每日彙整: 2018-08-05

樹莓派、樹莓派之學習、樹莓派之教育

STEM 隨筆︰古典力學︰轉子【五】《電路學》 五【電感】 III‧阻抗‧B

十二因緣

《緣起經》玄奘譯

佛言，云何名緣起初義？謂：依此有故彼有，此生故彼生。所謂：無明緣行，行緣識，識緣名色，名色緣六處，六處緣觸，觸緣受，受緣愛，愛緣取，取緣有，有緣生，生緣老死，起愁、歎、苦、憂、惱，是名為純大苦蘊集，如是名為緣起初義。

 

『邏輯學』上說『有 □ 則有 ○ ，無 ○ 則無 □』，既已『有 □ 』又想『無 ○ 』，哪裡能夠不矛盾的啊！過去魏晉時『王弼』講︰一，數之始而物之極也。謂之為妙有者，欲言有，不見其形，則非有，故謂之妙；欲言其無，物由之以生，則非無，故謂之有也。斯乃無中之有，謂之妙有也。

假使用『恆等式』 1 - x^n = (1 - x)(1 + x + \cdots + x^{n-1})

來計算 \frac{1 + x + \cdots + x^{m-1}}{1 + x + \cdots + x^{n-1}}，

將等於 \frac{1 - x^m}{1 - x^n} = (1 - x^m) \left[1 + (x^n) + { (x^n) }^2 + { (x^n) } ^3 + \cdots \right] = 1 - x^m + x^n - x^{n+m} + x^{2n} - \cdots，

那麼 1 - 1 + 1 - 1 + \cdots 難道不應該『等於』 \frac{m}{n} 的嗎？

一七四三年時，『伯努利』正因此而反對『歐拉』所講的『可加性』說法，『同』一個級數怎麼可能有『不同』的『和』的呢？？作者不知如果在太空裡，乘坐著『加速度』是 g 的太空船，在上面用著『樹莓派』控制的『奈米手』來擲『骰子』，是否一定能得到『相同點數』呢？難道說『牛頓力學』不是只要『初始態』是『相同』的話，那個『骰子』的『軌跡』必然就是『一樣』的嗎 ？？據聞，法國籍義大利裔大數學家『約瑟夫‧拉格朗日』伯爵 Joseph Lagrange 倒是有個『說法』︰事實上，對於『不同』的 m,n 來講， 從『幂級數』來看，那個 = 1 - x^m + x^n - x^{n+m} + x^{2n} - \cdots 是有『零的間隙』的 1 + 0 + 0 + \cdots - 1 + 0 + 0 + \cdots，這就與 1 - 1 + 1 - 1 + \cdots 『形式』上『不同』，我們怎麼能『先驗』的『期望』結果會是『相同』的呢！！

假使我們將『幾何級數』 1 + z + z^2 + \cdots + z^n + \cdots = \frac{1}{1 - z} ，擺放到『複數平面』之『單位圓』上來『研究』，輔之以『歐拉公式』 z = e^{i \theta} = \cos \theta + i\sin \theta，或許可以略探『可加性』理論的『意指』。當 0 < \theta < 2 \pi 時，\cos \theta \neq 1 ，雖然 |e^{i \theta}| = 1，我們假設那個『幾何級數』會收斂，於是得到 1 + e^{i \theta} + e^{2i \theta} + \cdots = \frac{1}{1 - e^{i \theta}} = \frac{1}{2} + \frac{1}{2} i \cot \frac{\theta}{2}，所以 \frac{1}{2} + \cos{\theta} + \cos{2\theta} + \cos{3\theta} + \cdots = 0 以及 \sin{\theta} + \sin{2\theta} + \sin{3\theta} + \cdots = \frac{1}{2} \cot \frac{\theta}{2}。如果我們用 \theta = \phi + \pi 來『代換』，此時 -\pi < \phi < \pi，可以得到【一】 \frac{1}{2} - \cos{\phi} + \cos{2\phi} - \cos{3\phi} + \cdots = 0 和【二】 \sin{\phi} - \sin{2\phi} + \sin{3\phi} - \cdots = \frac{1}{2} \tan \frac{\phi}{2}。要是在【一】式中將 \phi 設為『零』的話，我們依然會有 1 - 1 + 1 - 1 + \cdots = \frac{1}{2} ；要是驗之以【二】式，當 \phi = \frac{\pi}{2} 時，原式可以寫成 1 - 0 - 1 - 0 + 1 - 0 - 1 - 0 + \cdots = \frac{1}{2}。如此看來 s = 1 + z + z^2 + z^3 + \cdots = 1 +z s 的『形式運算』，可能是有更深層的『關聯性』的吧！！

