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Nguyen Huu Tam Ton ECE 586 Lab 3/26/ Lab 7: Frequency Modulation "Frequency Modulation (FM) is a type of modulation in which changes in the carrier wave frequency correlate directly to changes in the baseband signal." FM is considered an analog kind of modulation since the baseband signal is typically an analog waveform with no discrete, digital values." There are a few things to note regarding the FM signal. First, its envelopes are flat - remember that FM does not fluctuate the amplitude of the carrier. Second, its period (and thus frequency) changes as the amplitude of the message changes. Finally, as the message alternates above and below 0V, the signal's frequency alternates above and below the carrier's frequency. Part A:
Question 1: Draw the message and the modulated signal in time domain. Question 2: Why does the frequency of the carrier change? The amplitude change of the message signal changes the carrier signal's frequency. Part B: Check 20KHz step. Question 3: Using the scope, measure the magnitude of the FM signal, and calculate its power. Vpeak = 1.025 V, P = 𝑉𝑟𝑚𝑠^2 = 0.492 W Magnitude and Power Ground modulated.
2kHz sine modulated. Question 4: Using the scope, measure the magnitude of the FM signal, and calculate its power. Vpeak = 1.025 V, P = 𝑉𝑟𝑚𝑠^2 = 0. 503 W Question 5: Is the FM modulated signals with messages, respectively, 2kHz sine wave, and 0V DC have the same power? They have equal authority based on the answers to questions 3 and 4.
Frequency Component (Hz) Magnitude for each Frequency Component Power in each Frequency Component fc – 2fm 15400 Hz 0.141 0.01 W fc – fm 18000 Hz 0.374 0.07 W fc 20100 Hz 0.616 0.19 W fc + fm 22200 Hz 0.49 0.12 W fc + 2fm 24300 Hz 0.141 0.01 W
Question 6: Based on the measurements in your table, using Bessel table, can be found in textbook or on BB, find β (beta). P(0) = 0.5Ac^2 * J 0 (β)^ J 0 (β) = √(P( 0 ) / ( 0. 5 𝐴𝑐^2 )) J 0 (β) = √ 0. 19 / ( 0. 5 ( 0. 616 ) 2 ) 𝛽 = 1 Question 7: Calculated total power in all the frequency components in your table. Calculate the ratio of the obtained power to the FM power calculated in question 4. Ptotal = 0.4 W r = 0.4/0.49 = 82% Question 8: Measure the bandwidth of the FM signal. BW = fc + 2fm - (fc - 2fm) = 4fm Question 9: Calculate the ratio of the FM signal BW to the original signal BW. Can you
justify Carson’s rule in this case? Ratio = FM/Org = 4 : 1 => Carson’s rule applied. Question 10: Using Carson’s rule and the answer from Question 8, calculate β again and compare the answer in Question 6, comment the difference if any. BWc = 2(β + 1) x Wm β = BWc/(2Wm – 1 ) β = 4fm / 2fm - 1 β = 1 Question 11: How does the FM signal’s bandwidth compare to an AM signal’s bandwidth for the same message? FM bandwidth is significantly larger than AM bandwidth.
