howto:hambasics:sections:test
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howto:hambasics:sections:test [2021/01/03 13:16] – [The Complex Plane] 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: | ||
- | {{ggb>/ | + | <WRAP center round info 80%> |
You can move the point around to look at other complex numbers on the plane. | 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, | 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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This offers another interpretation of the Euler identity as the algebraic conversion between Cartesian and Polar coordinates: | This offers another interpretation of the Euler identity as the algebraic conversion between Cartesian and Polar coordinates: | ||
- | ^ | + | ^ |
^ Graphical |\$$(a, b) \$$ |\$$ (r\angle \theta) \$$ | | ^ Graphical |\$$(a, b) \$$ |\$$ (r\angle \theta) \$$ | | ||
^ Algebraic |\$$z = a + ib\$$ |\$$ z = re^{i\theta} \$$ | | ^ Algebraic |\$$z = a + ib\$$ |\$$ z = re^{i\theta} \$$ | | ||
- | This now allows us to simplify a lot of difficult mathematics. | + | This now allows us to simplify a lot of difficult mathematics. |
\$$ | \$$ | ||
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\$$ | \$$ | ||
- | The weird part about looking at a simple number like 8 in polar coordinates on the complex plane is that it can be viewed as many different points piled up on top of one another depending on how many times we go around | + | 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 | ||
+ | ^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