Fv = linspace(0, 1, fix(length(fft)/2)+1)*Fn; % One-Sided 'fft' Iv = 1:numel(Fv); % Index Vector The fft result to be plotted would then use ‘Fv’ and ‘Iv’ to create an equal-length fft vector. Use the ffts
x = linspace(0,1,20); t = linspace(0,0.5,5); sol = pdepe(m,@pdex1pde,@pdex1ic,@pdex1bc,x,t); % Extract the first solution component as u. u = sol(:,:,1); *** 답변 (0개) 이 질문에 답변하려면 로그인하십시오. ...
>> t=linspace(0,2*pi,100); >> y=5*t.*(sin(t).^2)-t.^2*(cos(t).^2); Error using * Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the number of rows in the second matrix. To perform elementwise multiplication, use...
y = linspace(-0.1,0.3,5) y = -0.1000 0 0.1000 0.2000 0.3000 x == y ans = 1 1 0 0 1 So, what happens when you do A = 5.8:0.1:6? How are the numbers created? And how can the following be explained? ThemeCopy A = 5.8:0.1:6; B = 5.8:0.1:6.1; A(2)-B(2) ans = 8....
Star Strider has a point too. However with the values you have, linspace does not actually pass very close to -2 or 2. (Try linspace(-10, -0.05)). At least on my system there are no out of bounds values involved. However if I use:
Fv = linspace(0, 1, NFFT/2+1)*Fn; Iv = 1:numel(Fv); TrFn = FTfiltered ./ FToriginal; figure plot(FV, abs(TrFn)) grid xlabel('Frequency') ylabel('Magnitude') title('Filter Transfer Function') The cutoff frequencies will be the points at whi...
I appreciate your help. I removed the . after sqrt, and I apologise for the insufficient info, I'm still new to this. The main body above is a function of v (labelled mainFunc) and I'm calling fzero in another script, using:
MATLAB Online에서 열기 1) Y=linspace(dy,dy*yNum*1000000,yNum); 2) X=linspace(dx,dx*xNum*1000000,xNum); 3) X=rot90(X,3); 댓글 수: 0 댓글을 달려면 로그인하십시오. 답변 (1개)
MATLAB Online에서 열기 Ran in: The‘rect’funciton does not exist in MATLAB, so I userectpulshere instead — %A1.2 T = 0.002; t = -0.01 : 0.0002 : 0.01;%---> I am not sure those time intervals... fc = 6000; g
When you called "ode45", you were only passing in a scalar value for the initial condition. However, it looks like your "odefun" expects a 3x1 solution vector at each time step and returns a 3x1 "dydt" vector. I don't know the details of the ODEs...