My question is similar to this question but I believe it to be more general.
I use Matlab's symbolic math toolbox to solve an equation:
MAZ = 0.5;
MAU = 1.0;
XI = 1.0;
ALPHA = 2.0;
DRG = 0.5;
SRG = 1.0;
PHI = 1 / (2 * MAU);
syms L;
f = 1 - DRG - sqrt(1 + (ALPHA * XI - L / (2 * XI * PHI) ) ^ 2 ) / ...
sqrt(1 + (ALPHA * XI) ^ 2) + L / (4 * PHI * SRG * sqrt(1 + (ALPHA * XI)^2));
a = solve(f,L,'Real',true);
The answer is:
a =
5^(1/2)/3 + (10*((4*5^(1/2))/25 + 6/25)^(1/2))/3 + 8/3
5^(1/2)/3 - (10*((4*5^(1/2))/25 + 6/25)^(1/2))/3 + 8/3
How do I automatically convert these expressions for a - which do not contain any symbolic expressions - to floats so that I can access them later in my code?
Matlab's symbolic math toolbox contains a function called double.
This function converts the result of the solve function - which is a sym - to a double:
double(a)
ans =
5.98921078320145
0.834834535131742
While double is correct, it is limited to the precision of 64bit floating points. Use vpa if a higher precision is needed:
>> vpa(a)
ans =
5.9892107832014511063435699584181
0.83483453513174202459587915406938
>> digits(99)
>> vpa(a)
ans =
5.98921078320145110634356995841813862213621375395128614574627036653958858547362556353272837962692249
0.834834535131742024595879154069378868157531819123064337100994463734092031618244369410214559292265698
Related
I have a problem in which I have to implement the following question in Matlab.
i(t) = A2 * sin(wr*t) * exp(-alpha*t); for t [0, 10] with step 0.5s
My approach is as follows
clc;
clear;
% Given Data
Vs = 220;
L = 5e-3;
C = 10e-6;
R = 22;
Vo = 50;
% a)
alpha = R / (2 * L);
omega_not = 1 / sqrt(L*C);
omega_r = sqrt( omega_not^2 - alpha^2 );
A2 = Vs / (omega_r * L);
t = 1:0.5:10;
i = A2 * sin( omega_r * t ) .* exp(-alpha * t);
% b)
t1 = pi / omega_r;
% c)
plot(t, i);
But it yields all the values of current equal to zero. Please help me solve the problem.
I think the problem is this part of the expression:
exp(-alpha * t)
When I run your code, -alpha equals -2200. The exponential for such a large negative number is so small that the code returns zero.
>> exp(-200)
ans =
1.3839e-87
>> exp(-1000)
ans =
0
I have seen that there was an interest in custom interpolation kernels for resize (MATLAB imresize with a custom interpolation kernel). Did anyone implemented the parametric Mitchell-Netravali kernel [1] that is used as default in ImageMagick and is willing to share the Matlab code? Thank you very much!
[1] http://developer.download.nvidia.com/books/HTML/gpugems/gpugems_ch24.html
// Mitchell Netravali Reconstruction Filter
// B = 0 C = 0 - Hermite B-Spline interpolator
// B = 1, C = 0 - cubic B-spline
// B = 0, C = 1/2 - Catmull-Rom spline
// B = 1/3, C = 1/3 - recommended
float MitchellNetravali(float x, float B, float C)
{
float ax = fabs(x);
if (ax < 1) {
return ((12 - 9 * B - 6 * C) * ax * ax * ax +
(-18 + 12 * B + 6 * C) * ax * ax + (6 - 2 * B)) / 6;
} else if ((ax >= 1) && (ax < 2)) {
return ((-B - 6 * C) * ax * ax * ax +
(6 * B + 30 * C) * ax * ax + (-12 * B - 48 * C) *
ax + (8 * B + 24 * C)) / 6;
} else {
return 0;
}
}
Here I got another approach with vectorization; according to my tests with upscaling (1000x1000 -> 3000x3000) this is faster than the standard bicubic even with a large Mitchell radius = 6:
function [outputs] = Mitchell_vect(x,M_B,M_C)
outputs= zeros(size(x,1),size(x,2));
ax = abs(x);
temp = ((12-9*M_B-6*M_C) .* ax.^3 + (-18+12*M_B+6*M_C) .* ax.^2 + (6-2*M_B))./6;
temp2 = ((-M_B-6*M_C) .* ax.^3 + (6*M_B+30*M_C) .* ax.^2 + (-12*M_B-48*M_C) .* ax + (8*M_B + 24*M_C))./6;
index = find(ax<1);
outputs(index)=temp(index);
index = find(ax>=1 & ax<2);
outputs(index)=temp2(index);
end
I got the following proposal for the Mitchel kernel called by imresize with the parameters B and C and a kernel radius using for-loops (and preallocation):
img_resize = imresize(img, [h w], {#(x)Mitchell(x,B,C),radius});
function [outputs] = Mitchell(x,B,C)
