Laplace transform ( "find x(t)" though ? )

[Color_of_Cyan]

Mod note: Please don't tinker with SIZE tags. Things are perfectly readable without them.

Homework Statement

Find x(t) = L^-1[(4e^-4s - 3)/(s² + 6s + 25)].

Homework Equations

L(x(t)) = _∞∫^∞x(t)e^-stdt
L^-1(x(s)) = (1/2π)_{(σ - ∞j)}∫^{(σ + ∞j)}[x(s)e^st]ds, "But you want to avoid this integral."
Laplace Table:
f(t), t > 0 ::::::::::::::::::::::::::::::::::::::::::::::: F(s)
_______________________________________
σ(t) :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1
u(t) :::::::::::::::::::::::::::::::::::::::::::::::::::::: (1/s)
t :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: (1/s²)
e^-at :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s + a)
e^at :::::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s - a)
te^-at ::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s+a)²
sin(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: B/(s² + B²)
cos(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: s/(s² + B²)
_______________________________________________________
Properties:
L[tf(t)] = -dF(s)/ds
L[x(at)] = (1/|a|) X (s/a), (unsure what this means at the moment though)
L[₀∫^tf(T)dT] = F(s)/s
Other properties

The Attempt at a Solution

It seems this problem is asking for doing the inverse Laplace transform on the equation
(4e^-4s - 3)/(s² + 6s + 25) but it seems I can't just do the integral. From what I hear "Laplace tables" are pretty normal to use, but I seem to be missing how they are supposed to help here, do they apply to Inverse Laplace transforms too?

 

[Mark44]

Homework Statement

Find x(t) = L^-1[(4e^-4s - 3)/(s² + 6s + 25)].

Homework Equations

L(x(t)) = _∞∫^∞x(t)e^-stdt
L^-1(x(s)) = (1/2π)_{(σ - ∞j)}∫^{(σ + ∞j)}[x(s)e^st]ds, "But you want to avoid this integral."
Laplace Table:
f(t), t > 0 ::::::::::::::::::::::::::::::::::::::::::::::: F(s)
_______________________________________
σ(t) :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1
u(t) :::::::::::::::::::::::::::::::::::::::::::::::::::::: (1/s)
t :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: (1/s²)
e^-at :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s + a)
e^at :::::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s - a)
te^-at ::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s+a)²
sin(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: B/(s² + B²)
cos(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: s/(s² + B²)
_______________________________________________________
Properties:
L[tf(t)] = -dF(s)/ds
L[x(at)] = (1/|a|) X (s/a), (unsure what this means at the moment though)
L[₀∫^tf(T)dT] = F(s)/s
Other properties

The Attempt at a Solution

It seems this problem is asking for doing the inverse Laplace transform on the equation
(4e^-4s - 3)/(s² + 6s + 25) but it seems I can't just do the integral.

I don't think they expect you to use the definition (evaluate the integral).

From what I hear "Laplace tables" are pretty normal to use, but I seem to be missing how they are supposed to help here, do they apply to Inverse Laplace transforms too?

Yes. If one column has the function, f(t), the other column has $\mathcal{L}(f(t)) = F(s)$, the Laplace transform of f. To find the inverse, match a function in the transform column with its counterpart on the same row.

Both the Laplace transform and inverse Laplace transform are linear, which means that the Laplace transform of a sum is the sum of the Laplace transforms. The same holds for the inverse Laplace transform. Break up your function into the sum of two functions, and work on those. You will also need to complete the square in the denominator to get it in the form of (s + <something>)² + <something>.

 

[rude man]

Break up the denominator into (s + a)(s + b), then use partial fraction expansion.
If the poles a and b turn out to be complex conjugates you also need to remember Euler's equation.
By no means try to do the integral. That's advanced math.
You know how to handle L^-1F(s)(e^-sT) given L^-1F(s) I hope ...

 

[Color_of_Cyan]

Sorry for getting back to this so late.

Break up the denominator into (s + a)(s + b), then use partial fraction expansion.
If the poles a and b turn out to be complex conjugates you also need to remember Euler's equation.
By no means try to do the integral. That's advanced math.
You know how to handle L^-1F(s)(e^-sT) given L^-1F(s) I hope ...

