Sketching Transfer function in time domain

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SUMMARY

The discussion focuses on applying the final and initial value theorems to the transfer function $$ y(s) = \frac {s}{(s+1)(2s+1)} u(s) $$, where $$ u(s) $$ is the step function $$ \frac {1}{s} $$. Participants clarify that the step function must be included when evaluating $$ \lim_{s \to 0} sY(s) $$ to find the output at both t=0 and t=infinity. The importance of distinguishing between the step function and the delta function is emphasized, as using the delta function would lead to incorrect interpretations of the system's response.

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gfd43tg
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Homework Statement


$$ y(s) = \frac {s}{(s+1)(2s+1)} u(s) $$
Where ##u(s)## is the step function ##\frac {1}{s}##

Find the output at t=0 and t= infinity

Homework Equations

The Attempt at a Solution


My question is kind of basic, so I know the final and initial value theorem

$$ \lim_{s \to 0} sY(s) = \lim_{t \to \infty} y(t) $$

But should I include the step function, or leave it out. Meaning, should I evaluate

$$ \lim_{s \to 0} \frac {s^{2}}{(s+1)(2s+1)} \frac {1}{s}$$
or rather,
$$ \lim_{s \to 0} \frac {s^{2}}{(s+1)(2s+1)} $$

The reason I am hesitating on this is because in the textbook example problem, they do not mention what the input function is, and proceed to solve without the step function. Then I solved a homework problem where they asked to match the transfer function output with a step function input, and at that time I did not even realize, so I was only doing limits of Y(s), not sY(s), and got them all right. So now my head is all jumbled up and I just want to get this thing cleared up!
 
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You need to include U(s) = 1/s.
If you used the initial & final value theorems without the U(s) input you'd be solving the problem with the input = kδ(t), the delta function aka impulse input, where k = 1 Volt-sec. If you're not familiar with the delta function, be careful with it. Unlike u(t) which is dimensionless, δ(t) has dimension 1/t.
 
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Makes sense why I got them all right then, the s and 1/s cancel anyways
 

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