The analog signal given below is sampled at 1200 samples per second. m(t) = 2 sin 120 pt + 3 sin 240 pt + 4 sin 360 pt + 5 sin 480 pt + 6 sin 600 pt + 7 sin 720 pt. The folding (maximum) frequency is __________ .

Where ? = conductivity

Given -> ? = 0.13 m²/v-s = 0.13 x 10⁴ cm²/V sec

P = 2.25 x 10¹⁵/cm³

We have, ? n_i = 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.

.

A_C cos?_ct

for envelope detection ? < 1 ? < 1 ? A_c should be at least-2.

Overall voltage gain will decrease as feedback signal comes into picture and since it is current-series feedback, input impedance increases.

= 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 .