I am trying to implement convolution in MATLAB without using built in command. I have obtained a code from google and some how understood it but still there is some confusion especially regarding the highlighted line and more specifically about the red encircled term, i know that we are using nested for loops for limits of integral but i am unable to understand the logic behind the red encircled term, how is this term relating to our formula of convolution. If i remove this term,my results are not valid any more
Also in typical theory formula of convolution we see input is multiplied with shifted impulse or shifted input is mulltiplied with impulse but here they are being multiplied without any apparent shift
My code is below:
clear
x=[1 2]
h=[3 4 5]
N = length(x);
M = length(h);
length_out=N+M-1;
y=zeros(1,length_out);
for i = 1:N
for k = 1:M
y(i+k-1) = y(i+k-1) + h(k)*x(i);
end
end
Related
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
In my problem the linear modal is defined as the first term in the series expansion of sin(x) so:
sin(x) = x is the linear modal.
So with this, I then have to write
as a system of x' and y', so:
I tried the linear modal in Euler's method, with initial conditions X(1) = 1 and V(1)=0 (V would be y then following the last equations):
for i = 1:1000
V(i+1) = V(i)-(1.*s) ;
X(i+1) = V(i);
end
Where s is the step size. But apparently I'm supposed to get a circle when I plot V with respect to X which makes sense, but all I get is a straight line.
If I change it to:
for i = 1:1000
V(i+1) = V(i)-(X(i).*s) ;
X(i+1) = V(i);
end
With s=0.8 I get a spiral, which looks like a development but I'm no closer to the circular shape that I am expecting. I think I just need a fresh pair of eyes to see where perhaps an obvious error lies.
You are coding something drastically different than you wrote down in the equations. If you wrote down the equations with the same order and variable names, you would probably already see your error.
Also, the following things should have tipped you off:
The step size does not appear as a linear factor in both equations.
In your first attempt, X has no impact on V.
Nonetheless, here is a direct translation of your equations to code (and the Euler method):
for i = 1:1000
x(i+1) = x(i) + y(i)*s;
y(i+1) = y(i) + sin(x(i))*s;
end
If you really want to use the linear approximation, just replace sin(x(i)) by x(i). However, the main point of the linear approximation is that it’s nice for analytical solutions and theoretical analysis. Since you are solving the problem numerically, I do not see a need for it.
Finally, note that since you are integrating a Hamiltonian system, you will very quickly see the problems of the Euler method, i.e., you will always get a blatant spiral. Just imagine what would happen if you simulate the motion of the earth around the sun with tangential vectors:
I have the following differential equation which I'm not able to solve.
We know the following about the equation:
D(r) is a third grade polynom
D'(1)=D'(2)=0
D(2)=2D(1)
u(1)=450
u'(2)=-K * (u(2)-Te)
Where K and Te are constants.
I want to approximate the problem using a matrix and I managed to solve
the similiar equation: with the same limit conditions for u(1) and u'(2).
On this equation I approximated u' and u'' with central differences and used a finite difference method between r=1 to r=2. I then placed the results in a matrix A in matlab and the limit conditions in the vector Y in matlab and ran u=A\Y to get how the u value changes. Heres my matlab code for the equation I managed to solve:
clear
a=1;
b=2;
N=100;
h = (b-a)/N;
K=3.20;
Ti=450;
Te=20;
A = zeros(N+2);
A(1,1)=1;
A(end,end)=1/(2*h*K);
A(end,end-1)=1;
A(end,end-2)=-1/(2*h*K);
r=a+h:h:b;
%y(i)
for i=1:1:length(r)
yi(i)=-r(i)*(2/(h^2));
end
A(2:end-1,2:end-1)=A(2:end-1,2:end-1)+diag(yi);
%y(i-1)
for i=1:1:length(r)-1
ymin(i)=r(i+1)*(1/(h^2))-1/(2*h);
end
A(3:end-1,2:end-2) = A(3:end-1,2:end-2)+diag(ymin);
%y(i+1)
for i=1:1:length(r)
ymax(i)=r(i)*(1/(h^2))+1/(2*h);
end
A(2:end-1,3:end)=A(2:end-1,3:end)+diag(ymax);
Y=zeros(N+2,1);
Y(1) =Ti;
Y(2)=-(Ti*(r(1)/(h^2)-(1/(2*h))));
Y(end) = Te;
r=[1,r];
u=A\Y;
plot(r,u(1:end-1));
My question is, how do I solve the first differential equation?
