How do i find a difference equation?
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I have this first order differential equation given:
i need to find a difference equation using MATLAB. I would appreciate if someone could help me with that. Thanks
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Azzi Abdelmalek
2013-1-24
Choose your sample time Te
Te=0.1 % Your sample time
N=1.5
D=[1 -0.5]
Modelc=tf(N,D) % Your continue transfer function
Modeld=c2d(Modelc,Te) % Your discrete transfer function
[Nd,Dd]=tfdata(Modeld,'v');
% From Y(z)/U(z)=Nd(z)/Dd(z) you can find the relation between y{n] and u[n]
10 个评论
LeBron
2013-1-24
Azzi Abdelmalek
2013-1-24
No, when you discretize a continuous system you have to take in account the two blocks ADC and DAC. If H(p) is the transfer function of your continuous system then, if you discretize, using a zoh method: the discrete transfer function G(z) will be
G(z)=(z/(z-1))*Ztransform(inverse_transform(H(p)/p))
Azzi Abdelmalek
2013-1-24
LeBron, I gave you the code which allows to find your discrete transfer function G(z). Do you know how to find the difference equation from G(z)? Other thing
G(z)=(z/(z-1))*Ztransform(inverse_transform(H(p)/p))
is not a code, it's what the above code is doing
Roger Stafford
2013-1-24
编辑：Azzi Abdelmalek
2013-1-25
I would advise you to look at the article
It gives an explanation of various Runga-Kutta methods of approximating the solution to ordinary differential equations of the kind you have. The discussion of RK4 shows you one method which is a fourth order approximation wherein it is assumed you can sample your u(t) at every h/2 interval with a step size of h in t. The f(t,y) they describe would be
f(t,y) = 1.5*u(t)+.5*y
in your case, and the corresponding difference equations are given there.
The article also describes other possible kinds of difference equations that one can use with differing amounts of accuracy.
Azzi Abdelmalek
2013-1-25
编辑：Azzi Abdelmalek
2013-1-25
He is asking for a difference equation, not a numeric resolution.
LeBron
2013-1-25
Azzi Abdelmalek
2013-1-25
For example
1 Y(z)
G(z)=----- = ------
z + 2 U(z)
Then
Y(z)(z+2)=U(z)
zY(z)+2Y(z)=U(z)
y[n+1]+2y[n]=u[n]
Roger Stafford
2013-1-25
Azzi, the Runga-Kutta method does give a set of difference equations, in fact a very respectable set indeed.
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