# Plotting output response of system ot the input stimulus given

#### Brian Moughtin

Hi

I need to "plot the output response of the system shown to the input stimulus given"

I suppose it would help if i knew where to start, but I'll try and explain as best I can.

There are two drawings.

Drawing 1 - shows a rectangle with 1 / 1 +0.5s inside and a line drawn horizontally from the centre of the left vertical side labelled Input, another line is drawns from the centre of the right vertical side saying Output.

Drawing 2 - Shows a graph and a curve starting at 10 on the Y axis with the figure 10e^-0.2t the curve flattens at 3.27 seconds on the X axis.

Now being the worlds worst student of maths, this is what I've got so far.

f(t) = 10e^-0.2t : 0 < t 3.27

f(t) = 5.19 : t>3.27

( Found 5.19 by 10e^(-0.2 x 3.27) = 5.19

Therefore f(t) = 10e^-0.2t - 10e^0.2t H(t-3.27) + 5.19H(t-3.27)

e^-3.27 = 0.038

5.19 x 0.038 = 0.19722

Thus

f(t) = 10e^0.2t - 0.197H(t-3.27) e^-(ta-3.27) +0.197H(t-3.27)

To give

F(s) = 10/(S + 1) - 0.197 x (e^3.27s) /(S+1) = (0.197 x e^3.27s)/S

So far this is probably easily understandable to some but makes me want to put a round in the chamber !
I've no confidence that what I've done is right so far, so would appreciate any comments and if anyone is an expert in this stuff ...the answer

As I've no access to any maths software and have never used it a plain english step by step explanation would be really appreciated!

Thanks

Brian

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#### AKG

Homework Helper
This is quite a puzzler, largely because you've made it very hard to understand what you're trying to do. I spent a minute trying to figure out what your diagram 1 looks like, but I don't think you refer to it anywhere in your problem. Are you just trying to simplify or re-express an expression in some desired form? Anyways, if f is supposed to be the function whose graph appears in figure 2, then you've given it correctly here:

f(t) = 10e^-0.2t - 10e^0.2t H(t-3.27) + 5.19H(t-3.27)

Next, although it's not entirely clear what you're doing, it appears you've neglected the "0.2":

f(t) = 10e^-0.2t - 10e^0.2t H(t-3.27) + 5.19H(t-3.27)

In fact, something is definitely wrong, since the expression on the right hand of your first equation has an $e^{-0.2t}$ term in it, and your second has no such term, so you have expressed the same f as two linearly independent functions, which is impossible. Also, although the coefficient of your Heaviside term goes from 5.19 to 0.197. If I knew what you were trying to do, I could do more than say what I think is wrong, I could try to suggest ways to fix it, but as I you've never told us what you're trying to do, I don't know what more to say. What's capital S by the way, and is it different from little s?

The rectangle with the lines coming out is your system. You put something in and it gives you something out. The 1/(1+0.5s) inside is called the transfer function or system function, which I'll call Z(s). To find the output Y(s) of the system, you just multiply the input by the system function.

So, Y(s) = Z(s)*F(s), assuming F(s) is the input. Then you need to do an inverse Laplace transform on Y(s) to turn it into y(t).

I think you have some problems when going from f(t) to F(s). Remember that

$$e^{-\alpha t}H(t) \rightarrow \frac{1}{s+\alpha}$$

#### Brian Moughtin

Heaviside

I've found an example in my college notes and followed it to add what I can, hope it makes sense.

This is where the drawing of the curve is shown

a) Define the function and hence find the Laplace.

f(t) = 10e^-0.2t : 0 < t 3.27

f(t) = 5.19 : t>3.27

( Found 5.19 by 10e^(-0.2 x 3.27) = 5.19

Therefore f(t) = 10e^-0.2t - 10e^0.2t H(t-3.27) + 5.19H(t-3.27)

e^-3.27 = 0.038

5.19 x 0.038 = 0.19722

Thus

f(t) = 10e^0.2t - 0.197H(t-3.27) e^-(t-3.27) +0.197H(t-3.27)

To give

F(s) = 10/(S+1) - (0.197 e^3.27s) /(S+1) + (0.197 e^3.27s)/S

b)

This input stimulus is applied to a system which has a transfer function of
1/ (1+0.5s) derive an expression for the output, and hence plot the response

i.p .....1/(1+0.5s)......o.p

o.p / i.p = 1/(1+0.5)

Therefore o.p. = 1 /(1+0.5s) x i.p.

op = 1/(1+0.5s) { 10/(s+1) - 0.197e^3.27s / (s+1) + 0.197e^3.27s / s }

= 0 / (0.5s +1) - 0.197e^3.27s / (0.5s +1) + 0.197e^3.27s / (s (0.5s+1))

T.P.F.

The line below is shown in my example from college but I don't know if it is the same for this question.

{ A/(s+1) + B /(s+1)^2 } + { C/(s+1) + D/(s+1)^2 } + { E /s + F/ (s+1) }

from this some values are worked out giving F(s)

and the example mentions converting back and shows F(t) = and some heaviside .

Hope that makes a bit more sense to someone than it does to me !

f(t) is the input to the system. You want to find the output y(t). You are given the transfer function Z(s). The transfer function is defined as the output divided by the input in the Laplace domain. Therefore, Y(s) = Z(S)*F(S). So you first need to convert f(t) to F(s), then then when you find Y(s), you need to transform it back into y(t).

I still think there is something wrong when you go from f(t) to F(s).

#### Brian Moughtin

Still struggling !

Thanks so far, I think bits are falling into place ! but I would appreciate any guidance on the next step,

"The line below is shown in my example from college but I don't know if it is the same for this question.

{ A/(s+1) + B /(s+1)^2 } + { C/(s+1) + D/(s+1)^2 } + { E /s + F/ (s+1) }"

#### Brian Moughtin

Partial Fractions

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