# CU-Boulder ECEN 2250 - Problem Set 6 (2 pages)

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## Problem Set 6

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## Problem Set 6

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Lecture Notes

Pages:
2
School:
University of Colorado at Boulder
Course:
Ecen 2250 - Introduction to Circuits and Electronics
##### Introduction to Circuits and Electronics Documents
• 2 pages

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ECEN 2250 Introduction to Circuits Electronics 10 01 11 Fall 2011 P Mathys Problem Set 6 Solutions are due Fri 10 07 11 Note For all problems you have to explain how you arrived at the solution legibly and in a logical order This includes i Statement of problem ii strategie s for solution iii execution of the strategy identifying each step and including a check of the solution s and iv logical and legible presentation of the solution 1 Use node voltage analysis to determine vA and vB in the following circuit 8R vA iS 5 vO 12R vB RL vO 2 Problem 8 10 in the book 3 Problem 8 28 in the book Use LTspice instead of OrCAD 4 Problem 8 42 in the book 5 The following RL circuit with input vS t VS can be used as a filter R vR t vS t vO t L a Determine the output vO t VO in the time and phasor domains assuming that vS t A cos t Let VO H j VS 1 denote the frequency response of the circuit Sketch and label the magnitude H j and the phase H j in degrees of the frequency response for 0 4000 rad sec i e f 0 2000 Hz when the time constant of the circuit is L R 1000 1 s Is this a lowpass LPF or a highpass HPF filter b Repeat a for the case when the filter output is the voltage vR t across the resistor instead of vO t 6 The following circuit is operated in the sinusoidal steady state i t Z1 vS t Z3 Z2 v t Z4 When vS t 20 cos 300000t V i t and v t are measured as i t 200 cos 300000t 15 mA and v t 10 cos 300000t 60 V Find elements R L C and values for Z1 Z4 such that this i t and this v t are produced in response to the given vS t The values for resistors capacitors and inductors in your solution must be in the range 10 100 0 01 1 F and 0 01 1 mH respectively Verify your solution using LTspice Is the solution unique c 2001 2011 P Mathys Last revised 10 02 11 PM 2

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