Sketching Transfer function in time domain

AI Thread Summary
To find the output at t=0 and t=infinity for the given transfer function, it is crucial to include the step function U(s) = 1/s in the calculations. The initial and final value theorems require evaluating the limit of sY(s), which incorporates the input function. Omitting U(s) would incorrectly imply using an impulse input instead of a step input. The confusion arises from differing examples in textbooks, but for accurate results, always consider the input function. Understanding the distinction between the step function and delta function is essential for proper application of these theorems.
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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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