howto:hambasics:sections:test
Differences
This shows you the differences between two versions of the page.
Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
howto:hambasics:sections:test [2021/01/03 20:27] – [Under Construction] va7fi | howto:hambasics:sections:test [2021/02/13 19:14] (current) – [Cartesian vs Polar] va7fi | ||
---|---|---|---|
Line 7: | Line 7: | ||
\$$ \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 ===== | ||
Line 24: | Line 24: | ||
\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 48: | Line 50: | ||
... 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 \$$ | | ||
Line 60: | Line 64: | ||
Note that very often, we use radians instead of degrees for the angle. | Note that very often, we use radians instead of degrees for the angle. | ||
* Imagine a circle. | * Imagine a circle. | ||
- | * Now imagine the length from the centre to the circle (along the ``radius" | + | * Now imagine the length from the centre to the circle (along the "radius" |
* Take that length and lay it down on the perimeter of the circle. | * 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). | * 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) | + | * That's why a circle has 2π radians |
==== Roots ==== | ==== Roots ==== | ||
Line 76: | Line 80: | ||
< | < | ||
{{ 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 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. | To find them, we first notice that the three solutions are spread out evenly around the circle, that is they are 120° apart. | ||
Line 140: | Line 144: | ||
===== 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 \$$ | | ||
Line 161: | Line 165: | ||
^ 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. |
\$$ | \$$ | ||
Line 192: | Line 196: | ||
\$$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) \$$ | \$$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, | + | Imagine having to add, subtract, multiply, or divide these together. |
< | < | ||
Line 206: | Line 210: | ||
\\ | \\ | ||
- | 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, | + | 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: | ||
Line 318: | Line 322: | ||
<WRAP center round info 80%> | <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. | + | * 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. | + | * 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'' | \$$ \dot{x}(t) = x'(t) = \frac{dx}{dt} \quad \text{and} \quad \ddot{x}(t) = x'' | ||
</ | </ | ||
Line 348: | Line 352: | ||
- | ===== Under Construction ===== | + | |
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. | 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: | + | 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} \$$ | \$$ \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. | + | Notice how the absolute value under the square root ensures that \$\beta\$ is always real. |
\$r\$ then: | \$r\$ then: | ||
- | \$$ r = \left\{ \begin{array}{rl} -\alpha \pm \beta & \text{if } b^2 - 4ac > 0,\\ | + | \$$ r = \left\{ \begin{array}{ll} -\alpha \pm \beta & \text{if } b^2 - 4ac > 0,\\ |
| | ||
Let's examine both of these cases in more detail. | 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