howto:hambasics:sections:test
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howto:hambasics:sections:test [2021/01/02 20:47] – [The Complex Plane] va7fi | howto:hambasics:sections:test [2021/02/13 19:02] – va7fi | ||
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- | ~~NOTOC~~ | ||
====== Optional Math of Waves ====== | ====== Optional Math of Waves ====== | ||
- | This is a brief survey of the math required to analyze waves at the first or second year university level. | + | This is a brief survey of the math required to analyze waves at the first or second year university level. If you did well in grade 12 high school math, you'll probably be able to follow this and learn some new and really cool math. |
===== Real numbers ===== | ===== Real numbers ===== | ||
- | If \$ (5) \times (5) = 25 \$ and \$ (-5) \times (-5) = 25 \$, what number | + | If \$ (5)^2 = 25 \$ and \$ (-5)^2 = 25 \$, what number can you put in the box so that: |
- | \$$ \Box \times \Box = -25 \$$ | + | \$$ \Box ^2 = -25 \$$ |
- | + | ||
- | It turns out that nowhere on the //real// number line is there a number such that when you multiply it by itself you get a negative number since two positive numbers give a positive number, and two negative numbers also give a positive number. | + | |
- | + | ||
- | But could we //invent// one? | + | |
+ | It turns out that there is no //real// number such that when you multiply it by itself you get a negative number. | ||
===== Complex Numbers ===== | ===== Complex Numbers ===== | ||
- | Let's create an imaginary number called \$ i \$ such that: | + | Let's create an //imaginary// number called \$ i \$ such that: |
+ | <WRAP center box 30em> | ||
\$$ i = \sqrt{-1} \qquad \Rightarrow \qquad i^2 = -1 \$$ | \$$ i = \sqrt{-1} \qquad \Rightarrow \qquad i^2 = -1 \$$ | ||
+ | </ | ||
- | Even though \$ i \$ is nowhere on the real line (in math, we say that: \$ i \not\in \mathbb{R} \$), we can non-the-less perform interesting mathematical operations with it: | + | Even though \$ i \$ is nowhere on the real line (in math, we say that: \$ i \not\in \mathbb{R} \$), we can none-the-less perform interesting mathematical operations with it: |
* We can add it to a real number and create a //complex// number: | * We can add it to a real number and create a //complex// number: | ||
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\begin{align*} | \begin{align*} | ||
- | (1+i)^2 &= 1 + 2i + i^2 \\ | + | (1+i)^2 |
+ | &= (1+i)\cdot(1+i) \\ | ||
+ | &= 1 + 2i + i^2 \\ | ||
&= 1 + 2i - 1 \\ | &= 1 + 2i - 1 \\ | ||
&= 2i | &= 2i | ||
Line 38: | Line 38: | ||
\end{equation*} | \end{equation*} | ||
- | === A Little Philosophy === | + | ==== A Little Philosophy |
- | If these weird numbers follow all of the algebra rules without inconsistencies, | + | 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), | ||
===== The Complex Plane ===== | ===== The Complex Plane ===== | ||
- | To represent a complex number graphically, | + | In the same way that we can represent |
- | You can move the point around to see what the polar representation | + | {{ numberline.png }} |
+ | |||
+ | ... 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, | ||
{{ggb>/ | {{ggb>/ | ||
+ | |||
+ | <WRAP center round info 80%> | ||
+ | You can move the point around to look at other complex numbers on the plane. | ||
+ | </ | ||
+ | |||
To convert between the Cartesian \$(a,b) \$ and the Polar \$ (r \angle \theta) \$ representations, | To convert between the Cartesian \$(a,b) \$ and the Polar \$ (r \angle \theta) \$ representations, | ||