As TroyHaskin pointed out in comments, one can determine D up to a constant factor, and that constant factor cancels out in D'/D anyway. Put another way: we can assume that D(1)=1 (a convenient number), since D can be multiplied by any constant. Now it's easy to find the coefficients (done with Wolfram Alpha), and the polynomial turns out to be
D(r) = -2r^3+9r^2-12r+6
with derivative D'(r) = -6r^2+18r-12. (There is also a smarter way to find the polynomial by starting with D', which is quadratic with known roots.)
I would probably use this information right away, computing the coefficient k of the first derivative:
r = a+h:h:b;
k = 1+r.*(-6*r.^2+18*r-12)./(-2*r.^3+9*r.^2-12*r+6);
It seems that k is always positive on the interval [1,2], so if you want to minimize the changes to existing code, just replace r(i) by r(i)/k(i) in it.
By the way, instead of loops like
for i=1:1:length(r)
yi(i)=-r(i)*(2/(h^2));
end
one usually does simply
yi=-r*(2/(h^2));
This vectorization makes the code more compact and can benefit the performance too (not so much in your example, where solving the linear system is the bottleneck). Another benefit is that yi is properly initialized, while with your loop construction, if yi happened to already exist and have length greater than length(r), the resulting array would have extraneous entries. (This is a potential source of hard-to-track bugs.)
I am trying to fit a line to some data without using polyfit and polyval. I got some good help already on how to implement this and I have gotten it to work with a simple sin function. However, when applied to the function I am trying to fit, it does not work. Here is my code:
clear all
clc
lb=0.001; %lowerbound of data
ub=10; %upperbound of data
step=.1; %step-size through data
a=.03;
la=1482/120000; %1482 is speed of sound in water and 120kHz
ep1=.02;
ep2=.1;
x=lb:step:ub;
r_sq_des=0.90; %desired value of r^2 for the fit of data without noise present
i=1;
for x=lb:step:ub
G(i,1)= abs(sin((a/la)*pi*x*(sqrt(1+(1/x)^2)-1)));
N(i,1)=2*rand()-1;
Ghat(i,1)=(1+ep1*N(i,1))*G(i,1)+ep2*N(i,1);
r(i,1)=x;
i=i+1;
end
x=r;
y=G;
V=[x.^0];
Vfit=[x.^0];
for i=1:1:1000
V = [x.^i V];
c = V \ y;
Vfit = [x.^i Vfit];
yFit=Vfit*c;
plot(x,y,'o',x,yFit,'--')
drawnow
pause
end
The first two sections are just defining variables and the function. The second for loop is where I am making the fit. As you can see, I have it pause after every nth order in order to see the fit.
I changed your fit formula a bit, I got the same answers but quickly got
a warning that the matrix was singular. No sense in continuing past
the point that the inversion is singular.
Depending on what you are doing you can usually change out variables or change domains.
This doesn't do a lot better, but it seemed to help a little bit.
I increased the number of samples by a factor of 10 since the initial part of the curve
didn't look sampled highly enough.
I added a weighting variable but it is set to equal weight in the code below. Attempts
to deweight the tail didn't help as much as I hoped.
Probably not really a solution, but perhaps will help with a few more knobs/variables.
