pcshw4_soln

# pcshw4_soln - THE STATE UNIVERSITY OF NEW JERSEY RUTGERS...

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THE STATE UNIVERSITY OF NEW JERSEY RUTGERS College of Engineering Department of Electrical and Computer Engineering 332:322 Principles of Communications Systems Spring 2004 Problem Set 4 1. Phase Locked Loop Demodulator (quickie): (a) Using what you know about phase locked loops (PLLs), show that the PLL can be used as a FM demodulation circuit. SOLUTION: Just use the ˙ φ t output – it’s the derivative of whatever was inside the argument of that incoming cos2 π f c t ϖ t . That is, ˙ φ t 2 π f c ˙ ϖ t . So you filter the D.C., and throw in a few multiplicative constants and you’re done. (b) How can you modify your PLL to obtain a PM demodulation circuit? SOLUTION: Integrate the output of your demodulator above. But note, you’ve GOT to get rid of the DC offset, otherwise you’ll have a lawsuit on your hands when 2 π f c t exceeds about ten kilovolts – how fast would that happen if the listener were tuned to 99.9FM? :) 2. Hilbert Transform Stuff for Culture: The Hilbert-transform of a Fourier transformable signal m t , denoted by m H t , is defined by m H t 1 π m t t τ d τ In the frequency domain, we have M H f j sgn f M f where m t , M f and m H t , M H f are Fourier transform pairs and sgn f is the signum function. Using the definition of the Hilbert transform, show that a single-sideband modu- lated signal resulting from the message signal m t and carrier cos 2 π f c t of unit amplitude is given by s t 1 2 m t cos 2 π f c t m H t sin 2 π f c t where the minus sign corresponds to the transmission of upper sideband and plus to the transmission of lower sideband. SOLUTION: Consider first the modulated signal s t 1 2 m t cos 2 π f c t m H t sin 2 π f c t Applying Fourier transform to the above equation, we get S f 1 4 M f f c M f f c 1 4 j M H f f c M H f f c 1

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From the defenition of Hilbert Transform, we have M H f j sgn f M f where sgn f is the signum function. Equivalently, we may write 1 j M H f f c sgn f f c M f f c 1 j M H f f c sgn f f c M f f c From the definition of signum function, we note the following for f 0 and f f c sgn f f c sgn f f c 1 Correspondingly, the equation for S f reduces to, S f 1 4 M f f c M f f c 1 4 M f f c M f f c 1 2 M f f c In words, the above result means that, except for a scaling factor, the spectrum of the modu- lated signal s t is the same as that of a DSB-SC modulated signal for f f c . For f 0 and f f c , we have sgn f f c 1 sgn f f c 1 Correspondingly, S f reduces to, S f 1 4 M f f c M f f c 1 4 M f f c M f f c 0 In words, for f f c , the modulated signal s t is zero.
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