Troubleshooting Log Formula for ILS DDM Calculation | Expert ATS Assistance

In summary, the conversation is about a student studying an ATS equipment called ILS and trying to use a formula to determine the DDM. However, when using the "log-1" function on a calculator, they received a Math Error. The conversation then discusses the meaning of "log-1" and how to format it. It is also mentioned that the logarithm is not defined for negative numbers. Finally, the conversation ends with the student asking for clarification on the formula and receiving help from another member.
  • #1
mcalves
10
0
Hello

Im studying an ATS equipment known as an ILS. One of the formulas to determine the DDM is :

E2 = E1 log-1(db/20)




When I try entering log-1, the calculator displays Math Error. What am I doing wrong?

Thanks for the help in advance
 
Physics news on Phys.org
  • #2
The logarithm is not defined for negative numbers.
 
  • #3
Are you sure that "log-1" means log(-1)?? Look more closely at the formula.
 
  • #4
Its log to the negative power of 1, I don't know how to format that below.


the funny thing is that there are examples in the book with the answer. For example:


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

Ecs = 1.0log-1(-3/20) = .707

Ess = 0.155log-1(-3/20) = 0.109
 
  • #5
mcalves said:
Its log to the negative power of 1, I don't know how to format that below.

[tex]\log^{-1}x = \frac{1}{\log{x}}
[/tex]

Click on the LaTeX to see the code.
 
Last edited:
  • #6
thanks neutrino, but what about that (-3/20) that will give me a negative. Logs don't work that way right?
 
  • #7
They don't. Are you sure db is negative?
 
  • #8
its interesting you say that because the its starts out with a 3db then they change it to a -3. I'll type the whole thing below...


E2 = E1 log-1(db/20)

Where:
E2 = before attenuation
E1 = after attenuation

The upper attenna signals exiting Z6 at output 2 are attenuated by 3db. Letting "A" equal .31, the Ess signal in the upper antenna would be .5A or .155 prior to Z6. So the equation becomes:

E2 = .155log-1(-3/20) = .109

What the heck does that mean ^^
 
  • #9
Sorry, I'm clueless. Maybe someone familiar with this topic (at the Electrical Engineering forum) will be able to answer your question.
 
  • #10
Thanks neutrino. I'll give it try. :)
 

What is a logarithm?

A logarithm is a mathematical function that helps us solve problems involving exponential growth and decay. It is the inverse of the exponential function and tells us the power to which a base number must be raised to get a certain value.

What is the log formula?

The log formula is logb(x) = y, which means the base b raised to the power of y is equal to x. This is the basic form of a logarithm and can be used to solve for any of the three variables (b, x, or y) given the other two.

How do I solve a log formula?

To solve a log formula, you need to first identify the base, or the number that is being raised to a power. Then, determine which variable (b, x, or y) is missing and use algebraic manipulation to isolate it on one side of the equation. Finally, use the properties of logarithms to simplify the equation and solve for the missing variable.

Can I use a calculator to solve log formulas?

Yes, most scientific calculators have a button or function for logarithms. Make sure to input the correct base and value to get an accurate answer. However, it is also important to understand the steps involved in solving a log formula manually.

Why are logarithms useful?

Logarithms are useful because they help us convert complicated exponential equations into simpler forms that are easier to solve. They are also used in many real-world applications, such as measuring decibels in sound, the pH levels of solutions, and the Richter scale for earthquakes.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
11
Views
3K
  • Aerospace Engineering
2
Replies
35
Views
3K
Replies
6
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
29K
  • STEM Academic Advising
Replies
13
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
2K
  • Classical Physics
Replies
22
Views
6K
  • General Discussion
Replies
2
Views
2K
  • Beyond the Standard Models
Replies
24
Views
7K
  • Atomic and Condensed Matter
Replies
4
Views
6K
Back
Top