jjy158158 发表于 2020-10-7 17:29

sin4423 发表于 2020-10-7 17:23
CD我早就不碰了,麻烦。解码器我觉得差别都没有那么巨大。功放有较大影响是真的。线材我就懒得说了,这个 ...

解码器没影响?那你有的玩了,慢慢折腾吧,一变压器都可以改变很多

sin4423 发表于 2020-10-7 17:59

jjy158158 发表于 2020-10-7 17:29
解码器没影响?那你有的玩了,慢慢折腾吧,一变压器都可以改变很多

请您仔细看,不是没影响,是没那么大的影响。

agnostic 发表于 2020-10-7 18:45

音乐之贼 发表于 2020-10-6 23:39
你测出符合标准了? 你喜欢用机线没人拦着你。




我只是好奇啊。不要回避问题吗?难道你真是用万用表测的???

音乐之贼 发表于 2020-10-7 19:35

agnostic 发表于 2020-10-7 18:45
我只是好奇啊。不要回避问题吗?难道你真是用万用表测的???


https://shop.sommercable.com/en/Cable/Bulk-Cable-Audio/
这是SOMMER官网

https://shop.sommercable.com/en/Cable/Bulk-Cable-Audio/Mikrofonkabel-SC-Primus-200-0151.html#tab_tender

这是其中随意一条线Primus的数据

我想你应该有能力看懂英文官网数据。

https://www.lulian.cn/product/701-cn.html绿联官网

可以用SOMMER去对比绿联山泽之类的USB数据线,没有对比就没有伤害。SOMMER数据清清楚楚,绿联数据在哪? 合格的USB线阻抗90欧+-15%,我可以看数据选择SOMMER,绿联的数据呢?那看出它合格了?

nickdo 发表于 2020-10-7 23:00

jjy158158 发表于 2020-10-7 17:29
解码器没影响?那你有的玩了,慢慢折腾吧,一变压器都可以改变很多

层次不同... 解码器都感觉差不多... 超级搞笑. 这个是LOW FI吧. HIGH什么FI.

等于跟一个 森林土著 解说 量子力学 没区别.

agnostic 发表于 2020-10-8 01:05

音乐之贼 发表于 2020-10-7 19:35
https://shop.sommercable.com/en/Cable/Bulk-Cable-Audio/
这是SOMMER官网



你还是没能回答你是怎么测阻抗的啊???什么设备怎么测的呢???

Sommer链接里哪有90欧姆了?

agnostic 发表于 2020-10-8 01:12

本帖最后由 agnostic 于 2020-10-8 01:14 编辑

nickdo 发表于 2020-10-7 23:00
层次不同... 解码器都感觉差不多... 超级搞笑. 这个是LOW FI吧. HIGH什么FI.

等于跟一个 森林土著 解 ...
影响大小肯定看和什么比了。DAC的作用和音箱、功放比那肯定是小很多的。音箱或耳机的作用应该超过50%,功放、耳放有30%的影响,然后是音源+DAC差不多10%,线材之类可以忽略不计(前提是合格合标的线材)。而如果是音箱的话,环境和房间声学处理和硬件至少应该有三七开的关系。

紫坛里那几个牛逼哄哄的,动辄一根几万的电源线一耳朵提升,结果房间连声学处理都几乎没有,放置一对牛B大喇叭的房间居然有石膏吊顶。还有一位上百万的设备背后一个玻璃书橱..........,真心看不下去了。这些影响比DAC都大了去了。

音乐之贼 发表于 2020-10-8 01:19

agnostic 发表于 2020-10-8 01:05
你还是没能回答你是怎么测阻抗的啊???什么设备怎么测的呢???

Sommer链接里哪有90欧姆了?

你的大脑理解能力真的有问题,SOMMER那么多线,里面有标准75/110欧---90欧+-15%的,也有恰好95欧左右的,也有35欧的,也有120欧的,网站都发给你了,自己找阻抗合适的90欧+-15%做USB数据线都不会?有些线35欧,有些50欧,用来做模拟音频线更合适。

你要是觉得绿联阻抗是标准90欧+-15%,请把数据来源公示给大家。

我SOMMER的线材数据来源已经公示给你了。

不敢公布数据的线材就是垃圾一堆,你喜欢用没人拦着你。

agnostic 发表于 2020-10-8 01:26

音乐之贼 发表于 2020-10-8 01:19
你的大脑理解能力真的有问题,SOMMER那么多线,里面有标准75/110欧---90欧+-15%的,也有恰好95欧左右的, ...

我的问题很简单,你是怎么测试的?这个对任何人类的理解力都不存在挑战吧?我就看了你贴的那个,请问哪条是指USB线的阻抗值了,有标注吗????

