how to convert a Laplace transform into a transfer function value

>> A= tf([1 0],[1 1])

A =

s

-----

s + 1

Continuous-time transfer function.

>> p= feedback(A,1)

p =

s

-------

2 s + 1

Continuous-time transfer function.

>> laplace(t)

ans =

1/s^2

>> p= feedback(ans,1)

Error using feedback

Not enough input arguments.

[Suman]

a = [2];
b = [1 0 0]; 
f= tf(a,b)

f =
 
   2
  ---
  s^2
 
Continuous-time transfer function.

syms t;
d= laplace(2*t)

d = 

S= feedback(f,1)

S =
 
     2
  -------
  s^2 + 2
 
Continuous-time transfer function.

D= feedback(d,1)

Error using feedback
Not enough input arguments.

So, why the laplace tranform is not considered as a transfer function ???? If I have a time domain system, then how we can calculate the transfer function of that system.

Answer 1

[Star Strider]

The ‘d’ expression currently exists as a symbolic object. It needs to be transformed into a double value and then to a system object to work here —

a = [2];
b = [1 0 0]; 
f= tf(a,b)

f =
 
   2
  ---
  s^2
 
Continuous-time transfer function.

syms t;
d= laplace(2*t)                             % Symbolic Object

d = 

[dn,dd] = numden(d)                         % Get Numerator & Denominator

dn = 

2

dd = 

dnp = double(sym2poly(dn))                  % Convert To Polynomial Vectors & 'double' Values (From Symbolic Variables)

dnp = 2

ddp = double(sym2poly(dd))                  % Convert To Polynomial Vectors & 'double' Values (From Symbolic Variables)

ddp = 1×3

1 0 0

dtf = tf(dnp, ddp)                          % Create AS Control System Toolbos 'system' Object

dtf =
 
   2
  ---
  s^2
 
Continuous-time transfer function.

S= feedback(f,1)

S =
 
     2
  -------
  s^2 + 2
 
Continuous-time transfer function.

D= feedback(dtf,1)

D =
 
     2
  -------
  s^2 + 2
 
Continuous-time transfer function.

.

See the documentation on the relevant Symbolic Toolbox functions numden, sym2poly, and double for details.

.