Convert from transfer function to time domain
显示 更早的评论
How can I convert from transfer function to time domain?
R0=0.000605797;
Qr=147964.8608;
R1=0.000856205;
C1=0.049823238;
% X=[SOC;Vrc];
A=[0 0;0 -1/R1*C1];
B=[1/Qr;1/C1];
C=[1 1];
D= R0;
% u=Il;
% Equations: X dot= AX+Bu & y= CX+Du+bo
sys= ss(A,B,C,D);
[b,a] = ss2tf(A,B,C,D) ;
采纳的回答
Probably the easiest way is to just simulate it.
R0=0.000605797;
Qr=147964.8608;
R1=0.000856205;
C1=0.049823238;
% X=[SOC;Vrc];
A=[0 0;0 -1/R1*C1];
B=[1/Qr;1/C1];
C=[1 1];
D= R0;
% u=Il;
% Equations: X dot= AX+Bu & y= CX+Du+bo
sys= ss(A,B,C,D);
figure
impulse(sys)
figure
step(sys)
There is no reason to create a transfer function from it.
.
4 个评论
Hend Mostafa
2022-5-27
I do not understand what you mean by ‘needing it in the time domain’.
Guessing here —
R0=0.000605797;
Qr=147964.8608;
R1=0.000856205;
C1=0.049823238;
% X=[SOC;Vrc];
A=[0 0;0 -1/R1*C1];
B=[1/Qr;1/C1];
C=[1 1];
D= R0;
% u=Il;
% Equations: X dot= AX+Bu & y= CX+Du+bo
sys= ss(A,B,C,D);
tfsys = tf(sys)
n = tfsys.Numerator;
d = tfsys.Denominator;
syms s t
H(s) = poly2sym(n,s) / poly2sym(d,s)
H = vpa(H, 5)
h(t) = ilaplace(H)
h = vpa(h, 5)
figure
fplot(h)
grid
axis([0 0.1 0 20])
This creates the transfer function as a Laplace polynomial fraction, then inverts it. Note that the result is the same as the impulse result in my original post.
.
Hend Mostafa
2022-5-27
Star Strider
2022-5-27
As always, my pleasure!
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