Welp, that's even more math review for me then (please bear with me again):

I'm not even sure how to do partial fractions here. It seems I can't divide out (4e-4s - 3)/(s2 + 6s + 25) or factor out (s2 + 6s + 25) easily so I use the quadratic formula, and with that I got this:

s = [-6 +- (36 - 100)^1/2]/2

s = [-6 +- (-64)^1/2]/2

s = [-6 +- (-64)^1/2]/2

s = [-6 +- (8i)]/2

s = -3 + 4i and s = -3 - 4i (and I got the same thing completing the square).

Is partial fractions with it now this?:

4e^-4s-3 = A(-4i - 3) + B(4i - 3)

 

[rude man]

I'm not even sure how to do partial fractions here. It seems I can't divide out (4e-4s - 3)/(s2 + 6s + 25) or factor out (s2 + 6s + 25) easily so I use the quadratic formula, and with that I got this:

s = [-6 +- (36 - 100)^1/2]/2

s = [-6 +- (-64)^1/2]/2

s = [-6 +- (-64)^1/2]/2

s = [-6 +- (8i)]/2

s = -3 + 4i and s = -3 - 4i (and I got the same thing completing the square).

Is partial fractions with it now this?:

4e^-4s-3 = A(-4i - 3) + B(4i - 3)

You derived the partial fraction denominators correctly.
You next rewrite 3/(s2 + 6s + 25) in partial fractions and invert those into the time domain.

 

[Color_of_Cyan]

You derived the partial fraction denominators correctly.
You next rewrite 3/(s2 + 6s + 25) in partial fractions and invert those into the time domain.

Where did you get the 3 from? Ah, do you mean change -3 +- 4i to time domain as in convert to polar & phasor form (because it's rectangular form)?

θ = tan^-1(j/x)

mag = (j² + x²)^1/2

It would be this:

-3 - 4i = (5 ∠ 53.1)

3 + 4i = (5 ∠ -53.1)

 

[rude man]

Where did you get the 3 from? Ah, do you mean change -3 +- 4i to time domain as in convert to polar & phasor form (because it's rectangular form)?

No.
Your expression to invert is
(4e^-4s - 3)/(s2 + 6s + 25)
= (-3)/(s2 + 6s + 25) + 4e^-4s/(s2 + 6s + 25).
The "3" is the numerator of the first term above with a minus sign. Deal with this term first, then do the e^-4s part afterwards.

In general: if you have a complex-conjugate pole pair as you have in this instance, if
F(s) = A/(s + a + jb) + A/(s + a - jb)
then f(t) = Ae^{-(a + jb)t} + Ae^{-(a - jb)t}
and then use the Euler relation e^jx = cos(x) + j sin(x)
to get a real f(t).

This is just a simple and obvious expansion of 1/(s + a) → e^-at for complex-conjugate pole pairs.
You have already found the poles in your post 4, to wit, s = -3 + 4i and s = -3 - 4j.

 

[Color_of_Cyan]

I take it you would end up putting the "pole expression" where it is "a + jb" then? Not really sure what goes in place of a or jb. Unless you meant to take the terms from the expression

What would happen to the 4 in the second term?

Would it then be something like this?:

f(t) = -3e^{-(-3 +4j)t} - 4e^-4se^{-(-3 - 4j)t}.

What would the x be for the Euler equation?

 

[rude man]

Not really sure what goes in place of a or jb.

a = -3
b = 4

What would the x be for the Euler equation?

x = b or -b.
Study my post 7 some more, then get going with inverse-transforming A/(s+a+jb) + A/(s+a-jb).

 

[Color_of_Cyan]

Ok so it's this:

f(t) = -3e^{-(-3 +4j)t} - 4e^-4se^{-(-3 - 4j)t}.

a = -3
b = 4

But at the same time "s = -3 + 4i and s = -3 - 4j"? Or is this just for the imaginary (s) domain ?

