Where ? = conductivity
Given -> ? = 0.13 m2/v-s = 0.13 x 104 cm2/V sec
P = 2.25 x 1015/cm3
We have, ? ni = 1.5 x 1010
Also n.p. =
? n = /p
= (1.6 x 10-19 x 0.13 x 104 x 2.25 x 1015) x
= (0.468) (4.5 x 1015)
? = 2.106 x 1015 ?/cm
J = ?E
? Current density = 2.106 x 1015 x 3.620 x 10-19
= 7.6237 x 10-4 A/m2.
n = 8
Thus (SNR)dB = 1.76 + 6.02 x 8 = 49.92 dB.
.
AC cos?ct
for envelope detection ? < 1 ? < 1 ? Ac should be at least-2.
= 2H log2 V bits/sec
V - discrete levels (Here Binary i.e. 2)
H - Bandwidth
? Maximum data rate = 2H log2 V
= 2(3 KHz) log2 2 = 2 x 3 x 103 x 1
= 6 x 103 = 6000 bps.
Power gain = 10
Thus directivity .
Here,
a = 1.59'' = 40.386 mm = 4.04 cm
? .
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