Trouble understand IVT, FVT, unit step input.

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In summary: IVT and FVT are two methods that can be used to calculate the response of a system to an impulse or step input. By multiplying the input by 1/s, we are ensuring that the impulse or step input is applied to the system at a constant frequency. This will result in a consistent response across different input frequencies.
  • #1
vysero
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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:

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?
 
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  • #2
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:
Code:
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.
 
  • #3
I am confused, let me see if I can't clear up what I did.

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

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

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

Lets say I can use l'hopital rule then:

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

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:

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

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

>> step(b)
>>

by doing the following limits:

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

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

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:

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

\(\displaystyle \lim_{{s}\to{\infty}} = 0\)
 
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  • #4
Circuits said:
I am confused, let me see if I can't clear up what I did.

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

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

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.

Circuits said:
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)?

\(\displaystyle \lim_{{s}\to{\infty}} = 0\)

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 $sY(s)$.
 
Last edited:

FAQ: Trouble understand IVT, FVT, unit step input.

1. What is IVT and why is it important?

IVT stands for Intermediate Value Theorem. It states that if a function is continuous on a closed interval, then it must take on every value between the minimum and maximum values on that interval. This is important because it allows us to prove the existence of solutions to certain problems.

2. What is FVT and how is it used?

FVT stands for Fundamental Value Theorem. It states that the definite integral of a function over an interval can be evaluated by finding the antiderivative of the function and plugging in the endpoints of the interval. This is used in many applications, such as finding the area under a curve or calculating work done by a variable force.

3. What is a unit step input?

A unit step input is a function that is defined as 0 for all negative values and 1 for all non-negative values. It represents an instantaneous change in a system, often used in control systems and signal processing.

4. How do IVT and FVT relate to each other?

While they may seem unrelated, IVT and FVT are both fundamental theorems in calculus that deal with continuity and integration, respectively. They both help us understand and solve problems involving functions and their behavior on intervals.

5. Can IVT or FVT be applied to non-continuous functions?

No, IVT and FVT both require the function to be continuous in order to be applicable. If a function is not continuous, then these theorems cannot be used and other methods must be employed to solve the problem.

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