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FM demodulation - Essay Example

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Name of student: Instructor: FM demodulation using CD 4046 CMOS Phase-Locked Loop (PLL) integrated circuit. Introduction FM is obtained by the variation of the frequency components of the modulating signal voltage through a Voltage Controlled Oscillator so that the output varies according to the input signal.FM demodulation is the reverse process of modulation where the Voltage Controlled Oscillator is locked to the incoming modulated FM signal by the Phase-Locked Loop to recover the original input signal…
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FM demodulation
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FM demodulation

The PLL has a voltage-controlled oscillator, a phase detector, and a low pass filter connected in a feedback loop. The input voltage determines the frequency of oscillation at the output of the VCO (Sedra 1998). This fosc is equal to the intermediate frequency fi required to be 470 kHz in this case. The circuit is built on a proto board according to the block diagram shown fig1 bellow; The components required are; CD 4046 CMOS Phase-Lock Loop (PLL) integrated circuit Resistors of values; 1k, 10k, 18k (2). Capacitors of values; 0.1?F, 0.01?F, 3900pF and, Breadboard. The actual circuit is realized according to the circuit shown in fig 2 bellow. The circuit components in the PLL providing a centre frequency of 470 kHz are determined according to the equation, fosc = 1/2пRC. Therefore 470000Hz*2п = 1/RC and taking R = 100K then 1/C =470000* 2п*100000= 3.386pF and the preferred value is 3400nf. The capture range is given by 2fc=1/п (2пfL/R1)1/2 and for audio the maximum range is 20 kHz and R1 = 1k therefore the frequency of the filter is given by; (2*20000*п) 2 = 2пfL/R1 fL= (2*2000*п)2*1000/2п = 25.136 GHz and so the filter components are obtained as shown bellow; 25.136*109 = 1/2пRfCf and with Rf taken as 18K then Cf = 1/ (25.136*109*2п*18000) = 35176.4pF The lock in range is given by fmax – fmin and; fmin= 1/R2(C1+32PF) = 1/10000(3900+32)*10^-12 = 25432.35Hz fmax = 1/R1(C1+32PF) + fmin = 1/1000(3900+32)*10^-12 +25432.35 = 279755.85 Hz Hence lock in range = 254323.5 Hz. Also, capture range 2fc = (2K0fpVDD)1/2 K0 =VDD/2= 15/2 =7.5 2fc = (2*7.5*25432.5*15)1/2 = 7564.6Hz Discussion Did the loop demodulate NBFM? The loop demodulates narrow band FM that occupies the frequency range of 0-15kHz as this frequency range lies within the capture range and the lock range of the circuit. Could it be used to demodulate WBFM without any alteration? The loop demodulate the wide band FM as the low pass filter above has a value of 25.136 GHz, which is well above the capture range of WBFM, which is customarily around 10.7 MHz with a system bandwidth of 200 kHz. This bandwidth is within the centre frequency range of 470 kHz (Alencar 2005) What happened to the lock-in and capture range when the loop components were altered? The capture range is low if the cut off frequency of the filter is lower; this is achieved by varying the filter components i.e. using large filter capacitance, and resistor values. On the other hand, the capture range is made large by using lower values of filter components thereby increasing the cut off frequency (Carlson 2002). A wider capture range is desirable as it enables demodulation of WBFM while a small capture range is desirable as it enables the attenuation of high frequency components thereby improving the signal to noise ratio of the system. The lock-in range follows the above relation, as it is also directly proportional to the filter cut off frequency. How would you modify the PLL in order to demodulate WBFM? From the relation of WBFM, 2fc = (2K0fpVDD)0.5 it is seen that the capture range 2fc is directly ... Read More
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