So F(s) is likely this then?:

3/(s - 3 + 4j) + 4e^-4s/(s - 3 - 4j).

You have to do the inverse Laplace transform in F(s) first?

You know how to handle L^-1F(s)(e^-sT) given L^-1F(s) I hope ...

Afraid I may need another hint with this, I'll get on this again soon though. Noted Mark's post that F(s) in the table is also the Laplace transform of f(t)

 

[rude man]

Ok so it's this:

f(t) = -3e^{-(-3 +4j)t} - 4e^-4se^{-(-3 - 4j)t}.But at the same time "s = -3 + 4i and s = -3 - 4j"? Or is this just for the imaginary (s) domain ?

So F(s) is likely this then?:

3/(s - 3 + 4j) + 4e^-4s/(s - 3 - 4j).

You have to do the inverse Laplace transform in F(s) first?

Wrong.

As I said before, forget about the 4 e^-s term until you've finished with the -3 term.
You have not followed what I stated in my post 7.

 

[Color_of_Cyan]

It got confusing because you have A = -3 and also -3 as one of the pole terms.

F(s) = -3/(s - 3 + 4j) - 3/(s - 3 - 4j) then

f(t) = -3e^{-(-3 + 4j)t} - 3e^{-(-3 - 4j)t}

f(t) = -3e^{(3 - 4j)t} - 3e^{(3 + 4j)t}

f(t) = -3e^(3t)e^-4jt - 3e^(3t)e^4jt

I assume you're supposed to only use the Euler relation to swap out the e^jx term in f(t) then? Would it still be done with the t you have in the F(t) formula? Because it would be e^-4jt and e^4jt instead, wouldn't it?

Otherwise it would be this..

e^Jj = cos(J) + sin(J)j; so...

e^-4j = cos(-4) + sin(-4)j = -0.65364 + 0.7568j

e^4j = cos(4) + sin(4)j = -0.65364 - 0.7568j

Is this correct so far?

 

[rude man]

You have not yet correctly done the partial fraction expansion. In other words, we have

-3/(s^2 + 6s + 25) = A/(s + a + jb) + B/(s + a - jb). (sorry, I said only "A" before but it's this) & you need to solve for A and B. If you do it right you will find B = -A.

Also, I meant to say a = 3, not -3, in post 9. The two denominator factors are
(s + s1) = (s + a + jb)
(s + s2) = (s + a - jb)
a = 3, b = 4. You're right, it's a coincidence that "3" appears in 2 places.

You need to solve for A and B before proceeding to the Euler relation thing.
Try to stick to a and b, not 3 and 4, until the end please. I get muddled myself with all those numbers floating around.

 

[Color_of_Cyan]

No worries.

-3/(s² + 6s + 25) = A/(s + a +jb) + B/(s + a - jb)

Since you have this:

(s + s1) = (s + a + jb)
(s + s2) = (s + a - jb)

can I write this instead?

-3/(s² + 6s + 25) = A/(s + s1) + B/(s + s2)

Also for partial fraction expansion the numerator degree is lower than the denominator, but how do I solve for A and B now then? Don't you set A or B equal to 0 first then solve for each of them? What is 's' (while assuming s1 and s2 are the poles I solved for earlier)?

 

[rude man]

No worries.

-3/(s² + 6s + 25) = A/(s + a +jb) + B/(s + a - jb)

Since you have this:

(s + s1) = (s + a + jb)
(s + s2) = (s + a - jb)

can I write this instead?
-3/(s² + 6s + 25) = A/(s + s1) + B/(s + s2)
[

Yes. You should. Makes things easier down the road.

Also for partial fraction expansion the numerator degree is lower than the denominator, but how do I solve for A and B now then? Don't you set A or B equal to 0 first then solve for each of them?

Yes. Look this up somewhere if you have to. It's standard partial fraction expansion.

What is 's' (while assuming s1 and s2 are the poles I solved for earlier)?

You know what s1 and s2 are. Look above.
What is "s"? I can't believe you asked that question.

