0
\$\begingroup\$

I'm trying to plot a -20dB/dec line on MATLAB but can't seem to get it correct. I know that a linear line in a loglog plot is actually an exponential curve in the linear scale. enter image description here

So writing a y = mx + c where m is -20dB/decade is not quite right in MATLAB. The closest I could get to a straight line was using this code below, but as you can see it is not the correct slope or the correct magnitudes.

I have also tried to back calculate it, but haven't solved it. Please help!

Frx = logspace(1,8,100);
y = zeros(length(Frx),1);
for i = 1:length(Frx)
    y(i) = 10^(-0.01*(log10(Frx(i))));
end
figure;
loglog(Frx,y);grid on
\$\endgroup\$
3
  • 1
    \$\begingroup\$ Seems as though it's just \$y=\frac{10^{8.5}}{x}\$ (linear scale.) I'm not sure where all the difficulty is at. If \$y^{\:'}=20\log_{10}\left(y\right)\$ and \$x^{\:'}=\log_{10}\left(x\right)\$ and therefore your log-log scale formula is just \$y^{\:'}=170-20\cdot x^{\:'}\$, anyway. I don't know anything about MATLAB. Don't use it. But the algebra seems easy enough to me without it. \$\endgroup\$ Commented Oct 13, 2023 at 2:48
  • \$\begingroup\$ @periblepsis Perhaps you could elaborate how you got to that linear scale equation please? Thanks! \$\endgroup\$ Commented Oct 14, 2023 at 23:25
  • \$\begingroup\$ @SM32 Straight substitution and solving for y. I gave you the primed equation and the definition of the primed variables. Take that last equation, substitute the definitions I gave earlier, then solve for y. \$\endgroup\$ Commented Oct 14, 2023 at 23:46

1 Answer 1

Reset to default
2
\$\begingroup\$

A line on a semi-log plot is an exponential; a line on a log-log plot is a power law. 20 dB is a factor of 100, and 'decade' means a factor of ten, so you want the 'decibels' value to be inversely proportional to the square of the frequency.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.