How would you draw this in MATLAB?

3 Ansichten (letzte 30 Tage)
S. David
S. David am 4 Mär. 2012
Hi,
I have an equation for the absorption coefficient in underwater acoustic communication channel (UWA) as:
alpha=0.11.*(f.^2./(1+f.^2))+44.*(f.^2./(4100+f))+2.75*10^-4.*f.^2+0.003;
where f is the frequency in kHz, and the absorption coefficient is given by dB/Km. I have a paper that plots this relation versus the frequency. It is straightforward as it appears at first, but when I tried to do it myself I have not gotten the same curve, even though the vertical axis in the paper is in dB/Km, which means it is just a direct substitution in the above equation. The question is why?
Thanks in advance
  3 Kommentare
Rick Rosson
Rick Rosson am 4 Mär. 2012
Please post your MATLAB code.
Jan
Jan am 4 Mär. 2012
All we know is the formula and that you do not get the "same curve". It might be a typo in the formula, another scaling of log-scaling, different colors or line styles, a post-script problem in the paper, etc. I do not see a chance to guess the source of the differences yet. Please post the necessary details.

Melden Sie sich an, um zu kommentieren.

Antworten (4)

S. David
S. David am 5 Mär. 2012
This is my code:
clear all;
clc;
f=10^3.*(0:1000);
alpha=(0.11.*((f.^2)./(1+f.^2))+44.*(f.^2./(4100+f))+(2.75.*10^(-4)).*(f.^2)+0.003);
plot(f,alpha)
Thanks

Jan
Jan am 5 Mär. 2012
Unfortunately you still did not describe, which difference you mean. I guess it is the small wobble near to 60Hz.
When I read the explanation about the Figure 1 in this paper, I find at first the above formula and then:
This formula is generally valid for frequencies above a
few hundred Hz. For lower frequencies, the following
formula may be used:
10 log a(f) = 0.002+ 0.11*(f.^2 ./ (1 + f.^2)) + 0.011 * f.^2
I guess, that the author used this formula for the low frequency part.
[EDITED] Check your formula again. It is "4100+f.^2" instead of "4100+f".
  4 Kommentare
S. David
S. David am 5 Mär. 2012
Not much difference when I wrote f.^2 instead of f. Now, I need the program to work. Later I can consider more efficient one.
Jan
Jan am 5 Mär. 2012
Ok, not much, but at least enough?
I suggest the standard notation of numbers mainly, because this is faster to write and to read. This is more important than the runtime.

Melden Sie sich an, um zu kommentieren.


G A
G A am 5 Mär. 2012
This gives the same picture as in the paper:
clear;
f=0:1000;%kHz
alpha=0.11.*f.^2./(1+f.^2)+44.*f.^2./(4100+f.^2)+2.75e-4.*f.^2+0.003;%dB/km
figure(1)
clf
plot(f,alpha,'-b')
  3 Kommentare
G A
G A am 5 Mär. 2012
multiplying f-array by 1e3 is not necessary
Honglei Chen
Honglei Chen am 5 Mär. 2012
In your original question, it says f is in kHz so that's why you don't need to multiply 1e3.

Melden Sie sich an, um zu kommentieren.


Saed
Saed am 5 Mär. 2012
That is great. What did you do? I mean I tried all of these stuff. Anyway, it is fine now. Thanks a lot.
  2 Kommentare
G A
G A am 5 Mär. 2012
1)I have used equation given in the paper. Jan has spotted your mistake. 2)I have removed frequency multiplication by 1e3 you have done.
S. David
S. David am 6 Mär. 2012
Thanks

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu MATLAB finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by