I am trying to solve a hysteresis problem. Considering hysteresis effect, the moisture content is history dependent: Consider a block of wood, which has initial moisture content w0, which is in equilibrium with initial environment of phi0; then you put the block into environment of phi1; now you need (w0, phi0 and phi1) to determine w1 corresponding to phi1; then you put the block into environment of phi2, to determine w2 corresponding to phi2, you need (w1, phi1 and phi2);....
In Modelica, you can only use the state variable at current time step; how can I use phi0 and phi1 simultaneously? not the time derivative of phi, der(phi). when w1 is determined, the values of w1 and phi1 will be transferred to w0 and phi0.
http://vbn.aau.dk/ws/files/13648994/Selected_Constitutive_Models_for_Simulating_the_Hygromechanical_Response_of_Wood
sorry I cannot upload an image, I upload the link of a paper, please check the Fig.6 and equation 17 in page 65 (90/135)
Sorry, I don't know whether it is clear or not. I want to implement sorption hysteresis in Modelica. Hysteresis is history dependent, I need the previous state as input to calculate current state variables. The lower continuous line is adsorption curve, upper continuous line is desorption curve, these two curves are function of relative humidity h, they are known; s(h)=(w[h]-wads[h])/(wdes[h]-wads[h]). z0 (h0,m0) is initial/previous moisture state (can be given by initial condition or calculated results from previous time step.) if the current h is bigger than h0, moisture state at zd0(relative humidity hd0) will be calculated from moisture state z0, then the curve along zd0-z0-right arrow will be defined as a function of hd0; now I want to calculate moisture content at current h (h>h0) on the defined curve zd0-z0-right arrow; but I cannot provide both h0 and h in the Modelica code. I define h as current state, I need the moisture state at previous time step h0. And every time step, when the moisture state at h is calculated (m at h are obtained, the values of m and h will be transferred to m0 and h0 for calculation in next time step.)
Related
I am in the midst of implementing a Kalman filter based AHRS in C++. There's something rather strange to me in the equations of the filter.
I can't find the part where the P (covariance) matrix is actually updated to represent uncertainty of predictions.
During the "predict" step P estimate is calculated from its previous value, A and Q. From what I understand A (system matrix) and Q (covariance of noise) are constant. Then during "Correct" P is calculated from K, H and predicted P. H (observation matrix) is constant, so the only variable that affects P is K (Kalman gain). But K is calculated from predicted P, H and R (observation noise) that are either constants or the P itself. So where is the part of the equations that makes P relate to x? To me it seems like P is recursively looping here depending only on the constants and initial value of P. This doesn't make any sense. What am I missing?
You are not missing anything.
It can come as a surprise to realise that, indeed, the state error covariance matrix (P) in a linear kalman filter does not depend on the the data (z). One way to lessen the surprise is to note what the covariance is saying: it is how uncertain you should be in the estimated state, given that the models you are using (effectively A,Q and H,R) are accurate. It is not saying: this is the uncertainty. By judicious tweaking of Q and R you could change P arbitrarily. In particular you should not interpret P as a 'quality' figure, but rather look at the observation residuals. You could, for example, make P smaller by reducing R. However then the residuals would be larger compared with their computed sds.
When the observations come in at a constant rate, and always the same set of observations, P will tend to a steady state that could, in principal, be computed ahead of time.
However there is no difficulty in applying the kalman filter when you have varying times between observations and varying sets of observations at each time, for example if you have various sensor systems with different sampling periods. In this case you will see more variation in P, though again in principal this could be computed ahead of time.
Further the kalman filter can be extended (in various ways, eg the extended kalman filter and the unscented kalman filter) to handle non linear dynamics and non linear observations. In this case because the transition matrix (A) and the observation model matrix (H) have a state dependency, so too will P.
As part of a course in signal processing at university, we have been asked to write an algorithm in Matlab to calculate the single sided spectrum of our signal using DFT, without using the fft() function built in to matlab. this isn't an assessed part of the course, I'm just interested in getting this "right" for myself. I am currently using the 2018b version of Matlab, should anyone find this useful.
I have built a signal of a 1 KHz and 2KHz sinusoid, phase shifted by 135 degrees (2*pi/3 rad).
then using the equations in 9.1 of Discrete time signal processing (Allan V. Oppenheim) and Euler's formula to simplify the exponent, I produce this code:
%%DFT(currently buggy)
n=0;m=0;
for m=1:DFT_N-1 %DFT_Fmin;DFT_Fmax; %scrolls through DFT m values (K in text.)
for n=1:DFT_N-1;%;(DFT_N-1);%<<redundant code? from Oppenheim eqn. 9.1 % eulers identity, K=m and n=n
X(m)=x(n)*(cos((2*pi*n*m)/DFT_N)-j*sin((2*pi*n*m)/DFT_N));
n=n+1;
end
%m=m+1; %redundant code?
end
This takes x as the input, in this case the signal mentioned earlier, as well as the resolution of the transform, as represented by the DFT_N, which has been initialized to 100. The output of this function, X, should be something in the frequency domain, but plotting X yields a circular plot slightly larger than the unit circle, and with a gap on the left hand edge.