我也不跟你绕圈子了,以你自己DIY的情形来看,我估计你根本不知道怎么按照USB标准测试所谓90欧姆的阻抗。如果你确实知道,那么很简单,请给出你的测试设备的照片并描述测试具体步骤。

野有蔓草 发表于 2020-10-8 02:23

sin4423 发表于 2020-10-7 17:00
怎么总是有人偷换概念,或者说按照自己的想法去理解。
我说的是符合人的口味,相当于做菜某些人喜欢咸点 ...

支持线材无用论的 大部分都不敢自己去实践
靠别人的理论或者自己的高中物理知识 得出这个结论

支持线材有用论的
反倒大部分都自己去实践了才得出这样的结论

请问按照科学来说 那个更符合科学观呢?
不能实践的科学 叫科学?

嗯嗯 明白了不花钱的改变才是好改变
另外要掏钱去尝试的改变 都是大骗子


说白了 不想花钱

野有蔓草 发表于 2020-10-8 02:31

本帖最后由 野有蔓草 于 2020-10-8 02:34 编辑

sin4423 发表于 2020-10-7 16:13
我记得有篇文章,是关于几个6.5寸监听音箱对比评测的。其中有点很有意思,就是不同的监听音箱品牌之间, ...
几个不同品牌的监听箱 单元都不同 要是盲听很难听出区别
一种可能前端太烂了
另外种就是写文章的人耳朵不行 在瞎扯淡
还不如直接说 只要是音箱 单元差不多 声音都是差不多的

如果你在上海 自己去橙音 听听看
就劲浪自己不同系列监听箱声音的差别有多大
去相信自己耳朵一回 好不好


野有蔓草 发表于 2020-10-8 02:36

本帖最后由 野有蔓草 于 2020-10-8 02:47 编辑

ang1206 发表于 2020-10-7 15:14
监听音箱声音更接近原声这是肯定的,这是录音工作的性质决定的,应该说所有的监听音箱不管什么牌子,发出 ...
录音师要毛好音箱
他们要的是麦克风 要的是多轨录音机

混音师才要监听音箱
但混音师要追求毛的原音 要原音 他们还赚什么钱
看这职业的名字 请读一遍
另外监听箱目的是还原?挑刺才是监听箱的目的


额 请别把别人的职业想当然






音乐之贼 发表于 2020-10-8 03:01

本帖最后由 音乐之贼 于 2020-10-8 03:35 编辑

agnostic 发表于 2020-10-8 01:26
我的问题很简单,你是怎么测试的?这个对任何人类的理解力都不存在挑战吧?我就看了你贴的那个,请问哪条 ...
你有什么技术能力去测试标准90欧+-15%?


展现给大家看看。贴图视频都可以。其实问题很简单。

1:要么你能够给大家展示你的测量技术让人相信 2:如果测量没设备有难处,请你给出厂家正规数据来源。


你不能一不测量或者二不给数据来源就在这胡搅蛮缠把。


-----------------------------------------------------------------------

当我测量遇到短板时候,我相信SOMMER,FURUTECH这样的专业大厂生产的线基,数据清清楚楚。毕竟工作在10MHZ的阻抗没有设备测不了。


绿联之流什么时候把数据说清楚了?


你一直回避说明不同线材的数据来源对比,请你好好研究下不同大厂的数据线DATASHEET,看看人家有没有说谎,再看看绿联之流的数据呢?

https://www.vhaudio.com/furutech-gt2-usb-cable.pdf


这是古河GT2的数据单PDF,你自己研究下。 然后请你找出绿联类似的数据单PDF给我看看。


agnostic 发表于 2020-10-8 13:28

音乐之贼 发表于 2020-10-8 03:01
你有什么技术能力去测试标准90欧+-15%?




按照你的回复和反复回避,可以推断其实你根本不懂USB标准里所谓的90欧姆是什么定义,自然也根本不懂如何测量,用什么仪器测量。也因此可以推定你所谓的USB线好坏完全是基于错误理解的,没有根据的瞎扯。
很明白了。

agnostic 发表于 2020-10-8 13:30

野有蔓草 发表于 2020-10-8 02:23
支持线材无用论的 大部分都不敢自己去实践
靠别人的理论或者自己的高中物理知识 得出这个结论



科学需要实践,但是实践了不等于就是科学的。这个是很基本的逻辑,我有点怀疑你能不能理解。顺便说一句,上次你要扯流体力学的问题,百度到伯努利公式的计算了吗?当年大学期末考试这部分内容是挂很多人的。

