howto:hambasics:sections:test
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howto:hambasics:sections:test [2021/01/02 08:55] – [Forme Polaire] va7fi | howto:hambasics:sections:test [2021/02/13 19:14] (current) – [Cartesian vs Polar] va7fi | ||
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- | ~~NOTOC~~ | + | ====== |
- | ====== | + | This is a brief survey of the math required to analyze waves at the first or second year university level. |
- | ===== Les nombres réels ===== | + | |
- | * \$ (5) \times (5) = \Box \$ | + | |
- | * \$ (-5) \times (-5) = \Box \$ | + | |
- | * \$ \Box \times \Box = -25 \$ Il n'existe pas de nombre qu'on peut multiplier par lui même et obtenir un résultat négatif, mais pourrait-on en // | + | |
+ | ===== Real numbers ===== | ||
+ | If \$ (5)^2 = 25 \$ and \$ (-5)^2 = 25 \$, what number can you put in the box so that: | ||
- | ===== Les nombres complexes ===== | + | \$$ \Box ^2 = -25 \$$ |
- | \$ i = \sqrt{-1} \qquad \Rightarrow \qquad i^2 = -1 \$ | + | |
- | Même si \$ i \not\in \mathbb{R} \$, on peut quand même faire des opérations mathématiques assez intéressantes avec: | + | It turns out that there is no //real// number such that when you multiply it by itself you get a negative number. |
- | * On peut multiplier: \\ \$ \begin{align*} (1+i)^2 &= 1 + 2i + i^2 \\ &= 1 + 2i - 1 \\ &= 2i \end{align*} \$ | + | |
- | * On peut trouver des racines: \\ \$ \begin{equation*} z^4 = 16 \Rightarrow z^2 = \left\{ \begin{array}{rl} 4 \Rightarrow z &= \pm 2 \\ -4 \Rightarrow z &= \pm 2i \end{array} \right. \end{equation*} \$ ou \\ \$\begin{equation*} z^8 = 256 \Rightarrow z^4 = \left\{ \begin{array}{rl} 16 \Rightarrow z^2 &= \left\{ \begin{array}{rl} 4 \Rightarrow z &= \pm 2 \\ -4 \Rightarrow z &= \pm 2i \end{array} \right. \\ -16 \Rightarrow z^2 &= \left\{ \begin{array}{rl} 4i \Rightarrow z &= \pm \sqrt{2}(1+i) \\ -4i \Rightarrow z &= \pm \sqrt{2}(1-i) \end{array} \right. \\ \end{array} \right. \end{equation*} \$ | + | |
- | === Un peu de philosophie... | + | ===== Complex Numbers ===== |
- | Si on peut soumettre ces nombres bizarres à toutes les règles d'algèbres qu'on connaît sans arriver à d' | + | Let's create an //imaginary// number called \$ i \$ such that: |
+ | <WRAP center box 30em> | ||
+ | \$$ i = \sqrt{-1} \qquad \Rightarrow \qquad i^2 = -1 \$$ | ||
+ | </ | ||
- | ===== Plan complexe ===== | + | Even though \$ i \$ is nowhere on the real line (in math, we say that: \$ i \not\in \mathbb{R} |
- | Pour représenter un nombre complexe graphiquement: | + | * We can add it to a real number and create a //complex// number: |
- | FIXME: geogebra | + | \$$(1 + i) \$$ |
- | Les points noires: \$ z = a + ib \$ sont des solutions de l' | + | * We can multiply complex numbers together: |
- | * ... | + | |
- | * ... | + | |
- | * ... | + | |
- | === Exercices === | + | \begin{align*} |
- | * Graphiquement, | + | (1+i)^2 |
- | * Comment pourraient-on calculer ces nombres algébriquement? | + | &= (1+i)\cdot(1+i) |
- | * Algébriquement, | + | &= 1 + 2i + i^2 \\ |
+ | &= 1 + 2i - 1 \\ | ||
+ | &= 2i | ||
+ | \end{align*} | ||
- | FIXME: Geogebra | + | * We can find roots: |
- | ===== Forme Polaire ===== | + | \begin{equation*} |