outputs= zeros(size(x,1),size(x,2));
for i = 1 : size(x,1)
for j = 1 : size(x,2)
ax = abs(x(i,j));
if ax < 1
outputs(i,j) = ((12-9*B-6*C) * ax^3 + (-18+12*B+6*C) * ax^2 + (6-2*B))/6;
elseif (ax >= 1) && (ax < 2)
outputs(i,j) = ((-B-6*C) * ax^3 + (6*B+30*C) * ax^2 + (-12*B-48*C) * ax + (8*B + 24*C))/6;
else
outputs(i,j) = 0;
end
end
end
end
2nd ODE to solve in MATLAB:
( (a + f(t))·d²x/dt² + (b/2 + k(t))·dx/dt ) · dx/dt - g(t) = 0
Boundary condition:
dx/dt(0) = v0
where
t is the time,
x is the position
dx/dt is the velocity
d2x/dt2 is the acceleration
a, b, v0 are constants
f(t), k(t) and h(t) are KNOWN functions dependent on t
(I do not write them because they are quite big)
As an example, using symbolic variables:
syms t y
%% --- Initial conditions ---
phi = 12.5e-3;
v0 = 300;
e = 3e-3;
ro = 1580;
E = 43e9;
e_r = 0.01466;
B = 0.28e-3;
%% --- Intermediate calculations ---
v_T = sqrt(((1 + e_r) * 620e6) /E) - sqrt(E/ro) * e_r;
R_T = v_T * t;
m_acc = pi * e * ro *(R_T^2);
v_L = sqrt (E/ro);
R_L = v_L * t;
z = 2 * R_L;
E_4 = B * ((e_r^2)* B * (0.9^(z/B)-1)) /(log(0.9));
E_1 = E * e * pi * e_r^2 * (-phi* (phi - 2*v_T*t)) /16;
E_2 = pi * R_T^2 * 10e9;
E_3 = pi * R_T^2 * 1e6 * e;
%% Resolution of the problem
g_t = -diff(E_1 + E_2 + E_3, t);
f(t,y)=(g_t - (pi*v_T*e*ro/2 + E_4) * y^2 /(y* (8.33e-3 + m_acc))];
fun=matlabFunction(f);
[T,Y]=ode45(fun,[0 1], v0]);
How can I rewrite this to get x as y=dx/dt? I'm new to Matlab and any help is very welcome !
First, you chould use subs to evaluate a symbolic function. Another approach is to use matlabFunction to convert all symbolic expressions to anonymous functions, as suggested by Horchler.
Second, you're integrating the ODE as if it is 1st order in dx/dt. If you're interested in x(t) as well as dx/dt(t), then you'll have to modify the function like so:
fun = #(t,y) [y(2);
( subs(g) - (b/2 + subs(k))*y(2)*y(2) ) / ( y(2) * (a + subs(f))) ];
and of course, provide an initial value for x0 = x(0) as well as v0 = dx/dt(0).
Third, the absolute value of the parameters is hardly ever a real concern. IEEE754 double-precision floating point format can effortlessly represent numbers between 2.225073858507201e-308 and 1.797693134862316e+308 (realmin and realmax, respectively). So for the coefficients you gave (O(1014)), this is absolutely not a problem. You might lose a few digits of precision if you don't take precautions (rescale to [-1 +1], reformulate the problem in different units, ...), but the relative error due to this is more than likely to be tiny and insignificant compared to the algorithmic error made by ode45.
<RANDOM_OPINIONATED_RANT>
Fourth, WHY do you use symbolic math for this purpose?! You are doing a numerical integration, meaning, there is no analytic solution anyway. Why bother with symbolics then? Doing the integration with symbolics (through vpa even) is going to be dozens, hundreds, yes, often even thousands of times slower than keeping (or re-implementing) everything numerical (which some would argue is already slow in MATLAB compared to a bare-metal approach).
Yes, of course, for this specific, individual, isolated use case it may not matter much, but for the future I'd strongly advise you to learn to:
use symbolics for derivations, proving theorems, simplifying expressions, ...
use numerics to implement any algorithm or function from which actual numbers are expected.
In other words, symbolics for drafting, numerics for crunching. And exactly zero symbolics should appear in any good implementation of any algorithm.
Although it's possible to mix them to some extent, that does not mean it is a good idea to do so. In fact, that's almost never. And the few isolated cases where it is the only viable option are not a vindication of the approach.
They are rare, isolated cases after all, far from the abundant norm.
For me it bears resemblance with the evil eval, with similar reasons for why it Should. Be. Avoided.