- | ^ \$$ a + ib \rightarrow r\angle \theta \$$ ^ \$$ r\angle \theta \rightarrow a+ib \$$ | | + | ^ \$$ (a, b) \rightarrow |
| \$$ r^2 = a^2 + b^2 \$$ | \$$ a = r\cos\theta \$$ | | | \$$ r^2 = a^2 + b^2 \$$ | \$$ a = r\cos\theta \$$ | | ||
| \$$ \tan \theta = \dfrac{b}{a} \$$ | \$$ b = r\sin\theta \$$ | | | \$$ \tan \theta = \dfrac{b}{a} \$$ | \$$ b = r\sin\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 (because the circumference is 2πr) | ||
==== Roots ==== | ==== Roots ==== | ||
- | The complex | + | The complex |
{{ggb>/ | {{ggb>/ | ||
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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? | 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. | + | Now move \$ w = 8 \$ and \$n = 3 \$ to have a look at the solutions |
< | < | ||
{{ complexroots.png? | {{ complexroots.png? | ||
- | So z = 2 was to be expected since \$ 2^3 = 8 \$ but it looks like there' | + | So \$z = 2\$ was to be expected since \$ 2^3 = 8 \$ but it looks like there' |
To find them, we first notice that the three solutions are spread out evenly around the circle, that is they are 120° apart. | 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, | + | So in polar coordinates, |
We can now convert them to Cartesian: | We can now convert them to Cartesian: | ||
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Let's check that the second solution works: | Let's check that the second solution works: | ||
\begin{align*} | \begin{align*} | ||
- | (-1 + i\sqrt{3})^3 &= (-1 + i\sqrt{3})(-1 + i\sqrt{3})(-1 + i\sqrt{3}) \\ | + | (-1 + i\sqrt{3})^3 &= (-1 + i\sqrt{3})\cdot(-1 + i\sqrt{3})(-1 + i\sqrt{3}) \\ |
- | &= (-1 + i\sqrt{3})\big(1 -2i\sqrt{3} +(i^2)(3)\big) \\ | + | &= (-1 + i\sqrt{3})\cdot\big(1 -2i\sqrt{3} +(i^2)(3)\big) \\ |
- | &= (-1 + i\sqrt{3})\big(1 -2i\sqrt{3} +(-1)(3)\big) \\ | + | &= (-1 + i\sqrt{3})\cdot\big(1 -2i\sqrt{3} +(-1)(3)\big) \\ |
- | &= (-1 + i\sqrt{3})(-2 -2i\sqrt{3}) \\ | + | &= (-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 | &= 8 | ||
\end{align*} | \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.)): | ||
- | ===== Forme Polaire ===== | + | <WRAP center box 30em> |
- | Comme pour les vecteurs, les nombres complexes peuvent être exprimés en coordonnées cartésiennes ou polaires. | + | \$$ e^{i \theta} |
- | FIXME: Geogebra | + | </ |
+ | Let's use two different ways to verify that this mysterious identity is true. | ||
- | === Exercices | + | ==== The Derivatives ==== |
- | * Convertir | + | If we separate this identity into two functions and take their derivatives, |
- | * Convertir | + | \begin{align*} |
+ | && f(\theta) &= e^{i \theta} &\text{ & }&& g(\theta) &= \cos \theta + i \sin \theta \\ | ||
+ | \Rightarrow && f' | ||
+ | \Rightarrow && f'(\theta) &= i \cdot f(\theta) &\text{ & }&& g' | ||
+ | \end{align*} | ||
+ | We know that there' | ||
- | ===== Formule de Euler ===== | + | ==== Taylor |
- | La formule de Euler expose une relation très profonde entre les fonctions trigonométriques et les fonctions exponentiels ((Note, obtenir "la plus belle équation du monde", | + | Another method to verify the Euler identity is to use Taylor series: |
- | \$$ e^{i \theta} = \cos \theta + i \sin \theta \$$ | + | |
- | Vérifions cette identité mystérieuse de deux façons. | + | \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*} | ||
- | === La dérivée === | + | - Replace |
- | Si on sépare cette identité en deux fonctions et qu'on prend la dérivées de ces deux fonctions, on remarque que: | + | - Replace |
- | \$ \begin{align*} | + | |