- | Comme pour les vecteurs, les nombres complexes peuvent être exprimés en coordonnées cartésiennes ou polaires. | + | z^4 = 16 \Rightarrow z^2 = \left\{ \begin{array}{rl} 4 \Rightarrow z &= \pm 2 \\ |
+ | -4 \Rightarrow z &= \pm 2i \end{array} \right. | ||
+ | \end{equation*} | ||
- | FIXME: Geogebra | + | ==== A Little Philosophy ==== |
+ | If these weird numbers follow all of the algebra rules without inconsistencies, | ||
+ | In a certain way, negative numbers are just as weird as complex numbers: after all, we know what 5 cars look like, but what does −5 cars mean? And yet, in certain context (like temperature), | ||
- | Pour convertir d'une représentation à l' | + | ===== The Complex Plane ===== |
- | |<100% - >| | + | In the same way that we can represent real numbers by a point on the real number line...((Number line picture from [[wp> |
- | ^ \$ a + ib & | + | |
- | | \$ r^2 &= a^2 + b^2 \$ | \$ a &= r\cos\theta \$ | | + | |
- | | \$ \tan \theta &= \dfrac{b}{a} \$ | \$ b &= r\sin\theta \$ | | + | |
- | === Exercices | + | {{ numberline.png }} |
- | * Convertir | + | |
- | * Convertir | + | ... we can also represent a complex number graphically on a complex plane by using the horizontal axis for the real part and the vertical axis for the imaginary part. For example, \$ (1 + i) \$ would be represented as a point 45° up the horizontal axis and \$ \sqrt{2} \$ away from the origin: |
+ | |||
+ | <WRAP center round info 80%> | ||
+ | You can move the point around to look at other complex numbers on the plane. | ||
+ | </ | ||
+ | |||
+ | {{ggb>/ | ||
+ | |||
+ | To convert between the Cartesian \$(a,b) \$ and the Polar \$ (r \angle \theta) \$ representations, | ||
+ | |||
+ | ^ \$$ (a, b) \rightarrow (r\angle \theta) \$$ ^ \$$ (r\angle \theta) \rightarrow (a, b)\$$ | | ||
+ | | \$$ r^2 = a^2 + b^2 \$$ | \$$ a = r\cos\theta \$$ | | ||
+ | | \$$ \tan \theta | ||
+ | |||
+ | Note that very often, we use radians instead of degrees for the angle. | ||
+ | * Imagine a circle. | ||
+ | * Now imagine the length from the centre to the circle (along the " | ||
+ | * Take that length and lay it down on the perimeter of the circle. | ||
+ | * The angle that this length covers is 1 radian (because of the length of the radius on the circle). | ||
+ | * That's why a circle has 2π radians (because the circumference is 2πr) | ||
+ | |||
+ | ==== Roots ==== | ||
+ | The complex plane has many useful applications, | ||
+ | |||
+ | {{ggb>/ | ||
+ | |||
+ | Without using the graph above, what do you expect the solution(s) to \$ z^3 = 8 \$ will be? That is, what number(s), when multiplied by itself three times gives 8? | ||
+ | |||
+ | Now move \$ w = 8 \$ and \$n = 3 \$ to have a look at the solutions graphically, | ||
+ | |||
+ | < | ||
+ | {{ complexroots.png? | ||
+ | So \$z = 2\$ was to be expected since \$ 2^3 = 8 \$ but it looks like there are two more solutions. \\ \\ | ||
+ | |||
+ | To find them, we first notice that the three solutions are spread out evenly around the circle, that is they are 120° apart. | ||
+ | So in polar coordinates, | ||
+ | |||
+ | We can now convert them to Cartesian: | ||
