I am sorry If I say some thing silly.Please forgive me:
I am trying to convert Matlab code(given below) to VHDL code,using HDL coder.It contains a function called sum.But when I try to convert the code it gives me error :
code generation only supports SumModes 'SpecifyPrecision' and
'KeepLSB' for 'SUM' when the size of the inputs can vary at run-time.
But the thing is I have never used functions before.Can any one please help me on it.How should change my code to convert it to VHDL.It would be really nice!
function y = fcn(n,y1,y2)
n=10;
x1c=zeros(2*n-1,1);
for i=1:2*n-1
if(i>n)
j1=1;
k1=2*n-i;
j2=i-n+1;
k2=n;
else
j1=n-i+1;
k1=n;
j2=1;
k2=i;
end
x1c(i)=sum((y1(j1:k1)).*y2(j2:k2));
end
x1c=flipud(x1c).';
y=x1c;
This is cross correlation of y1 and y2. y1 and y2 are two vectors of same length and n is length of y1. I am really stuck please help!
Thanks in advance!
#Haider: Take a look at
n=10;
y1 = 1:n;
y2 = n:-1:1;
x1c=zeros(1,2*n-1);
for i=1:2*n-1
y1_temp = zeros(1,2*n-1);
y2_temp = zeros(1,2*n-1);
if(i>n)
j1=1;
k1=2*n-i;
j2=i-n+1;
k2=n;
else
j1=n-i+1;
k1=n;
j2=1;
k2=i;
end
y1_temp(j1:k1) = y1(j1:k1);
y2_temp(j1:k1) = y2(j2:k2);
x1c(i)=sum(y1_temp.*y2_temp);
end
I compared the result with the Matlab xcorr function, and it seems the vector is reversed. Does this solve the error?
Related
I am trying to solve a particular system of ODE's dF/dt = A*F, F_initial = eye(9). Being a Matlab novice, I am trying to somehow use the implemented ode45 function, and I found useful advises online. However, all of them assume A to be constant, while in my case the matrix A is a function of t, in other words, A changes with each time step.
I have solved A separately and stored it in a 9x9xN array (because my grid is t = 0:dt:2, N=2/dt is the number of time-steps, and A(:,:,i) corresponds to it's value at the i-th time step). But I can't implement this array in ode45 to eventually solve my ODE.
Any help is welcomed, and please tell me if I have missed anything important while explaining my problem. Thank you
First of all, F must be a column vector when using ode45. You won't ever get a result by setting F_initial = eye(9), you'd need F = ones(9,1).
Also, ode45 (documentation here, check tspan section) doesn't necessarily evaluate your function at the timesteps that you give it, so you can't pre-compute the A matrix. Here I'm going to assume that F is a column vector and A is a matrix which acts on it, which can be computed each timestep. If this is the case, then we can just include A in the function passed to ode45, like so:
F_initial = ones(9,1);
dt = 0.01;
tspan = 0:2/dt:2;
[t, F] = ode45(#(t,F) foo(t, F, Ainput), tspan, F_initial);
function f = foo(t, F, Ainput)
A = calculate_A(t, Ainput);
f = A*F;
end
function A = calculate_A(t, Ainput)
%some logic, calculate A based on inputs and timestep
A = ones(9,9)*sqrt(t)*Ainput;
end
The #(x) f(x,y) basically creates a new anonymous function which allows you to treat y as a constant in the calculation.
Hope this is helpful, let me know if I've misunderstood something or if you have other questions.
Can anybody help me with this assignment please?
I am new to matlab, and passing this year depends on this assignment, i don't have much time to explore matlab and i already wasted alot of time trying to do this assignment in my way.
I have already wrote the equations on the paper, but transfering the equations into matlab codes is really hard for me.
All i have for now is:
syms h
l = (0.75-h.^2)/(3*sqrt((5*h.^2)/4)); %h is h_max
V_default = (h.^2/2)*l;
dv = diff(V_default); %it's max. when the derivative is max.
h1 = solve( dv ==0);
h_max = (h1>0);
l_max = (0.75-h_max.^2)/(3*sqrt((h_max/2).^2+(h_max.^2)));
V_max = ((h_max.^2)./(2.*l_max));
but it keep give me error "Error using ./
Matrix dimensions must agree.
Error in triangle (line 9)
V_max = ((h_max.^2)./(2.*l_max)); "
Not really helping with the assignment here, but with the Matlab syntax. In the following line:
l_max = (0.75-h_max.^2)/(3*sqrt((h_max/2).^2+(h_max.^2)));
you're using / that is a matrix divide. You might want to use ./ which will divide the terms element by element. If I do this
l_max = (0.75-h_max.^2) ./ (3*sqrt((h_max/2).^2+(h_max.^2)));
then your code doesn't return any error. But I have no idea if it's the correct solution of your assignment, I'll leave that to you!
