Chap6StudentSolutions

The dividing time td between these two parts is set

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Unformatted text preview: ) 2500 I(x ) 0 0.0004 -0.2 0.2 x 0 (b) Now let the slit be replaced by two slits each 0.1 mm in width, separated by 1 mm (center-to-center) and centered on the optical axis. Plot the intensity of light at the viewing screen if the other parameters are the same as in part (a). Similar to (a) I(x ) 0 1.6e-05 -1 x0 1 42. In Figure 42-1 is a circuit diagram of a half-wave rectifier followed by a capacitor to smooth the response voltage. Model the diode as ideal and let the excitation be a cosine at 60 Hz with an amplitude of 120 2 volts. Let the RC time constant be 0.1 seconds. Then the response voltage will look as illustrated in Figure E42-2 . Find and plot the magnitude of the CTFT of the response voltage. + v i (t) + R C - vo (t) - Figure E42-1 A half-wave rectifier with a capacitive smoothing filter Solutions 6-35 M. J. Roberts - 8/16/04 175 vo(t) vi (t) 0.05 t -175 Figure E42-2 Excitation and response voltages The response voltage has two parts, the exponential decay time and the cosinusoidal charging time. The dividing time, td , between these two parts is set by the intersection of the cosine and the exponential decay. The peak of the cosine is 120 2 . The decay time constant is 0.1 seconds. Therefore the dividing time is the solution of 120 2 cos(120πtd ) = 120 2e − td 0.1 or cos(120πtd ) = e − td 0.1 This is a transcendental equation best solved numerically. This equation is simple enough that a trial-and-error method converges very quickly to a solution. That solution is td = 15.23906 ms . Therefore the description of the response voltage over one period is − 0t.1 , 0 < t < 15.23906 ms e v o1 ( t) = 120 2 2 cos(120πt) , 15.23906 < t < 16 ms 3 or − 0t.1 t − 0.00761953 t − 0.01595286 v o1 ( t) = 120 2 e rect + cos(120πt) rect 0.01523906 0.001427607 The CTFT of the response is the CTFT of this voltage convolved with a comb to make it periodically repeat. The CTFT of one period is 0.01523906 − t − j 2πft dt ∫ e 0.1e 0 Vo1 ( f ) = 120 2 1 + [δ ( f...
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This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.

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