+ | * \$z = 2\angle 0^\circ = \big(2\cos(0^\circ), | ||
+ | * \$z = 2\angle 120^\circ = \big(2\cos(120^\circ), | ||
+ | * \$z = 2\angle 240^\circ = \big(2\cos(240^\circ), | ||
+ | |||
+ | Let's check that the second solution works: | ||
+ | \begin{align*} | ||
+ | (-1 + i\sqrt{3})^3 &= (-1 + i\sqrt{3})\cdot(-1 + i\sqrt{3})(-1 + i\sqrt{3}) \\ | ||
+ | &= (-1 + i\sqrt{3})\cdot\big(1 -2i\sqrt{3} +(i^2)(3)\big) \\ | ||
+ | &= (-1 + i\sqrt{3})\cdot\big(1 -2i\sqrt{3} +(-1)(3)\big) \\ | ||
+ | &= (-1 + i\sqrt{3})\cdot(-2 -2i\sqrt{3}) \\ | ||
+ | &= 2 + 2i\sqrt{3} - 2i\sqrt{3} -(2)i^2(3) \\ | ||
+ | &= 2 -(2)(-1)(3) \\ | ||
+ | &= 2 + 6 \\ | ||
+ | &= 8 | ||
+ | \end{align*} | ||
+ | |||
+ | </ | ||
+ | \\ | ||
+ | |||
+ | ===== The Euler Identity ===== | ||
+ | The Euler identity exposes a deep relationship between trigonometric and exponential functions((Note, | ||
+ | \$$ e^{i \pi} = \cos (\pi) + i \sin (\pi) = -1 + 0i = -1\$$ | ||
+ | \$$ \Rightarrow e^{i \pi} = -1\$$ | ||
+ | which is amazingly beautiful because it relates \$ e = 2.71828... \$, \$i = \sqrt{-1} \$, \$\pi = 3.14159... \$, and \$-1\$ in the most surprising and elegant way.)): | ||
+ | |||
+ | <WRAP center box 30em> | ||
+ | \$$ e^{i \theta} = \cos \theta + i \sin \theta \$$ | ||
+ | |||
+ | </ | ||
+ | |||
+ | Let's use two different ways to verify that this mysterious identity is true. | ||
+ | |||
+ | ==== The Derivatives ==== | ||
+ | If we separate this identity into two functions and take their derivatives, | ||
+ | \begin{align*} | ||
+ | && f(\theta) &= e^{i \theta} &\text{ & }&& g(\theta) &= \cos \theta + i \sin \theta \\ | ||
+ | \Rightarrow && f' | ||
+ | \Rightarrow && f' | ||
+ | \end{align*} | ||
+ | |||
+ | We know that there' | ||
+ | |||
+ | ==== Taylor ==== | ||
+ | Another method to verify the Euler identity is to use Taylor series: | ||
+ | |||
+ | \begin{align*} | ||
+ | e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots \\ | ||
+ | \sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \\ | ||
+ | \cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots | ||
+ | \end{align*} | ||
+ | |||
+ | - Replace \$ x \$ with \$ i \theta \$ in the Taylor series of \$ e^x \$ | ||
+ | - Replace \$ x \$ with \$ \theta \$ in the Taylor series for \$ \sin x \$ and \$ \cos x \$ | ||
+ | - Add the Taylor series of \$ \cos \theta \$ and \$ i \sin \theta \$ together and you'll get the Taylor series for \$ e^{i \theta} \$ | ||
+ | |||
+ | |||
+ | ===== Euler Identity and Polar-Cartesian Representations ===== | ||
+ | |||
+ | In the previous section, we saw that a complex number \$z = a + ib \$ could be represented as a point \$(a, b)\$ on the complex plane, which could also be viewed in polar coordinates as (\$r\angle | ||
+ | |||
+ | ^ \$$ (a,b) \rightarrow (r\angle \theta) \$$ ^ \$$ (r\angle \theta) \rightarrow (a,b)\$$ | | ||
+ | | \$$ r^2 = a^2 + b^2 \$$ | \$$ a = r\cos\theta \$$ | | ||
+ | | \$$ \tan \theta = \dfrac{b}{a} \$$ | \$$ b = r\sin\theta \$$ | | ||
+ | |||
+ | This means that: | ||
+ | |||
+ | \begin{align*} | ||
+ | z &= a + ib \\ | ||
+ | &= r\cos\theta + i r\sin\theta \\ | ||
+ | &= r\big( \cos\theta + i \sin\theta \big) \\ | ||
+ | &= r e^{i\theta} | ||