In line 5, the result h1 is a vector of two values but the variable itself remains symbolic, from the Symbolic Math Toolbox. MATLAB treats such variables slightly different. For that reason, the line h_max = (h1>0) doesn't really do what you expect. As I think from this point, you are interested in one value h_max, I would convert h1 to a regular MATLAB variable and change your code to the following:
h1 = double(solve( dv ==0)); % converts symbolic to regular vectors
h_max = h1(h1>0); % filters out all negative and zero values
l_max = (0.75-h_max.^2)/(3*sqrt((h_max/2).^2+(h_max.^2)));
V_max = ((h_max.^2)./(2.*l_max));
EDIT.
If you still have error, it means solve( ...) returns more than 1 positive values. In this case, as suggested, use dotted operations, such as ./ but the results in l_max and V_max will not be a single value but vectors of the same size as h_max. Which means you don't have one max Volume.
For a project I need to understand a matlab code, but as I am quite new I dont really understand what is happening. I have a function file and a script file.
Function:
function dxdt = sniffer_ode(t,x,par,tu)
X = x(1);
R = x(2);
k1 = par(1);
k2 = par(2);
k3 = par(3);
k4 = par(4);
S = interp1(tu(:,1),tu(:,2),t);
dxdt(1) = k3*S-k4*X;
dxdt(2) = k1*S-k2*X*R;
dxdt = dxdt(:); %dxdt should be column
and the script file:
%sniffer
close all
%initial conditions:
X0=0; R0=0;
x0=[X0 R0];
%parameters:
k1=1; k2=1; k3=1; k4=1;
par=[k1 k2 k3 k4];
%input:
tu=[ 0 , 0
1 , 0
1.01, 1
20 , 1];
[t,x] = ode45(#sniffer_ode,[0 20],x0, [],par,tu);
plot(t,x);
So the question is: What is happening? I also need to plot S in the same figure as X and R. How do I do this?
I appreciate your help!
This is a really basic Matlab question. There is tons of information about your requested topic. I think these slides will help you on the right path.
However, a quick explanation; the first code you provide is the function which describes your ordinary differential equation. This function always has to be of the form x' = f(t,x,...). Herein t is the time and x is the state. After the state (on the place of the dots ...) you can define other input parameters, such as is being done in your ode function. Furthermore, the interp1 function interpolates the data provided.
The second code you provide is the code you start within Matlab. Parameters are defined, after which the ordinary differential equation is solved and plotted.
If you have any further questions I would recommend that you first try to find your answer using a search engine.
I have been going round and round in circles trying to get Matlab to solve the resonator circuit equation with a time varying input voltage. It works just fine as long as all the in arguments of the derivative functions are scalar values.
I have gone through others questions and found answers suggesting anonymous functions or using interpolation. When I try to implement these suggestions, I get a return error that tells me I do not have enough initial conditions to match the output of ode function or the matrices being concatenated do not match..
Here is my code:
%% Define Parameters
f0=1494.72e6;
Q=80;
R=1;
L=(Q*R)/(2*pi*f0);
C=1/((2*pi*f0)^2*L);
tp=71e-9;
N=2^16;
n=(0:N-1);
TT=1e-6;
h=TT/N;
t=n*h;
start_pulse=1;
end_pulse=round(tp/h);
V=zeros(size(t));
%% Create Voltage Pulse
for ii=1:length(n);
if ii>=start_pulse && ii<=end_pulse
V(ii)=sin(2*pi*f0*t(ii));
end
end
%%
tspan=[0 TT];
x0=[0 0];
sol=ode45(#ode,tspan,x0,[],V);
int=(0:h:TT);
sint=deval(sol,int);
plot(int,sint*C);
MY ode funtion is the following:
function [ dx ] = ode( t,x,V)
f0=1494.72e6;
Q=80;
R=1;
L=(Q*R)/(2*pi*f0);
C=1/((2*pi*f0)^2*L);
dx1 = x(2);
dx2 =((-x(1)./(L*C))-(R*x(2)./L)-(V./(L*C)));
dx = [dx1; dx2];
end
As you can see, L,C,and R all are scalar values. If I replace 'V' in dx2 with '1', the program runs just fine. I need to change V to be the matrix defined above.
Any help at all would be greatly appreciated!!! Thanks in advance!!! :)
I've got a silly problem, I'm looking to take a few data points, fit a polynomial function through it and then differentiate that function to get that particular functions optimal point. As such I have done some reading online and I've used the Matlab 'spline' function. Here is some code:
a = [50; 100; 150;200;250;300;350]
b = [56;23;22;18;14;15;21]
y = spline(a,b)
But when I used diff(y) I get the following error:
??? Error using ==> diff
Function 'diff' is not supported for class 'struct'.
I'm not too familiar with Matlab, so any help would really be appreciated
As per comments:
y = polyfit(a,b,2)
syms x
g = y(1)*x^2 + y(2)*x + y(3)
diff(g)
and you get the derivative of the function g. Much thanks to the guys in the comment section!