I am struggling to see how I am supposed to convert this to the stem() plots as given by the in-built DFT algorithm.
Many thanks, J.
This is your bug:
replace X(m)=x(n)*(cos.. with X(m)=X(m)+x(n)*(cos..
For a given m, it does not integrate over the variable n, but overwrites X(m) only the last calculation for n = DFT_N-1.
Notice that integrating over n=1:DFT_N-1 omits one harmonic, i.e., the first one, exp(-j*2*pi). Replace
n=1:DFT_N-1 with n=1:DFT_N to include that. I would also replace m=1:DFT_N-1 with m=1:DFT_N for plotting concerns.
Also replace any 2*pi*n*m with 2*pi*(n-1)*(m-1) to get the phase right, since the first entry of X should correspond to zero-frequency, yielding sum_n x(n) * (cos(0) + j sin(0)) = sum_n x(n). If your signal x is real-valued then the zero-frequency component X(1) should be real-valued, angle(X(1))=0.
Last remark, don't forget to shift zero-frequency component to the center of the spectrum for better visibility, X = circshift(X,floor(size(X)/2));
If you are interested in the single-sided spectrum only, than you can just calculate X(m) for m=1:DFT_N/2 since X it is conjugate symmetric around m=DFT_N/2, i.e., X(DFT_N/2+m) = X(DFT_N/2-m)', due to exp(-j*(pi*n+2*pi/DFT_N*m)) = exp(-j*(pi*n-2*pi/DFT_N*m))'.
As a side note, for a given m this program calculates an inner product between the array x and another array of complex exponentials, i.e., exp(-j*2*pi/DFT_N*m*n), for n = 0,1,...,N-1. MATLAB syntax is very convenient for such calculations, and you can avoid this inner loop by the following command
exp(-j*2*pi/DFT_N*m*(0:DFT_N-1)) * x
where x is a column vector. Similarly, you can avoid the first loop too by expanding your complex exponential vector row-wise for every m, i.e., build the matrix exp(-j*2*pi/DFT_N*(0:DFT_N-1)'*(0:DFT_N-1)). Then your DFT is simply
X = exp(-j*2*pi/DFT_N*(0:DFT_N-1)'*(0:DFT_N-1)) * x
For single-sided spectrum, instead use
X = exp(-j*2*pi/DFT_N*(0:floor((DFT_N-1)/2))'*(0:DFT_N-1)) * x
The task is to create a cone hat in Matlab by creating a developable surface with numerical methods. There are 3 parts of which I have done 2. My question is regarding part 3 where I need to calculate the least rectangular paper surface that can contain the hat. And I need to calculate the material waste of the paper.
YOU CAN MAYBE SKIP THE LONG BACKGROUND AND GO TO LAST PARAGRAPH
BACKGROUND:
The cone hat can be created with a skewed cone with its tip located at (a; 0; b) and with a circle-formed base.
x = Rcos u,
y = Rsin u
z = 0
0<_ u >_2pi
with
known values for R, a and b
epsilon and eta ('n') is the curves x- and y- values when the parameter u goes from 0 to 2pi and alpha is the angle of inclination for the curve at point (epsilon, eta). Starting values at A:
u=0, alhpa=0, epsilon=0, eta=0.
Curve stops at B where the parameter u has reached 2pi.
1.
I plotted the curve by using Runge-Kutta-4 and showed that the tip is located at P = (0, sqrt(b^2 + (R-alpha)^2))
2.
I showed that by using smaller intervals in RK4 I still get quite good accuracy but the problem then is that the curve isn't smooth. Therefore I used Hermite-Interpolation of epsilon and eta as functions of u in every interval to get a better curve.
3.
Ok so now I need to calculate the least rectangular paper surface that can contain the hat and the size of the material waste of the paper. If the end angle alpha(2pi) in the template is pi or pi/2 the material waste will be less. I now get values for R & alpha (R=7.8 and alpha=5.5) and my task is to calculate which height, b the cone hat is going to get with the construction-criteria alpha(2pi)=pi (and then for alpha(2pi)=pi/2 for another sized hat).
