By L. A. A. Warnes (auth.)
ISBN-10: 0333580001
ISBN-13: 9780333580004
ISBN-10: 0333638034
ISBN-13: 9780333638033
ISBN-10: 1349137499
ISBN-13: 9781349137497
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Extra info for Solutions to Problems: Electronic and Electrical Engineering
Sample text
Without going into too much mathematics, we can write a = 100 - a (a < < 1) and then vo(t) = (100- a)Vexp(-100 + a)t _ lOOVexp(-lOOt) -a = (100 - a) Vexp ( -lOOt)exp (at) -a -a lOOVexp( -lOOt) -a Solutions to problems = (100- a)Vexp(-lOOt)(l +at) _ 100Vexp(-100t) -a -a = (100 + lOOat- a) Vexp( -lOOt) - lOOVexp( -lOOt) -a = Vexp( -lOOt) - lOOVexp( -lOOt) 47 where terms in dl and higher are ignored. This agrees with the previous problem's answer. 5647 Vat t = 20 ms. 4353 exp( -t/35), moving the time origin along 20 ms for convenience.
17b we see that the j 5 0 is in parallel with the 2 0 giving an impedance of Zz = j 10/(2 + j 5) 0, which is in series with -j 4 0. This impedance is '7. _1_L_5_. 91 o = 4 . 15. s. s. [T/2- sinwl] T ~! s. value is Irmr. s. value is A square wave can be treated as DC + AC for power purposes only if the time spent at the peak value is the same as the time spent at the trough. If the square wave is shifted by D. s. s. value is the same as the square root of sum of the squares of the alternating and direct components.
1b, if the op amp is ideal, /in= 0 and Zm = oo. 1c, /in = (Vin - V0 )IR. 1d, lin = (Vin - V 0 )IR. The voltage at node A is Vin and then Substituting for Vo in the equation for V in gives V in - V in(l + lljwCR) ... R = -v. Rl + jwCI 1 + jwCR2 ... CR~> or wCR2 > > 1 and wCR 1 < < 1; Zin = jwCR 1R2 • The circuit is an active inductor of inductance Leq = CR 1R2 = 10- 7 x 500 X 107 = 500 H. 5, V8 = Vo and the sum of the outgoing currents is V 0 /R + (V0 => - VA)jwC = 0 VA = V 0 (1 + lljwCR) Then at node A the sum of the outgoing currents is (VA- Vu)jwC +(VA- VjjwC +(VA- Vj/R = 0 VAU2wC + 1/R) - V 0 (11R + jwC) = Vm,iwC Substituting for VA leads to VJ(1 + 1/jwCR)(2 + 1ljwCR) - (1 + 1/jwCR)] = V in vo vin 1 1 = ---------- = ---------------1- llw2 C 2R 2 + 2/jwCR (1 + lljwCR) 2 = a?
Solutions to Problems: Electronic and Electrical Engineering by L. A. A. Warnes (auth.)
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