Circle-trig6.svg

複數平面之單位圓

300px-Unit_circle_angles_color.svg

220px-Periodic_sine

假使我們將【二】式對 \phi 作『逐項微分』得到 \cos{\phi} - 2\cos{2\phi} + 3\cos{3\phi} - \cdots = \frac{1}{4} \frac{1}{{(\cos \frac{\phi}{2})}^2}，此時令 \phi = 0，就得到 1 - 2 + 3 - 4 + 5 - \cdots = \frac{1}{4}。如果把【一】式改寫成 \cos{\phi} - \cos{2\phi} + \cos{3\phi} - \cdots = \frac{1}{2} 然後對 \phi 作『逐項積分』 \int \limits_{0}^{\theta} ，並將變數 \theta 改回 \phi 後得到 \sin{\phi} - \frac{\sin{2\phi}}{2} + \frac{\sin{3\phi}}{3} - \cdots = \frac{\phi}{2}；再做一次 作『逐項積分』 \int \limits_{0}^{\theta} ，且將變數 \theta 改回 \phi 後將得到 1 - \cos{\phi} - \frac{1 - \cos{2\phi}}{2^2} + \frac{1 - \cos{3\phi}}{3^2} - \cdots = \frac{\phi^2}{4}，於是當 \phi = \pi 時，1 + \frac{1}{3^2} + \frac{1}{5^2} + \cdots = \frac{\pi^2}{8}。然而 1 + \frac{1}{3^2} + \frac{1}{5^2} + \cdots = [1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots] - [\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \cdots] =[1 - \frac{1}{4}][1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \cdots] ，如此我們就能得到了『巴塞爾問題』的答案 \sum \limits_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}。那麼

S= \ \ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + \cdots 減
4S=\ \ \ \ \ \ 4 + \ \ \ \ \ 8 + \ \ \ \ \ 12 + \cdots 等於
-3S= 1 - 2 + 3 - 4 + 5 - 6 + \cdots = \frac{1}{4}，所以 S = - \frac{1}{12}。

但是這樣的作法果真是有『道理』的嗎？假使按造『級數的極限』 之『定義』，如果『部份和』 S_n = \sum \limits_{k=0}^{n} a_n 之『極限』 S = \lim \limits_{n \to \infty} S_n 存在， S 能不滿足 S = a_0 + a_1 + a_2 + a_3 + \cdots = a_0 + (S - a_0) 的嗎？或者可以是 \sum \limits_{n=0}^{\infty} k \cdot a_n \neq k \cdot S 的呢？即使又已知 S^{\prime} = \sum \limits_{n=0}^{\infty} b_n ，還是說可能會發生 \sum \limits_{n=0}^{\infty} a_n + b_n \neq S + S^{\prime} 的哩！若是說那些都不會發生，所謂的『可加性』的『概念』應當就可以看成『擴大』且包含『舊有』的『級數的極限』 的『觀點』的吧！也許我們應當使用別種『記號法』來『表達』它，以免像直接寫作 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + \cdots = - \frac{1}{12} 般的容易引起『誤解』，畢竟是也存在著多種『可加法』的啊！至於說那些『可加法』的『意義詮釋』，就看『使用者』的吧！！

─── 《【SONIC Π】電聲學之電路學《四》之《 V!》‧下》

 

傳說過去有位哲學家講︰

牛頓『運動定律』 \vec{F} = m \cdot \vec{a} ，不過是『因果律』之案例耳。

如果以此立論，莫非

曾經自然不但忌真空，而後厭運動，所以阻抗生耶？

假使考察

\frac{Cause}{Effect} = Impedance

這個特殊形式，可以用來計算 \frac{\vec{F}}{\vec{a}} ，得到 m 乎？

要是

\frac{d}{dt} \vec{p} = \frac{d}{dt} m \cdot \vec{v} = \frac{d \ m}{dt} \cdot \vec{v} + m \cdot \frac{d \ \vec{v}}{dt} = \vec{F}

又該怎麼論呢？？

倘若不知『電抗』意指︰

阻抗

阻抗（electrical impedance）是電路中電阻、電感、電容對交流電的阻礙作用的統稱。阻抗是一個複數，實部稱為電阻，虛部稱為電抗；其中電容在電路中對交流電所起的阻礙作用稱為容抗，電感在電路中對交流電所起的阻礙作用稱為感抗，容抗和感抗合稱為電抗 。阻抗將電阻的概念加以延伸至交流電路領域，不僅描述電壓與電流的相對振幅，也描述其相對相位。當通過電路的電流是直流電時 ，電阻與阻抗相等，電阻可以視為相位為零的阻抗。阻抗的概念不僅存在與電路中，在力學的振動系統中也有涉及。

阻抗通常以符號 \displaystyle Z 標記。阻抗是複數，可以用相量 \displaystyle Z_{m}\angle \theta 或 \displaystyle Z_{m}e^{j\theta } 來表示；其中， \displaystyle Z_{m} 是阻抗的大小， \displaystyle \theta 是阻抗的相位。這種表式法稱為「相量表示法」。