野有蔓草 发表于 2020-10-8 13:42

本帖最后由 野有蔓草 于 2020-10-8 13:48 编辑

agnostic 发表于 2020-10-8 13:30
科学需要实践,但是实践了不等于就是科学的。这个是很基本的逻辑,我有点怀疑你能不能理解。顺便说一句, ...
你的意思 大家坛搞的这么多届盲听比赛 还是样本不够多不能证明线材是有用的?或者说是有差别的?
反倒你自己也没对比过 按照自己的知识 就能判断出线材无用?或者说是没差别的?
敢问你是材料专业还是电声专业的或者冶金专业?
我倒认识个冶金专业的 在搞线材
网上吵这么多 你就不能掏个千把块钱 买个好点的线材或者押金试听个好点的线材
然后拍个照片 说 有毛差别 你看 我都对比过了 你们都是胡扯
不舍得千把块钱上可能的当
但押金不至于没有吧 快递费不至于出不起吧
你这么有常识 怎么就没勇气踏出实践的这一步呢
花点快递费 就能打我们这些线材有用论的人脸 多划算

音乐之贼 发表于 2020-10-8 15:18

本帖最后由 音乐之贼 于 2020-10-8 15:20 编辑

agnostic 发表于 2020-10-8 13:28
按照你的回复和反复回避,可以推断其实你根本不懂USB标准里所谓的90欧姆是什么定义,自然也根本不懂如何 ...
你但凡上点心,就会去研究下什么是标准典型阻抗了,要么就自己有设备有能力测量10MHZ下的阻抗值,要么就能给出厂家正规数据,也就不会和我胡搅蛮缠了
麻烦你后面回复上点干货行么?要么给出技术建议要么给出真实数据。

Characteristic impedance
From Wikipedia, the free encyclopedia
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This article is about impedance in electrical circuits. For impedance of electromagnetic waves, see Wave impedance. For characteristic acoustic impedance, see Acoustic impedance.

A transmission line drawn as two black wires. At a distance x into the line, there is current phasor I(x) traveling through each wire, and there is a voltage difference phasor V(x) between the wires (bottom voltage minus top voltage). If {\displaystyle Z_{0}}Z_{0} is the characteristic impedance of the line, then {\displaystyle V(x)/I(x)=Z_{0}}V(x)/I(x)=Z_{0} for a wave moving rightward, or {\displaystyle V(x)/I(x)=-Z_{0}}V(x)/I(x)=-Z_{0} for a wave moving leftward.

Schematic representation of a circuit where a source is coupled to a load with a transmission line having characteristic impedance {\displaystyle Z_{0}}Z_{0}.
The characteristic impedance or surge impedance (usually written Z0) of a uniform transmission line is the ratio of the amplitudes of voltage and current of a single wave propagating along the line; that is, a wave travelling in one direction in the absence of reflections in the other direction. Alternatively, and equivalently, it can be defined as the input impedance of a transmission line when its length is infinite. Characteristic impedance is determined by the geometry and materials of the transmission line and, for a uniform line, is not dependent on its length. The SI unit of characteristic impedance is the ohm.

The characteristic impedance of a lossless transmission line is purely real, with no reactive component. Energy supplied by a source at one end of such a line is transmitted through the line without being dissipated in the line itself. A transmission line of finite length (lossless or lossy) that is terminated at one end with an impedance equal to the characteristic impedance appears to the source like an infinitely long transmission line and produces no reflections.


Contents
1      Transmission line model
2      Derivation
2.1      Using telegrapher's equation
2.2      Alternative approach
3      Lossless line
4      Surge impedance loading
5      Practical examples
6      See also
7      References
7.1      Sources
8      External links
Transmission line model
The characteristic impedance {\displaystyle Z(\omega )}{\displaystyle Z(\omega )} of an infinite transmission line at a given angular frequency {\displaystyle \omega }\omegais the ratio of the voltage and current of a pure sinusoidal wave of the same frequency travelling along the line. This definition extends to DC by letting {\displaystyle \omega }\omegatend to 0, and subsists for finite transmission lines until the wave reaches the end of the line. In this case, there will be in general a reflected wave which travels back along the line in the opposite direction. When this wave reaches the source, it adds to the transmitted wave and the ratio of the voltage and current at the input to the line will no longer be the characteristic impedance. This new ratio is called the input impedance.

The input impedance of an infinite line is equal to the characteristic impedance since the transmitted wave is never reflected back from the end. It can be shown that an equivalent definition is: The characteristic impedance of a line is that impedance which, when terminating an arbitrary length of line at its output, produces an input impedance of equal value. This is so because there is no reflection on a line terminated in its own characteristic impedance.


Schematic of Heaviside's model of an infinitesimal segment of transmission line.
Applying the transmission line model based on the telegrapher's equations as derived below, the general expression for the characteristic impedance of a transmission line is:

{\displaystyle Z_{\text{o}}={\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}}{\displaystyle Z_{\text{o}}={\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}}
where

{\displaystyle R}R is the resistance per unit length, considering the two conductors to be in series,
{\displaystyle L}L is the inductance per unit length,
{\displaystyle G}G is the conductance of the dielectric per unit length,
{\displaystyle C}C is the capacitance per unit length,
{\displaystyle j}j is the imaginary unit, and
{\displaystyle \omega }\omegais the angular frequency.
A surge of energy on a finite transmission line will see an impedance of {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}} prior to any reflections returning; hence surge impedance is an alternative name for characteristic impedance. Although an infinite line is assumed, since all quantities are per unit length, the “per length” parts of all the units cancel, and the characteristic impedance is independent of the length of the transmission line.