+ | \end{align*} | ||
+ | |||
+ | This offers another interpretation of the Euler identity as the algebraic conversion between Cartesian and Polar coordinates: | ||
+ | |||
+ | ^ | ||
+ | ^ Graphical |\$$(a, b) \$$ |\$$ (r\angle \theta) \$$ | | ||
+ | ^ Algebraic |\$$z = a + ib\$$ |\$$ z = re^{i\theta} \$$ | | ||
+ | |||
+ | This now allows us to simplify a lot of difficult mathematics. | ||
+ | |||
+ | \$$ | ||
+ | z^3 = 8 = \left\{ | ||
+ | \begin{array}{c} | ||
+ | 8e^{0 i} \\ | ||
+ | 8e^{2\pi i} \\ | ||
+ | 8e^{4\pi i}\\ | ||
+ | \vdots | ||
+ | \end{array} | ||
+ | \right. | ||
+ | \$$ | ||
+ | |||
+ | \$$ | ||
+ | \Rightarrow z = \left\{ | ||
+ | \begin{array}{lcl} | ||
+ | \left(8e^{0 i}\right)^{\frac{1}{3}} &=& 8^{\frac{1}{3}} e^{\frac{0}{3}} = 2\\ | ||
+ | \left(8e^{2\pi i}\right)^{\frac{1}{3}} &=& 8^{\frac{1}{3}} e^{\frac{2\pi}{3}i} = 2 \left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right) = 2 \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3} \\ | ||
+ | \left(8e^{4\pi i}\right)^{\frac{1}{3}} &=& 8^{\frac{1}{3}} e^{\frac{4\pi}{3}i} = 2 \left(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}\right) = 2 \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = -1 - i\sqrt{3} \\ | ||
+ | & | ||
+ | \end{array} | ||
+ | \right. | ||
+ | \$$ | ||
+ | |||
+ | If we had used more than 3 numbers, the roots would have started repeating. | ||
+ | |||
+ | ===== Cartesian vs Polar ===== | ||
+ | Which is the best representation: | ||
+ | |||
+ | \$$z_1 = 1 + i = \sqrt{2}e^\left(i\frac{\pi}{4}\right) \quad \text{and} \quad z_2 = -1 + i = \sqrt{2}e^\left(i\frac{3\pi}{4}\right) \$$ | ||
+ | |||
+ | Imagine having to add, subtract, multiply, or divide these together. | ||
+ | |||
+ | < | ||
+ | |< 100% - 50% 50% >| | ||
+ | ^Operation | ||
+ | ^Addition | ||
+ | ^Subtraction | ||
+ | ^Multiplication | Moderate: \$$ z_1 \cdot z_2 = (1 + i) \cdot (-1 + i) \$$ \$$ = (1)(-1) + (1)(i) + (i)(-1) + (i)(i) \$$ \$$ = -1 + i - i -1 = -2 \$$ | Easiest: | ||
+ | ^Division | Tedious: \$$ \frac{z_1}{z_2} = \frac{(1 + i)}{(-1 + i)} = \frac{(1 + i)(-1 - i)}{(-1 + i)(-1 - i)} \$$ \$$ = \cdots = \frac{-2i}{2} = -i \$$ | Easiest: | ||
+ | ^Exponentiation | The bigger the exponent, the more tedious: \$$z_1^{100} = (1 + i)^{100} = \cdots \$$ | Easy no matter how big the exponent: | ||
+ | ^Roots | ||
+ | </ | ||
+ | \\ | ||
+ | |||
+ | The lesson here is that since the polar representation uses exponents, and exponents turn multiplication into addition((\$ x^a \cdot x^b = x^{a+b} \$)), the polar representation is easiest for multiplication, | ||
+ | ===== Important Algebraic Results ===== | ||
+ | * Use the Euler identity to get the following two useful results: | ||
+ | |||
+ | <WRAP center box 30em> | ||
+ | \$$ \cos \theta = \dfrac{e^{i \theta} + e^{-i \theta}}{2} \qquad \text { & } \qquad \sin \theta = \dfrac{e^{i \theta} - e^{-i \theta}}{2i} \$$ | ||
+ | </ | ||
+ | |||
+ | < | ||
+ | * Modify the Euler identity to see what happens when the angle is \$ (-\theta) \$ | ||
+ | |||
+ | \begin{align*} | ||
+ | && e^{i \theta} | ||
+ | \Rightarrow && e^{i (-\theta)} &= \cos (-\theta) + i \sin (-\theta) \\ | ||