 

[LCKurtz]

Since using partial fractions doesn't seem to be getting anywhere with the OP, I have another suggestion. Your basic problem is to calculate$$
{\mathcal L}^{-1}\frac 1 {s^2+6s + 9}$$If you know the formula$$
{\mathcal L}^{-1} F(s+a) = e^{-at}{\mathcal L}^{-1}F(s)$$all you have to do is complete the square on the denominator, use that formula, and look at your sine and cosine transforms. You don't need partial fractions.

 

[rude man]

Since using partial fractions doesn't seem to be getting anywhere with the OP, I have another suggestion. Your basic problem is to calculate$$
{\mathcal L}^{-1}\frac 1 {s^2+6s + 9}$$If you know the formula$$
{\mathcal L}^{-1} F(s+a) = e^{-at}{\mathcal L}^{-1}F(s)$$all you have to do is complete the square on the denominator, use that formula, and look at your sine and cosine transforms. You don't need partial fractions.

Sure. Very elegant! But that requires availability of the sine and cosine transforms. Partial fractions is more fundamental, requiring only the pair 1/(s+a) ↔ exp(-at). (Higher-order numerators can be dealt with by f'(t) ↔ sF(s) etc.)

You might as well then just go one step further and look up the inverse transform of the entire original function in one fell swoop, which as a matter of fact I did to double-check my p.f. derivation. Which, in fact, is what a practicing EE would do automatically of course, since he/she would have an extensive table to work with. Mine has 157 transform pairs.

But IMO the beginning student is I think better advised to familiarize him/herself with the partial fraction expansion method, including the realization that the method covers complex-conjugate poles as well as real ones.

 

[LCKurtz]

Of course, I understand your point. But given that the OP in fact gave table entries, my assumption would be that the exercise in question was for practice in using the tables.

 

[rude man]

Of course, I understand your point. But given that the OP in fact gave table entries, my assumption would be that the exercise in question was for practice in using the tables.

You're right, I missed seeing those on the OP's 1st post. But it still requires use of your special relation which I doubt he/she has encountered (anyway, not on his/her list.). Your approach is really clever - I hadn't thought of it - but perhaps a bit advanced for the OP's level of expertise. Thanks for your inputs!

 

[Color_of_Cyan]

I need to find a way to put more time into this, sorry. I'm a desperate guy by the way (if that matters, heh).

You know what s1 and s2 are. Look above.
What is "s"? I can't believe you asked that question.

s = -3 + 4j and s = -3 - 4j, and that's added to 's1' and 's2' in the denominator (which is it supposed to be)?

Actually wait, I think I see you just keep 's' as the variable PLUS either of the previous pole terms (s1 and s2)? It's really been some time seen I've seen any algebra II equation like this (if at all, simplifying something with partial fractions involving imaginary terms). How do you solve for A easily then?

But let me see..

-3/(s2 + 6s + 25) = A/(s + s1) + B/(s + s2)

The way I think it goes, (s - 3 + 4i)(s - 3 - 4i) composes (s² + 6s + 25)?

B = 0, then maybe s1 = -3 + 4i (while keeping +s in the denominator)

-3/(s2 + 6s + 25) = A/(s - 3 + 4i)

A = -3(s - 3 + 4i)/[(s2 + 6s + 25)]

A = -3(s - 3 + 4i)/[(s - 3 + 4i)(s - 3 - 4i)]

A = -3/(s - 3 - 4i) ?

Unless there's a mistake somewhere.

 

[rude man]

I need to find a way to put more time into this, sorry. I'm a desperate guy by the way (if that matters, heh).

Actually wait, I think I see you just keep 's' as the variable PLUS either of the previous pole terms (s1 and s2)? It's really been some time seen I've seen any algebra II equation like this (if at all, simplifying something with partial fractions involving imaginary terms). How do you solve for A easily then?

But let me see..

-3/(s2 + 6s + 25) = A/(s + s1) + B/(s + s2)

The way I think it goes, (s - 3 + 4i)(s - 3 - 4i) composes (s² + 6s + 25)?

That's right.

B = 0, then maybe s1 = -3 + 4i (while keeping +s in the denominator)

That's wrong. Show us how you solved for A and B. Study up on partial fraction expansion if you have to.