The voltage and current phasors on the line are related by the characteristic impedance as:

{\displaystyle {\frac {V_{(+)}}{I_{(+)}}}=Z_{\text{o}}=-{\frac {V_{(-)}}{I_{(-)}}}}{\displaystyle {\frac {V_{(+)}}{I_{(+)}}}=Z_{\text{o}}=-{\frac {V_{(-)}}{I_{(-)}}}}
where the subscripts (+) and (−) mark the separate constants for the waves traveling forward (+) and backward (−).

Derivation
Using telegrapher's equation

Consider one section of the transmission line for the derivation of the characteristic impedance. The voltage on the left would be V and on the right side would be V + dV . This figure is to be used for both the derivation methods.
The differential equations describing the dependence of the voltage and current on time and space are linear, so that a linear combination of solutions is again a solution. This means that we can consider solutions with a time dependence {\displaystyle e^{j\omega t}}e^{j\omega t} – doing so is functionally equivalent of solving for the Fourier coefficients for voltage and current amplitudes at some fixed angular frequency {\displaystyle \omega }\omega . Doing so causes the time dependence to factor out, leaving an ordinary differential equation for the coefficients, which will be phasors, dependent on position (space) only. Moreover, the parameters can be generalized to be frequency-dependent.

Let

{\displaystyle V(x,t)\equiv V(x)\ e^{+j\omega t}}{\displaystyle V(x,t)\equiv V(x)\ e^{+j\omega t}}
and

{\displaystyle I(x,t)\equiv I(x)\ e^{+j\omega t}}{\displaystyle I(x,t)\equiv I(x)\ e^{+j\omega t}}
Take the positive direction for {\displaystyle V}V and {\displaystyle I}I in the loop to be clockwise.

We find that

{\displaystyle {\text{d}}V=-(R+j\omega L)\ I\ dx=-Z\ I\ {\text{d}}x}{\displaystyle {\text{d}}V=-(R+j\omega L)\ I\ dx=-Z\ I\ {\text{d}}x}
and

{\displaystyle {\text{d}}I=-(G+j\omega C)\ V\ {\text{d}}x=-Y\ V\ {\text{d}}x}{\displaystyle {\text{d}}I=-(G+j\omega C)\ V\ {\text{d}}x=-Y\ V\ {\text{d}}x}
or

{\displaystyle {\frac {{\text{d}}V}{{\text{d}}x}}=-Z\ I\qquad }{\displaystyle {\frac {{\text{d}}V}{{\text{d}}x}}=-Z\ I\qquad } and {\displaystyle \qquad {\frac {{\text{d}}I}{{\text{d}}x}}=-Y\ V}{\displaystyle \qquad {\frac {{\text{d}}I}{{\text{d}}x}}=-Y\ V}
where

{\displaystyle Z\equiv R+j\omega L\qquad }{\displaystyle Z\equiv R+j\omega L\qquad } and {\displaystyle \qquad Y\equiv G+j\omega C\ }{\displaystyle \qquad Y\equiv G+j\omega C\ }.
These two first-order equations are easily uncoupled by a second differentiation, with the results:

{\displaystyle {\frac {{\text{d}}^{2}V}{{\text{d}}x^{2}}}=ZY\ V}{\displaystyle {\frac {{\text{d}}^{2}V}{{\text{d}}x^{2}}}=ZY\ V}
and

{\displaystyle {\frac {{\text{d}}^{2}I}{{\text{d}}x^{2}}}=ZY\ I}{\displaystyle {\frac {{\text{d}}^{2}I}{{\text{d}}x^{2}}}=ZY\ I}
Notice that both {\displaystyle V}V and {\displaystyle I}I satisfy the same equation.