</RANDOM_OPINIONATED_RANT>
With the full code, it's easy to come up with a complete solution:
% Initial conditions
phi = 12.5e-3;
v0 = 300;
x0 = 0; % (my assumption)
e = 3e-3;
ro = 1580;
E = 43e9;
e_r = 0.01466;
B = 0.28e-3;
% Intermediate calculations
v_T = sqrt(((1 + e_r) * 620e6) /E) - sqrt(E/ro) * e_r;
R_T = #(t) v_T * t;
m_acc = #(t) pi * e * ro *(R_T(t)^2);
v_L = sqrt (E/ro);
R_L = #(t) v_L * t;
z = #(t) 2 * R_L(t);
E_4 = #(t) B * ((e_r^2)* B * (0.9^(z(t)/B)-1)) /(log(0.9));
% UNUSED
%{
E_1 = #(t) -phi * E * e * pi * e_r^2 * (phi - 2*v_T*t) /16;
E_2 = #(t) pi * R_T(t)^2 * 10e9;
E_3 = #(t) pi * R_T(t)^2 * 1e6 * e;
%}
% Resolution of the problem
g_t = #(t) -( phi * E * e * pi * e_r^2 * v_T / 8 + ... % dE_1/dt
pi * 10e9 * 2 * R_T(t) * v_T + ... % dE_2/dt
pi * 1e6 * e * 2 * R_T(t) * v_T ); % dE_3/dt
% The derivative of Z = [x(t); x'(t)] equals Z' = [x'(t); x''(t)]
f = #(t,y)[y(2);
(g_t(t) - (0.5*pi*v_T*e*ro + E_4(t)) * y(2)^2) /(y(2) * (8.33e-3 + m_acc(t)))];
% Which is readily integrated
[T,Y] = ode45(f, [0 1], [x0 v0]);
% Plot solutions
figure(1)
plot(T, Y(:,1))
xlabel('t [s]'), ylabel('position [m]')
figure(2)
plot(T, Y(:,2))
xlabel('t [s]'), ylabel('velocity [m/s]')
Results:
Note that I've not used symbolics anywhere, except to double-check my hand-derived derivatives.
Upon request by Martin here is the basic problem. There is a function M(x) which is supposed to be minimized over the interval [lb, ub].
M = #(x) (a_1 * x + b_1) * (log((a_1 * x + b_1)/P_1) + X_u)...
+ (a_2 * x + b_2) * (log((a_2 * x + b_2)/P_2) + X_m)...
+ x * (log(x / P_3) + X_d);
lb = max(0, -b_1 / a_1);
ub = -b_2 / a_2;
where the inputs are:
P_1 = 0.6;
P_2 = 0.2;
P_3 = 0.2;
a_1 = 0.7071;
a_2 = -1.7071;
b_1 = 0.0245;
b_2 = 0.9755;
X_u = 44;
X_m = 2.9949;
X_d = 0;
The other option would be to solve for the root of the equation m_dash:
m_dash = #(x) log(((a_1 .* x + b_1).^a_1) .* ((a_2 .* x + b_2).^a_2) .* x)...
- log((P_1.^a_1) .* (P_2.^a_2) .* P_3) + a_1 .* X_u + a_2 .* X_m + X_d;
Any help would be greatly appreciated.
If you want to minimize a function over a certain interval, you can use the fminbnd function from the Optimization Toolbox. If you don't have that toolbox installed, you can either try a free alternative, or instead coerce the built-in function fminsearch to only return results from the interval:
rlv = 1e12; % ridiculously large value
M_hacked= #(x) rlv*((x < lb) + (x > ub)) + M(x);
x_min = fminsearch(M_hacked, (lb + ub)/2)
I introduced a new function, M_hacked, which returns ridiculously large values for x outside of the interval.
This is not be the most elegant solution, but it should do for your problem.
My question today is related to this previous question. I am following this research paper. I am trying to duplicate figure 8 located on page 20. I have a screenshot:
I'm confused on how to plot the left figure this in MATLAB because now a instead of having time varying we have the treatment varying. Here's what I have from the previous question:
function dX = CompetitionModel(~,X)
bs = 8e-3;
bl = 4e-3;
bh = 6.4e-3;
N = sum(X);
K = 1e8;
m1 = 2e-5;
m2 = 9e-9;
p = 5e-13;
I = 1e-3;
T = 1e-3; % Treatment
a = 0;
dX = [X(1) * (bs * (1 - N/K) - I - T - m1) - p * X(1) * (X(2) + X(3));
X(2) * (bl * (1 - N/K) - I - a*T - m2) + m1 * X(1) + p * X(2) * (X(1) - X(3));
X(3) * (bh * (1 - N/K) - I - a*T) + m2 * X(2) + p * X(3) * (X(1) + X(2))];
end
To plot my equations in the previous question, I typed the following in the command window:
>> [t,Y] = ode45(#CompetitionModel, [0 4.5e4], [1e4 0 0]);
>> plot(t,X(:,1), t,X(:,2), t,X(:,3))
In my function file, I have Treatment already defined. I'm guessing that it shouldn't be anymore. So what can I do so that I have Treatment varying instead of time? I hope my question makes sense.
You still solve the equation in regards to the time - but solely plot the value at the time t = 1 month.
To vary the treatment you need an additional loop around the ode45 call and pass the current treatment-value to the function dX
for treatment = 10^-4:10^-5:10^-3
[t,Y] = ode45(#CompetitionModel, [0 4.5e4], [1e4 0 0], [] , treatment);
plot(treatment,Y(end,1), 'x')
plot(treatment,Y(end,2), 'kx')
plot(treatment,Y(end,3), 'rx')
hold on
end
the function dX now has to be changed to accept the treatment input:
function dX = CompetitionModel(~,X, T)
Finally, comment your old treatment assignment in the function dX: %T = 1e-3; % Treatment