Is There Another Way to Rearrange This Formula for ILS Glideslope?

[mcalves]

Hello all, this is a formula that calculates DDM (depth of modulation) at certain outputs of the ILS's Glideslope. I posted this question some time ago but I think I left out some important details. Any help is really appreciated.


This is basic info on the ILS
http://en.wikipedia.org/wiki/Instrument_Landing_System

db = 20log(E1/E2)

E1 is the input voltage before attenuation
E2 is the output voltage after attenuation

E1 and db seems like is always given. E2 is what needs to be solved.

My course book rearranges the formula to this: E2 = E1log-1(db/20)

-that makes NO sense to me. Log to negative power - or the reciprical?

this is one of the examples they have:

db = -3
E1 = .155

E2 = .155log-1(-3/20) = 0.109

more examples:

2.5log-1(-3/20) = 1.77

1.0log-1(-3/20) = 0.707

.494log-1(-3/20) = 0.342

Is there another way to rearrange the formula that makes sense?

Thank you.

 

[Dick]

They mean by log^(-1) the inverse function of log. I'll admit, that's a funny notation. But if you are dealing with base 10 logs, then log^(-1)(x)=10^x. Since log(10^x)=x.

 

[mcalves]

Dick, I'm not real big on math, I am not sure what you mean.

Basically

db = 20log(E1/E2)

db = -3
E1 = .155

How would you solve for E2

 

[mcalves]

by the way yes, the way its written in the book its

log^-1

thanks

 

[Dick]

Dick, I'm not real big on math, I am not sure what you mean.

Basically

db = 20log(E1/E2)

db = -3
E1 = .155

How would you solve for E2

I would write db/20=log(E1/E2), then take 10 to the power of both sides (this is log^(-1)). So 10^(db/20)=E1/E2.

 

[Sleek]

Yes, it is Antilog. If the base is 10, then you will find Antilogs for them in logarithmic tables. Base e can be converted into 10 if necessary (by multiplying roughly with 2.303).

Regards,
Sleek.

 

[Dick]

If it's talking about decibels, then the base of the log is 10.

 

[mcalves]

So take this example that's in the book:

E2 = .155log^-1(-3/20) = 0.109

The answer is already given, but how would I enter this in the calculator? Every way I tried has yielded a syntex or math error.

thanks in advance

 

[Hootenanny]

So take this example that's in the book:

E2 = .155log^-1(-3/20) = 0.109

The answer is already given, but how would I enter this in the calculator? Every way I tried has yielded a syntex or math error.

thanks in advance

The argument of a logarithm must be positive, so in this case E2 is undefined.

 

[Dick]

The argument of a logarithm must be positive, so in this case E2 is undefined.

It's NOT a log, it's an antilog. On the calculator I'm looking at above the 'log' key is a second function labelled $$10^x$$. That's the 'antilog'.

 

[mcalves]

okay using the antilog key finally worked. I get that 0.109 answer above with no errors.

Thanks Dick, Sleek, and Hootenanny for responding. Much appreciated!

 

[Hootenanny]

It's NOT a log, it's an antilog. On the calculator I'm looking at above the 'log' key is a second function labelled $$10^x$$. That's the 'antilog'.

Indeed 10^x gives the correct result, I've never seen that notation before.