Since {\displaystyle ZY}{\displaystyle ZY} is independent of {\displaystyle x}x and {\displaystyle t}t, it can be represented by a single constant {\displaystyle -k^{2}}-k^{2}. That is:

{\displaystyle -k^{2}\equiv Z\ Y\ }{\displaystyle -k^{2}\equiv Z\ Y\ }
so

{\displaystyle j\ k=\pm {\sqrt {Z\ Y\ }}}{\displaystyle j\ k=\pm {\sqrt {Z\ Y\ }}}
The minus sign is included for later convenience. Because of it, we can write the above equation as

{\displaystyle k=\pm \omega {\sqrt {\left(L-jR/\omega \right)\left(C-jG/\omega \right)\ }}=\pm \omega {\sqrt {L\ C\ }}{\sqrt {\left(1-j{\frac {R}{\omega L}}\right)\left(1-j{\frac {G}{\omega C}}\right)\ }}}{\displaystyle k=\pm \omega {\sqrt {\left(L-jR/\omega \right)\left(C-jG/\omega \right)\ }}=\pm \omega {\sqrt {L\ C\ }}{\sqrt {\left(1-j{\frac {R}{\omega L}}\right)\left(1-j{\frac {G}{\omega C}}\right)\ }}}
which is correct for all transmission lines. And for typical transmission lines, that are built to make wire resistance loss {\displaystyle R}R small and insulation leakage conductance {\displaystyle G}G low, and further, with high frequencies, the inductive reactance {\displaystyle \omega L}{\displaystyle \omega L} and the capacitive admittance {\displaystyle \omega C}{\displaystyle \omega C} will both be large, so the constant {\displaystyle k}k is very close to being a real number:

{\displaystyle k\approx \pm \omega {\sqrt {LC\ }}.}{\displaystyle k\approx \pm \omega {\sqrt {LC\ }}.}
Further, with this definition of {\displaystyle k}k the position- or {\displaystyle x}x-dependent part will appear as {\displaystyle \ \pm j\ k\ x\ }{\displaystyle \ \pm j\ k\ x\ } in the exponential solutions of the equation, similar to the time-dependent part {\displaystyle \ +j\ \omega \ t\ }{\displaystyle \ +j\ \omega \ t\ }, so the solution reads

{\displaystyle V(x)=v_{(+)}\ e^{-jkx}+v_{(-)}e^{+jkx}}{\displaystyle V(x)=v_{(+)}\ e^{-jkx}+v_{(-)}e^{+jkx}}
where {\displaystyle v_{(+)}}{\displaystyle v_{(+)}} and {\displaystyle v_{(-)}}{\displaystyle v_{(-)}} are the constants of integration for the forward moving (+) and backward moving (−) waves, as in the prior section. When we recombine the time-dependent part we obtain the full solution:

{\displaystyle V(x,t)\quad =\quad V(x)\ e^{+j\omega t}\quad =\quad v_{(+)}\ e^{-jkx+j\omega t}+v_{(-)}e^{+jkx+j\omega t}}{\displaystyle V(x,t)\quad =\quad V(x)\ e^{+j\omega t}\quad =\quad v_{(+)}\ e^{-jkx+j\omega t}+v_{(-)}e^{+jkx+j\omega t}}
Since the equation for {\displaystyle I}I is the same form, it has a solution of the same form:

{\displaystyle I(x)=i_{(+)}\ e^{-jkx}+i_{(-)}\ e^{+jkx}}{\displaystyle I(x)=i_{(+)}\ e^{-jkx}+i_{(-)}\ e^{+jkx}}
where {\displaystyle i_{(+)}}{\displaystyle i_{(+)}} and {\displaystyle i_{(-)}}{\displaystyle i_{(-)}} are again constants of integration.

The above equations are the wave solution for {\displaystyle V}V and {\displaystyle I}I. In order to be compatible, they must still satisfy the original differential equations, one of which is

{\displaystyle {\frac {{\text{d}}V}{{\text{d}}x}}=-Z\ I}{\displaystyle {\frac {{\text{d}}V}{{\text{d}}x}}=-Z\ I}
Substituting the solutions for {\displaystyle V}V and {\displaystyle I}I into the above equation, we get

{\displaystyle {\frac {\text{d}}{{\text{d}}x}}\left=-(R+j\omega L)\left[\ i_{(+)}\ e^{-jkx}+i_{(-)}\ e^{+jkx}\right]}{\displaystyle {\frac {\text{d}}{{\text{d}}x}}\left=-(R+j\omega L)\left[\ i_{(+)}\ e^{-jkx}+i_{(-)}\ e^{+jkx}\right]}
or

{\displaystyle -jk\ v_{(+)}\ e^{-jkx}+jk\ v_{(-)}\ e^{+jkx}=-(R+j\omega L)\ i_{(+)}\ e^{-jkx}-(R+j\omega L)\ i_{(-)}\ e^{+jkx}}{\displaystyle -jk\ v_{(+)}\ e^{-jkx}+jk\ v_{(-)}\ e^{+jkx}=-(R+j\omega L)\ i_{(+)}\ e^{-jkx}-(R+j\omega L)\ i_{(-)}\ e^{+jkx}}
Isolating distinct powers of {\displaystyle e}e and combining identical powers, we see that in order for the above equation to hold for all possible values of {\displaystyle x}x we must have:

