How to calculate gradient from semilogy plot graph?

I have data and already plotted in semilogy graph,

I need to know the gradient from that graph

Any one can help me, please?

I use Matlab 2016b versiaon

Answer 1

[Torsten]

In the left part, your functional equation is approximately

f(x) = 10^(-3.2*x+70)

I don't know from your question whether you mean the gradient of your original function f(x) or that of log10(f(x)) that is shown in the plot.

[christina widyaningtyas]

y-axis is pressure data, and x-axis is time

I want to calculate the gradient or pressure

[Torsten]

Then the gradient is

syms t
f = 10^(-3.2*t+70);
gradf = diff(f,t)

gradf = 

whatever the units for pressure and time in the plot might be.

[christina widyaningtyas]

But, there is a error warning

"Error using ^

Inputs must be a scalar and a square matrix.

To compute elementwise POWER, use POWER (.^) instead."

I tried to fix it : f = 10.^(-3.2*t+70);

but it still error with this note :

Error using diff

Difference order N must be a positive integer scalar.

There are negative value, do you have any advice to fix it?

[Torsten]

I don't know what you try to do.

The linear part of the semilogy plot can be approximated by g(t) = 70-3.2*t.

Thus the original pressure function is

p(t) = 10^g(t) = 10^(70-3.2*t).

Now I just calculated its gradient symbolically:

syms t
p = 10^(70-3.2*t)

p = 

gradp = diff(p,t)

gradp = 

Or maybe you don't have a licence for the Symbolic Toolbox ?

Try

license checkout Symbolic_Toolbox

ans = 1

[Torsten]

I think I made a mistake in the calculation of your pressure function.

I assumed that in the semilogy plot, log(p) is plotted against t. But that's not true - it's also p that is shown.

The following code should be correct:

b = log10(70);
a = log10(6/70)/30;

a = -0.0356

ans = 1.0853

syms t
p = 10^(a*t+b);
double(subs(p,t,0))

ans = 70.0000

double(subs(p,t,30))

ans = 6.0000

gradp = diff(p,t);
p = matlabFunction(p);
gradp = matlabFunction(gradp);
t = 0:0.1:40;
plot(t,p(t))

plot(t,gradp(t))

Answer 2

[BELDA]

1° question n°1 :

b = log10(70);

a = log10(6/70)/30;

Vous avez élévé la fonction f(x) en log base 10 c a d en échelle semilog alors que x est en échelle linéaire afin de mieux définir ma fonction f(x);

pourquoi pour a=log10(6/70)/30 comment on trouve (6/70)/30 pour -3.2x

D'avance merci .

[Torsten]

You search for the constants of a linear function a*x+b such that (from the graph)

p(0) = 10^(a*0+b) = 70

p(20) = 10^(a*20+b) = 6 (I don't know how I came up with p(30) = 6 as done above).

The equations are thus

70 = 10^b, thus b = log10(70) and

6 = 10^(a*20+b) -> log10(6) = a*20+log10(70) -> a = (log10(6)-log10(70))/20 = log10(6/70) / 20.

The original pressure curve is thus approximately

p(t) = 10^(log10(6/70) / 20 * t + log10(70))