具體而言，阻抗定義為電壓與電流的頻域比率[1]。阻抗的大小 \displaystyle Z_{m} 是電壓振幅與電流振幅的絕對值比率，阻抗的相位 \displaystyle \theta 是電壓與電流的相位差。採用國際單位制，阻抗的單位是歐姆（Ω），與電阻的單位相同。阻抗的倒數是導納，即電流與電壓的頻域比率。導納的單位是西門子 (單位)（舊單位是姆歐）。

英文術語「impedance」是由物理學者奧利弗·黑維塞於1886年發表論文《電工》給出[2][3]。於1893年，電機工程師亞瑟·肯乃利（Arthur Kennelly）最先以複數表示阻抗[4]。

 

實藉『相量』V, I, Z 將

v(t) = R \cdot i(t)

i(t) = C \cdot \frac{d}{dt} v(t)

v(t) = L \cdot \frac{d}{dt} i(t)

推廣為

\frac{V}{I} = Z 者，恐易輕忽

Resistance vs reactance

Resistance and reactance together determine the magnitude and phase of the impedance through the following relations:

\displaystyle {\begin{aligned}|Z|&={\sqrt {ZZ^{*}}}={\sqrt {R^{2}+X^{2}}}\\\theta &=\arctan {\left({\frac {X}{R}}\right)}\end{aligned}}

In many applications, the relative phase of the voltage and current is not critical so only the magnitude of the impedance is significant.

Resistance

Main article: Electrical resistance

 

Resistance \displaystyle \scriptstyle R is the real part of impedance; a device with a purely resistive impedance exhibits no phase shift between the voltage and current.

\displaystyle \ R=|Z|\cos {\theta }\quad

Reactance

Main article: Electrical reactance

 

Reactance \displaystyle \scriptstyle X is the imaginary part of the impedance; a component with a finite reactance induces a phase shift \displaystyle \scriptstyle \theta between the voltage across it and the current through it.

\displaystyle \ X=|Z|\sin {\theta }\quad

A purely reactive component is distinguished by the sinusoidal voltage across the component being in quadrature with the sinusoidal current through the component. This implies that the component alternately absorbs energy from the circuit and then returns energy to the circuit. A pure reactance will not dissipate any power.

Capacitive reactance

Main article: Capacitance

 

A capacitor has a purely reactive impedance which is inversely proportional to the signal frequency. A capacitor consists of two conductors separated by an insulator, also known as a dielectric.

\displaystyle X_{C}=-(\omega C)^{-1}=-(2\pi fC)^{-1}\quad

The minus sign indicates that the imaginary part of the impedance is negative.

At low frequencies, a capacitor approaches an open circuit so no current flows through it.

A DC voltage applied across a capacitor causes charge to accumulate on one side; the electric field due to the accumulated charge is the source of the opposition to the current. When the potential associated with the charge exactly balances the applied voltage, the current goes to zero.

Driven by an AC supply, a capacitor will only accumulate a limited amount of charge before the potential difference changes sign and the charge dissipates. The higher the frequency, the less charge will accumulate and the smaller the opposition to the current.

Inductive reactance

Main article: Inductance

 

Inductive reactance \displaystyle \scriptstyle {X_{L}} is proportional to the signal frequency \displaystyle \scriptstyle {f} and the inductance \displaystyle \scriptstyle {L}\scriptstyle {L} .

\displaystyle X_{L}=\omega L=2\pi fL\quad

An inductor consists of a coiled conductor. Faraday’s law of electromagnetic induction gives the back emf \displaystyle \scriptstyle {\mathcal {E}} (voltage opposing current) due to a rate-of-change of magnetic flux density \displaystyle \scriptstyle {B} through a current loop.

\displaystyle {\mathcal {E}}=-{{d\Phi _{B}} \over dt}\quad

For an inductor consisting of a coil with \displaystyle N loops this gives.

\displaystyle {\mathcal {E}}=-N{d\Phi _{B} \over dt}\quad

The back-emf is the source of the opposition to current flow. A constant direct current has a zero rate-of-change, and sees an inductor as a short-circuit (it is typically made from a material with a low resistivity). An alternating current has a time-averaged rate-of-change that is proportional to frequency, this causes the increase in inductive reactance with frequency.

Total reactance

The total reactance is given by

\displaystyle {X=X_{L}+X_{C}} (note that \displaystyle X_{C} is negative)

so that the total impedance is

\displaystyle \ Z=R+jX

 

文本裡的『the opposition to』敘述，怕難了

Variable impedance

In general, neither impedance nor admittance can be time varying as they are defined for complex exponentials for –∞ < t < +∞. If the complex exponential voltage–current ratio changes over time or amplitude, the circuit element cannot be described using the frequency domain. However, many systems (e.g., varicaps that are used in radio tuners) may exhibit non-linear or time-varying voltage–current ratios that appear to be linear time-invariant (LTI) for small signals over small observation windows; hence, they can be roughly described as having a time-varying impedance. That is, this description is an approximation; over large signal swings or observation windows, the voltage–current relationship is non-LTI and cannot be described by impedance.

 

言何事呦★