MATLAB step response of a signal

[gmish27]

to evaluate the step response of a transfer function, say Ts, we write in matlab: step(Ts)

Mathematically this is same as multiplying the transfer function Ts with 1/s and then plotting the inverse laplace transform of (1/s)*Ts.

I am trying the same in matlab. Have a look at the code:

Ts=tf[1,[1 6 9]);
step(Ts) %step response using step function
----------------------

syms s;
Tt=ilaplace(1/(s^3+6*(s^2)+9*s)); %product of Ts and 1/s
ezplot(Tt)
----------------------

In both cases am not getting the same plot. Can anyone provide help me with an explanation/solution???

 

[barry]

There's an initial condition associated with the inverse Laplace. Could that be the problem?

 

[gmish27]

There's an initial condition associated with the inverse Laplace. Could that be the problem?

well the initial conditions are all zeros. You can proceed with that...

 

[barry]

Maybe it's a sampling-time issue. How different are the two results? Have you looked at the MATLAB website for answers?

 

[gmish27]

Maybe it's a sampling-time issue. How different are the two results? Have you looked at the MATLAB website for answers?

Here is the image using both the methods:
using tf
using ilaplace

 

[barry]

Well, the first thing that jumps out at me is that your first plot, although having the right shape has a maximum of 0.1. It should be 1.0. Next, the time scale in the second plot starts at -6 seconds. What you've got there is a time machine.

BUT, I think your main problem is that you've got a bad transfer function. 1/(s^2 +6s +9)==> 1/(s+3)(s+3). Your step response is then: 1/(s(s+3)(s+3)), which is of the form: 1/s(s+a)(s+b). If you look at any table of Laplace transforms you'll see that the inverse transform of your step response is: 1/(a+b) * {1-[(be^-at)-(ae^-bt)]/(b-a)}. You've got a denominator of zero, which makes everybody unhappy.

 

[gmish27]

Well, the first thing that jumps out at me is that your first plot, although having the right shape has a maximum of 0.1. It should be 1.0. Next, the time scale in the second plot starts at -6 seconds. What you've got there is a time machine.

BUT, I think your main problem is that you've got a bad transfer function. 1/(s^2 +6s +9)==> 1/(s+3)(s+3). Your step response is then: 1/(s(s+3)(s+3)), which is of the form: 1/s(s+a)(s+b). If you look at any table of Laplace transforms you'll see that the inverse transform of your step response is: 1/(a+b) * {1-[(be^-at)-(ae^-bt)]/(b-a)}. You've got a denominator of zero, which makes everybody unhappy.

the ilaplace of Y(s)=H(s)*X(s), that i am getting is: 1/9 - t/(3*exp(3*t)) - 1/(9*exp(3*t))
what's the problem with this.

And why output should be 1.0??? ( in case of step response)

 

[barry]

First, I was wrong about 1.0; it SHOULD be 1/9. If you look at your transfer function , if you remember that s=jw, and assume jw=0 for a dc signal, your transfer function reduces to 1/9. The inverse Laplace that I get is:

1/9u(t) - (1/9)*exp(-3t) + (1/3)*t*exp(-3t) ; it looks like you've moved numerator and denominator terms. This also approaches a final value of 1/9. The step response and inverse Laplace plots are very similar, although I get a weird overshoot at the start of the inverse Laplace plot. Being mathematically challenged I don't understand some of this.

If you use the transform pair I posted previously... 1/(a+b) * {1-[(be^-at)-(ae^-bt)]/(b-a)}... everything goes to zero. That can't be right. I used partial fraction expansion to get my transform...maybe there's something wrong there. Here's my plots (hope I can figure out how to import pictures...)


https://obrazki.elektroda.pl/0_1314378798.jpg