Question

3. Show that the circuit below can be used as a DSB demodulator. Specify the properties of the filter in the out. O a(..+b(.) m(t)cos(2n fet) -cos(21 fc1) Filter a.)+b(.) Figure 2:

Answer 1

There is little sign mistake in the question.either at the summer both the previous output must be adding in nature.other wise they will cancel each other and nothing is obtained . Hence I assume that both the. Previous output are adding.
Solution Vlt) altbl 1² Y(+) me), cos(zated) -Cos (hatt) + filter mit X Haltbl 12 Vla by seeing the above Diagram: This is simply Synchronous detection, Here the DSB-sc signal is first multiplied carrier signal with locally generated low pass cos (2afet) and then pass through filter (LF. Hence, at of a low the Original modulating Signal 18 the output pass filter (LPF), recovered, -DSB - SC Vlt) Low Balance Modulator Enct) cos(aktet) 7 M& Pass filter cosleafet) A fm

or Phe Balanced modulator (or multiplier Product modulator) output can be given as.. V(+): (m(+). Cosenfet.cos (Ratet VIH = [mu). Cos?(2x+c+)] VH = mut 1+ cos curfen) 2 ULH mith mets. cos Cuartet) 2 tes) ] 2 Now, As there you can see in the digram Summer LUCE) & ves] yet YH 2V(2) mit) cos(unto + 18 (044) 2 yet: 2 met (441 MW) + mth cos (un fet) passed Now, output Signal, yet manecer through a Low pass filter Double frequency Signal 2fc. Which rejects the SAF mith, and Hence, modela ing Signal is recovered ine Spectrum of the cufut of Balance modulator ! UCH = met 2 + m(H). Coelenter] (m (+-2fe) MLH) 2 mt-efc) & mett 2fc) T (f+ 2

met) cos(2ktet) EL, [MCF- Mcf-te) & mif&te) a ted 2 VE) 1 m(o) 2 mo) I mo) 2fe for arte 2f+fm-fm ton Efe-fom 24 eft fm ine spectrum output of eff 18 of the VU) I mo) 7f - fm fm filter properties filrer. Hence pass Low a Since eliminare Higher it's main work like 2fc. frequency component