School of Engineering Department of Electrical and Computer Engineering 332:221 Principles of Electrical Engineering I Fall 2012 Differential Equations + Euler + Phasors Christopher Rose ABSTRACT You have a network of resistors, capacitors and inductors. They are driven by voltage and current sources. You are asked to find a given output. You are freaking out because unlike resistive networks, everything is TIME VARYING! 1. Get a paper bag and place it over your head to stop hyperventilating. 2. Chill out because they’re only linear constant coefficient differential equations and those are EASY to solve if you understand the trick. This writeup is about that trick. We also (re)introduce the Euler expansion in the service of talking about sinusoidal steady state response and phasors. 1 LCC Differential Equations A linear constant coefficient differential equation is of the form N X (k) akX = V (t) k=0 where X(k) is the kth derivative of X and V (t) is a “drive term” (something you apply to the “system” which the differential equations describe. As you already know from class (and from the book), all of those circuits with inductors and capacitors result in such differential equations. N is AT MOST the number of capacitors and inductors in the circuit (but it could be fewer depending on the arrangement of those elements – inductors in series act like a single inductor, for instance). 1.1 The Basics So, how to solve the beast?!!? It starts with a theorem that I’ll not prove. Theorem 1 Picard’s Theorem Any solution to an Nth order linear constant coefficient differential equation is composed of two parts: 1 • The Particular solution – ANY Xp(t) that satisfies N X (k) akXp = V (t) k=0 and • The Homogeneous solutions – ANY set of N “linearly independent” Xh(t) that satisfy N X (k) akXh = 0 k=0 h = 1; 2; :::N The term linear independence simply means that you can never find a set of coefficients fchg such that N X chXh(t) = 0 k=0 Put another way, if a set of functions is linearly independent, you can’t “build” any one of the functions by linearly combining the other functions. Now let’s get cute. In class (and here) we’re going go GUESS the homogeneous solutions by snt saying they must be of the form Ane where An is some constant (possibly complex) and sn is a (possibly complex) number. For later convenience we’ll say sn = σn + j!n p where σn and !n are real and j = −1. So with that guess, we proceed just like in class to recognize that dkA esnt n = A sk est dtk n n snt So, using the solution Ane in our homogeneous (unforced – i.e., V (t) = 0) equation we have N " N # X k snt snt X k akAnsne = Ane aksn = 0 k=0 k=0 snt Since we can’t guarantee that An and e are zero 8t, our only recourse is to find fsng such that the bracketed POLYNOMIAL in sn is zero. That polynomial is called the characteristic equation and the satisfying sn are called its roots. You’ll see these again and again and again and again and again in your ECE lives – and beyond. snt This means that there are N homogeneous solutions, fAne g, n = 1; 2; :::N, to our differen- tial equation. Thus, the general solution is N X snt X(t) = Xp(t) + Ane n=1 (And yes, Virginia, there can be repeated roots, but we’re going to ignore that case for now :) Well ... if you MUST know, if you have Nm repeated roots sm, then the Nm homogeneous solutions are proportional to tNm−1esmt – try it if you’re really curious. :) ) 2 The last thing we need to do is determine those constants fAng. We can get these from “initial conditions.” For instance, in a circuit, perhaps the inductor has current through it and capacitors charge on them at t = 0 and we want the solution for t > 0. For an Nth order differential equation, we’ll need N initial conditions to determine those fAng. Interested in learning more? Too bad. Why? Read on. :) 1.2 Short Shrift for Transient Response Because most of you have not already HAD a differential equations course, we’re NOT going to worry TOO much about these homogeneous solutions except to say that for all the circuits we’ll consider, all of those homogeneous solutions will have sn with NEGATIVE REAL PARTS. What does that mean? It means that as time goes, on all the esnt ! 