+ | \Rightarrow && e^{-i \theta} &= \cos \theta - i \sin \theta | ||
+ | \end{align*} | ||
+ | |||
+ | * Add the original Euler identity to the new one for \$ (-\theta) \$ and simplify: | ||
+ | \begin{align*} | ||
+ | && e^{i \theta} | ||
+ | + && \underline{e^{-i \theta}} &= \underline{\cos \theta - i \sin \theta} \\ | ||
+ | \Rightarrow && e^{i \theta} + e^{-i \theta} &= 2\cos \theta \\ | ||
+ | \Rightarrow && \cos \theta &= \frac{e^{i \theta} + e^{-i \theta}}{2} \\ | ||
+ | \end{align*} | ||
+ | |||
+ | * To get the second equation, take the original Euler identity and subtract the new one for \$ (-\theta) \$ from it and simplify: | ||
+ | \begin{align*} | ||
+ | && e^{i \theta} | ||
+ | - && \underline{e^{-i \theta}} &= \underline{\cos \theta - i \sin \theta} \\ | ||
+ | \Rightarrow && e^{i \theta} - e^{-i \theta} &= 2i\sin \theta \\ | ||
+ | \Rightarrow && \sin \theta &= \frac{e^{i \theta} - e^{-i \theta}}{2i} \\ | ||
+ | \end{align*} | ||
+ | </ | ||
+ | \\ | ||
+ | |||
+ | * Use the Euler identity to show that: | ||
+ | |||
+ | <WRAP center box 30em> | ||
+ | \$$ | ||
+ | \cos (\theta + \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi\\ | ||
+ | \sin (\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi | ||
+ | \$$ | ||
+ | </ | ||
+ | |||
+ | < | ||
+ | * Multiply the Euler identities for \$\theta\$ and \$\phi\$ and simplify. | ||
+ | |||
+ | \begin{align*} | ||
+ | && e^{i \theta} | ||
+ | \times && \underline{e^{i \phi}} | ||
+ | \Rightarrow && e^{i \theta} e^{i \phi} &= \big(\cos \theta + i \sin \theta\big)\cdot \big(\cos \phi + i \sin \phi \big) \\ | ||
+ | \Rightarrow && e^{i\theta + i\phi} | ||
+ | \Rightarrow && e^{i (\theta + \phi)} | ||
+ | \Rightarrow && e^{i (\theta + \phi)} | ||
+ | \Rightarrow && e^{i (\theta + \phi)} | ||
+ | \end{align*} | ||
+ | |||
+ | But the Euler identity for angle \$ (\theta + \phi) \$ is: | ||
+ | \$$ | ||
+ | e^{i (\theta + \phi)} = \color{green}{\cos(\theta + \phi)} + i\color{blue}{\sin(\theta + \phi)} | ||
+ | \$$ | ||
+ | |||
+ | |||
+ | So comparing the <fc # | ||
+ | \$$ | ||
+ | \color{green}{\cos(\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi} \\ | ||
+ | \color{blue}{\sin(\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi} | ||
+ | \$$ | ||
+ | </ | ||
+ | \\ | ||
+ | |||
+ | |||
+ | * Use the two earlier results to show that: | ||
+ | |||
+ | <WRAP center box 30em> | ||
+ | \$$ \sin (\theta + \Delta \theta) + \sin (\theta - \Delta \theta) = 2 \cos \Delta \theta \sin \theta \$$ | ||
+ | </ | ||
+ | |||
+ | < | ||
+ | * Use the last result for \$ \sin(\theta + \phi) \$ but replace \$ \phi \$ with \$ \Delta \theta \$ | ||
+ | |||
+ | \$$ | ||
+ | \sin (\theta + \Delta \theta) = \cos \theta \sin \Delta \theta + \sin \theta \cos \Delta \theta | ||
+ | \$$ | ||
+ | |||
+ | Similarily, | ||
+ | \begin{align*} | ||
+ | && \sin (\theta - \Delta \theta) & | ||
+ | \Rightarrow && \sin (\theta - \Delta \theta) & | ||
+ | \end{align*} | ||
+ | |||
+ | Adding both of these together, we get: | ||
+ | \begin{align*} | ||
+ | \sin (\theta + \Delta \theta) + \sin (\theta - \Delta \theta) & | ||
+ | & \quad - \cos \theta \sin\Delta \theta + \sin \theta \cos \Delta \theta \\ | ||