 

[Color_of_Cyan]

That's right.

That's wrong. Show us how you solved for A and B. Study up on partial fraction expansion if you have to.

Upon some more review:

-3/(s² + 6s + 25) = A/(s - 3 + 4i) + B/(s - 3 - 4i)

-3(s² + 6s + 25) = [A/(s - 3 + 4i)][(s - 3 + 4i)(s - 3 - 4i)] + [B/(s - 3 - 4i)][(s - 3 + 4i)(s - 3 - 4i)]

becomes this:

-3(s² + 6s + 25) = A(s - 3 - 4i) + B(s - 3 + 4i)

s = (3 + 4i) ----> A = 0:

-3/[(3 + 4i)² + 6(3 + 4i) + 25) = B(8i)

-3/(9 + 24i + 16i² + 18 + 24i + 25) = B(8i)

-3/(9 + 24i + -16 + 18 + 24i + 25) = B(8i)

-3/(9 + 48i + 27) = B(8i)

-3/[8i(9 + 48i + 27)] = B

s = (3 - 4i) ----> B = 0:

-3/[(3 - 4i)² + 6(3 - 4i) + 25] = A(-8i)

-3/[(3 - 4i)² + 6(3 - 4i) + 25] = A(-8i)

-3/[9 - 24i +16i² + 18 - 24i + 25] = A(-8i)

-3/[9 - 24i - 16 + 18 - 24i + 25] = A(-8i)

-3/[9 - 48i + 27] = A(-8i)

-3/[-8i(9 - 48i + 27)] = AIs there something still wrong?

 

[rude man]

Upon some more review:

-3/(s² + 6s + 25) = A/(s - 3 + 4i) + B/(s - 3 - 4i)

Wrong.
Correct is -3/(s² + 6s + 25) = A/(s + 3 + 4j) + B/(s+ 3 - 4j).
I would further prefer A/(s + 3 + 4j) + B/(s+ 3 - 4j) = A/(s + s1) + B/(s + s2)
where s1 = a + jb, s2 = a - jb.
Later, a = 3 and b = 4.
Makes the math easier.
If you do it right you will get purely imaginary numbers for A and B.
BTW since this is the engineering forum you should use j instead of i for
√(-1). i is current.

 

[Color_of_Cyan]

In post 21 you said (s - 3 + 4i)(s - 3 - 4i) composing (s2 + 6s + 25) was right. But you don't actually use them as the partial fraction denominators for A and B here? Is it only because it's "imaginary"?

I thought if you have (algebra) say (x + 2)(x - 3), then x = 3 and x = -2 (from setting them equal to 0). Here it doesn't look like it's the same way in this case (where it seems you went without flipping the sign for the pole (or variable?), for the denominator. IE s1 = 3 + 4j and s2 = 3 - 4j.)I'll take your word for it in the meantime though:

-3/(s2 + 6s + 25) = A/(s + 3 + 4j) + B/(s+ 3 - 4j)

s = -3 - 4j ----> A = 0:

-3/[(-3 - 4j)² + 6(-3 - 4j) + 25] = B/(-8j)

-3/[9 + 12j + 12j + 16j² - 18 - 24j + 25] = B/(-8j)

-3/[9 + 12j + 12j -16 - 18 - 24j + 25] = B/(-8j)

-3/[9 -16 - 18 + 25] = B/(-8j)

-3/0 = B/(-8j).

B = 24j/0 ? This is an imaginary term? Was / is there another way to find B? Or is something wrong?

s = -3 + 4j ----> B = 0:

-3/[9 - 12j - 12j + 16j² + 6(3 + 4j) + 25] = A/(8j)

-3/[9 - 12j - 12j + 16j² + 18 + 24j + 25] = A/(8j)

-3/[9 - 12j - 12j - 16 + 18 + 24j + 25] = A/(8j)

-3/[9 - 16 + 18 + 25] = A/(8j)

-3/(36) = A/(8j)

A = -24j/36

A = -2j/3

 

[rude man]

In post 21 you said (s - 3 + 4i)(s - 3 - 4i) composing (s2 + 6s + 25) was right. But you don't actually use them as the partial fraction denominators for A and B here? Is it only because it's "imaginary"?