...
step=.01; %step-size through data
...
x=r;
y=G;
t=x.*sqrt(1+x.^(-2));
t=log(t);
V=[ t.^0];
w=ones(size(t));
for i=1:1:1000
% Trying to solve for value of c
% c that
% yhat = V*c approximates y
% or y = V*c
% V'*y = V'*V * c
% c = (V'*V) \ V'*y
V = [t.^i V];
c = (V'*diag(w.^2)*V ) \ (V'*diag(w.^2)*y) ;
yFit=V*c;
subplot(211)
plot(t,y,'o',t,yFit,'--')
subplot(212)
plot(x,y,'o',x,yFit,'--')
drawnow
pause
end
It looks like more of a frequency estimation problem, and trying to fit a unknown frequency
with polynomial tends to be touch and go. Replacing the polynomial basis with a quick
sin/cos basis didn't seem to do to bad.
V = [sin(t*i) cos(t*i) V];
Unless you specifically need a polynomial basis, you can apply your knowledge of the problem domain to find other potential basis functions for your fit, or to attempt to make the domain in which you are performing the fit more linear.
As dennis mentioned, a different set of basis functions might do better. However you can improve the polynomial fit with QR factorisation, rather than just \ to solve the matrix equation. It is a badly conditioned problem no matter what you do however, and using smooth basis functions wont allow you to accurately reproduce the sharp corners in the actual function.
clear all
close all
clc
lb=0.001; %lowerbound of data
ub=10; %upperbound of data
step=.1; %step-size through data
a=.03;
la=1482/120000; %1482 is speed of sound in water and 120kHz
ep1=.02;
ep2=.1;
x=logspace(log10(lb),log10(ub),100)';
r_sq_des=0.90; %desired value of r^2 for the fit of data without noise present
y=abs(sin(a/la*pi*x.*(sqrt(1+(1./x).^2)-1)));
N=2*rand(size(x))-1;
Ghat=(1+ep1*N).*y+ep2*N;
V=[x.^0];
xs=(lb:.01:ub)';
Vfit=[xs.^0];
for i=1:1:20%length(x)-1
V = [x.^i V];
Vfit = [xs.^i Vfit];
[Q,R]=qr(V,0);
c = R\(Q'*y);
yFit=Vfit*c;
plot(x,y,'o',xs,yFit)
axis([0 10 0 1])
drawnow
pause
end
The following is Exercise 3 of a Numerical Analysis task I have to do as part of my university course on the subject.
Find an approximation of tomorrow's temperature based on the last 23
values of hourly temperature of your city ( Meteorological history for
Thessaloniki {The city of my univ} can be found here:
http://freemeteo.com)
You will approximate the temperature function with a polynomial of
2nd, 3rd and 4th degree, using the Least Squares method. Following
that, you will find the value of the function at the point that
interests you. Compare your approximations qualitatively and make a
note to the time and date you're doing the approximation on.
Maybe it's due to fatigue due to doing the first two tasks without break, or it's my lack of experience on numerical analysis, but I am completely stumped. I do not even know where to start.
I know it's disgusting to ask for a solution without even showing signs of effort, but I would appreciate anything. Leads, tutorials, outlines of the things I need to work on, one after the other, anything.
I'd be very much obliged to you.
NOTE: I am not able to use any MATLAB in-built approximation functions.
In general, if y is your data vector belonging to the times t, and c is the coefficient vector you're interested in, then you need to solve the linear system
Ac = y
in a least-squares sense, where
A = bsxfun(#power, t(:), 0:n)
In MATLAB you can do this with mldivide:
c = A\y(:)
Example:
>> t = 0 : 0.1 : 20; %// Define some times
>> y = pi + 0.8*t - 3.2*t.^2; %// Create some synthetic data
>> y = y + randn(size(y)); %// Add some noise for good measure
>>
>> n = 2; %// The order of the polynomial for the fit
>> A = bsxfun(#power, t(:), 0:n); %// Design matrix
>> c = A\y(:) %// Solve for the coefficient matrix
c =
3.142410118189416e+000
7.978077631488009e-001 %// Which works pretty well
-3.199865079047185e+000
But since you are not allowed to use any built-in functions, you can use this simple solution only to check your own outcomes. You'll have to write an implementation of the equations given on (for example) Wolfram's MathWorld.