Question

1. To transmit an AM signal, how would you minimize power requirement. If an 150 kW AM transmitter modulates at 95%, calculate the required power for one of the side band signals and determine its ratio with the AM transmitter power. [10] 2. Sketch and describe a simple AM Modulator circuit, and explain how it generates the AM signal with the signal status at each stage, starting from the mixture of information signal and carrier signal. [10] 3. Sketch and describe a AM De-modulator circuit with an RC low pass filter and explain how one would retrieve the original information signal. 4. Explain how would our derive the AM modulated signal, and write a mathematical expression for the e(t), with plots in spectral diagram. [10]

Answer 1

we will minimize the by sanding only single power tauirement side-band. لطمطم هاكا کوه - اد ptotel - Pellty P = 150x103 Hoggy? E 103.359kw Ptofel a Pc & Pistol - PSB = P total -Pc b. b4 kW lous-sideband . PB a 96.64 kW 23,32kw

classmate Date Page modulation ! Definition ! It is the process in of carrier signal i.e. whe varies accord. amplitude variation which one of the parameter amplitude , tregn or phase to message signal's Amplitude variation modulation : - let mit) - message signal carover sigmal (CH: AC Cos27. fet General expr at AM s/y = (SAM = Aclit ka mlt)] Cosenfell ..Ka. Amplitude sensitivity of AM modulator SAM(A) . ACos 27 fet + Ac mlt). Ka Coszafct Cagouier modulated Samit: cit)t ka mit)clt) Because of wasted but extra carrier tom demodulation becomes power will be simple. spectrum . analysis of Multi-tone ! LAM: ___ -mith R TMTEJ BWL-D -CS Hz SAMlt): A cos anfitt Ackqm Lt) Cosan fet USB KAM LSB LB A USB_ ds Scanned fittor CamScanner -tc fet w 6 few fe fcto 3 SE

Demodulation of AM:- Envelope detector synchronous detr or or Diode coherent det detr (451) (any value of la) el Envelop Detector :- SAMLE) [ { RL MIE) E.D. is very simple and cheaper . LED extracts positive envelope of applied signal For proper envelope detection , Rec a should be possibly minimg. RSC high i UUUU 0 proper choice of RLC r proper choice / (1) RC L L X (ii) 1 RLC C 1 fessfühed Bite > fbingana) clipping ) - Scanner

Single Tone AM : - let mlt) Am cos 20 fimt SAM(t) = A (it ka Am Cos. 21 fmt) Cosan fet __ Ka Am = a + modulation index Cills | Max [ Ka mlu] uxloo % of modulation depth of modulation r : 43 Demod simple u<l - Under mod" Mal - Coritical min us! 7. over imod? f Santt): Ac[it_u Cos 27 fm t) cos enfet. I SAM (t) = A Cos zx fet & ACM cosan fet Cos zx fmt | Sambue Ac Cosza fet + Açu cos 28 (fet.fm) Ezt Açucas 204te-tant Courie USB SB Scanned with CamScanner

classmate Date Page - -रि-4 4 ACH -- - - - - - L LSB USB - - -fc-fim fe fetfm o te-am fc fut fm AM BW = (fot fm) - ( fc- fm) = afm = 2 x msg. frequ. canned with FamScanner
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