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Sec. 5–15 Study-Aid Examples 399
SA5–3 FM Transmitter with Frequency Multipliers As shown in Fig. 5–43, an FM transmitter
consists of an FM exciter stage, a *3 frequency multiplier, an up-converter (with a bandpass
filter), a *2 frequency multiplier, and a *3 frequency multiplier. The oscillator has a frequency of
80.0150 MHz, and the bandpass filter is centered around the carrier frequency, which is located at
approximately 143 MHz. The FM exciter has a carrier frequency of 20.9957 MHz and a peak
deviation of 0.694 kHz when the audio input is applied. The bandwidth of the audio input is
3 kHz. Calculate the carrier frequency and the peak deviation for the FM signals at points B, C, D,
E, and F. Also, calculate the bandwidth required for the bandpass filter and the exact center
frequency for the bandpass filter.
Solution As shown in Sec. 4–12, a frequency multiplier produces an output signal at the nth
harmonic of the input, and it increases any PM or FM variation that is on the input signal by a
factor of n. That is, if the input signal has an angle variation of u (t), the output signal will have a
variation of nu (t), as shown by Eq. (4–73). Thus, the peak deviation at the output of a frequency
multiplier is (∆F) out = n(∆F) in , because ∆F = (12 p)
du(t)dt. The FM exciter output has a
carrier frequency ( f c ) A = 20.9957 MHz and a peak deviation of (∆F) A = 0.694 kHz. Thus, the FM
signal at point B has the parameters
The mixer (multiplier) produces two signals—a sum frequency term and a difference frequency
term at point C—with the carrier frequencies
and
(f c ) B = 3(f c ) A = 62.9871 MHz and (¢F) B = 3(¢F) A = 2.08 kHz
(f c ) C sum = f 0 + (f c ) B = 143.0021 MHz
(f c ) C diff = f 0 - (f c ) B = 17.0279 MHz
Because the mixer output signal has the same complex envelope as the complex envelope at its
input (see Sec. 4–11), all the modulation output parameters at the mixer output are the same as
those for the input. Thus, the sum and difference carrier frequencies are frequency modulated, and
the peak deviation for each is (∆F) C = (∆F) B = 2.08 kHz. The bandpass filter passes the 143-MHz
term. Consequently, the FM signal at point D has the parameters
(f c ) D = (f c ) C sum = 143.0021 MHz and (¢F) D = (¢F) C = 2.08 kHz
The FM signals at points E and F have the parameters
(f c ) E = 2(f c ) D = 286.0042 MHz and (¢F) E = 2(¢F) D = 4.16 kHz
(f c ) F = 3(f c ) E = 858.0126 MHz and (¢F) E = 3(¢F) E = 12.49 kHz
m(t)
Audio
input
FM
exciter
A × 3 B C D × 2 E × 3
Bandpass
F s(t)
Frequency
Frequency Frequency
filter
multiplier
multiplier multiplier FM
output
Oscillator
Figure 5–43
FM transmitter.