By L. A. A. Warnes (auth.)
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Additional info for Solutions to Problems: Electronic and Electrical Engineering
Sample text
1. + s2 Io.. 9 dB. When w > > 1
The exact response is 3 dB down at the comer frequency as usual. 4. 41 ( •) 180 20 10 dB 0 exact _:;>--, 2 --- 165 ' 150 4 exact j 135 120 -10 105 -20 90 log 10 (J) 0 2 3 .... 5a, VB = V0 and at node A, VA = VB Vo and no current flows in the feedback loop, nor from the input: Vo = VIn• the amplitude response is 0 dB at low frequencies and the phase response is oo. At high frequencies R > > 1/jwC, so that VB =VA = 0- the capacitor is a short-circuit to ground. The circuit is then a feedback amplifier with unity gain and 180° phase shift: Vo = -VIn.
5 + 1/s = -12s s2 + 5s + 4 i = - s x -12/s s+ 2 z -12s (s + 4)(s + 1) Then we find A(s + 1) + B(s + 4) = -12 (s + 2)Z A = s+4 B +-s+ 1 = -12s Equating coefficients of s: A + B = -12 Equating constant terms: A + 4B = 0 From which we find A = -16 and B = 4, so that i(t) = [4exp( -t) - 16exp ( -4t)] u(t) A The transformed voltage is given by = A Proceeding as before we find A(s + 1) + B(s + 4) ... 0. 15s) + 45s 2 Hence i ... 1 exp( -1 OOOt)]u(t) A. 362 mV. Q ! 5) 250, so A = [20 A - ;!!! 5t)]u(t) A.
