Q4.B: For the synchronous normal AM demodulator shown in figure 4, If the 2-sided channel noise PSD is n/2 (W/Hz) : BAM (t) + n(t) BPF center = fe BW=2f. S. DC LPF foam=f block N. i) Prove that SNR, m(t)? nfm ii) Then Prove that you can write: VZ coswt m(t)? Figure 4 Si A2 + m(t)² nfm SNRO = 9AM (t) = V2[A+m(t)]cos w̟t |COS

Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Question
Q4.B: For the synchronous normal
AM demodulator shown in figure 4,
If the 2-sided channel noise PSD is
n/2 (W/Hz) :
S.
DC
Ыock
Si
BAM(t) + n(t)
ВРF
LPF
center = fe
%3D
BW=2f,
Ni
m(t)?
i) Prove that SNR, =
nfm
ii) Then Prove that you can write:
Vz coswt
Figure 4
m(t)?
Si
SNR.
A2 + m(t)² nfm
gAH(t) = v2[A +m(t)]cos w̟t
Transcribed Image Text:Q4.B: For the synchronous normal AM demodulator shown in figure 4, If the 2-sided channel noise PSD is n/2 (W/Hz) : S. DC Ыock Si BAM(t) + n(t) ВРF LPF center = fe %3D BW=2f, Ni m(t)? i) Prove that SNR, = nfm ii) Then Prove that you can write: Vz coswt Figure 4 m(t)? Si SNR. A2 + m(t)² nfm gAH(t) = v2[A +m(t)]cos w̟t
Expert Solution
Step 1

For the synchronous normal AM demodulator shown in the figure. 

Electrical Engineering homework question answer, step 1, image 1

Given : gAM(t)=2[A+m(t)]cos ωct

If the two-sided channel PSD is η2( W/Hz).

1. We need to prove that SNR0=m(t)2¯ηfm

2. Then we need to prove that SNR0=m(t)2¯A2+m(t)2¯Siηfm

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