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I need to generate two signals which in the end I want to connect. The problem is that the end condition of the 1st signal can be quite different compared to the initial conditions in my 2nd signal. Subsequently it can result in a sudden and unrealistic jump in my final signal. Final signal is the 2 connected signals.
How could I smooth the connection in my final signal?
Thanks!
What about some sort of cross-fading:
S1 = rand(1000,1);
S2 = rand(1000,1) + 1;
%\\ cross-fade over last 200 elements
n = 200;
W = linspace(1,0,n)'; %'
S1(end-n+1:end) = S1(end-n+1:end).*W;
S2(1:n) = S2(1:n).*(1-W);
S12 = zeros(size(S1,1) + size(S2,1) - n, 1);
S12(1:size(S1,1)) = S1;
S12(end-size(S1,1)+1:end) = S12(end-size(S1,1)+1:end) + S2;
This was using a linear weighting for the fading, you might choose something else but I think this will sort of work.
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I need to write a MATLAB function that coverts from RBG to ych1ych2.
Y = 0.299\*R + 0.587\*G +0.114\*B
Ch1 = R - (G+B)/2
Ch2 = (√(3)/2) \* (B-G)
To convert RGB to the luma, chroma cyan-red, chroma green-blue system of Carron (YCh1Ch2), the conversion matrix you need is
rgb2ych1ch2 = [ 0.299 0.587 0.114 ; 1 -0.5 -0.5 ; 0 -sqrt(3)/2 sqrt(3)/2 ];
(This is just the three equations you present converted directly to matrix form.)
You can then convert by multiplying this matrix with your rgb values:
ych1ch2 = rgb2ych1ch2 * rgb;
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I have this trigonometric equation;
cos(2*pi*50*t)+cos(2*pi*100*t)
I want to graphic of equation and I want to find field for a period. How can I do?
Graph:
>> f = #(t) cos(2*pi*50*t) + cos(2*pi*100*t);
>> x = linspace(0, 1/50, 100);
>> y = f(x);
>> plot(x,y)
Area over 1 period:
>> integral(f, 0, 1/50)
or just do it manually:
∫ ( cos(2π·50t) + cos(2π·100t) ) dt =
-1/2π·( 1/50·sin(2π·50t) + 1/100·sin(2π·100t) )
which, evaluated between 0 and 1/50, equals 0.
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I'd like to write the Matlab code for the following equation:
z(k)=lamda*x(k)+(1-lamda)*z(k-1)
lamda can be of any value. x is a 1000x22 matrix. Z(0)=0.
Can anyone help me please?
You can use iteration function.same thing like this
function z = itrationFunctio(k,x,lambda)
if(k == 0)
z = 0;
else
z = lambda*x+(1-lambda)*itrationFunctio((k-1),x,lambda);
end
and in your code just call itrationFunctio(k,x,lambda).
Does that vectorized solution work for you?
% parameters
x = rand(10,1);
lambda = 2;
% init z, normalize with (1-lambda)
z(2:numel(x)) = lambda/(1-lambda)*x(2:end);
% cumsum, denormalize with (1-lambda)
z = cumsum(z)*(1-lambda)
However I don't get why your x is a matrix and not a vector. What is z supposed to be, in what dimension works k? So in case xrepresents a couple of vectors you want to calculate in parallel, that could work:
% parameters
x = rand(1000,22);
lambda = 2;
% init z, normalize with (1-lambda)
z(2:size(x,1),:) = lambda/(1-lambda)*x(2:end,:);
% cumsum, denormalize with (1-lambda)
z = cumsum(z,1)*(1-lambda)
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I have a function (the hill function) when X and K are the same (x==k) the output should be 0.5, when I test the function it gives a result of 0.5, when I try to plot it, I do not get 0.5 for my Y. Can anyone explain what I am doing wrong?
n = 1;
k = 1;
x = [0:0.01:2].';
y = (x.^n)/((x.^n)+(k.^n));
plot(x,y);
n = 1;
k = 1;
x = [0:0.01:2].';
y = (x.^n)./((x.^n)+(k.^n));
plot(x,y);
You missed a dot before the division.
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Given the following code:
t = -5:.1:5;
w = pi;
x = zeros(101,1);
Xt = zeros(101,1);
for i = 1 : 101;
x = exp((-3*t)+(-1i*w*t));
Xt = trapz(t, x);
end
disp (length(x))
disp (length(Xt))
disp (Xt)
The values of Xt do not change, which is a problem.
How should this be coded in order for Xt to change when x is changed?
Side note:
Xt(i) = trapz(t,x);
Reduces the vector from length 101 to length 1 and therefore cannot be used.
i am not sure if this is what you wanted. anyways while working on imaginary numbers it is always a good idea to not use i and j as common variables, just to avoid confusion (IMO)
xt = zeros(101,1);
x = exp((-3.*t)+(-1i*w.*t));
for k=2:101
xt(k)=trapz(t(1:k),x(1:k));
end