Edit: MATLAB.
I'm trying to create a function which returns an array of the velocity per second for a free falling object. The input argument for the function is the height h. Also, if the object hits the ground before 1 second, it should return the velocity for the time it hits the ground.
Example: If the object falls from 80 meters, the function should return
v =
9.8 19.6 29.4 39.2
My attempt looks like this:
function freefall(h)
g = 9.8; % gravity acceleration
h_t = linspace(0,h); % the height should start from 0 to input height h
t = sqrt(2.*h_t/g); % time for the free fall
n=0;
if t < 1
velocity = sqrt(2*g*h)
disp(velocity)
else
for time = n+1 % An attempt to try making t into integers(tried floor(t) but didn't work)
v = g.*time
while time <= t % Don't want the time to exceed the amount of time it takes for object to land
disp(v)
end
end
end
end
The output just becomes v = 9.82, and I'm out of ideas to try and make this work.
Is this all you are trying to do?
tfinal = sqrt(2*h/g); % time to fall to ground
if tfinal < 1
velocity = g*tfinal; % velocity when hitting ground is less than 1 sec
else
velocity = g*(1:tfinal); % velocities at 1 sec intervals, not including hitting ground
end
If you always wanted to include the final velocity in the output, you could do this instead:
tfinal = sqrt(2*h/g); % time to fall to ground
t = 1:tfinal; % range of integer times spaced by 1 sec
if tfinal ~= fix(tfinal) % if final time is not an integer
t = [t,tfinal]; % append final time to time array
end
velocity = g*t; % velocities at integer times plus final time
I don't have matlab or scilab at the moment so I fiddled around using online scilab to write some test code. Here what I came up with:
function freefall(h)
g = 9.8;
t = floor(sqrt(2*h/g));
if t < 1
t = 1;
t_mat = linspace(1, t, t);
t_mat = g * t_mat;
return t_mat;
end
You can then call that function this way:
disp(freefall(80));
What's missing on your code is rather to limit the increments of linspace, for some odd reason you tried to compare a vector to a scalar (that if t < 1 on your code), and finally your function did not return an answer, instead it prints it.
¯\(ツ)/¯ correct me if I'm wrong, I haven't write any matlab code for like 6 years.
Note: that online scilab doesn't seem to support function so, you have to use bare code without the function block. You can replace that return with disp(t_mat).
Related
I have some synthetic signal data to cross correlate before moving on with real data but the xcorr function is returning an odd result.
I have added a delay onto either the start (delay =100) or end of the respective data sets which are the same size (2101 x 1 double) and put them through xcorr which has returned an array (4201 x 1 double).
The result has NaNs between rows 101 and 2205 and then a ring down effect after that start at an order of x10^5. I have now attached pictures of the result
Can anyone please offer some suggestions on how to correct what I am attempting to do? I am expecting to be able to plot the result and see a spike at the delay that I have set.
Thanks
EDIT:
Short example code
X = [-4.99 -0.298 4.95 12.06 15.76 18.86 19.00 17.82 14.35 11.77 6.71 0.80 -5.07 -11.79 -15.34 -18.60 -18.56 -19.31 -14.37 -11.51 -5.04];
Y = [14.13 18.48 7.53 -3.41 -8.41 -13.40 -15.37 -17.34 -16.83 -16.33 -12.21 -8.09 -8.80 -9.52 3.90 17.31 17.52 17.72 17.73 17.75 16.90];
N = length(X);
delay = 2;
% set up two new arrays which will have random noise at ends
Xx = zeros(N+delay,1);
Yx = zeros(N+delay,1);
for i=1:N+delay
if i<=delay
Xx(i) = rand;
elseif i>delay
Xx(i) = X(i-delay);
end
end
for i=1:N+delay
if i<=N
Yx(i) = Y(i);
elseif i>N
Yx(i) = rand;
end
end
C = xcorr(Xx,Yx);
I'm trying to calculate the DT value from a model I set up on Sim4Life. Firstly, i'd like to say that I am a complete beginner and I am trying to understand how programming works in general.
