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Unformatted text preview: Lecture S9 Muddiest Points General Comments In this lecture, we began learning how to find the time response of RC circuits to initial conditions. To do this, we used the node method, and used the constitutive relations for capacitors and resistors. This led to coupled, first order differential equations, which we can solve by guessing that the solution is exponential, proportional to e st . We then find that solutions are possible only for some exponentials. This gives the characteristic values, s , of the system. Responses to MuddiestPartoftheLecture Cards (57 cards) 1. Why do we assume that the guess we made ( e st ) is right and go ahead as if it could not be wrong? (1) We assume the exponential as a trial answer, and show that it indeed works. That is, the trial solution does indeed solve the differential equations, and you can choose constants (the characteristic vector) to solve the initial conditions. It could have been wrong, and we would have found out if it were. 2. When doing row reduction, can you alter columns at all? (1) The row reduction approach seems to have no rules, i.e., you can do anything to the rows. The legal row operations are (1) scaling of a row; (2) exchanging rows; and (3) adding the multiple of one row to another row. You can’t do column operations. 3. Why is it necessary to do row reduction when solving for the [characteristic vectors]? (1) Row reduction is the most direct way to find solutions to the equation M ( s ) E = 0 (8) 4. When solving for characteristic vectors, why can we choose an arbitrary value for E 3 ? (2) If the characteristic vector...
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This note was uploaded on 01/28/2012 for the course AERO 16.01 taught by Professor Markdrela during the Fall '05 term at MIT.
 Fall '05
 MarkDrela

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