Dimpulse Function in Matlab - matlab

Im working with Signals and i need to determine the impulse of a Discrete Signal using Matlab dimpulse function.
For some reason i dont have dimpulse function on my Matlab...
Does anyone knows if its really an Matlab function or a custom made function? if its an Matlab one which toolbox is it?
i tried the commands which dimpulse and it says it doesn't find it....I need this function for Discret Signals...
I´m running Matlab 2012b on a Mac if that helps ...
Thanks ...

Dimpulse is not an official Matlab function. However, Google turned up a few universities that have written the function, such as here:
function [yout,x,n] = dimpulse(a,b,c,d,iu,n)
%DIMPULSE Impulse response of discrete-time linear systems.
% DIMPULSE(A,B,C,D,IU) plots the response of the discrete system:
%
% x[n+1] = Ax[n] + Bu[n]
% y[n] = Cx[n] + Du[n]
%
% to an unit sample applied to the inputs IU. The number of
% points is determined automatically.
%
% DIMPULSE(NUM,DEN) plots the impulse response of the polynomial
% transfer function G(z) = NUM(z)/DEN(z) where NUM and DEN contain
% the polynomial coefficients in descending powers of z.
%
% DIMPULSE(A,B,C,D,IU,N) or DIMPULSE(NUM,DEN,N) uses the user-
% supplied number of points, N. When invoked with left hand
% arguments,
% [Y,X] = DIMPULSE(A,B,C,D,...)
% [Y,X] = DIMPULSE(NUM,DEN,...)
% returns the output and state time history in the matrices Y and X.
% No plot is drawn on the screen. Y has as many columns as there
% are outputs and X has as many columns as there are states.
%
% See also IMPULSE, STEP, INITIAL, LSIM.
% J.N. Little 4-21-85
% Revised CMT 7-31-90, ACWG 5-30-91, AFP 10-1-94
% Copyright 1986-2002 The MathWorks, Inc.
% $Revision: 1.15 $ $Date: 2002/04/04 15:15:36 $
%warning(['This calling syntax for ' mfilename ...
% ' will not be supported in the future: use IMPULSE(SYS,...) instead.'])
ni = nargin;
no = nargout;
if ni==0,
eval('exresp(''dimpulse'')')
return
end
error(nargchk(2,6,ni))
Ts = -1;
% Determine which syntax is being used
switch ni
case 2
if size(a,1)>1,
% SIMO syntax
a = num2cell(a,2);
den = b;
b = cell(size(a,1),1);
b(:) = {den};
end
sys = tf(a,b,Ts);
n = [];
case 3
% Transfer function form with time vector
if size(a,1)>1,
% SIMO syntax
a = num2cell(a,2);
den = b;
b = cell(size(a,1),1);
b(:) = {den};
end
sys = tf(a,b,Ts);
n = c;
case 4
% State space system without iu or time vector
sys = ss(a,b,c,d,Ts);
n = [];
otherwise
% State space system, with iu but w/o time vector
if min(size(iu))>1,
error('IU must be a vector.');
elseif isempty(iu),
iu = 1:size(d,2);
end
sys = ss(a,b(:,iu),c,d(:,iu),Ts);
if ni<6,
n = [];
end
end
if no==1,
yout = impulse(sys,n);
yout = yout(:,:);
elseif no>1,
[yout,t,x] = impulse(sys,n);
yout = yout(:,:);
x = x(:,:);
n = length(t);
else
impulse(sys,n)
end
% end dimpulse

