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0

In your circuit Vc(s) is impressed across each of the 5 branches. You can write a node equation on that top node fairly easily. The current leaving that node in the leftmost branch is Vc(s)/20. Look carefully at the second branch from left and you will see that the current flowing down that branch is [Vc(s) + 1V]/(20+2s). You should be able to work it ...

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Just put some formula onto the various nodes: - Can you see how this works and how E(s) becomes: - $$R(s) - Y(s)$$ Therefore: - $$Y(s) = (R(s) - Y(s)) \cdot \left( \dfrac{K}{(s+3)(s+5)}\cdot\dfrac{1}{s}\right)$$ Just rearrange to group Y(s) terms and solve. Can you take it from here (it should be fairly straightforward)?

0

Yes, you can picture it by looking at the time index (n in your case) as going from -infinity on the left to infinity on the right. What does time reversal mean? Your function values (e.g., step function values) will reverse, as time "moves" from right to left instead (for example, instead of having the value 0 from -infinity to -1 and then having 1 from 0 ...

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By definition, if you replace n with k-n, then it's time reversed. What value you use for k seems to me to be a matter of convention. It affects the time shift of the signal. K will depend on what time datum you reverse it about.

5

Great question. Of course, transforms are the best way to solve this, which you already know. But maybe when you start with a bunch of numbers from an oscilloscope instead of a nice, neat formula, your perspective changes a bit. Here is another way to approach the problem numerically. It is really best suited if you have some a priori info to bring to the ...

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