How do I evaluate this equation? (dB to THD)

[TonyRPH]

There are two equations given on this page

One converts % THD to dB and I understand this one and have a working example in Excel.

THD to dB = 20*LOG(0.0000178/100) = -134.99 dB

But the dB to THD calculation (top one in the attached image) I just cannot get my head around.

(a) is "Distortion attenuation" in dB and (k) is "Distortion factor THD" in %

I'm reading it as 10*(-134/20)*100 - however I'm somewhat confused by what looks like 10*ak over 2D? (ak) is the distortion attenuation in dB.

But if that's 2D in the equation (as against 20) then I'm stumped!

Thanks in advance!

[TonyRPH]

jeremyc said:

here's your problem. You should read it as:
10 to the power of (-134/20) all multiplied by 100.

It's a simple re-arrangement of the other equation to solve for k in the following steps:
1. Divide both sides by 20.
2. Take the antilog of both sides.
3. Multiply both sides by 100

Thanks! But you make it seem so simple.

Now I just need to establish how I raise 10 to a power in Excel lol...

You can tell I didn't do very well in maths at school...!

[TonyRPH]

Thank you all for bearing with this numbskull.

In Excel I put: =POWER(-134/20,10)*100 - and instead of getting 0.0000178 - I get 18228378046?

If I do it this way: =10^(-134/20)*100 - I get something closer - 0.00001995 - but it should be 0.0000178 ?

Even if I calculate -134/20 separately (-6.7) and then put 10^(-6.7)*100 it's still the wrong answer so can't be the order of calculation?

What am I doing wrong?

[TonyRPH]

Mr Pointy said:

Haven't you got the number & the power the wrong way round?

=POWER(10,(-134/20))*100

=0.000019952623

Edited by Mr Pointy on Thursday 22 April 15:29

I did yes - but the answer should be 0.0000178 ?

[TonyRPH]

DOH!

I've just demonstrated what an idiot I really am....

Working it back, 0.00001995 = -134dB

It's all about the decimal places and rounding lol.

Thanks all.