0. Formally, lim esnt = lim eσnt+j!nt = 0; n = 1; 2; ··· ;N t!1 t!1 because we assume all the σn < 0. So in the STEADY STATE, we won’t worry about the “tran- sients” because they’ll all “die out” with time. That said, I MIGHT give you a problem on a quiz where the transients DO NOT die out – and if you’re currently awake you’ll imediately realize that any such circuit will either have to have NO resistors (which dissipate energy) or have to have an active element (like a transistor or an op amp) in it since passive circuits with resistors, capacitors and inductors cannot have undying transients (the resistors eventually turn the initial condition energies into heat and the corresponding circuit state variables to zero). Now, let’s look at some useful math. 2 USEFUL MATH 2.1 The Euler Expansion The Euler expansion is a cute identity that you should tatoo on your retinas: ejx = cos x + j sin x It comes from Taylor series (well, McLauren series actually). From your calculus days you should remember that 1 X xk ex = k! k=0 You should also remember that 1 k X k x cos x = (−1) 2 k! k=0;even and 1 k X k−1 x sin x = (−1) 2 k! k=1;odd 3 So, if we replace x by jx we have 1 k 1 k 1 k jx X k x X k x X k−1 x e = j = (−1) 2 + j (−1) 2 k! k! k! k=0 k=0;even k=1;odd The first series you’ll recognize as 1 k X k x cos x = (−1) 2 k! k=0;even and the second one is 1 k X k−1 x j sin x = j (−1) 2 k! k=1;odd So, once again, with feeling, ejx = cos x + j sin x It’s also useful to note here that the complex conjugate (inverting the sign of the imaginary portion of a complex number) of ejx is ejx∗ = (cos x + j sin x)∗ = cos x − j sin x = e−jx where the ∗ notation implies complex conjugation. The euler expansion will be really important because we’re going to quickly start using it to represent sinusoids as in ejx + e−jx cos x = 2 cos x = <fejxg and ejx − e−jx sin x = 2j sin x = =fejxg Also note that sin !t = cos(!t − π=2) = < e−jπ=2+j!t = < e−jπ=2ej!t = < −jej!t Yes, that’s a subtle hint that you need to be facile deriving/understanding/memorizing things like j = ejπ=2 and the like. :) 4 2.2 Complex Numbers and Phasors In a previous version of these notes, I directed you to the text section on phasors – thinking you folks already knew ALL about complex numbers from some other course (in high school or col- lege). Woe to us all (and to quizlettes 8&9 AND the last part of problem 1 on Quiz II!) the moment I made that assumption! So this next bit is a SHORT attempt to lay phasors all out in a way that will allow you to see the equivalence between phasors and complex numbers. First off, I KNOW you all know what a complex number is C = a + bj p where j = −1 (again, mathematicians step off! i/I is reserved for CURRENT). The righthand side of the definition is often called “standard form.” Now, what’s interesting about C is that it has two “orthogonal” components – a real (<) part and an imaginary (=) part. This allows us to think of C in the “complex plane” as in FIGURE 1 That C is a PHASOR with magnitude jCj and angle b |C| C θ a Figure 1: Complex number vector (phasor) representation. θ. Combined with euler’s expansion, FIGURE 1 has ALL THE THINGS YOU NEED TO UNDERSTAND PHASORS AND COMPLEX NUMBERS!!!!! First, as soon as we draw C in the complex plane as a point (a; b), wep can see that the associated vector (nose at (a; b) and tail at the origin) has magnitude jCj = a2 + b2 and angle 6 C = arctan(a; b). Now, here’sp where it gets COOLIO!p (passe though he is :) ). Look at the figure. Obviously cos θ = a= a2 + b2 and sin θ = b= a2 + b2. However, here’s where the MAGIC happens p a + bj = a2 + b2ejθ 5 Why? ejθ = cos θ + j sin θ as you know from Euler’s expansion. But looky here! We then have p p a b a2 + b2ejθ = a2 + b2 p + j p = a + bj a2 + b2 a2 + b2 p So, the equivalence between a + bj and a2 + b2ejθ is EXACT!!!! This latter representation is called phasor notation. Now, some basic mechanics are useful too. For instance, we define the COMPLEX CONJU- GATE of a complex number as C∗ = (a + bj)∗ = a − bj That is, we just multiply the imaginary part by −1 to get the complex conjugate.