB has already been
assigned to the LNA. Taking into account the 18 dB gain of the LNA, the required noise figure
for the quadrature down-converter becomes 25.6 dB. However, in order to get some margin, the
noise figure specification of the receive path has been set to 5 dB, requiring a noise figure less
than 18.6 dB from the quadrature down-converter. For a single down-converter this boils down
to a required noise figure of 21.6 dB, provided that all the output noise between DC and 200 kHz
is taken into account. The detailed proof is given in Subsection A.3.2. ”
我在subsection A.3.2里找的公式，算出来，对于8.87dB的总NF, 来说，down converter的NF 应该为16dB多一点，也不该是25.6dB啊。
如果按照 拉扎维的 射频微电子 一书中，找的计算级联系统的NF公式来算的话，也得不到25.6dB啊。
我非常想知道，是我算错了，还是我有地方没考虑清楚。或者是作者的一点小失误啊。但作者是鲁文大牛，我也不能轻易怀疑。
所以，我想请各位读过这本书的朋友看下。 教下我。在此谢谢了~
不懂。
我算过了，书上给的数据没有问题。这个问题当然用级联噪声系数公式来计算，不过要注意的是，级联公式中的各个数据不是取对数的值，而是绝对数值。
然也。
如果是对数，作除法似乎就不好解释了。
dingyie xian
NF=10logF，级联公式用的是F来计算，这里的Gain也是绝对值