tv = linspace(0, L-1, L)/Fs; % Time Vector One way to compute the frequency vector: 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
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:
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 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
Hi, I need to draw the Second Order Response function for Zeta = 0.5, 1 ,2 and tau Tau = 0.5: The functions are defined as: Here is what I did: loadxival.txt; tau=0.5; v=xival; n=length(v); t=linspace(-10,10); fori=1:n ...
Open in MATLAB Online w1=104; s=0.3; u=1; >> w2=w1*(1-(s^2))^(1/2) w2 = 99.2097 >> y=(s)/(1-(s^2)) y = 0.3297 >> t=linspace(0,0.2,100); >> x=u*(1-exp(-s*w1*t)*(cos(w2*t)-y*sin(w2*t)));
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...