Question

Hi everyone! Can anyone please tell me how they expanded f(t) please? I posted the work. Thanks! :o)

Answer 1

@rational @Michele_Laino Thanks! :O)

Answer 2

Hi Michele! :o)

Answer 3

Hi!

Answer 4

Do you think you can help me figure out how the got their final expansion of f(t) ?

Answer 5

I am totally fine with the 2nd to the last f(t), but I can't for the life of me get the last expansion!

Answer 6

have you tried to make the Fourier transform of both sides?

Answer 7

we don't know how to do that...this is just using laplace transform to solve differential equations is all

Answer 8

I just need to understand how they got the last f(t) so I can finish the problem

Answer 9

I am stuck at this point

Answer 10

is u(t) the step function, right?

Answer 11

yes

Answer 12

I think that the functions listed below have to be substituted into your differential equation, and then you have solve it

Answer 13

oops...you have to solve it

Answer 14

yes they do...I don't need help with that

Answer 15

I just need to know how they did the algebra between f(t) = t-tU(t-1) and the last f(t)

Answer 16

does that make sense what I need?

Answer 17

ok! I understand!

Answer 18

I mean we know the "t" function is "on", then shuts "off" hence t-tU(t-1)
but then the function is actually at zero for t>1
don't know where they got all the extra junk

Answer 19

please you have to change variable, like this:
\[\Large \tau = t - 1\quad \to \quad t = \tau + 1\]
where \tau is the new variable

Answer 20

there are no "tau's"...we only used those in "convolution" this section has no use of "taus"

Answer 21

let me show you another example and then it will make sense what I am asking...one sec

Answer 22

it is a change of variable only, we have the subsequent steps:
\[\Large \begin{gathered}
t - tu\left( {t - 1} \right) = \left( {\tau + 1} \right) - \left( {\tau + 1} \right)u\left( \tau \right) = \hfill \\
= \tau + 1 - \tau u\left( \tau \right) - u\left( \tau \right) = \hfill \\
= t - \left( {t - 1} \right)u\left( {t - 1} \right) - u\left( {t - 1} \right) \hfill \\
\end{gathered} \]

Answer 23

okay, I think I understand...please do me a favor, look at this new post to see an earlier problem dealing only with 1's and zero's and look at the f(t)
then maybe you can see why I am confused and why I don't understand why they are changing variables...one sec, look at this

Answer 24

it is the same post

Answer 25

sorry

Answer 26

see how easy that one is compared to the first one :o)

Answer 27

they are three expressions of the same function. Your teacher has wrote them in order to let you know how to get the last expression for the function f(t). Namely the third expression is the last step, whereas the first and second expressions are the first and second steps respectively, needed to write the third step

Answer 28

or, in other words, the last expression, is a more compact form for function listed in the first or the second step

Answer 29

well if I am stuck at t-tU(t-1)
cam you please tell me the very next step I have to do?

Answer 30

If you prefer, you can use the function listed above, namely:
\[\Large f\left( t \right) = \left\{ {\begin{array}{*{20}{c}}
{t,\quad 0 \leqslant t < 1} \\
{t - t = 0,\quad t \geqslant 1}
\end{array}} \right.\]

Answer 31

that's fine...okay, go ahead

Answer 32

whereas for second question, you can use this expression, for your function f(t):
\[\Large f\left( t \right) = \left\{ {\begin{array}{*{20}{c}}
{1,\quad 0 \leqslant t < 1} \\
{ - 1,\quad t \geqslant 1}
\end{array}} \right.\]

Answer 33

I need to know how to physically write down the process of going from t-tU(t-1) to
t-(t-1)U(t-1)-U(t-1)
can you help me with each baby step step?

Answer 34

I know you said they changed variables or something but I don't understand what that is

Answer 35

as I wrote before you have to change variable, for example:
\[ \Large t - 1 = \tau \] so I can write this:
\[\Large t = \tau + 1,\] and substituting into the formula for f(t), I get:
\[\Large \begin{gathered}
t - tu\left( {t - 1} \right) = \left( {\tau + 1} \right) - \left( {\tau + 1} \right)u\left( \tau \right) = \hfill \\
\hfill \\
= \tau + 1 - \tau u\left( \tau \right) - u\left( \tau \right) \hfill \\
\end{gathered} \]
Now I make this substitution:
\[\Large t - 1 = \tau \]
and I can write:
\[\Large \tau + 1 - \tau u\left( \tau \right) - u\left( \tau \right) = t - \left( {t - 1} \right)u\left( {t - 1} \right) - u\left( {t - 1} \right)\]

Answer 36

do you know why you have to change the variables when on the other easier problem I showed you, you don't have to do any of that?

Answer 37

why don't you have to change variables on the easier problem?

Answer 38

I changed variable, in order to show you how to do to going from this expression:

f(t)=t-tu(t-1)

to this one:

f(t)=t-(t-1)u8t-1)-u(t-1)

Answer 39

and that is what they did also?

Answer 40

is there an easier way?

Answer 41

no, I think that it is the only way

Answer 42

oops..the unique way

Answer 43

oops.. another typo, here is the right expression
f(t)=t-(t-1)u(t-1)-u(t-1)

Answer 44

how did you go from (T+1)-(T+1)U(T) to the next step?
did you factor out (T+1) or something because I can't get what you got

Answer 45

you have to do the multiplications indicated, namely:
(T+1)-(T+1)U(T) =T+1-T U(T) - U(T)

Answer 46

ok, i see it now..then after that you change variables again?

Answer 47

then you have to return to old variable t, namely:
t=T+1, so T=t-1, and we have to replace T with t-1, like this:
t-1+1-(t-1)U(t-1)-U(t-1)

Answer 48

I got it! :o)
Thanks so much Michele! :o)

Answer 49

Thank you! :) @SinginDaCalc2Blues