- | \text{Si} && f(\theta) & | + | |
- | \Rightarrow && f'(\theta) &= i e^{i \theta} &\text{ et }&& g' | + | |
- | \Rightarrow && f' | + | |
- | \end{align*} \$ | + | |
- | On sait qu'il n'y a qu'une fonction \$ h(x) \$ qui satisfasse l' | ||
- | === Taylor | + | ===== Euler Identity and Polar-Cartesian Representations ===== |
- | Une autre méthode de vérifier la formule d'Euler est d' | + | |
- | \$ \begin{align*} | + | In the previous section, we saw that a complex number |
- | e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \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*} | + | |
- | Si on remplace | + | ^ \$$ (a,b) \rightarrow (r\angle \theta) |
+ | | \$$ r^2 = a^2 + b^2 \$$ | \$$ a = r\cos\theta \$$ | | ||
+ | | \$$ \tan \theta | ||
- | === Exercices === | + | This means that: |
- | * Utiliser la formule de Euler pour obtenir ces deux résultats qui seront très pratique: | + | |
- | \$$ \cos \theta = \dfrac{e^{i \theta} + e^{-i \theta}}{2} \qquad \text { et } \qquad \sin \theta = \dfrac{e^{i \theta} - e^{-i \theta}}{2i} \$$ | + | |
- | * Utiliser la formule de Euler pour démonter d'un seul coup que: | ||
- | \$ | ||
\begin{align*} | \begin{align*} | ||
- | \cos (\theta + \phi) &= \cos \theta | + | z &= a + ib \\ |
- | \sin (\theta + \phi) & | + | |
+ | & | ||
+ | &= r e^{i\theta} | ||
\end{align*} | \end{align*} | ||
- | \$ | ||
+ | This offers another interpretation of the Euler identity as the algebraic conversion between Cartesian and Polar coordinates: | ||
- | * Utiliser les deux résultats de la formule de Euler pour démontrer que: | + | ^ |
+ | ^ 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 ===== | ||
+ | | ||
+ | |||
+ | <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 \$$ | \$$ \sin (\theta + \Delta \theta) + \sin (\theta - \Delta \theta) = 2 \cos \Delta \theta \sin \theta \$$ | ||
- | (ce qui sera très important quand nous ferons de l' | + | </ |
+ | < | ||
+ | * Use the last result for \$ \sin(\theta + \phi) \$ but replace \$ \phi \$ with \$ \Delta \theta \$ | ||
- | ===== Équations différentiels ===== | + | \$$ |
+ | \sin (\theta + \Delta \theta) | ||
+ | \$$ | ||
- | Dans la physique des ondes, on aura bientôt à trouver la solutions d'une équation différentiel de ce type: | + | Similarily, |
- | \$$ a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \$$ | + | \begin{align*} |
+ | && \sin (\theta - \Delta \theta) & | ||
+ | \Rightarrow && \sin (\theta - \Delta \theta) &= -\cos \theta \sin\Delta \theta + \sin \theta \cos \Delta \theta \\ | ||
+ | \end{align*} | ||
- | Dans nos applications, les paramètres | + | Adding both of these together, we get: |
+ | \begin{align*} | ||
+ | \sin (\theta + \Delta \theta) + \sin (\theta - \Delta \theta) &= \cos \theta \sin \Delta \theta + \sin \theta \cos \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/ | ||
- | \$ \begin{align*} | + | |
+ | ===== 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} \\ | & x (t) = e^{rt} \\ | ||
\Rightarrow \qquad & \dot{x}(t) = r e^{rt} \\ | \Rightarrow \qquad & \dot{x}(t) = r e^{rt} \\ | ||
\Rightarrow \qquad & \ddot{x}(t) = r^2 e^{rt} | \Rightarrow \qquad & \ddot{x}(t) = r^2 e^{rt} | ||
- | \end{align*} | + | \end{align*} |
- | et notre équation différentiel devient: | + | When we put these into the differential equation, we get: |
- | \$ \begin{align*} | + | \begin{align*} |
& a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \\ | & 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 & 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 & e^{rt} (a r^2 + b r + c ) = 0 \\ | ||
