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## AD835ARZ Datasheet(HTML) 12 Page - Analog Devices

 12 / 16 page AD835Rev. D | Page 12 of 16SQUARING AND FREQUENCY DOUBLINGAmplitude domain squaring of an input signal, E, is achievedsimply by connecting the X and Y inputs in parallel to producean output of E2/U. The input can have either polarity, but theoutput in this case is always positive. The output polarity can bereversed by interchanging either the X or Y inputs.When the input is a sine wave E sin ωt, a signal squarer behavesas a frequency doubler because()(ωt )UEUωtE2cos12sin22−=(6)While useful, Equation 6 shows a dc term at the output, whichvaries strongly with the amplitude of the input, E.Figure 24 shows a frequency doubler that overcomes thislimitation and provides a relatively constant output over amoderately wide frequency range, determined by the timeconstant R1C1. The voltage applied to the X and Y inputs isexactly in quadrature at a frequency f = ½πC1R1, and theiramplitudes are equal. At higher frequencies, the X input becomessmaller while the Y input increases in amplitude; the oppositehappens at lower frequencies. The result is a double frequencyoutput centered on ground whose amplitude of 1 V for a 1 Vinput varies by only 0.5% over a frequency range of ±10%.Because there is no squared dc component at the output, suddenchanges in the input amplitude do not cause a bounce in the dc level.AD83512348VOLTAGEOUTPUTR297.6ΩR1R3301ΩX2VPWY1X1Y2+5VC1–5VVNZ765VGFigure 24. Broadband Zero-Bounce Frequency DoublerThis circuit is based on the identityθ=θθ2sin21sincos(7)At ωO = 1/C1R1, the X input leads the input signal by 45° (and isattenuated by √2, while the Y input lags the input signal by 45°and is also attenuated by √2. Because the X and Y inputs are 90°out of phase, the response of the circuit is()()(tUEtEtEUWω=°+ω°−ω=2sin245sin245sin212)(8)which has no dc component, R2 and R3 are included to restorethe output to 1 V for an input amplitude of 1 V (the same gainadjustment as previously mentioned). Because the voltage acrossthe capacitor (C1) decreases with frequency, while that acrossthe resistor (R1) increases, the amplitude of the output variesonly slightly with frequency. In fact, it is only 0.5% below its fullvalue (at its center frequency ωO = 1/C1R1) at 90% and 110% ofthis frequency.