+ | &= 2\sin \theta \cos \Delta \theta | ||
+ | &= 2\cos \Delta \theta \sin \theta | ||
+ | \end{align*} | ||
+ | </ | ||
+ | \\ | ||
+ | |||
+ | This last result is the basis behind why [[howto/ | ||
+ | |||
+ | |||
+ | ===== Differential Equations ===== | ||
+ | |||
+ | In the physics of wave, we often have to find solutions to the following type of differential equations: | ||
+ | <WRAP center box 30em> | ||
+ | \$$ a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \$$ | ||
+ | </ | ||
+ | |||
+ | <WRAP center round info 80%> | ||
+ | * A differential equation is an equation that relates a function to its derivatives in some ways and the question is: given some information about the system, what's the function (or family of functions) that satisfy the differential equation. | ||
+ | |||
+ | * In physics we often use a dot above the function to indicate a derivative with respect to time, where as in math, we'll often use an apostrophe. | ||
+ | \$$ \dot{x}(t) = x'(t) = \frac{dx}{dt} \quad \text{and} \quad \ddot{x}(t) = x'' | ||
+ | </ | ||
+ | |||
+ | |||
+ | In our applications, | ||
+ | |||
+ | The next step is to try this ``test function'' | ||
+ | |||
+ | \begin{align*} | ||
+ | & x (t) = e^{rt} \\ | ||
+ | \Rightarrow \qquad & \dot{x}(t) = r e^{rt} \\ | ||
+ | \Rightarrow \qquad & \ddot{x}(t) = r^2 e^{rt} | ||
+ | \end{align*} | ||
+ | |||
+ | When we put these into the differential equation, we get: | ||
+ | |||
+ | \begin{align*} | ||
+ | & a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \\ | ||
+ | \Rightarrow \qquad & a (r^2 e^{rt}) + b (r e^{rt}) + c (e^{rt}) = 0 \\ | ||
+ | \Rightarrow \qquad & e^{rt} (a r^2 + b r + c ) = 0 \\ | ||
+ | \Rightarrow \qquad & a r^2 + b r + c = 0 \\ | ||
+ | \Rightarrow \qquad &r = - \dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} | ||
+ | \end{align*} | ||
+ | |||
+ | So what does that result mean? Remember, what we're looking for is the function \$x(t)\$ that satisfies the differential equation \$ a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \$ | ||
+ | |||
+ | |||
+ | |||
+ | What we've go so far says that our test function \$x(t) = e^{rt}\$ will satisfy the differential equation if \$r\$ is given by above equation. | ||
+ | |||
+ | To simplify the notation, let's define \$\alpha\$ and \$\beta\$ as: | ||
+ | \$$ \alpha = \dfrac{b}{2a} \qquad \text{and} \qquad \beta = \dfrac{\sqrt{|{b^2 - 4ac}|}}{2a} \$$ | ||
+ | |||
+ | Notice how the absolute value under the square root ensures that \$\beta\$ is always real. | ||
+ | |||
+ | \$r\$ then: | ||
+ | |||
+ | \$$ r = \left\{ \begin{array}{ll} -\alpha \pm \beta & \text{if } b^2 - 4ac > 0,\\ | ||
+ | | ||
+ | |||
+ | Let's examine both of these cases in more detail. | ||
+ | |||
+ | ==== Case 1: Over Damped Oscillation ==== | ||
+ | When \$ b^2 - 4ac > 0 \$ , \$ r \$ is real and the general solution is: | ||
+ | |||
+ | \begin{align*} | ||
+ | x(t) &= A_1 e^{r_1 t} + A_2 e^{r_2 t} \\ | ||
+ | &= A_1 e^{( -\alpha + \beta) t} + A_2 e^{( -\alpha - \beta) t} \\ | ||
+ | &= A_1 e^{-\alpha t} e^{\beta t} + A_2 e^{-\alpha t} e^{-\beta t} | ||
+ | \end{align*} | ||
+ | |||
+ | \$$ x(t) = e^{-\alpha t} ( A_1 e^{\beta t} + A_2 e^{-\beta t} ) \$$ | ||
