how can i get a slope in the loglog plot?

i enter the data in "ID"

ID = [ 2.33879000000000e-19,

7.68704000000000e-17,

.....

0.00865008100000000,

0.00892525100000000,

0.00920535600000000 ] ( for ID(1,1)~ID(600,1))

loglog(ID);

i get a loglog plot for data ID,

and i want to get a slope of this loglog plot (at all point)

and i want to get a graph for this slope data

how can i do this?

Answer 1

[Torsten]

ID = ...;
dID = gradient(log10((1:600).'),log10(ID));
loglog(ID)
hold on
plot(log10((1:600).'),dID)

[준형 박]

i cant get the graph i want , help

ex) in the plot, loglog(ID), ID.1 (570, 2.35232) , ID.2(121, 0.00139276)

The regime between the two points appears linear in loglog plot, and calculating the slope with a calculator shows about 4.795.

but in plot(log((1:600).'),dID), dID's value is about 0.2

how can i get 4.795?

[Torsten]

Do you want the slope in the original scale

slope = gradient((1:100).',ID) = (ID(i+1)-ID(i-1))/2 (i=2,...,length(ID)-1)

or in the loglog scale

slope = gradient(log((1:100).'),log(ID)) = (log(ID(i+1))-log(ID(i-1)))/(log(i+1)-log(i-1)) (i=2,...,length(ID)-1)

And I think log10 instead of log is required:

slope = gradient(log10((1:100).'),log10(ID))

if you want the latter (see above).

?

[준형 박]

thank you for answer

i want to get loglog plot slope

when I get 4.795, I use this equation >> log(2.35232/0.00139276) / log(570/121) = 4.795 (use calculator)

but

when i use this

slope = gradient(log10((1:100).'),log10(ID));

i get a slope value about 0.1 ~ 0.2 .. why?

(log(ID(i+1))-log(ID(i-1)))/(log(i+1)-log(i-1))

I think this equation is the same as my intention

but i can't get slope data

[Torsten]

Why log(570/121) ?

If you use loglog(ID), you get a graph with x-coordinates log10(i) and y-coordinates log10(ID(i)).

The slope in a single x-coordinate is thus approximately log10(ID(i+1)/ID(i-1))/log10((i+1)/(i-1)).

But the difference between 570 and 121 is not 2 ( (i+1) - (i-1) = 2).

Or do you have an independent x-vector and you use

loglog(x,ID)

?

Because in your first contribution, you wrote loglog(ID) as plot command.

If you just want to calculate d(log10(ID))/d(log10(x)) between two given points as I suspect, then

i1 = 3;
i2 = 54;
slope_i2_i1 = log10(ID(i2)/ID(i1))/log10(i2/i1)

is the way to go.

But this is not the pointwise slope (thus an approximation to the loglog derivative).

[준형 박]

OK I understand

I think i misunderstood about x-coordinate

thank you for the reply.