In post 13 I corrected myself.

-3/(s² + 6s + 25) = A/(s + 3 + 4j) + B/(s+ 3 - 4j)

Correct. Also = A/(s+s1) + B/(s+s2) where
s1 = a + jb
s2 = a - jb
and a = 3, b = 4.

s = -3 - 4j ----> A = 0:

Wrong. Go

-3/(s² + 6s + 25)
= -3/(s+s1)(s+s2)] =A/(s+s1) + B/(s+s2)
Multiply both sides of the last equation by (s+s1), then let s = -s1 on both sides again.
What do you get for A? Show your work.

 

[Color_of_Cyan]

s1 = a + jb
s2 = a - jb

s1 = 3 + 4j
s2 = 3 - 4j

-3/[(s+s1)(s+s2)] =A/(s+s1) + B/(s+s2)

-3[s + s1]/[(s+s1)(s+s2)] =A[s + s1]/(s+s1) + B[s + s1]/(s+s2)

-3/(s+s2) = A + B[s + s1]/(s + s2)

-3/(s + 3 - 4j) = A + B[s + 3 + 4j]/(s + 3 - 4j)

s = -s1 = -3 - 4j; B ----> 0

= -3/(-3 - 4j + 3 -4j) = A + B[-3 - 4j + 3 + 4j]/(s + 3 - 4j)

A = -3/8j

Next:
-3/[(s+s1)(s+s2)] =A/(s+s1) + B/(s+s2)

-3[s + s2]/[(s+s1)(s+s2)] =A[s + s2]/(s+s1) + B[s + s2]/(s+s2)

-3/(s+s1) = A[s + s2]/(s + s1) + B

= -3/(s + 3 + 4j) = A[s + 3 - 4j]/(s + 3 + 4j) + B

s = -s2 = -3 + 4j; A ---> 0

-3/8j = B

Slightly different way (just forgot to multiply both sides instead of just 1):

-3/(s² + 6s + 25) = A/(s + 3 + 4j) + B/(s + 3 - 4j)

-3/(s² + 6s + 25) = A/(s + 3 + 4j) + B/(s + 3 - 4j)

[-3/(s + 3 + 4j)(s + 3 - 4j)][(s + 3 - 4j)(s + 3 + 4j)] = [A/(s + 3 + 4j)][(s + 3 - 4j)(s + 3 + 4j)] + [B/(s + 3 - 4j)][(s + 3 - 4j)(s + 3 + 4j)]

-3 = A(s + 3 - 4j) + B(s + 3 + 4j)

s = -s2 ---> (-3 + 4j); A ---> 0

-3 = B(-3 + 4j + 3 + 4j)

B= -3/8j

 

[rude man]

s1 = a + jb
s2 = a - jb

s1 = 3 + 4j
s2 = 3 - 4j

-3/[(s+s1)(s+s2)] =A/(s+s1) + B/(s+s2)

-3[s + s1]/[(s+s1)(s+s2)] =A[s + s1]/(s+s1) + B[s + s1]/(s+s2)

-3/(s+s2) = A + B[s + s1]/(s + s2)

OK up to here, but now let s = -s1. And please don't revert to using nunbers for s1 and s2 again. B=0 and A=-j3/8 are wrong.

 

[Color_of_Cyan]

-3/(-s1 + s2) = A

What now? If I substitute back now it would be this:

s1 = a + jb
s2 = a - jb

s1 = 3 + 4j
s2 = 3 - 4j

-3(-3 - 4j + 3 - 4j)

A = -3/8j (same thing I just got...)

Did you mean clear the denominator of any 'j' ?

A = [-3/8j][j/j]

A = [-3j/8j²]

A = 3j/8

 

[rude man]

-3/(-s1 + s2) = A

What now? If I substitute back now it would be this:

s1 = a + jb
s2 = a - jb

s1 = 3 + 4j
s2 = 3 - 4j

-3(-3 - 4j + 3 - 4j)

A = -3/8j (same thing I just got...)