For the co-efficients of {\displaystyle e^{-jkx}\quad {\text{ : }}\quad -j\ k\ v_{(+)}=-(R+j\omega L)\ i_{(+)}}{\displaystyle e^{-jkx}\quad {\text{ : }}\quad -j\ k\ v_{(+)}=-(R+j\omega L)\ i_{(+)}}
For the co-efficients of {\displaystyle e^{+jkx}\quad {\text{ : }}\quad +j\ k\ v_{(-)}=-(R+j\omega L)\ i_{(-)}}{\displaystyle e^{+jkx}\quad {\text{ : }}\quad +j\ k\ v_{(-)}=-(R+j\omega L)\ i_{(-)}}
Since {\displaystyle jk={\sqrt {(R+j\omega L)(G+j\omega C)\ }}}{\displaystyle jk={\sqrt {(R+j\omega L)(G+j\omega C)\ }}}

{\displaystyle +{\frac {v_{(+)}}{i_{(+)}}}={\frac {R+j\omega L}{jk}}={\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\equiv Z_{\text{o}}}{\displaystyle +{\frac {v_{(+)}}{i_{(+)}}}={\frac {R+j\omega L}{jk}}={\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\equiv Z_{\text{o}}}
{\displaystyle -{\frac {v_{(-)}}{i_{(-)}}}={\frac {R+j\omega L}{jk}}={\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\equiv Z_{\text{o}}}{\displaystyle -{\frac {v_{(-)}}{i_{(-)}}}={\frac {R+j\omega L}{jk}}={\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\equiv Z_{\text{o}}}
hence, for valid solutions require

{\displaystyle v_{(+)}=+Z_{\text{o}}\ i_{(+)}\quad {\text{ and }}\quad v_{(-)}=-Z_{\text{o}}\ i_{(-)}}{\displaystyle v_{(+)}=+Z_{\text{o}}\ i_{(+)}\quad {\text{ and }}\quad v_{(-)}=-Z_{\text{o}}\ i_{(-)}}
It can be seen that the constant {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}}, defined in the above equations has the dimensions of impedance (ratio of voltage to current) and is a function of primary constants of the line and operating frequency. It is called the “characteristic impedance” of the transmission line, and conventionally denoted by {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}}.

{\displaystyle Z_{\text{o}}\quad =\quad {\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\quad =\quad {\sqrt {\ {\frac {\ L\ }{C}}\ }}{\sqrt {{\frac {\ 1-j\left({\frac {R}{\omega L}}\right)\ }{\ 1-j\left({\frac {G}{\omega C}}\right)\ }}\ }}}{\displaystyle Z_{\text{o}}\quad =\quad {\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}\quad =\quad {\sqrt {\ {\frac {\ L\ }{C}}\ }}{\sqrt {{\frac {\ 1-j\left({\frac {R}{\omega L}}\right)\ }{\ 1-j\left({\frac {G}{\omega C}}\right)\ }}\ }}}
for any transmission line, and for well-functioning transmission lines, with {\displaystyle R}R and {\displaystyle G}G both very small, or {\displaystyle \omega }\omegavery high, or all of the above, we get

{\displaystyle Z_{\text{o}}\approx {\sqrt {{\frac {L}{C}}\ }}}{\displaystyle Z_{\text{o}}\approx {\sqrt {{\frac {L}{C}}\ }}}
hence the characteristic impedance is typically very close to being a real number (see also the Heaviside condition.)

Alternative approach
We follow an approach posted by Tim Healy. The line is modeled by a series of differential segments with differential series {\displaystyle (R{\text{d}}x,L{\text{d}}x)}{\displaystyle (R{\text{d}}x,L{\text{d}}x)} and shunt {\displaystyle (C{\text{d}}x,G{\text{d}}x)}{\displaystyle (C{\text{d}}x,G{\text{d}}x)} elements (as shown in the figure above). The characteristic impedance is defined as the ratio of the input voltage to the input current of a semi-infinite length of line. We call this impedance {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}}. That is, the impedance looking into the line on the left is {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}}. But, of course, if we go down the line one differential length {\displaystyle {\text{d}}x}{\text{d}}x, the impedance into the line is still {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}}. Hence we can say that the impedance looking into the line on the far left is equal to {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}} in parallel with {\displaystyle C{\text{d}}x}{\displaystyle C{\text{d}}x} and {\displaystyle G{\text{d}}x}{\displaystyle G{\text{d}}x}, all of which is in series with {\displaystyle R{\text{d}}x}{\displaystyle R{\text{d}}x} and {\displaystyle L{\text{d}}x}{\displaystyle L{\text{d}}x}. Hence:

