F07 314 HW 06 all

# F07 314 HW 06 all - EECS 314 Fall 2007 HW 06 Problem 1...

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EECS 314 Fall 2007 HW 06 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) © 2007 Alexander Ganago This problem was given on exam 2 in EECS 314 Winter 2005 The circuit shown on the diagram below is fed with an input signal, which is a square wave at a low frequency. Which of the output waveforms is observed? A. Waveform A B. Waveform B C. Waveform C D. Waveform D E. None of the above. In your homework paper, justify your answer (based on KVL equations, differential equations, and the continuity requirements) and prove that other answers are wrong. Hints: 1. The voltage abruptly changes twice per period T of the input square wave, at time = 0 and T/2. In practice, the voltage cannot change instantaneously but it can change very fast. For example, if the rise time of the square wave input is a few nanoseconds and the period of the square wave is in the microsecond or millisecond range, the changes of input voltage are practically instantaneous. 2. From the flat tops of the output waveforms one can conclude that DC steady-state conditions (zero time derivatives) are reached before the voltage changes, that is at time t = 0- and at time t = (T/2)-. 3. Assume that the 1 Vppk input square wave has the voltage that varies from 0 V to +1 V. 4. Write the KVL equations and consider the continuity requirements at time t = 0- (before the input voltage jumps from low to high) and at time t = 0+ (just after the voltage has jumped from low to high). 5. Also, write the KVL equations and consider the continuity requirements at time t = (T/2)- (before the input voltage jumps from high to low) and at time t = (T/2)+ (just after the voltage has jumped from high to low).

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EECS 314 Fall 2007 HW 06 Problem 2 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) © 2007 Alexander Ganago The big picture The circuit shown on this diagram is an integrator: its output voltage is V OUT ( t ) = K " V IN ( t ) " dt # Here the constant K does not depend on time and voltage, but it depends on R and C in the circuit. Note that, in order to match the units on the both sides of the above equation, the constant K must have units of (sec) -1 . Also note that there is a typo on page 27 of Chapter 13 in Making sense of EE. Problem Part 1 For the circuit shown on the diagram, derive the equation for the coefficient K (be careful with its sign!). Your answer: K = _________________ For the circuit shown on the diagram, assume R = 15 k Ω , C = 4 nF, and calculate the numerical value of K (be careful with its sign!). Your answer: K = _________________ Part 2 Suppose that the input signal is a constant equal to +20 mV, and that the output voltage in the circuit above (with component values as in Part 1 ) was set to zero at time t = 0 . Obtain the equation for the output voltage as function of time. Since the integral of a constant is a linear function, the output voltage would eventually outgrow the source
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F07 314 HW 06 all - EECS 314 Fall 2007 HW 06 Problem 1...

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