Question

r(t)

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1 | ________
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0 1 2 3 4
t


I dont know how to read step functions.. How do I read this graph/find the equation of this graph????? PLEASE HELP

Answer 1

Came out messed up.

Point is,

The switch starts at t=1 and ends at t=2.

Answer 2

I want to know how to write out the equation for these graphs

Answer 3

r(t)=u(t-1)-u(t-2)

Answer 4

u is the unit step func.

Answer 5

ooh
i see i see.. and for example...

let's say i wanted to graph 2t-2u(t-1).. would it become...

(drawing it on paint)

Answer 6

I think i've done the step function wrong... how would i do it?

Answer 7

If I had a million guesses i'd never be able to do that... How come it goes up at a gradient like that?

Answer 8

see u(t-1) starts at t=1
before t=1
your graph will be like y=2t...

Answer 9

at t=1
r(t)=2*1-2=0
after t=1 ur u(t-1) has a const. value 1
so then ur equation r(t)=2(t)-2

Answer 10

But why does the 2u(t-1) go up like that? I thought it was meant to be straight line :S Man that's confusing lol

Answer 11

no...man ....u have a term 2t...that give u st. line...

Answer 12

hello ggaborne...........

Answer 13

Yep Thanks alot.. Helps!

Answer 14

I have another question...

Answer 15

I want to find the Inverse Laplace of

\[(e ^{-t} - (e ^{-2t}) + (t^2))/(t)(t^2 -3t + 2)\]

I was going to use partial fraction decomposition... but I can't get it working.

Answer 16

now i'm really confused....but still it appears to me that partial fraction would work...

Answer 17

Constantly getting the wrong answer... The question is...

y'' - 3y' + 2y = r(t)

y(0) = 1, y'(0) = 3

And r(t) is the graph i drew in the first post... The one where dipan said it is u(t-1) - u(t-2)

Answer 18

how did u approached?

Answer 19

I took the laplace of

y'' - 3y' + 2y = u(t-1) - u(t-2)

This gave me (t^2 - 3t + 2)Y(t) - t = (e^-t - e^-2t)/t

Rearranging to get Y(t) on one side...
I get the \[(e ^{-t} −(e ^{-2t})+(t^2))/(t)(t^2−3t+2)\]

Then I get using partial..
\[[(e ^{-t} −(e ^{-2t})+(t^2))\] = A(t-1)(t-2) + B(t)(t-2) + C(t)(t-1)

and i'm not sure how to find my unknowns A, B and C

Answer 20

But even if i do find these constants.. A B and C.... it will turn out like the answer in the back of the book..

Answer 21

hey...wait wait... u can't write [(e^−t−(e^−2t)+(t^2))
= A(t-1)(t-2) + B(t)(t-2) + C(t)(t-1)

better to take this in the form of \[\frac{e^{-at}}{(t+b)^p}\]

Answer 22

i'm confused...

Answer 23

t(t^2-3t+2)=t(t-1)(t-2)
\[\frac{e^{-t}-e^{-2t}+t^2}{t(t-1)(t-2)}=\frac{e^{-t}}{t(t-1)(t-2)}-\frac{e^{-2t}}{t(t-1)(t-2)}+\frac{t^2}{t(t-1)(t-2)}\]

Answer 24

And then...? Find the partial decomposition of each one individually?

Answer 25

yes but dont make any thing with the exponential terms...

Answer 26

and \[\frac{e^{-at}}{(t+b)^p}\]gives u a direct form to solve it u can go to wiki for the details of that formula...

Answer 27

Could you please give me the link from wiki? I don't know where I'd find it..