Question

Q: One type of modulators is shown below: y(t) Rs 10 Bandpass filter von Vc(t) 1- What is the name of this modulator, and what is it used for? 2- If m (-0.1 cos (2000 ) and Ve(t)=0.3 cos (40000 ) are the input of the diode, find the output of the diode if the input output characteristics of the diode is: Vo(t) = 0.3*Vi (t) + 0.2*Vi?(t) Then, find Vo(t), if the center frequency of the band pass filter is (ta = 20kHz), and the bandwidth of the filter is 2 kHz

Answer 1

velt) VCD Ś Bandpass | filter lult 1. It is called as Square Law Modulator. It consists of the following: - a. A non-linear Device Diode. b. A bandpass filter c. A carrier source and modulating signal ne velt) and mLt) respectively. It is used for generating Amplikido Modulation Waves 2. Giren, message signal mLt) = o.1 Cos (2000 tt) frequency In = 1000 Hz Carrier source v (t) = 0.3 cos (40000 Tt) frequency to - 20000 Hz They are connected in series with each other and their sum V; Lt) is applied at the input of the non-linear device such as

diode Vilt) - mLt) + velt) = 0.1 cos(2000nt) + 0.3 cos (40000nt) Lyu And ginen, input output characteristics of the diode is (2) V Lt) = 0.3 * V. (t) +0. 2* V² (t) Now, substituting Eg (1) in Eq (2) v (t) = 0.3 0.1 cos (2000 OTT4172 - 0.3 Oolcos (20007t) +0.3 cos (40000 tt) / + 0.2 0.1 Cos (2000 t) +0.3 cos (40000 ) VID - 0.03.COS (200015 t) +0.09 cos(40000T 0.2 0.01 COS² (2000nt) +0.06os (2000Tt) cost 40000 t)+0.09 cos? ( 400007+)] - 0.03 cos (20001 t) +0.09 cos ( 40000 t) + 0.00205 220001T +)+0.012 00s (200017+)* Cos (400001 t) +0.018 cos2 (400001t) Modulating signal :- 0.03 cos (200071 t) Carrier segnal :-0.09 cos(40000Tt) Am wave with only side bands:- 0.01260S (20001 t) Squared modulating signal :-0.002.Cos? (2000110) Squared carrier :-0.018 COS? (4000 OTT t )

consider only carrier signal and AM wave with only side bands terms. Output after band filter pass filter ; - The circuit (Le tuned circuit) is turned to frequency, fo=20kHz and its bandwidth is 2 KHZ. BW = 2 fm .fm = 1000 Hz = 2kHz Hence, the output contains only two terms. Nolt) = 0.09cos (40000Tt) + 0.012 * cos (20001Tt) * cos (HOOOOTE) = 6:09 + 0.012 C05 (20001 +)(40K TIC) V,Lt) = 0.09[1+0:012 (08(2000Mt)]cas (4 OKTt) Comparing the above volt) with standard AM wave equation le. Slt) - Ec [ 14m alt)] cos (2017et) 0.09