Now, I have a function with some constants and two variables, the one being time Dt (starting from 1 sec to 900 secs) and the other being the initial DT_i value. I want to calculate the increase of temperature for every second and create a loop that replaces the DT_i value with the DT_1_i value and also calculates the increased temperature DT_i_1. The function looks like this: DT_1_i=DT_i+Dt.
I know it is a very simple problem but I couldn't work my way through other similar questions. Any help would be appreciated.
Temperature variation:
You need initial temperature variation , I used 0
T(i+1) stands for Next temperature variation
T(i) stands for present temperature variation
i stands for time step, dt
Read through comment in my code
Time
Use for loop to set the time for i = 1 : 900 %Temperature increase end
i =1:900 just means
first run use time = 1s,
second run time = 1+1 = 2
so on till 900
The code is as follow
% Initial Temperature variation is set to zero, unless you have some data
d = 1.3;
c = 3.7;
S_i = 3*10^3;
t_reg = 900;
%Time
t = 1:900;
% Length oftime to help me know the size of the variable needed to
% initialize
l = length(t);
% Initialize variable that used to store DT it helps speed up
% comutation
% Initial Temperature variation is set to zero, unless you have some data
DT = zeros(1, l);
for i = 1:900
% the value of i represent dt, first run i = 1, dt = 1, second run
% i = 2 and dt = 2 so on
if i == 900
%do nothing already reached the last index 900, i+1 = 901 will be
%out of range
else
DT(i+1) = DT(i) + (i./t_reg).*(d.*sqrt(c*S_i)-DT(i+1));
end
end
I have a mathematical equation that describes a dynamical system as
The parameters are defined as follows
k1=1; S=1; Kd=1; p=2; tau=10; k2=1; ET=1; Km=1;
I coded the system as
y(1) = 1; % based on the y-axes starting point in the last figure
y(2) = y(1) + k1*S*Kd^p/(Kd^p + y(1)^p) - k2*ET*y(1)/(Km + y(1)); % to avoid errors
for t=1:100
y(t+1) = y(t+1) + (k1*S*Kd^p/(Kd^p + y(t)^p) - k2*ET*y(t+1)/(Km + y(t+1)));
end
plot(y);
Note that I did not use tau=10 for simplicity and instead used a delayed version by 1 instead of 10 (because I am not sure how to insert a delay of 10)
And obtained the following result
However, I need to obtain this
Can anyone help me rectify the mistake in my code?
Thanking you in advance.
If we assume that for Y(t) = 0 for t < 0 then you're code could be modified to produce a similar plot. However, it looks like the plot you are looking to generate uses different initial conditions. If you're just looking to measure Tc then it appears that the signal stabilizes with the period you're looking for.
k1=1; S=1; Kd=1; p=2; tau=10; k2=1; ET=1; Km=1;
% time step size (tau MUST be divisible by dt to ensure proper array indexing)
dt = 0.01;
% time series
t = -10:dt:100;
% initialize y to all zeros so that y(t)=0 for all t<0 (initial condition)
y = zeros(size(t));
% Find starting and ending indexes to iterate from t=0 to t=100-dt
idx0 = find(t == 0);
idx1 = numel(t)-1;
% initial condition y(0) = 1
y(idx0) = 1;
for n = idx0:idx1
% The indexing used here ensures the following equivalences.
% y(n+1) = y(t+dt)
% y(n) = y(t)
% y(n - round(tau/dt)) = y(t-tau)
%
% Note that (y(t+dt)-y(t))/dt is approximately y'(t)
% Solving for y(t+dt) we get the following formula
y(n+1) = y(n) + dt*((k1*S*Kd^p/(Kd^p + y(n - round(tau/dt))^p) - k2*ET*y(n)/(Km + y(n))));
end
% plot y(t) for t > 0
plot(t(t>0),y(t>0));
Result
Seeing as things stabilize we can take the values in one of the periods and use those for the initial conditions and we get.
Edit: To elaborate, the function contains a delay of 10 which means that instead of just a single initial condition at y(0), we also need to initialize all values from t=-10 to 0. In the code posted in this answer I arbitrarily assumed that y(t) = 0 for t < 0 and y(0) = 1 because I don't know otherwise. Once we run the code and see that the signal becomes periodic we can borrow the values from one of these periods to use those as the initial conditions.
