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总谐波失真系数

(2012-09-10 14:28:04)
标签：

杂谈

 分类： 电子硬件线路

总谐波失真（英文全称Total Harmonic Distortion，THD ）是指用信号源输入时，输出信号比输入信号多出的额外谐波成分。功放工作时，由于电路 不可避免的振荡或其他谐振 产 生的二次，三次谐波与实际输入信号叠加， 在输出端输出的信号就不单纯是与输入信号完全相同的成分， 而是包括了谐波成分的信号， 这些多余出来的谐波成分与实际输入信号(基频)的对比， 用百分比来表示就称为总谐波失真。

 

一般来说， 总谐波失真在1000赫兹附近最小， 所以大部分功放表明总谐波失真是用1000Hz信号做测试， 但有些更严格的厂家也提供20－ 20000赫兹范围内的总谐波失真数据。总谐波失真在1％以下， 一般耳朵分辨不出来，超过10％就可以明显听出失真的成分。 这个总谐波失真的数值越小，音色就更加纯净。 一般产品的总谐波失真都小于1％@1kHz，但这个数值越小， 表明产品的品质越高。

 

 

One of the ways of expressing the "goodness" of an amplifier or other device is to use a number, based on measurements at a given power output level, expressing its "Total Harmonic Distortion." If an amplifier or other device is given a pure sine wave (i.e. just one frequency) at its input, the signal at the output will never be an exact copy of the input. There will always be some deviation in the shape of the waveform, which can be expressed as a series of "harmonics" of the fundamental frequency. This number indicates the RMS voltage equivalent of total harmonic distortion power , as a percentage of the total output RMS voltage. 
Clear as mud? Let's say you've got a harmonic analyser or other way of determining the amount of each harmonic (such as an FFT spectrograph program on your computer). How do you convert these readings into a single THD figure, for comparison with other devices? 
Before we go much further, it bears pointing out that there is considerable (and very valid) criticism of the practise of a single THD number to define the goodness of an amplifier. There are several weaknesses to this approach: not all harmonics are equally discordant; even harmonics tend to be much less dissonant then odd ones; higher harmonics are generally considered more dissonant; and a percentage number does not necessarily relate to how the ear perceives sound and distortion. Furthermore, it only describes one "kind" of distortion, and says nothing about intermodulation, phase errors, or other anomalies in the audio signal. There have been attempts to redefine THD to give more "weight" to higher harmonics, and other refinements, but as far as I know the industry has not agreed on any of these modified specs. 
All that being said, a THD figure can be useful when combined with other specifications, such as IMD (intermodulation distortion), etc. Just don't assume that an amplifier with lower THD than another will necessarily sound better. 
Let's first do it the wrong way, even though it appears intuitive. (Follow along with your calculator if you really want to understand what's going on here.) You've got a sine-wave generator at, say, 400 Hz. at the input of your amplifier, and set it so that it outputs 4 volts RMS into an 8-ohm load. You correctly compute the power into the load (using the formula P=E2 /R) as 4*4/8 = 2 watts. Let's say you measure the second harmonic at 0.3 volts RMS, and the third harmonic at 0.5 volts. For the sake of this discussion, let's assume that all other harmonics are way in the mud and can be neglected. 
It doesn't take long to add 0.3 and 0.5 (=0.8), divide by 4 volts, and multiply by 100 to come up with an apparent THD of 20% . However, this answer is incorrect . The reason is that we have to add Power and convert to an equivalent voltage. Let's do it the proper way, as summarized in the table below: 
Harmonic Voltage Power % Distortion
Fundamental 4.0 volts 2 watts
2nd Harm. 0.3 volts .01125 W 0.56225 %
3rd Harm. 0.5 volts .03125 W 1.5625 %
Total
.0425 W 2.12475 %

"Hey, neat," you may be saying. "Only 2.12475% !" But we're not there yet; 2.12475% is another wrong answer. While this apparent THD figure is, indeed, representative of the total power of the harmonic content, we still have to convert this power to anequivalent RMS voltage in order to come up with a THD figure according to standard methodology. 
So the correct answer is obtained by taking total harmonic power (0.0425 watts) and calculating the equivalent RMS voltage , given by the square root of power times output impedance, or about 0.583 volts in the example. i.e. SQRT(0.0425 watts * 8 ohms). So the THD value, as specified would actually be 100 times 0.583 volts divided by 4 volts(基频电压) = about 14.6% (correct answer).  注意这里的电压值都是rms value!如果是峰值的话要乘以0.707哦

 

To generalize this as an equation, 
THD(%) = 100 * SQRT[(P2 + P3 + P4 + ... + Pn) * Zout] / Vt

where TDH(%) is total harmonic distortion, P represents the power of each harmonic, Zout equals the load impedance, and Vt is the fundamental voltage .

Another way of getting the correct answer would be to take the square root of the sum of the squares of harmonic component voltages. This is because power is proportional to the square of the voltage. Let's try it. SQRT(0.32 + 0.52) = SQRT(.09 + .25) = SQRT(0.34) = 0.583. Multiply by 100 (to get percent) and divide by 4 (fundamental voltage) gives the same correct result, approx. 14.6%. 


In equation form, 

THD(%) = 100 * SQRT[(V22 + V32 + V42 + ... + Vn2)] / Vt

where TDH(%) is total harmonic distortion, V represents the RMS voltage of each harmonic, and Vt is the fundamental voltage. 
Note that Zout is not present in this variant of the equation; 
since P = V2/Z, the impedance terms cancel. 

Working from a Spectrogram
 

在labview中将harmonic distortion analyzer vi中的highest harmonic设置为5，这个很重要！否则运行的结果用19项谐波计算而用显示的前5项结果计算会导致结果不一致！下面将数据在EXCEL中算了下THD=5.25% 与Labview计算的一致，也验证了THD公式的正确性。

peak value rms value rms squred value
Fundamental 0.9778 0.691409011 0.47804642 THD= 0.0525 THD公式SQRT(SUM(D3:D6))/C2
2nd 0.0222 0.015697771 0.00024642
3rd 0.0406 0.028708535 0.00082418
4th 0.0101 0.007141778 0.000051005
5th 0.0198 0.014000714 0.00019602

 

You can have a look at the formulae in the "Power" columns and "Totals" rows if you want to understand what's going on here. I hope that some of you will find this useful.


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