Related

Obtaining steady state solution for spring mass dashpot system

I'm trying to solve the following problem using MATLAB but I faced multiple issues. The plot I obtained doesn't seem right even though I tried to obtain the steady-state solution, I got a plot that doesn't look steady.
The problem I'm trying to solve
The incorrect plot I got.
and here is the code
% system parameters
m=1; k=1; c=.1; wn=sqrt(k/m); z=c/2/sqrt(m*k); wd=wn*sqrt(1-z^2);
% initial conditions
x0=0; v0=0;
%% time
dt=.001; tMax=8*pi; t=0:(tMax-0)/999:tMax;
% input
A=1
omega=(2*pi)/10
F=A/2-(4*A/pi^2)*cos(omega*t); Fw=fft(F);
F=k*A*cos(omega*t); Fw=fft(F);
% normalize
y = F/m;
% compute coefficients proportional to the Fourier series coefficients
Yw = fft(y);
% setup the equations to solve the particular solution of the differential equation
% by the method of undetermined coefficients
N=1000
T=10
k = [0:N/2];
w = 2*pi*k/T;
A = wn*wn-w.*w;
B = 2*z*wn*w;
% solve the equation [A B;-B A][real(Xw); imag(Xw)] = [real(Yw); imag(Yw)] equation
% Note that solution can be obtained by writing [A B;-B A] as a scaling + rotation
% of a 2D vector, which we solve using complex number algebra
C = sqrt(A.*A+B.*B);
theta = acos(A./C);
Ywp = exp(j*theta)./C.*Yw([1:N/2+1]);
% build a hermitian-symmetric spectrum
Xw = [Ywp conj(fliplr(Ywp(2:end-1)))];
% bring back to time-domain (function synthesis from Fourier Series coefficients)
x = ifft(Xw);
figure()
plot(t,x)
Your forcing function doesn't look like the triangle wave in the problem. I edited the %% time section of your code into the following and appeared to give a steady state response.
%% time
TP = 10; % forcing time period (10 s)
dt=.001;
tMax= 3*TP; % needs to be multiple of the time period
t=0:(tMax-0)/999:tMax;
% input
A=1; % Forcing amplitude
omega=(2*pi)/TP;
% forcing is a triangle wave
% generate a triangle wave with min/max values of 0/1.
F = 0*t;
for i = 1:length(t)
if mod(t(i), TP) <= TP/2
F(i) = mod(t(i), TP)/(TP/2);
else
F(i) = 2 - mod(t(i), TP)/(TP/2);
end
end
F = F*A; % scale triangle wave by amplitude
% you can also use MATLAB's sawtooth() function if you have the signal
% processing toolbox

MATLAB's lsim() vs for-loop Simulation // Different results for the same system

I've spent quite some time trying to simulate a simple SISO system using two approaches:
1) Using lsim() in MATLAB
2) By writing down the difference equations myself and iterate over them in a loop.
I was never able to get the same simulation results from both approaches, and I have no idea what I am doing wrong.
I stacked my code in a single m-file so it's easier to follow. Here is the code:
function main()
clear all
clc
simulateUsing_lsim()
simulateUsing_loop()
end
%%%%%% Simulating using lsim %%%%%%%
function simulateUsing_lsim()
% Define the continuous-time closed-loop system
P = getContPlant();
[Kp,Ki,Kd] = get_PIDgains();
C = pid(Kp,Ki,Kd);
clSys_cont = feedback(C*P,1);
% Define the discrete-time closed-loop system
hk = get_sampling_time();
clSys_disc = c2d(clSys_cont,hk);
% Generate the reference signal and the time vector
[r,t] = getReference(hk);
%% Simulate and plot using lsim
figure
lsim(clSys_disc,r,t)
%% Finding and plotting the error
y = lsim(clSys_disc,r);
e = r - y;
figure
p = plot(t,e,'b--');
set(p,'linewidth',2)
legend('error')
xlabel('Time (seconds)')
ylabel('error')
% xlim([-.1 10.1])
end
%%%%%% Simulating using loop iteration (difference equations) %%%%%%%
function simulateUsing_loop()
% Get the cont-time ol-sys
P = getContPlant();
% Get the sampling time
hk = get_sampling_time();
% Get the disc-time ol-sys in SS representation
P_disc = ss(c2d(P,hk));
Ad = P_disc.A;
Bd = P_disc.B;
Cd = P_disc.C;
% Get the PID gains
[Kp,Ki,Kd] = get_PIDgains();
% Generate the reference signal and the time vector
[r,t] = getReference(hk);
%% Perform the system simulation
x = [0 0]'; % Set initial states
e = 0; % Set initial errors
integral_sum = 0; % Set initial integral part value
for i=2:1:length(t)
% Calculate the output signal "y"
y(:,i) = Cd*x;
% Calculate the error "e"
e(:,i) = y(:,i) - r(i);
% Calculate the control signal vector "u"
integral_sum = integral_sum + Ki*hk*e(i);
u(:,i) = Kp*e(i) + integral_sum + (1/hk)*Kd*(e(:,i)-e(:,i-1));
% Saturation. Limit the value of u withing the range [-tol tol]
% tol = 100;
% if abs(u(:,i)) > tol
% u(:,i) = tol * abs(u(:,i))/u(:,i);
% else
% end
% Calculate the state vector "x"
x = Ad*x + Bd*u(:,i); % State transitions to time n
end
%% Subplots
figure
plot(t,y,'b',t,r,'g--')
%% Plotting the error
figure
p = plot(t,e,'r');
set(p,'linewidth',2)
legend('error')
xlabel('Time (seconds)')
ylabel('error')
end
function P = getContPlant()
s = tf('s');
P = 1/(s^2 + 10*s + 20);
end
function [Kp,Ki,Kd] = get_PIDgains()
Kp = 350;
Ki = 300;
Kd = 50;
end
function hk = get_sampling_time()
hk = 0.01;
end
function [r,t] = getReference(hk)
[r,t] = gensig('square',4,10,hk);
end
I got the plant model P and its PID controller from this page (see equation 10), where the system is simulated against a step reference and the result looks pretty much exactly like the lsim() result (just for a single step peak).
However, the result of simulating the system using lsim() is this:
whereas, using the for loop, I got this performance:
I would highly appreciate any help or clarification why I am getting different results.

