Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C

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Step of 7 18.023P Refer to Figure 17 11 in the text book The transfer function of the filter normalised to the centre frequency is, = ww = Substitute Q=20 20 - T(ja) 2 Step 7 (a) Represent the square wave of amplitude V and frequency in series. The series contains no even The even harmonics relative to the fundamental amplitude is 0 Thus, the second harmonic relative to the fundamental amplitude Step (b) Determine the magnitude of third harmonic 1 20 3 3 1 1 60 = 3 9 3600 1 0.016667 = 3 ) =6.25x10⁻³ Thus, the magnitude of third harmonic is Step (c) Determine the magnitude of fifth 1 20 5 5 1 1 100 = 5 1 25 10000 1 0.01 = 5 =2.08x10⁻³ Thus, the magnitude of fifth harmonic is Step (d) The magnitude relative to 8th and 10th harmonics is (zero). Determine the magnitude of seventh 1 1 20 7 7 1 140 7 19600 0.0071 7 ) Step of 7 Determine the magnitude of ninth harmonic 1 1 20 9 9 1 20 9 1 180 = 9 -80 1 81 32400 1 0.00556 = 9 ) Step Determine the rms of harmonics to the = = /39.0625+4.3264+1.071225+0.39125025x10 Thus, the rms of harmonics to the tenth is