+ | |||
+ | It's normal to have two constants of integration since our differential equation has a second degree derivative in it. To find these constants, we'd need to know more about the system' | ||
+ | |||
+ | ==== Case 2: Under Damped ==== | ||
+ | When \$ b^2 - 4ac < 0 \$ , \$ r \$ is complex and we'll be using the Euler identity to simplify our solutions | ||
+ | |||
+ | \begin{align*} | ||
+ | x(t) &= A_1 e^{r_1 t} + A_2 e^{r_2 t} \\ | ||
+ | &= A_1 e^{( -\alpha + i \beta) t} + A_2 e^{( -\alpha - i \beta) t} \\ | ||
+ | &= A_1 e^{-\alpha t} e^{i \beta t} + A_2 e^{-\alpha t} e^{-i \beta t} \\ | ||
+ | &= e^{-\alpha t} ( A_1 e^{i \beta t} + A_2 e^{-i \beta t} ) \\ | ||
+ | &= e^{-\alpha t} \Big( A_1 \big(cos( \beta t) + i \sin( \beta t) \big) + A_2 \big(cos( -\beta t) + i \sin( -\beta t) \big) \Big) \\ | ||
+ | &= e^{-\alpha t} \Big( A_1 \big(cos(\beta t) + i \sin(\beta t) \big) + A_2 \big(\cos(\beta t) - i \sin(\beta t)\big)\Big) \\ | ||
+ | &= e^{-\alpha t} \Big( (A_1 + A_2) \cos(\beta t) + i (A_1 - A_2) \sin(\beta t) \Big) \\ | ||
+ | &= e^{-\alpha t} \Big( a_1 \cos(\beta t) + a_2 \sin(\beta t) \Big) \\ | ||
+ | &= e^{-\alpha t} \Big( A \sin \phi \cos(\beta t) + A \cos \phi \sin(\beta t) \Big) \\ | ||
+ | &= Ae^{-\alpha t} \Big(\sin \phi \cos(\beta t) + \cos \phi \sin(\beta t) \Big) \\ | ||
+ | \end{align*} | ||
+ | |||
+ | In the last three lines, we've redefined the constants of integration a few times so that: | ||
+ | |||
+ | \begin{align*} | ||
+ | a_1 &= A_1 + A_2 & , a_2 &= i(A_1 - A_2) \\ | ||
+ | a_1 &= A \sin \phi & , a_2 &= A \cos \phi | ||
+ | \end{align*} | ||
+ | |||
+ | And we finally use one of the trig identities we proved earlier to write the solution as: | ||
+ | \$$ x(t) = A e^{-\alpha t} \sin(\beta t + \phi) \$$ | ||
+ | |||
+ | ==== Case 3: Critically Damped ==== | ||
+ | When \$ b^2 - 4ac = 0 \$ , \$ r = -\frac{b}{2a} \$ is real and negative but our test solution is under determined. | ||
+ | \begin{align*} | ||
+ | && | ||
+ | \Rightarrow && \dot x(t) &= re^{rt}(A + Bt) + Be^{rt} \\ | ||
+ | && | ||
+ | && | ||
+ | \Rightarrow && | ||
+ | && | ||
+ | && | ||
+ | \end{align*} | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==== Exemple ==== | ||
+ | Nous avons donc deux types de solutions complètement différents qui dépendent de trois paramètres \$ a, b, c \$. Pour voir comment ces paramètres affectent le graphique, imaginons qu'une de nos conditions initiales est \$ \phi = \frac{\pi}{2} \$ . Ça veut dire que: | ||
+ | |||
+ | \$ \begin{align*} | ||
+ | & a_1 = A \sin \pi/2 = A & & a_2 = A \cos \pi/2 = 0 \\ | ||
+ | \Rightarrow \qquad & A_1 + A_2 = A & & A_1 - A_2 = 0 \\ | ||
+ | \Rightarrow \qquad & A_1 = A/2 & & A_2 = A/2 | ||
+ | \end{align*} \$ | ||
+ | |||
+ | Dans ce cas particulier, | ||
+ | |||
+ | \$ \begin{equation*} x(t) = \left\{ \begin{array}{rl} A e^{-\alpha t} \dfrac{e^{\beta t} + e^{-\beta t}}{2} & \text{si } b^2 - 4ac > 0 ,\\ | ||
+ | A e^{-\alpha t} \cos(\beta t) & \text{si } b^2 - 4ac < 0 ,\\ | ||
+ | \end{array} \right. \end{equation*} \$ | ||
+ | |||
+ | FIXME Geogebra | ||
howto/hambasics/sections/test.txt · Last modified: 2021/02/13 19:14 by va7fi