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MIT18_03S10_c17

# MIT18_03S10_c17 - 18.03 Class 17 Linearity and time...

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18.03 Class 17 , March 12, 2010 Linearity and time invariance [1] RLC [2] Superposition III [3] Time invariance [4] Review of solution methods [1] We've spent a lot of time with mx" + bx' + cx = q(t) . There are many other systems modeled by this equation. For example here is a series RLC circuit: ____ / \ || ------( )-----/\/\/\/-----------------------------||------ | \____/ || | | | | | |_______________________________________________________________| I = current; the same everywhere V = voltage increase across the power source V_R = voltage drop across the resistor V_C = voltage drop across the capacitor Input signal: V System response: I KVL: V_R + V_C = V Component behavior: V_R = R I V_C' = (1/C) I Differentiate KVL : V_R' + V_C' = V' so R I' + (1/C) I = V' In Lecture 3 we offered this as an example of a first order LTI system. Now let's add another component, an inductor. ____ / \ __ ___ __ || ------( )-----/\/\/\/--------/ X X \----------||------ | \____/ V V || | | | | | |_______________________________________________________________|

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The voltage drop across an inductor depends not on the current but rather on the derivative of the current: V_L = L I' so V_L' = L I" . KVL now says V_L + V_R + V_C = V so L I" + R I' + (1/C) I = V' (*) The system serves as an OPERATOR: L D^2 + R D + (1/C) Id takes the current I , as a function of time, and gives the derivative of the impressed voltage. (I have to use a different symbol for the identity operator here, so it doesn't get confused with current. I chose "Id.") [2] Suppose you want to solve u" - 4u = cosh(2t) Remember, I can write the left hand side as p(D) u where p(s) = s^2 - 4 . First you have to know what cosh(t) is : cosh(x) = ( e^x + e^{-x} ) / 2 sinh(x) = ( e^x - e^{-x} ) / 2 The right hand side is a linear combination of exponentials, each of which we know very well how to deal with.
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