So I took the first equation above (the expression containing b) and rewrote it like an integral:
TO THE QUESTION
What I understand is that I need to solve this integral in matlab and then choose b so that alpha(2pi)-pi=0 (using the given criteria above).
The values for R and alpha is given and t is defined as an interval earlier (in part 1 where I did the RK4). So when the integral is solved I get f(b) = 0 which I should be able to solve with for example the secant method? But I'm not able to solve the integral with the matlab function 'integral'.. cause I don't have the value of b of course, that is what I am looking for. So how am I going to go about this? Is there a function in matlab which can be used?
You can use the differential equation for alpha and test different values for b until the condition alpha(2pi)=pi is met. For example:
b0=1 %initial seed
b=fsolve(#find_b,b0) %use the function fsolve or any of your choice
The function to be solved is:
function[obj]=find_b(b)
alpha0=0 %initual valur for alpha at u=0
uspan=[0 2*pi] %range for u
%Use the internal ode solver or any of your choice
[u,alpha] = ode45(#(u,alpha) integrate_alpha(u,alpha,b), uspan, alpha0);
alpha_final=alpha(end) %Get the last value for alpha (at u=2*pi)
obj=alpha_final-pi %Function to be solved
end
And the integration can be done like this:
function[dalpha]=integrate_alpha(u,alpha,b)
a=1; %you can use the right value here
R=1; %you can use the right value here
dalpha=(R-a*cos(u))/sqrt(b^2+(R-a*cos(u))^2);
end
Here is a mass-spring-damper system with an impulse response, h and an arbitrary forcing function, f (cos(t) in this case). I am trying to use Matlab's FFT function in order to perform convolution in the frequency domain. I am expecting for the output (ifft(conv)) to be the solution to the mass-spring-damper system with the specified forcing, however my plot looks completely wrong! So, i must be implementing something wrong. Please help me find my errors in my code below! Thanks
clear
%system parameters
m=4;
k=256;
c=1;
wn=sqrt(k/m);
z=c/2/sqrt(m*k);
wd=wn*sqrt(1-z^2);
w=sqrt(4*k*m-c^2)/(2*m);
x0=0; %initial conditions
v0=0;
%%%%%%%%%%%%%%%%%%%%%%%%%
t=0:.01:2*pi ;%time vector
f=[cos(t),zeros(1,length(t)-1)]; %force f
F=fft(f);
h=[1/m/wd*exp(-z*wn*t).*sin(wd*t),zeros(1,length(t)-1)]; %impulse response
H=fft(h);
conv=H.*F; %convolution is multiplication in freq domain
plot(0:.01:4*pi,ifft(conv))
To see what is expected run this code. Enter in cos(t); 4; 1; 256 for the inputs. You can see that it reaches a steady state at an amplitude much different than the plot generated from the above FFT code.
%%%FOR UNDERDAMPED SYSTEMS
func=input('enter function of t---> ','s');
m=input('mass ');
c=input('c ');
k=input('k ');
z=c/2/sqrt(k*m);
wn=sqrt(k/m);
wd=wn*sqrt(1-z^2);
dt=.001;
tMax=20;
x0=0;
v0=0;
t0=0;
t=0:dt:tMax;
X(:,1)=[x0;v0;t0];
functionForce=inline(func);
tic
for i=1:length(t)-1
a=([0, 1, 0; -wn^2, -2*z*wn, 0; 0,0,0]*[X(1,i);X(2,i);X(3,i)]+[0;functionForce(X(3,i));0]);
Xtemp=X(:,i)+[0;0;dt/2] + a*dt/2;
b=([0, 1, 0; -wn^2, -2*z*wn, 0; 0,0,0]*[Xtemp(1);Xtemp(2);Xtemp(3)]+[0;functionForce(X(3,i));0]);
Xtemp=Xtemp+[0;0;dt/2] + b*dt/2;
c=([0, 1, 0; -wn^2, -2*z*wn, 0; 0,0,0]*[Xtemp(1);Xtemp(2);Xtemp(3)]+[0;functionForce(X(3,i));0]);
Xtemp=Xtemp + [0;0;dt] +c*dt;
d=([0, 1, 0; -wn^2, -2*z*wn, 0; 0,0,0]*[Xtemp(1);Xtemp(2);Xtemp(3)]+[0;functionForce(X(3,i));0]);
X(:,i+1)=X(:,i)+(a+2*b+2*c+d)*dt/6+[0;0;dt];
end
toc
figure(1)
axis([0 tMax min(X(1,:)) max(X(1,:))])
plot(t,X(1,:))
The initial transient will appear in the FFT method, so you will need to increase the time span (eg t=0:0.01:15*pi) to ensure the result includes the steady state. Incidentally since you truncate h after the same duration, increasing the time span also makes your impulse response h a better approximation to the actual infinite length impulse response.
