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Detector for Amplitude Shift Keyed Signal

IP.com Disclosure Number: IPCOM000034862D
Original Publication Date: 1989-May-01
Included in the Prior Art Database: 2005-Jan-27
Document File: 2 page(s) / 37K

Publishing Venue

IBM

Related People

Maddox, RA: AUTHOR [+2]

Abstract

Digital information is carried by an amplitude shift keyed, or off-on keyed, signal through the presence or absence of a particular carrier frequency waveform in successive intervals of time, or time slots. Detection of the amplitude shift keyed (ASK) signal may be accomplished by heterodyning the ASK signal with a local oscillator signal to produce an intermediate frequency signal, which is then detected. Two such detection circuits are the subject of this article. In the first detector (Fig. 1), the received ASK signal E1 and the local signal E2 are multiplied together to form a signal E3. The signal E3 is passed through a low pass filter to produce a signal E4 including a signal component at the difference frequency between the received signal E1 and the local signal E2.

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Detector for Amplitude Shift Keyed Signal

Digital information is carried by an amplitude shift keyed, or off-on keyed, signal through the presence or absence of a particular carrier frequency waveform in successive intervals of time, or time slots. Detection of the amplitude shift keyed (ASK) signal may be accomplished by heterodyning the ASK signal with a local oscillator signal to produce an intermediate frequency signal, which is then detected. Two such detection circuits are the subject of this article. In the first detector (Fig. 1), the received ASK signal E1 and the local signal E2 are multiplied together to form a signal E3. The signal E3 is passed through a low pass filter to produce a signal E4 including a signal component at the difference frequency between the received signal E1 and the local signal E2. This signal takes the form: E3 = AB/2[cos[(l1+l2)t+d]+cos[(l1-l2)t+d]]
(1) In this expression, the signal E1 is equal to A cos (l1t+d), and the signal E2 is equal to B cos (l2t). The first low pass filter removes the l¶+lý term from the signal E3 in producing signal E4. The signal E4 is then squared to produce the signal E5: E5 = (AB)2/8[cos[(2l1-2l2)t+2d']+1] (2) In the expression for the signal E5, the phase angle d' differs from the original phase angle d due to phase shifts introduced by the first low pass filter. The signal E5 is passed through a second low pass filter, leaving only a DC term, which is independent of the phase difference between t...