From the diagram you posted we can use our intuition to guess that, before time 0, the signal probably looks something like the region highlighted in the figure below.
If, rather than using zero to initialize y at y < 0, we copy the values in the red highlighted region, then we get a plot that is more like what you desire.
To get the plot shown above I ran the script once, then found the indices in y for the part I wanted to use as initial conditions, then copied those into a new array.
init_cond = y(7004:8004);
Then I changed script to use this array as the initial condition and changed the initial y values to
y = zeros(size(t));
y(1:1001) = init_cond;
and ran the modified script again.
Edit 2: The built-in function dde23 appears to be applicable for your problem. To see an example run the command edit ddex1 in the command window.
So I am trying to go through a for loop that will increment .1 every time and will do this until the another variable h is less than or equal to zero. Then I am suppose to graph this h variable along another variable x. The code that I wrote looks like this:
O = 20;
v = 200;
g = 32.2;
for t = 0:.1:12
% Calculate the height
h(t) = (v)*(t)*(sin(O))-(1/2)*(g)*(t^2);
% Calculate the horizontal location
x(t) = (v)*(t)*cos(O);
if t > 0 && h <= 0
break
end
end
The Error that I keep getting when running this code says "Attempted to access h(0); index must be a positive integer or logical." I don't understand what exactly is going on in order for this to happen. So my question is why is this happening and is there a way I can solve it, Thank you in advance.
You're using t as your loop variable as well as your indexing variable. This doesn't work, because you'll try to access h(0), h(0.1), h(0.2), etc, which doesn't make sense. As the error says, you can only access variables using integers. You could replace your code with the following:
t = 0:0.1:12;
for i = 1:length(t)
% use t(i) instead of t now
end
I will also point out that you don't need to use a for loop to do this. MATLAB is optimised for acting on matrices (and vectors), and will in general run faster on vectorised functions rather than for loops. For instance, your equation for h could be replaced with the following:
O = 20;
v = 200;
g = 32.2;
t = 0:0.1:12;
h = v * t * sin(O) - 0.5 * g * t.^2;
The only difference is that you have to use the element-wise square (.^2) rather than the normal square (^2). This means that MATLAB will square each element of the vector t, rather than multiplying the vector t by itself.
In short:
As the error says, t needs to be an integer or logical.
But your t is t=0:0.1:12, therefore a decimal value.
O = 20;
v = 200;
g = 32.2;
for t = 0:.1:12
% Calculate the height
idx_t = 1:numel(t);
h(idx_t) = (v)*(t)*(sin(O))-(1/2)*(g)*(t^2);
% Calculate the horizontal location
x(idx_t) = (v)*(t)*cos(O);
if t > 0 && h <= 0
break
end
end
Look this question's answer for more options: Subscript indices must either be real positive integers or logical error
I'm trying to set up a piecewise transfer function of a filter in matlab to get its impulse response. I have the code below:
function H = H(w)
H = zeros(size(w)); % Preallocating enough memory for y
nd = 0;
region1 = (abs(w)<(pi/4)) & (abs(w)>(pi/8)) ; % First interval
H(region1) = exp((-(w(region1))*1i*nd));
region2 = (abs(w)<(7*pi/8)) & (abs(w)>(5*pi/8)); % Second interval
H(region2) = exp((-0.5*(w(region1))*1i*nd));
region3 = ~(abs(w)<(pi/4)) & (abs(w)>(pi/8)) & ~(abs(w)<(7*pi/8)) & (abs(w)>(5*pi/8)) ; % Third interval
H(region3) = 0;
But it gives me this error, when I try to run:
In an assignment A(I) = B, the number of elements in B and I must be the same.Error
in H (line 9)
H(region2) = exp((-0.5*(w(region1))*1i*nd));
Am I going about this the right way or is there an easier way to do something like this?
I think the problem is that:
H(region2) = exp((-0.5*(w(region1))*1i*nd));
Should be:
H(region2) = exp((-0.5*(w(region2))*1i*nd));
Where region1 is corrected as region2.
Also, nd is always 0.
You ask if you're going about it the right way, seems decent enough to me as long as you realize the frequency response between the points you specify could be all over the place, or not depending on the transitions.