How to debug this Matlab code to model glider descent?

I have this Matlab project but for some reason I just cannot stop thinking about it because I could not get it to work.
Objective:
This is a MATLAB script that would calculate the change of pressure, temperature and density of a glider that is being dropped from 10000 feet. As it falls, we want to use those new values calculated and then plugged in a function that has 4 equations that need to be differentiated at every point using ode45 as well as the new values of P T and Rho.
Here is the main code:
% HouseKeeping:
clc
clear all
close all
% Constants:
S = 232; % ft^2
Cd0 = 0.02;
K = 0.07;
W = 11000; % lbf
Cl_max = sqrt(Cd0/K);
Cd_max = 2*K*Cl_max^2;
Rho_10000 = .001756; % slugs/ ft^3
%Initial conditions:
t = 0; % Sec
x = 0; % ft
h = 10000; % ft
v = sqrt((2*W)/(Rho_10000*S*Cl_max)); %ft/s
gamma = -Cd_max/Cl_max;
% Find Endurance:
V_RD= sqrt((2*W)/(S* Rho_10000* sqrt(3*Cd0/K)));
RD= V_RD/((sqrt(3*Cd0/K))/(2*Cd0)) ; % ft/s
Endurcance= h/RD; % 958.3515 sec
% Sea Level values:
TSL = 518.69; % Rankine
PSL = 2116.199414; % lbs/ft^2
RhoSL = 0.0023769; % slugs/ft^3
while h > 0
tspan = [t t+1];
i=1;
X = [x;h;v;gamma;Rho_10000];
Time(i)= t;
% Calling ODE45:
[F] = ode45(# D,tspan, X)
% Hight Varying Parameters:
T = TSL - 0.00356616*h;
P = (1.137193514E-11)*(T)^5.2560613;
Rho = (RhoSL * TSL / PASL)*(P / T);
a = 49.0214 * (T)^.5;
H_Del(i) = (-Cd_max/Cl_max)*(plotted_x(i))+10000;
V_Del(i) = sqrt((2*W)/(Rho*S*Cl_max));
Gamma_Del(i) = -Cd_max/Cl_max;
i= i+1;
X = [ x ; H_Del(i); V_Del(i); Gamma_Del(i); Rho];
end
% Plots:
%1
figure (1)
plot(F(:,1),'-r',F(:,3),'-b')
title('Velocity vs Distance');
xlabel('x (ft)');
ylabel('v (ft/s)');
%2
Figure (2)
plot(F(:,1),'-r',F(:,2),'-b')
title('Altitude vs Distance ');
xlabel('x (ft)');
ylabel('h (ft)');
%3
figure (3)
plot(F(:,1),'-r',F(:,4),'-b')
title('Gamma vs Distance');
xlabel('x (ft)');
ylabel('Gamma (rad)');
%4
figure (4)
plot(t,'-r',F(:,3),'-b')
title('Velocity vs Time');
xlabel(' t (s)');
ylabel('v (ft/s)');
%5
figure (5)
plot(t,'-r',F(:,1),'-b')
title(' Distance vs Time ');
xlabel('t (s)');
ylabel('x (ft)');
%6
figure (6)
plot(t,'-r',F(:,4),'-b')
title('Gamma vs Time ');
xlabel('t (s)');
ylabel('Gamma (rad)');
%7
figure (7)
plot(t,'-r',F(:,3),'-b')
title('Velocity vs Time');
xlabel('t (s)');
ylabel('V (ft/s)');
Here is the Function
function [F] = D(X)
% Constants:
S = 232; % ft^2
Cd0 = 0.02;
K = 0.07;
W = 11000; % lbf
Cl_max = sqrt(Cd0/K);
Cd_max = 2*K*Cl_max^2;
Rho_10000 = .001756; % slugs/ ft^3
% Initial conditions:
t = 0; % Sec
x = 0; % ft
h = 10000; % ft
v = sqrt((2*W)/(Rho_10000*S*Cl_max)); % ft/s
gamma = -Cd_max/Cl_max;
g= 32.2; % ft/s^2
% Equations:
X_Pr = X(3)*cos(X(4));
H_Pr = X(3)*sin(X(4));
V_Pr = (-32.2./W)*(0.5*X(5)*Cd_max*S*X(3)^2 + W*sin(X(4)));
G_Pr = (32.2./(W*X(3)))*(0.5*X(5)*Cl_max*S*X(3)^2 - W*cos(X(4)));
F = [X_Pr;H_Pr;V_Pr;G_Pr];
I am not very good with Matlab but I did my very best! I went to my professors for help but they said they were too busy. I even stalked every senior I knew and they all said they did not know how to do it. My next project will be assigned soon and I think that if I cannot do this then I would not be able to do the next one.
Your code produces the following error:
Error using main>D
Too many input arguments.
This means that ode45 tries to call your provided function D with too many input arguments. You should check the desired odefun format in the ode45 documentation: dydt = odefun(t,y)
So, you should change your function declaration of D to
function [F] = D(t, X)
This should solve your first problem, but a following error pops up:
D returns a vector of length 4, but the length of initial conditions
vector is 5. The vector returned by D and the initial conditions
vector must have the same number of elements.
Again, you should check the ode45 documentation. Your function should return the derivatives of all your input variables X: F = dX/dt. You should also return the derivate of the fifth element Rho_10000.
Next, I got some error about undefined variables such as PASL. Probably because you did not post your full code.
Besides of the errors, you should really check your code again. You have written an infinite while loop while h > 0. You never change h in the loop, nor you use the output of ode45 in your loop. Furthermore you always overwrite your i and X value at the beginning of the loop, which is probably not what you want.
This is not a full answer to your question, but I hope you will be able to continue and post smaller, well defined problems, instead of one big problem which is very difficult to answer completely.

