總諧波失真(英文全稱Total HarmonicDistortion,THD)是指用信號源輸入時,輸出信號比輸入信號多出的額外諧波成分。功放工作時,由于電路不可避免的振蕩或其他諧振產(chǎn)生的二次,三次諧波與實際輸入信號疊加,在輸出端輸出的信號就不單純是與輸入信號完全相同的成分,而是包括了諧波成分的信號,這些多余出來的諧波成分與實際輸入信號(基頻)的對比,用百分比來表示就稱為總諧波失真。
一般來說,總諧波失真在1000赫茲附近最小,所以大部分功放表明總諧波失真是用1000Hz信號做測試,但有些更嚴格的廠家也提供20-20000赫茲范圍內(nèi)的總諧波失真數(shù)據(jù)。總諧波失真在1%以下,一般耳朵分辨不出來,超過10%就可以明顯聽出失真的成分。這個總諧波失真的數(shù)值越小,音色就更加純凈。一般產(chǎn)品的總諧波失真都小于1%@1kHz,但這個數(shù)值越小,表明產(chǎn)品的品質(zhì)越高。
| One of theways of expressing the "goodness" of an amplifier or other deviceis to use a number, based on measurements at a given power outputlevel, expressing its "Total Harmonic Distortion." If an amplifieror 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 copyof the input. There will always be some deviation in the shape ofthe waveform, which can be expressed as a series of "harmonics" ofthe fundamental frequency. This numberindicatesthe RMS voltage equivalent of totalharmonic distortion power, as a percentage of thetotal output RMS voltage. Clear as mud? Let's say you've got a harmonic analyser or other wayof determining the amount of each harmonic (such as an FFTspectrograph program on your computer). How do you convert thesereadings into a single THD figure, for comparison with otherdevices? Before we go much further, it bears pointing out that there isconsiderable (and very valid) criticism of the practise of a singleTHD number to define the goodness of an amplifier. There areseveral weaknesses to this approach: not all harmonics are equallydiscordant; even harmonics tend to be much less dissonant then oddones; higher harmonics are generally considered more dissonant; anda percentage number does not necessarily relate to how the earperceives sound and distortion. Furthermore, it only describes one"kind" of distortion, and says nothing about intermodulation, phaseerrors, or other anomalies in the audio signal. There have beenattempts to redefine THD to give more "weight" to higher harmonics,and other refinements, but as far as I know the industry has notagreed on any of these modified specs. All that being said, a THDfigurecanbe useful whencombined with other specifications, such as IMD (intermodulationdistortion), etc. Just don't assume that an amplifier with lowerTHD than another will necessarily soundbetter. ![]() Let's first do it thewrongway,even though it appears intuitive. (Follow along with yourcalculator if you really want to understand what's going onhere.)You've got asine-wave generator at, say, 400 Hz. at the input of youramplifier, and set it so that it outputs4 voltsRMSinto an 8-ohmload.You correctly compute the power intothe load (using the formula P=E2/R) as 4*4/8 = 2watts.Let's say you measurethe second harmonic at 0.3 volts RMS, and the third harmonic at 0.5volts.For the sake of this discussion,let's assume that all other harmonics are way in the mud and can beneglected. 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 THDof20%. However, this answerisincorrect. The reason is thatwe have to addPowerand convertto an equivalent voltage. Let's do it the proper way, as summarizedin 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."Only2.12475%!" But we're notthere yet; 2.12475% isanotherwronganswer. While thisapparent THD figure is, indeed, representative of thetotalpowerof the harmoniccontent, we still have to convert this power to anequivalent RMSvoltagein order to come up with a THD figureaccording to standard methodology. |
| THD(%) = 100 * SQRT[(P2+ P3+ P4+ ... + Pn) * Zout] / Vt |
where TDH(%) is total harmonic distortion, P represents the powerof each harmonic, Zoutequals the loadimpedance, andVtis the fundamentalvoltage .
Another way of getting the correct answer would be to takethesquare root of the sum of the squares ofharmonic component voltages. This is because power isproportional 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 divideby4 (fundamentalvoltage)gives thesamecorrectresult,approx.14.6%.
In equation form,
| THD(%) = 100 * SQRT[(V22+ V32+ V42+ ... + Vn2)] / Vt |
where TDH(%) is total harmonic distortion, V represents the RMSvoltage of each harmonic, and Vtisthefundamentalvoltage.
Note that Zoutis not present inthis variant of the equation;
since P = V2/Z, the impedanceterms cancel.
| Working from aSpectrogram |
在labview中將harmonic distortion analyzervi中的highestharmonic設(shè)置為5,這個很重要!否則運行的結(jié)果用19項諧波計算而用顯示的前5項結(jié)果計算會導(dǎo)致結(jié)果不一致!下面將數(shù)據(jù)在EXCEL中算了下THD=5.25%與Labview計算的一致,也驗證了THD公式的正確性。
| 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'sgoing on here. I hope that some of you will find thisuseful.
愛華網(wǎng)


