1 Notes on Laplace circuit analysis

Transcription

1 Physics - David Kleinfeld - Spring 7 Notes on aplace circuit analysis Background We previously learned that we can transform from the time domain to the frequency domain under steady-state conditions thus solve algebraically for the transfer function between the input output of a circuit This analysis allowed us to replace indictors capacitors by their complex impedance, as derivatives in time were replaced by i integrals in time were replaced by /i The steady-state time dependence is then found by transforming back to the time domain The problem is that we often have a signal that turns on at a specific time, which we will take to be t with no loss of generality an we calculate the transient behavior with a transform method, as opposed to performing a convolutional integral in the time domain? et s recall the transform from the time domain to the frequency domain It is given by: Ṽ vt e it First, what happens when Ṽ cannot exist because the integral does not converge at t /or t? For example, suppose ftconstant or worse yet a polynomial in time? One way to deal is to add an integrating factor; we chose an exponential as this will suppress any polynomial Thus we add a factor exp a t to the integr, we can take a Second, what happens when the system is causal, so that V t for t <? Here we take the lower limit as t rather than t This is equivalent to multiplying the integr by a step function, denoted ut, where, t >, ut, otherwise, All of this leads to the aplace transform: V s vt e at e it 3 vt e st where s a + i vt is understood as vtut Keep in mind that the units of V s are Volts time The inverse transform is a bit more involved, but we will show how this can be readily done for any of the functions that arise in linear circuit analysis We have vtut ds V s e st, πi

2 which is a contour integral in the complex s-plane All we need to know about, at least to start analyzing the kind of circuits familiar to the class, are two rules dvt dvt e st s vt e st + vt e st sv s v 5 t dx vx t dx vx e st V s 6 s three transforms e st s 7 e at sint cost e at e st s + a sint e st d sint s 8 s + 9 s + s s + Note that the derivative transform includes initial conditions, as shown in the table: Application with step-to-constant input et s apply our new knowledge to a circuit that has a switch that closes at time t Thus the current at t + equals the current at t, which is I since the current through an inductor cannot change instantaneously The initial voltage across the capacitor however, may not be zero This V + V since the voltage across the capacitor cannot change instantaneously,

3 The equations are: Transforming, we get v o + dit + Rit + t it + V v o s We multiply through bys/ terms to get so that + sis + RIs + s Is + V s v o + s Is + R Is sis + + V 3 vo V Is s + R s + We let V simply to minimize the algebra in the following mathematics Thus: Is v o s + ks + o 5 where k R is a decay rate o is a resonant frequency The first thing we need to do is factor the denominator We have roots k ± i o k 6 Thus with Is v o s as a 7 a k + i o k 8 thus it v o e st ds πi s as a 9 In order to solve this we need a refresher on the residue theorem "Blitz refresher on auchy s Residue Theorem" Integrals in the complex plane, of the form used in linear circuit analysis, may be evaluated by dsf s πi Σ Residues When F s Any regular function Polynomial function with simple zeros qs ps the residue at each zero of ps, or pole of F s, is given by the expression Residue qs sspole ps s 3

4 For example, with qs rse st ps s as b s ys z, we have rse st dsf s ds s as bs c s ys z rse st πi s b s ys z rse st ] sa + + s as b s y sz rae at πi a b a ya z + + rze zt ] z az b z y Note that complex poles always appear as conjugate pairs Thus, for example, with F s e st s as a we find ft πi πi dsf s e st ds s as a πi πi e at a a + ea t a a e e Rea]t iima]t e iima]t ] iima] erea]t Ima] sinima]t ] which is just the form of our solution for the prior circuit application The auchy residue theorem for the inverse transform thus yields: it v o e Rea]t Ima] sinima]t v o e kt o k sin o k t Note that the shift in the natural frequency, from o to o k o + ], is quite clear When the loss is high, ie, k > o, the sine term becomes a hyperbolic sine the current just rises decays exponentially For the special case of k o, so called critical damping, it v o R kt e kt Note also that the current at very short times is limited by the highest impedance, which is the induc-

5 tance In particular it t v o t 3 Application with step-to-sinusoid tone input et s now move to a more interesting dynamics replace the source with a cosine that turns on at t, that is so that v o cost + dit v o t v o cost 3 + Rit + t it where, for simplicity, we take the initial voltage on the capacitor to be zero Transforming, we get Is i o k s s as a s i o k s s as a s is + i 5 where we use the same abbreviations as above We choose to use costut as this reverts to ut as The circuit will respond at both the driven frequency at the natural frequency We have, from the residue theorem, four terms that we will evaluate in pairs, ie it i o f t + f t] 6 where f t f t k s e st s a s + k s e st sa + s as + sa 7 k s e st s as a s + i k s e st si + s as a s i s i 8 Before we solder on with f t, let s calculate some of the algebraic terms that we will need, ie, a a i o k, 9 aa o, 3 a k o + i o k 3 a k o i o k 3 5

6 f t ke kt k o + ik o k o k i k + o + ik ei o k t k o ik o k o k k + o ik o k e i o k t 33 We rationalize the denominator, noting that k + o + ik o k k + o ik o k o + k, 3 k o + ik o k k + o ik o k o o k + ik o k k o ik o k k + o + ik o k o o k ik o k so that f t ke kt 37 o k o + k ] o o k ] e i o kt e i o k t + k o i k ei o kt + e i o k t ke kt o k o + k ] o o k ] sin o k t + k o k cos o k t oke kt o k o + k ] o o o k o + k sin o k t + The weighting factors for the sine cosine terms satisfy the right triangle rule o k o k o + k cos o k t o o k ] + k o k ] o o + k ] 38 so that with the definition k o φ atan k o o k 39 we have f t oke kt o k o + k ] cosφ ] sin o k t + sinφ ] cos o k t o ke kt o + k o k sin o k t + φ The first term is maximized for the choice of drive frequency o k, which is slightly lower than the natural response frequency of o k astly, as a sanity check, in the limit of, we recover the result for the response to a step input, ie f t ke kt o k sin o k t 6

7 et s now move on to the driven term f t We first note the evaluations: s as a si s + ks + o si o + ik, s as a s i s + ks + o s i o ik 3 o + ik] o ik] o + k Then k i e it f t o + ik]i + k i e it o 5 ik] i k e it o ik] e it o + ik] i o + k k k eit + e it o + k + o eit + e it i k o + k o sint + k cost k o o + k o + k sint + k o + k cost The weighting factors for the sine cosine terms satisfy the right triangle rule, so f t k o + k sinφ ]sint + cosφ ]cost 6 k o + k cos t φ where φ atan o k 7 it v o k f t + f t 8 v o o o e kt o o + k o k sin o k t + φ + cos t φ o At t +, the amplitude of the response is i + v o R k o + k 9 7