howto:hambasics:sections:test
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howto:hambasics:sections:test [2021/01/03 11:55] – [Euler Identity and Polar-Cartesian Representations] va7fi | howto:hambasics:sections:test [2021/02/13 19:03] – va7fi | ||
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====== 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. | ||
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\$$ \Box ^2 = -25 \$$ | \$$ \Box ^2 = -25 \$$ | ||
- | It turns out that there is no //real// number such that when you multiply it by itself you get a negative number, but could we invent an // | + | It turns out that there is no //real// number such that when you multiply it by itself you get a negative number. But could we invent an // |
===== Complex Numbers ===== | ===== Complex Numbers ===== | ||
Let's create an // | Let's create an // | ||
- | |||
<WRAP center box 30em> | <WRAP center box 30em> | ||
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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 | ||
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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), | 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 ===== | ||
In the same way that we can represent real numbers by a point on the real number line...((Number line picture from [[wp> | In the same way that we can represent real numbers by a point on the real number line...((Number line picture from [[wp> | ||
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... 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: | ... 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>/ | {{ggb>/ | ||
- | |||
- | You can move the point around to see what the polar representation \$ (r \angle \theta) \$ is. | ||
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 ==== | ||
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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 |
< | < | ||
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</ | </ | ||
\\ | \\ | ||
- | |||
===== The Euler Identity ===== | ===== The Euler Identity ===== | ||
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</ | </ | ||
- | Let's use two different ways to verify that this mysterious identity | + | Let's use two different ways to verify that this mysterious identity |
==== The Derivatives ==== | ==== The Derivatives ==== | ||
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===== Euler Identity and Polar-Cartesian Representations ===== | ===== 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 \theta \$. We saw that to convert between the Cartesian \$(a,b) \$ and the Polar \$ (r \angle \theta) \$ 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 \theta) \$. We saw that 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 \$$ | | ||
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\end{align*} | \end{align*} | ||
- | This offers another interpretation of the Euler identity: | + | This offers another interpretation of the Euler identity |
- | ^ | + | ^ |
- | ^ Algebraic | + | ^ Graphical |
- | ^ Graphical | + | ^ Algebraic |
+ | 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 ===== | ===== Important Algebraic Results ===== | ||
* Use the Euler identity to get the following two useful results: | * Use the Euler identity to get the following two useful results: | ||
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\\ | \\ | ||
- | ===== Équations différentiels ===== | + | This last result is the basis behind why [[howto/ |
- | Dans la physique des ondes, on aura bientôt à trouver la solutions d'une équation différentiel de ce type: | ||
- | \$$ a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \$$ | ||
- | Dans nos applications, | + | ===== 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 \$$ | ||
+ | </ | ||
- | \$ \begin{align*} | + | <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