So, updating your code to:
T=15*pi;
N=floor(T/.01);
t=[0:N-1]*0.01; ;%time vector
...
plot([0:2*N-2]*0.01, real(ifft(conv)));
axis([0 T]); % only show the duration where the driving force is active
would correspondingly show the following response:
which shows the initial transient, and eventually the steady state. You may notice that the plot is similar to your alternate implementation up to a scaling factor.
This difference in scaling comes from two factors. The first one is simply due to the fact that the convolution in your FFT based implementation computes a sum which isn't weight by the time step (as compare with the dt scaling used in your second implementation). The second one comes from the fact that the second implementation does not account for the mass m for the effect of the driving force.
After accounting for those two factors, you should get the following responses:
where the blue curve represents the FFT based implementation (same as the above curve but scaled down by dt=0.01), and the red curve represents your alternate implementation (with the functionForce divided by m).
I have attempted to write a code in order to solve the following coupled partial differential EM wave equations:
The code employs finite difference time domain using the Yee algorithm which can be read about in the following two online documents:
http://www.eecs.wsu.edu/~schneidj/ufdtd/ufdtd.pdf
http://www.eecs.wsu.edu/~schneidj/ufdtd/chap3.pdf
I want my source at the left hand side boundary to be a sinusoidal wave with parameters as such:
Ex(0,t) = E0 sin(2πft) for 0 ≤ t ≤ 1/f
The code updates the electric and magnetic field properties of the wave with each loop.
My initial code is as follows:
%FDTD Yee algorithm to solve coupled EM wave equations
clear
clc
G=50; %Specify size of the grid
f=10^3; %choose initial frequency of wave in hertz
e=1; %specify permitivity and permeability (normalised condition)
u=1;
Nt=150; %time steps
E0=1; %Electric Field initial amplitude
%Specify step sizes using corruant condition
c=3*10^8;
dx=0.01;
dt=dx/2*c;
%make constant terms
c1=-dt/(dx*e);
c2=-dt/(dx*u);
%create vgector place holders
Ex=zeros(1,G);
Hy=zeros(1,G);
%create updating loop
M=moviein(Nt);
for t=1:Nt
% Spatial Ex
for k=2:G-1
Ex(k)=Ex(k)+c1*(Hy(k)-Hy(k-1));
end
Ex(G)=0; %PEC boundary condition
%E Source at LHS boundary
Ex(1)=E0*sin(2*pi*f*t);
%Spatial Hy
for n=1:G-1
Hy(n)=Hy(n)+c2*(Ex(n)-Ex(n+1));
end
Hy(G)=0; %PMC boundary condition
plot(Ex);
M(:,t) = getframe;
end
movie(M,1);
Basically I want to create an updating movie which shows the sinusoidal wave propagating to the right hand side boundary coded as a perfect electrical conductor; therefore reflecting the wave, and then propagating back to the left hand side boundary coded as a perfect insulator; absorbing the wave.
The problems I have are as follows:
1) I'm not sure how to properly implement my desired source. It don't appear to be purely sinusoidal.
2) The wave I've coded begins to propagate but it quickly disappears for the majority of the simulation. I do not know why this is occurring
3) I do not know much about running a movie simulation and the plot oscillates as the solution is being run. How can I stop this?
Your wave attenuates because the diference equations are not correctly implemented; instead:
Ex(k)=Ex(k)+c1*(Hy(k)-Hy(k-1));
you should use
Ex1(k)=Ex(k)+c1*(Hy(k)-Hy(k-1));
and instead of:
Hy(n)=Hy(n)+c2*(Ex(n)-Ex(n+1));
you should use:
Hy1(n)=Hy(n)+c2*(Ex(n)-Ex(n+1));
and, in the end of the loop update the current "dataframe":
Hy = Hy1;
Ey = Ey1;
(you should take care also the boundary conditions).
If you want a fixed plot frame that doesn't change when your data changes, create first a axis where you can plot into, with a defined xmin/max and ymin/max, see http://www.mathworks.com/help/matlab/ref/axis.html
You should set the Courant number closer to 1 say 0.995. Thus delta_t = 0.995*delta_x/c.
Also assuming delta_x is in METRIC units then e and u should be in metric units.
I do not know about the specific coding language used but in c or c++ there is no need for intermediate variable Ey1 etc.
Also there should be at least 10 samples per wavelength for accuracy ( preferably 60). Thus wavelength = 60*delta_x and thus the frequency equals roughly of the order 10 to power of 9. Also, I think the sinesoidal source should be E0 * sin(2* pi * f * t * delta_t). You need to adjust your constants, and try it again