Solving ODEs with Matlab, with varying Parameters

Lets say I have a simple logistic equation
dx/dt = 2ax(1 - x/N)
where N is the carrying capacity, a is some growth rate, and both a and N are parameters I'd like to vary.
So what I want to do is to plot a 3D graph of my fixed point and the two parameters.
I understand how to find a fixed point of a single parameter.
Here is my sample code
function xprime = MyLogisticFunction(t,X) %% The ODE
% Parameters
N = 10 % Carrying Capacity
a = 0.5 % Growth Rate
x1prime = 2*a*X(1)*(1 - X(1)/N );
xprime = [x1prime ]';
end
Next my solver
% Initial Number
x0 = 0.4;
%Time Window
tspan=[0 100];
[t,x]=ode45(#MyLogisticFunction,tspan,x0);
clf
x(end,1) % This gives me the fixed point for the parameters above.
So my real question is, how do I put a for loop across two functions, that allows me to vary a and N, so that I can plot out a 3D graph of a and N and my fixed point x*.
I've tried combining both functions into one .m file but it does not seem to work
You need to pass the parameters to your function:
function xprime = MyLogisticFunction(t,X,a,N) %% The ODE
% Parameters (passed as function arguments)
% N = 10 % Carrying Capacity
% a = 0.5 % Growth Rate
x1prime = 2*a*X(1)*(1 - X(1)/N );
xprime = [x1prime ]';
end
and then when you call the ode solver:
% Initial Number
x0 = 0.4;
%Time Window
tspan=[0 100];
a = 0.1:0.1:1; % or whatever
N = 1:10; % or whatever
x_end = zeros(length(a),length(N));
for ii = 1:length(a)
for jj = 1:length(N)
[t,x]=ode45(#(t,X)MyLogisticFunction(t,X,a(ii),N(jj)),tspan,x0);
x_end(ii,jj) = x(end,1);
end
end

