Question

8. (15 points) Design an Armstrong indirect FM modulator to generate an FM signal with carrier frequency 97.3MHZ and Af = 10.24kHz. A NBFM generator of fe Af = 5Hz is available. Only frequency doublers can be used as multipliers. Additionally, a local oscillator (LO) with adjustable frequency between 400kHz and 500kHz is readily available for frequency mixing. 20KHZ and %3D

Answer 1

Step 1: First find the n₁ and n₂ by using the formula of ∆f₄=n_1n2∆f_{1 ,}

_{Step 2: Calculate output of first multiplier}

fc₂ =M₁fc₁

Calculate the input of second multiplier

fc₃=fc₄/M₂,

Step 3:Find the valid f_LO (Range 400 KHz to 500 KHZ)

mu mlt) IMBEM I Frequency or Faequency L multiplies multiplier Int (0) fa=20kHz Afi - SH3 ñ two (4= 97.3MH3 aty = 10.24kHZ The Locat Oscillator adjustable 400 KHz to SCOKH3. frequency range we know that the Frequency deviation is Afy a min2 Afi where in factor of Flys+ multiples n2 Factor of second multipies min2 = t4 10-24 X1000 - Afy 1024 X 1000 = - 5 = 2048 In the problem given, we have only doublers sort min2 2048 mine = 211 of mina= 25.26 11 = 25, 12:26 n1= 32 m2=64

The output of the First multiplies: fc2= nix for = 32 X 20 = 640 KH3 The input of the second multiplier f63 = = 97.3 64 a 1520.316142 so that the Frequency of the oscillatoy fc3 = $c2 + fro 1520-31= GIOKH3 I flo flo = 980.31 KHZ For Faequency consides oniy tve sign but fro Not im the range of local oscillatoz. so this design is not right. so now consider mine=21 = 24 27 ni = 24=16 72-27 = 128

NOLO the output of the first multiplier fc2= nixtes = 16X20 = 320KHZ The imput of the second multiplier 5c3 = fc4 = 97.3 X1000 -= 760.156kHa 128 N2 so that the frequency of the Oscillatoy f(3= fcz i fho 760.156 = 320flo flo = 440.156 KHZ consider only tve sign. flo- 440.156 KH3 Tie frequency of the local oscillator is lie in between 400 to soo kit2. So this designi's correct. So the i nie 16, 92-128, flo - 440.156 KH3 TO