- | \Rightarrow \qquad & a r^2 + b r + c = 0 | + | \Rightarrow \qquad & a r^2 + b r + c = 0 \\ |
- | \end{align*} | + | \Rightarrow \qquad &r = - \dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} |
+ | \end{align*} | ||
- | \$ \Rightarrow | + | So what does that result mean? Remember, what we're looking for is the function |
- | Puisque \$ r \$ contient une racine carré, il peut être réel, ou complexe. Pour simplifier la notation, disons que: \$ \alpha = \dfrac{b}{2a} \qquad \text{et} \qquad \beta = \dfrac{\sqrt{|{b^2 - 4ac}|}}{2a} \$ | ||
- | Nous pourrons donc dire que: \$ \begin{equation*} r = \left\{ \begin{array}{rl} -\alpha \pm \beta & \text{si } b^2 - 4ac > 0,\\ | ||
- | | ||
- | Examinons les deux cas en plus de détails. | + | 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. |
- | === Cas 1: === | + | To simplify the notation, let's define \$\alpha\$ and \$\beta\$ as: |
- | Quand \$ b^2 - 4ac > 0 \$ , \$ r \$ est réel et la solutions général sera: | + | \$$ \alpha = \dfrac{b}{2a} \qquad \text{and} \qquad \beta = \dfrac{\sqrt{|{b^2 - 4ac}|}}{2a} |
- | \$ \begin{align*} | + | 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} \\ | 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 + \beta) t} + A_2 e^{( -\alpha - \beta) t} \\ | ||
&= A_1 e^{-\alpha t} e^{\beta t} + A_2 e^{-\alpha t} e^{-\beta t} | &= A_1 e^{-\alpha t} e^{\beta t} + A_2 e^{-\alpha t} e^{-\beta t} | ||
- | \end{align*} | + | \end{align*} |
\$$ x(t) = e^{-\alpha t} ( A_1 e^{\beta t} + A_2 e^{-\beta t} ) \$$ | \$$ x(t) = e^{-\alpha t} ( A_1 e^{\beta t} + A_2 e^{-\beta t} ) \$$ | ||
- | C'est normal | + | It's normal |
- | === Cas 2: === | + | ==== Case 2: Under Damped ==== |
- | Quand \$ b^2 - 4ac < 0 \$ , \$ r \$ est complexe et nous utiliserons la formule de Euler pour simplifier notre solutions. | + | When \$ b^2 - 4ac < 0 \$ , \$ r \$ is complex and we'll be using the Euler identity to simplify our solutions |
- | \$ \begin{align*} | + | \begin{align*} |
x(t) &= A_1 e^{r_1 t} + A_2 e^{r_2 t} \\ | 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 + i \beta) t} + A_2 e^{( -\alpha - i \beta) t} \\ | ||
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&= 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 + 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_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) | + | &= e^{-\alpha t} \Big( A \sin \phi \cos(\beta t) + A \cos \phi \sin(\beta t) \Big) \\ |
- | \end{align*} | + | &= Ae^{-\alpha t} \Big(\sin \phi \cos(\beta t) + \cos \phi \sin(\beta t) \Big) \\ |
+ | \end{align*} | ||
- | Dans les deux dernière ligne, nous avons ré-écrit les constantes d'intégration: | + | In the last three lines, we've redefined the constants of integration a few times so that: |
- | \$ \begin{align*} | + | \begin{align*} |
a_1 &= A_1 + A_2 & , a_2 &= i(A_1 - A_2) \\ | a_1 &= A_1 + A_2 & , a_2 &= i(A_1 - A_2) \\ | ||
a_1 &= A \sin \phi & , a_2 &= A \cos \phi | a_1 &= A \sin \phi & , a_2 &= A \cos \phi | ||
- | \end{align*} | + | \end{align*} |
- | + | ||
- | Pour finalement pouvoir utiliser une identité trigonométrique: | + | |
+ | 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) \$$ | \$$ x(t) = A e^{-\alpha t} \sin(\beta t + \phi) \$$ | ||
- | === Exemple === | + | ==== 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: | 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: | ||
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FIXME Geogebra | FIXME Geogebra | ||
+ |
howto/hambasics/sections/test.txt · Last modified: 2021/02/13 19:14 by va7fi