OK, I guess I missed that. Yes, A is now correct. But try again with B.

 

[Color_of_Cyan]

-3[s + s2]/[(s+s1)(s+s2)] =A[s + s2]/(s+s1) + B[s + s2]/(s+s2)

-3/(s + s1) = A[s +s2]/(s+1) + B
-3/(-s2 + s1) = B

s1 = 3 + 4j
-s2 = -3 + 4j

-3/(-3 + 4j + 3 + 4j) = B

B = (-3/8j)

B = (3j/8)

 

[rude man]

B = (-3/8j)

B = (3j/8)

Right!
So now, given F(s) = (-j3/8)/(s+s1) + j3/8/(s+s2), where s1 = a + jb, s2 = a - jb, a = 3, b = 4,
what is f(t)?

 

[Color_of_Cyan]

That's what I was going to ask. Any hints on that, or just look at the Laplace table? What about the 4e^-4s term from the beginning?
Well just for the -3 term from the beginning (and looking back at the Laplace table) I'd say this:
f(t) ::::::::: F(s)
e^-at ::::::::: 1/(s + a)
__________________
F(s) = (-j3/8)/(s+s1) + j3/8/(s+s2),
and a = s1 and s2? What happens to the -+3j/8 ?
I think this? (Just for the -3 at the beginning though):
=-3j/8e^{(-3 - 4j)t} + 3j/8e^{(-3 + 4j)t}

 

[rude man]

That's what I was going to ask. Any hints on that, or just look at the Laplace table? What about the 4e^-4s term from the beginning?

For the n^th time, ignore that term until you've inverse-transformed the -3 term!

Well just for the -3 term from the beginning (and looking back at the Laplace table) I'd say this:
f(t) ::::::::: F(s)
e^-at ::::::::: 1/(s + a)
__________________
F(s) = (-j3/8)/(s+s1) + j3/8/(s+s2),

Perfect!

I think this? (Just for the -3 at the beginning though):
f(t) = -3j/8e^{(-3 - 4j)t} + 3j/8e^{(-3 + 4j)t}

This is 100% right!
Now, you know f(t) must be real, so what are you going to do with the above?
Hint: Euler relation! And remember from high school, e^{(x + y)} = e^xe^y.

 

[Color_of_Cyan]

For the n^th time, ignore that term until you've inverse-transformed the -3 term!
Hint: Euler relation! And remember from high school, e^{(x + y)} = e^xe^y.

Haha, okay sorry, I thought the '-3' was already done.
f(t) = -3j/8e^{(-3 - 4j)t} + 3j/8e^{(-3 + 4j)t}
= f(t) = (-3j/[8(e^-3t)(e^-4jt] + (3j/[8(e^-3t)(e^4jt)
Euler's relation:
e^xj = (cos x) + (sin x)j
(e^-4jt) = cos(-4)t + sin (-4)jt
e^4jt = cos(4)t + sin(4)jt
Do you just plug these back in and then that's -3 done?

 

[rude man]

Haha, okay sorry, I thought the '-3' was already done.
f(t) = -3j/8e^{(-3 - 4j)t} + 3j/8e^{(-3 + 4j)t}
= f(t) = (-3j/[8(e^-3t)(e^-4jt] + (3j/[8(e^-3t)(e^4jt)

At this point, factor out the e^-3t in both terms, then work with what's left using Euler.

Euler's relation:
(e^-4jt) = cos(-4)t + sin (-4)jt
e^4jt = cos(4)t + sin(4)jt

Rewrite as e^j4t = cos(4t) + j sin(4t) etc.

 

[Color_of_Cyan]

(-3e^3tj/8)[(cos(4t) + sin(4t)j) - 1/(cos(4t) + sin(4t)j)]

I feel more comfortable putting the 'j' after the terms not before them is that bad? Is this it for the -3 term now?

 

[rude man]

(-3e^3tj/8)[(cos(4t) + sin(4t)j) - 1/(cos(4t) + sin(4t)j)]

I feel more comfortable putting the 'j' after the terms not before them is that bad?

