Convert from transfer function to time domain

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) ;

Answer 1

[Star Strider]

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.

.

[Hend Mostafa]

I just have a state space function, my plan is to convert it to s domain then to time domain because I need it in time domain

[Star Strider]

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)

tfsys =
 
  0.0006058 s^2 + 20.11 s + 0.0003933
  -----------------------------------
             s^2 + 58.19 s
 
Continuous-time transfer function.

n = tfsys.Numerator;
d = tfsys.Denominator;

syms s t 
H(s) = poly2sym(n,s) / poly2sym(d,s)

H(s) = 

H = vpa(H, 5)

H(s) = 

h(t) = ilaplace(H)

h(t) = 

h = vpa(h, 5)

h(t) = 

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]

I understood, Thank you very much

[Star Strider]

As always, my pleasure!