How to do Z transform with Z^-1 format?
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Diego Marvid
il 27 Set 2020
Modificato: Walter Roberson
il 21 Gen 2024
I want to do the Z transform of
which theoretically is
.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/367285/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/367288/image.png)
However, Matlab's Z transform function (ztrans) gives it in Z format and not in Z^-1 format.
The equation that ztrans(f) gives me is:
.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/367291/image.png)
I want to know if there is a clean way to get the Z transform in Z^-1 format as shown above.
If the code is generic and not specific to this example will help a lot.
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Star Strider
il 27 Set 2020
The only option I can think of has nothing to do with the Symbolic Math Toolbox, and is instead the Variable argument in the Control System Toolbox. That will allow you to specify z^-1 as the variable.
Since it woulld be best to use the Control System Toolbox anyway (the Symbolic Math Toolbox is not designed for efficient transfer function analysis), that would be my option.
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Star Strider
il 27 Set 2020
The heaviside function may not be correct. The delta function would be more likely to be correct. Consider the Laplace and z-transform versions of both. (Those are the bases of my preferences.)
Paul
il 23 Feb 2023
The problem is that sym2poly returns the polynomial coefficient is descending powers of z. So the tf needs be created in z, and then changed to z^-1. Two step process:
syms x n
v = 1;
sympref('HeavisideAtOrigin', v);
x = heaviside(n-1);
H = ztrans(x)
[num, den] = numden(H);
H = tf(sym2poly(num), sym2poly(den), -1)
H.Variable = 'z^-1'
Più risposte (2)
Vinit
il 2 Nov 2023
Modificato: Walter Roberson
il 21 Gen 2024
clear all
close all
clc
V = 220;
R = 10;
L = 0.051;
dt = le-3;
t = 0:dt:0.05;
y = [0];
for k =1:length(t)-1
y (K+1)=(1-R*dt/L)*y(k)+dt*(1/L)*V;
end
figure(1)
plot(t,y)
grid on
xlabel ('time (s)')
ylabel ('current (Amp)')
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Vinit
il 2 Nov 2023
Modificato: Walter Roberson
il 21 Gen 2024
The solution of equation (5) & (8) is obtainedby using MATLABProgram to solve difference equations
%%Program to plot the difference equation to verify
%Results with time domain equations
%Considering the systemof RL Circuit
% The expression is V=Ri+L(di/dt) clear all
close all clc
%System Parameters
V=220; % Maximum valueof the voltage (Volt)
R=10; % Resistance Value(Ohm)
L=0.051; % Inductor value(Henry)
dt=1e-3; % Time period (T)
t=0:dt:0.05; % Samples, k
y=[0]; % Output stack
fork=1:length(t)-1
y(k+1)=(1-R*dt/L)*y(k)+dt*(1/L)*V; end
figure(1)plot(t,y) grid on
xlabel ('Time (s)') ylabel('Current(Amp)')
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