ECE 201 - Exam 2 - F06

The circuit below models the scenario where one is

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Unformatted text preview: 1 30 V + - t = 2 msec. 1 µF S2 1 KΩ + vC(t) - Answer = #1 3 Problem 4. The circuit below models the scenario where one is trying to display a square wave from a function generator on an oscilloscope in the laboratory. We are using a 50 Ohm square wave generator to create a ±5 V square wave v (t) with 3 diﬀerent frequencies. The frequency f0 of the square wave is related to its period T0 by f0 = 1/T0 . The output of the square wave generator is connected to the input of the oscilloscope using a section of coaxial cable. Models for each of the components of the system are connected as shown. Measurements of 3 diﬀerent frequencies are taken. The resulting oscilloscope voltages vscope (t) are shown as the three superimposed Traces in the ﬁgure below the circuit. Note that the oscilloscope time scale is adjusted for each measurement such that exactly 2.5 periods are displayed in each case. From the possibilities below pick the most reasonable answer for the frequencies corresponding to Traces 1, 2, and 3. f0,T race 1 = 500 MHz (1) f0,T race 2 = 50 MHz f0,T race 3 = 5 MHz f0,T race 1 = 500 KHz (2) f0,T race 2 = 50 KHz f0,T race 3 = 5 KHz f0,T race 1 = 5 MHz (3) f0,T race 2 = 50 MHz f0,T race 3 = 500 MHz f0,T race 1 = 50 MHz (4) f0,T race 2 = 500 MHz f0,T race 3 = 5 MHz f0,T race 1 = 5 KHz (5) f0,T race 2 = 50 KHz f0,T race 3 = 500 KHz f0,T race 1 = 50 KHz (6) f0,T race 2 = 500 KHz f0,T race 3 = 5 KHz f0,T race 1 = 50 GHz (7) f0,T race 2 = 500 GHz f0,T race 3 = 5 GHz + 50 Ω v(t) + - 30 pF 1 MΩ 20 pF vscope(t) model of square wave generator model of coaxial cable model of oscilloscope input 5 Trace 1 4 Trace 2 3 vscope(t) in Volts 2 Trace 3 1 0 −1 −2 −3 −4 −5 0 0.5 1 1.5 Time normalized by period, i.e., t/T0 Answer = #3 4 2 2.5 Problem 5. The circuit shown is in the steady state with the switch closed. At time t = 0 the switch is opened. Find the current i(t) for t > 0. (1) 4 − 6e−50t A (5) 2 + 10e−30t A (2) 10 − 2e−20t A (6) 4 + 6e−50t A (3) 3 + 2e−30t A (7) 10 − 5e−20t A 0.5 H 10 Ω t=0 15 Ω i(t) Answer = #6 5 + - 100 V (4) 3 − 2e−20t A...
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This note was uploaded on 09/14/2012 for the course ECE 201, 202 taught by Professor Capano during the Fall '11 term at Purdue.

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