Matlab plot in loop error

I am creating figures in a for loop. The figure is a 2D mesh plot, which is supposed to be updated every iteration. The value to be plotted in a 200x200 array.
My problem is: It seems the calculation is running every iteration, but the plot is always the first one created, no matter I just plot or save to file.
Here is my code:
x = 1:200;
y = x;
for i = 1:100000
c = calculate(stuff, c); % value to be created, nothing to do with x and y
h = figure;
mesh(x,y,c);
saveas(h, sprintf('FIG%d.jpg',i);
drawnow; % did not work with or without this command
close(h);
end
First, thank you for all your inputs and suggestions! I didn't expect to get so many help within such a short time!
Then, I can answer some of the confusions here.
To Daniel: yes the c is changing. The program is calculating c based on its previous value. And there is sufficient step for c to change.
To R.Schifini: I tried pause(.1) but it didn't help unfortunately
To Andrew: thanks for pointing it. The complete program is attached now. And as to Daniel, the program calculate the value of c based on its previous values.
To The-Duck: I tried clf(h, 'reset') but unfortunately it didn't help.
Complete code:
Main program: please refer to wikipedia for the physical equation if you are interested
http://en.wikipedia.org/wiki/Cahn%E2%80%93Hilliard_equation
% Program to calculate composition evolution for nucleation and growth
% by solving Cahn-Hilliard equation - Time dependent non-linear
% differential equation
% Parameter
sig = 0.1; % J/m^2
delta = 10E-9; % m
D = 1E-9; %m^2/s
A = 10*sig/delta; % J/m
K = 3*sig*delta; % J/m^3
M = D/(2*A); % m^2/s
N = 200; % mesh size
dt = 1E-12; %s
h = delta/10;
% Rng control
r = -1+2.*rand(N);
beta = 1E-3;
n = 10000;
% initialization
c0 = zeros(200);
c0 = c0+ 0.1+beta.*r;
c = c0;
x = h.*linspace(-N/2,N/2,N);
y=x;
% Iteration
for i = 1:n
LP_c = laplacian(c,h);
d_f = A*(4*(c.^3)-6*(c.^2)+2*c);
sub = d_f - (2*K)*LP_c;
LP_RHS = laplacian(sub,h);
RHS = M*LP_RHS;
c = c + dt.*RHS;
% Save image every 2000 steps
% if ( i==1000 || i==10000 || i==100000)
% h = mesh(x,y,c);
% pause(.1);
% saveas(h, sprintf('FIG%d.jpg',i));
% clf(h,'reset');
% end
end
%h = figure;
mesh(x,y,c);
Laplacian function:
function LP_c = laplacian(c,h)
v1 = circshift(c,[0 -1]);
v2 = circshift(c,[0 1]);
v3 = circshift(c,[-1 0]);
v4 = circshift(c,[1 0]);
LP_c = (v1+v2+v3+v4-4.*c)./(h^2);
end
Result:
You can see the commented part in main program is for plotting periodically. They all give the same plots for each iteration. I tried the current OR version, also tried if ( mod(i,2000) == 0) to plot more pics. There is no difference. Shown:
However, if I comment out the periodic plotting, just run the program for different values of n, I got different plots, and they obey physical laws (evolving structure), shown in time order
Therefore I excluded the possibility that c might not update itself. It has to be some misuse of the plotting function of matlab. Or maybe some memory issue?
An interesting point I discovered during edition session: If I put the command h = figure in front of the loop and plot after the loop end, like this:
h = figure;
% Iteration
for i = 1:n
LP_c = laplacian(c,h);
d_f = A*(4*(c.^3)-6*(c.^2)+2*c);
sub = d_f - (2*K)*LP_c;
LP_RHS = laplacian(sub,h);
RHS = M*LP_RHS;
c = c + dt.*RHS;
end
mesh(x,y,c);
It seems all value of c calculated during the loop will overlap and give a figure shown below: I guess this indicates some facts about the plotting function of matlab, but I am not sure
Btw, can I answer directly to each comment and high light the new added section in my post? Sorry I am not as familiar with Stack Overlow as I should have :)
I ran your routine and with the following changes it works for me:
% Iteration
for i = 1:n
LP_c = laplacian(c,h);
d_f = A*(4*(c.^3)-6*(c.^2)+2*c);
sub = d_f - (2*K)*LP_c;
LP_RHS = laplacian(sub,h);
RHS = M*LP_RHS;
c = c + dt.*RHS;
% Save image every 2000 steps
if ( mod(i,2000)==0)
h1 = mesh(x,y,c);
drawnow;
saveas(h1, sprintf('FIG%d.jpg',i));
end
end
The main change is the figure handle variable from h to h1. Why? You are already using variable h in your equations.
Regards,