{\displaystyle Z_{\text{o}}=(R+j\omega L){\text{d}}x+{\frac {1}{\ (G+j\omega C){\text{d}}x+{\frac {1}{Z_{\text{o}}}}\ }}}{\displaystyle Z_{\text{o}}=(R+j\omega L){\text{d}}x+{\frac {1}{\ (G+j\omega C){\text{d}}x+{\frac {1}{Z_{\text{o}}}}\ }}}
{\displaystyle Z_{\text{o}}=(R+j\omega L){\text{d}}x+{\frac {\ Z_{\text{o}}\ }{Z_{\text{o}}(G+j\omega C){\text{d}}x+1\ }}}{\displaystyle Z_{\text{o}}=(R+j\omega L){\text{d}}x+{\frac {\ Z_{\text{o}}\ }{Z_{\text{o}}(G+j\omega C){\text{d}}x+1\ }}}
{\displaystyle Z_{\text{o}}+Z_{\text{o}}^{2}(G+j\omega C){\text{d}}x=(R+j\omega L){\text{d}}x+Z_{\text{o}}(G+j\omega C){\text{d}}x(R+j\omega L){\text{d}}x+Z_{\text{o}}}{\displaystyle Z_{\text{o}}+Z_{\text{o}}^{2}(G+j\omega C){\text{d}}x=(R+j\omega L){\text{d}}x+Z_{\text{o}}(G+j\omega C){\text{d}}x(R+j\omega L){\text{d}}x+Z_{\text{o}}}
The {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}} terms cancel, leaving

{\displaystyle Z_{\text{o}}^{2}(G+j\omega C){\text{d}}x=(R+j\omega L){\text{d}}x+Z_{\text{o}}(G+j\omega C)(R+j\omega L)({\text{d}}x)^{2}}{\displaystyle Z_{\text{o}}^{2}(G+j\omega C){\text{d}}x=(R+j\omega L){\text{d}}x+Z_{\text{o}}(G+j\omega C)(R+j\omega L)({\text{d}}x)^{2}}
The first-power {\displaystyle {\text{d}}x}{\text{d}}x terms are the highest remaining order. In comparison to {\displaystyle {\text{d}}x}{\text{d}}x, the term with the factor {\displaystyle ({\text{d}}x)^{2}}{\displaystyle ({\text{d}}x)^{2}} may be discarded, since it is infinitesimal in comparison, leading to:

{\displaystyle Z_{\text{o}}^{2}(G+j\omega C){\text{d}}x=(R+j\omega L){\text{d}}x}{\displaystyle Z_{\text{o}}^{2}(G+j\omega C){\text{d}}x=(R+j\omega L){\text{d}}x}
and hence

{\displaystyle Z_{\text{o}}=\pm {\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}}{\displaystyle Z_{\text{o}}=\pm {\sqrt {{\frac {R+j\omega L}{G+j\omega C}}\ }}}
Reversing the sign on the square root has the effect of changing the direction of the flow of current.

Lossless line
The analysis of lossless lines provides an accurate approximation for real transmission lines that simplifies the mathematics considered in modeling transmission lines. A lossless line is defined as a transmission line that has no line resistance and no dielectric loss. This would imply that the conductors act like perfect conductors and the dielectric acts like a perfect dielectric. For a lossless line, R and G are both zero, so the equation for characteristic impedance derived above reduces to:

{\displaystyle Z_{\text{o}}={\sqrt {{\frac {L}{C}}~}}~.}{\displaystyle Z_{\text{o}}={\sqrt {{\frac {L}{C}}~}}~.}
In particular, {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}} does not depend any more upon the frequency. The above expression is wholly real, since the imaginary term j has canceled out, implying that {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}} is purely resistive. For a lossless line terminated in {\displaystyle Z_{\text{o}}}{\displaystyle Z_{\text{o}}}, there is no loss of current across the line, and so the voltage remains the same along the line. The lossless line model is a useful approximation for many practical cases, such as low-loss transmission lines and transmission lines with high frequency. For both of these cases, R and G are much smaller than ωL and ωC, respectively, and can thus be ignored.

The solutions to the long line transmission equations include incident and reflected portions of the voltage and current:

{\displaystyle V={\frac {V_{r}+I_{r}Z_{c}}{2}}e^{\gamma x}+{\frac {V_{r}-I_{r}Z_{c}}{2}}e^{-\gamma x}}{\displaystyle V={\frac {V_{r}+I_{r}Z_{c}}{2}}e^{\gamma x}+{\frac {V_{r}-I_{r}Z_{c}}{2}}e^{-\gamma x}}
{\displaystyle I={\frac {V_{r}/Z_{c}+I_{r}}{2}}e^{\gamma x}-{\frac {V_{r}/Z_{c}-I_{r}}{2}}e^{-\gamma x}}{\displaystyle I={\frac {V_{r}/Z_{c}+I_{r}}{2}}e^{\gamma x}-{\frac {V_{r}/Z_{c}-I_{r}}{2}}e^{-\gamma x}}
When the line is terminated with its characteristic impedance, the reflected portions of these equations are reduced to 0 and the solutions to the voltage and current along the transmission line are wholly incident. Without a reflection of the wave, the load that is being supplied by the line effectively blends into the line making it appear to be an infinite line. In a lossless line this implies that the voltage and current remain the same everywhere along the transmission line. Their magnitudes remain constant along the length of the line and are only rotated by a phase angle.
Surge impedance loading
In electric power transmission, the characteristic impedance of a transmission line is expressed in terms of the surge impedance loading (SIL), or natural loading, being the power loading at which reactive power is neither produced nor absorbed:

