MHB Trouble understand IVT, FVT, unit step input.

[vysero]

Here is my transfer function, G(s) = 1/(s+1). I know this will be a half parabola looking output with a unit step input whose initial value will be 0 and final value will be 1. I got the initial value by takeing the limit of the function at inf and the final value by taking the limit of the function at 0.

My first question is this: does the IVT and FVT only apply to a function given it has a unit step input? For instance, if I graph this function on my calculator I get an exponentially decaying graph that starts at one and goes to zero.

Say my transfer function is H(s) = (s-10)/(s^2 +20s). The limit @ inf is 1.

My second question (is a double): If I plug H(s) into MATLAB with the following code:

>> a = tf([1 -10 ],[1 20 0])

a =
 
    s - 10
  ----------
  s^2 + 20 s
 
Continuous-time transfer function.

>> step(a)

I get a ramp output that starts at zero and decays to -inf with a slope of -1. How can that be if my limit @ inf says the final value should be 1?

also, When I try to figure out the limit @ 0 I get the undefined answer -10/0 so how do I get around that? I don't believe l'hopital's rule applies to undefined forms like division by zero unless its 0/0 right?

 

[Dethrone]

1. IVT and FVT would apply to any transfer function where the poles are located in the left-hand plane (negative real parts) and not more than one pole at the origin. That is, for stable systems. For a transfer function like $G(s)=1/(s+1)$, we wouldn't normally need to apply IVT or FVT because the inverse laplace transform is pretty simple; it is just $g(t)=e^{-st}, t>0$ which as you've mentioned, starts at $1$ and tends to $0$.

If your transfer function is $H(s) = (s-10)/(s^2 +20s)$, then the denominator factors to $s(s+20)$ with poles at $0$ and $-20$, which satisfies the conditions to use IVT and FVT (you can find the conditions on wikipedia).

2. I think you're getting confused here. FVT and IVT apply to systems, not outputs. When we apply IVT and FVT to $H(s) = (s-10)/(s^2 +20s)$, we're only getting the initial and final behaviors of $h(t)=\mathcal{L}^{-1}\{H(s)\}$. It gives us a convenient way to find the behavior of time-domain functions in the frequency domain - nothing more. Btw, IVT gives me 1 and FVT gives me -0.5 (you might want to double check your calculus there). When you're finding the step response, what you're actually going to get is the inverse laplace transform of $H(s) = \frac{s-10}{s^2+20s}\cdot \frac{1}{s}$, and if you apply IVT and FVT to that, you'll get the same answer as the graph you generated.

Btw, a neat tool on MatLab that I recently discovered:

import sys.*

a = tf([1 -10 ],[1 20 0])
ltiview(a)

Then right click should give much more options like Step, Impulse, Bode Plots, etc. You should also be able to trace.

 

[vysero]

I am confused, let me see if I can't clear up what I did.

$$H(s) = \frac{s-10}{{s}^{2}+20s}$$

$$\lim_{{s}\to{\infty}}(H(s)) = 0$$

$$\lim_{{s}\to{0}}(H(s)) = \frac{-10}{0}$$ (undefined) <- Can I use l'hopital rule on this?

Lets say I can use l'hopital rule then:

$$\lim_{{s}\to{0}}(H(s)) = .05$$

This is the part that really confuses me because people have given me your same answer before and it doesn't make sense. If I have a simple function like G(s) = \frac{1}{s+10} I can pinpoint the initial value and final value as displayed in MATLAB using this code:

>> b=tf([1],[1 10])

b =
 
    1
  ------
  s + 10
 
Continuous-time transfer function.

>> step(b)
>>

by doing the following limits:

$$\lim_{{s}\to{inf}}(G(s)) = 0$$

$$\lim_{{s}\to{0}}(G(s)) = .1$$

So why do I need to multiply H(s) by 1/s to get the initial value and final value for the graph MATLAB displays if I don't have to do it for G(s)?

Also multiplying H(s) by 1/s and evaluating limits:

$$\lim_{{s}\to{0}}(H(s)\frac{1}{s}) = \frac{-10}{0}$$ <- still undefined even after using l'hopital rule

$$\lim_{{s}\to{\infty}} = 0$$

 

[Dethrone]

I am confused, let me see if I can't clear up what I did.

$$H(s) = \frac{s-10}{{s}^{2}+20s}$$

$$\lim_{{s}\to{\infty}}(H(s)) = 0$$

Careful here, we're applying the limit to $sH(s)$, not $H(s)$. Then, we will be able to apply l'Hopital's rule properly. Let me know if that resolves your problem.

I think you need to revisit what things like "impulse-response" or "step-response" refers to. Step-response is when the input is $u(t)$, with laplace transform of $1/s$. You might have heard that the $u(t)$ is an integrator. You might have also learned that convolution in time is multiplication in the frequency domain. So with input $1/s$, a system $G(s)$, we get the output as $G(s)/s$. This simple computation (rather than convolution) is one of the reasons why the frequency domain is so convenient.

So why do I need to multiply H(s) by 1/s to get the initial value and final value for the graph MATLAB displays if I don't have to do it for G(s)?

$$\lim_{{s}\to{\infty}} = 0$$

Sorry, you have to do it for the $G(s)$ too. The only time you don't is when you're dealing with the impulse response. Remember, IVT and FVT only relate an expression in frequency domain to time domain. That's all it does.

To summarize, given the transfer function $G(s)=1/(s+1)$ with a step-input, the step-response is $Y(s)=1/[s(s+1)]$. Then apply IVT and FVT to $sY(s)$.