Download Solutions to Problems: Electronic and Electrical Engineering by L. A. A. Warnes (auth.) PDF

By L. A. A. Warnes (auth.)

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Extra resources for Solutions to Problems: Electronic and Electrical Engineering

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At 9 rad/s and below the phase response will be 90°. 8 rad/s the phase response will be 90° - 45° = 45°. At 1 krad/s the phase response is 90° - 90o - oo = oo. At 11 krad/s the phase response is 90° - 90° - 45° = -45°. Finally, at 110 krad/s and above the phase response is 90° - 90° - 90° = -90°. 3 shows the Bode plot. ----:---=:=+ 2 Thus the comer frequency is we = 11CR2 = (20 x 10- 9 x 500 x lQJ)- 1 = 100 rad/s and the gain is R2 /R 1 = 100 or 20log 10 100 = 40 dB. 4, which is the Bode plot. The exact response is 3 dB down at the comer frequency as usual.

8 rad/s and 11 krad/s the response will be flat. It will then fall at 20 dB/decade above 11 krad/s. At 9 rad/s and below the phase response will be 90°. 8 rad/s the phase response will be 90° - 45° = 45°. At 1 krad/s the phase response is 90° - 90o - oo = oo. At 11 krad/s the phase response is 90° - 90° - 45° = -45°. Finally, at 110 krad/s and above the phase response is 90° - 90° - 90° = -90°. 3 shows the Bode plot. ----:---=:=+ 2 Thus the comer frequency is we = 11CR2 = (20 x 10- 9 x 500 x lQJ)- 1 = 100 rad/s and the gain is R2 /R 1 = 100 or 20log 10 100 = 40 dB.

619. 01t)]u(t) If the supply voltages are v ±15 V, then the time to saturate will be when vo(t) ~ =:: t =:: The exponential part is negligible. 11, v8 = 0 and then by KCL vA -vo fD.. 2 x 10 - 8()()2 (s + 800)2 + 56()()2 = which can be detransformed to vo(t) = -10exp( -800t)sin(5600t)u(t) V. 5 as would be obtained using wn 5 600 radls, though the difference is insignificant. 1538s + 10- s -14000 s + 1538s + 592000 = 2 4 2 -14000 (s + 769? which detransforms to V0 (t) = -14000texp( -769t)u(t) V. 061 x 10St)u(t) kA.

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