{\displaystyle {\mathit {SIL}}={\frac {{V_{\mathrm {LL} }}^{2}}{Z_{0}}}}{\mathit {SIL}}={\frac {{V_{\mathrm {LL} }}^{2}}{Z_{0}}}
in which {\displaystyle V_{\mathrm {LL} }}V_{\mathrm {LL} } is the line-to-line voltage in volts.

Loaded below its SIL, a line supplies reactive power to the system, tending to raise system voltages. Above it, the line absorbs reactive power, tending to depress the voltage. The Ferranti effect describes the voltage gain towards the remote end of a very lightly loaded (or open ended) transmission line. Underground cables normally have a very low characteristic impedance, resulting in an SIL that is typically in excess of the thermal limit of the cable. Hence a cable is almost always a source of reactive power.

Practical examples
Standard      Impedance
(Ω)      Tolerance
Ethernet Cat.5      100         ±5Ω
USB         90      ±15%
HDMI         95      ±15%
IEEE 1394      108         +3%
−2%
VGA         75         ±5%
DisplayPort      100      ±20%
DVI         95      ±15%
PCIe         85      ±15%
The characteristic impedance of coaxial cables (coax) is commonly chosen to be 50 Ω for RF and microwave applications. Coax for video applications is usually 75 Ω for its lower loss.

See also: Nominal impedance § 50 Ω and 75 Ω
See also
Ampère's circuital law
Characteristic acoustic impedance
Electrical impedance – The opposition of a circuit to a current when a voltage is applied
Maxwell's equations – Equations describing classical electromagnetism
Transmission line
Wave impedance
Space cloth – Hypothetical plane with resistivity of 376.7 ohms per square.
References
"The Telegrapher's Equation". mysite.du.edu. Retrieved 9 September 2018.
"Derivation of Characteristic Impedance of Transmission line". GATE ECE 2018. 16 April 2016. Archived from the original on 9 September 2018. Retrieved 9 September 2018.
"Characteristic Impedance". www.ee.scu.edu. Retrieved 2018-09-09.
"SuperCat OUTDOOR CAT 5e U/UTP" (PDF). Archived from the original (PDF) on 2012-03-16.
"Chapter 2 – Hardware". USB in a NutShell. Beyond Logic.org. Retrieved 2007-08-25.
"AN10798 DisplayPort PCB layout guidelines" (PDF). Retrieved 2019-12-29.
"Evaluation" (PDF). materias.fi.uba.ar. Retrieved 2019-12-29.
"VMM5FL" (PDF). pro video data sheets. Archived from the original (PDF) on 2016-04-02. Retrieved 2016-03-21.
Sources
Guile, A. E. (1977). Electrical Power Systems. ISBN 0-08-021729-X.
Pozar, D. M. (February 2004). Microwave Engineering (3rd ed.). ISBN 0-471-44878-8.
Ulaby, F. T. (2004). Fundamentals Of Applied Electromagnetics (media ed.). Prentice Hall. ISBN 0-13-185089-X.


agnostic 发表于 2020-10-8 19:08

野有蔓草 发表于 2020-10-8 13:42
你的意思 大家坛搞的这么多届盲听比赛 还是样本不够多不能证明线材是有用的?或者说是有差别的?
反倒 ...

我不知道你所说的盲听是怎么进行。严密的科学对比测试也是经历了很多年才成熟起来的。

线材当然有用,没线材你喇叭怎么出声音呢?

agnostic 发表于 2020-10-8 19:10

音乐之贼 发表于 2020-10-8 15:18
你但凡上点心,就会去研究下什么是标准典型阻抗了,要么就自己有设备有能力测量10MHZ下的阻抗值,要么就 ...

你抄一段维基百科完全就代表你明白USB阻抗是啥了?你不是号称自己亲自动手的吗?怎么让你拿个证据出来就那么困难呢?

sin4423 发表于 2020-10-8 20:03

agnostic 发表于 2020-10-8 19:08
我不知道你所说的盲听是怎么进行。严密的科学对比测试也是经历了很多年才成熟起来的。

线材当然有用, ...

线材的作用也就是个连接作用,只要合格,都不会是系统的瓶颈,换线说有改变可以。说什么作用多大,那是纯粹意淫。
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查看完整版本: 实践证明,USB线对声音影响很大