How to find total Steady-State Error due to reference and disturbance

I have this system and wanted to find the error for when R(s) is a 3 step input and the error when D(s) is a unit step input. I'm doing it this way but I think it's wrong. Can someone tell me if this is the right way?

km = 3.5;
tm = 0.5
K = 5;

s = tf('s');
G = km/(tm*s+1);
sys = feedback(G*K,1,-1);
SP=3; %input value
[y,t]=step(SP*sys); %response of the system to a step with amplitude SP
sserror=abs(SP-y(end)) %get the steady state error

Answer 1

[Paul]

Hi Vivianne,

If this problem is to be attacked using the Final Value Theorem, we can proceed either numerically or symbolically. Here's the former.

km = 3.5;
tm = 0.5;
K = 5;
s = tf('s');

Define the plant transfer function

G = km/(tm*s+1);

Closed loop transfer function from R(s) to E(s)

EoverR = feedback(1,G*K,-1)

EoverR =
 
   0.5 s + 1
  ------------
  0.5 s + 18.5
 
Continuous-time transfer function.

Before trying to find the steady state output, we need to verify that the system is stable.

pole(EoverR)

ans = -37

The single pole is in the left half plane, so the system is stable and we can proceed

SP = 3; %input value

E(s) in response to R(s) = SP/s

E = EoverR*SP/s

E =
 
     1.5 s + 3
  ----------------
  0.5 s^2 + 18.5 s
 
Continuous-time transfer function.

The Final Value Theorem tell us that the steady state value of e(t) is the limit as s ->0 of s*E(s)

s*E

ans =
 
   1.5 s^2 + 3 s
  ----------------
  0.5 s^2 + 18.5 s
 
Continuous-time transfer function.

We see that the free s in numeator cancels the free s in the denominator, so the limit is simply evaluating SP*EoverR at s = 0

ess = freqresp(EoverR*SP,0)

ess = 0.1622

You can use a similar procedure for the transfer function from D(s) to E(s).