Yes, sin(4t)j = sin(4tj) which is obviously not the same as j sin(4t). The j should really always come first, as in j100 instead of 100j. For another example, Euler is always written as e^jx = cos(x) + j sin(x).

(-3e3tj/8)[(cos(4t) + sin(4t)j) - 1/(cos(4t) + sin(4t)j)]
Is this it for the -3 term now?

No. It's (-3e^-3t), not (-3e^3t). Then, factor that out from the remaining expression.
Then, what's with the division sign all of a sudden?
You are not translating your -3t term from exponential form to sines and cosines correctly.

 

[Color_of_Cyan]

I thought I could do that because it was 1/e^-3 (for BOTH denominators), and I thought I could factor both out as just e^3, And then I divided out only one of the Euler relations.. so I can't do that or I did it wrong?

 

[rude man]

I thought I could do that because it was 1/e^-3 (for BOTH denominators), and I thought I could factor both out as just e^3, And then I divided out only one of the Euler relations.. so I can't do that or I did it wrong?

All wrong.
Look, you have F(s) = A/(s+s1) + B/(s+s2).
What is the inverse transform of F(s) for this?

 

[Color_of_Cyan]

The transform of F(s) = A/(s+s1) + B/(s+s2)

is Ae^-s1(t) + Be^-s2(t)

Can I try from post 35 one more time though?:

f(t) = (-3j/[8(e^-3t)(e^-4jt)]) + (3j/[8(e^-3t)(e^4jt)])

(1/e^-3t)[ (-j3/8(cos(-4t) + jsin(-4t))) + (j3/8(cos(4t) + jsin(4t))) ]

 

[rude man]

The transform of F(s) = A/(s+s1) + B/(s+s2)

is Ae^-s1(t) + Be^-s2(t)

No. Don't say "s₁(t). Say "s₁t" etc. So correct would be
"The inverse transform of F(s) = A/(s+s₁) + B/(s+s₂)
is Ae^-s₁t + Be^-s₂t ... (1)

Can I try from post 35 one more time though?:

f(t) = (-3j/[8(e^-3t)(e^-4jt)]) + (3j/[8(e^-3t)(e^4jt)])

This is almost correct. Fix up your parentheses.

(1/e^-3t)[ (-j3/8(cos(-4t) + jsin(-4t))) + (j3/8(cos(4t) + jsin(4t))) ]

How did you manage to get (1/e^-3t)?
Also, your parentheses don't look right. Too many of them!
Other than that this looks OK now.
So now combine (-j3/8)(cos(-4t) + jsin(-4t)) + (j3/8)(cos(4t) + jsin(4t)) using Euler to come up with a real function of time. In other words, all the j's have to disappear.

 

[rude man]

Since you are seemingly having a lot of trouble with this complex math maybe you should look at posts 2 and 16 again. It would avoid all the complex math if you do it that way. Let us know if you want this explained to you.

 

[LCKurtz]

Since you are seemingly having a lot of trouble with this complex math maybe you should look at posts 2 and 16 again. It would avoid all the complex math if you do it that way. Let us know if you want this explained to you.

I have been waiting for the OP to get done with this method to suggest that. But note that I had a typo in post #16 which I can't edit. It should have said complete the square on$$\frac 1 {s^2+6s+25}$$and use that shifting formula.

 

[rude man]

I have been waiting for the OP to get done with this method to suggest that. But note that I had a typo in post #16 which I can't edit. It should have said complete the square on$$\frac 1 {s^2+6s+25}$$and use that shifting formula.

Hadn't noticed that. Right.

EDIT: If the OP wants to investigate it that way, is it OK with you if I turn him over to you?

 

[Color_of_Cyan]

Since you are seemingly having a lot of trouble with this complex math maybe you should look at posts 2 and 16 again. It would avoid all the complex math if you do it that way. Let us know if you want this explained to you.

I'd be able to make it look better here on the forums if only I knew the way Kurtz posted his fractions (like on post 43 on here)...

$$\frac {1} {e^{-3t}} ( \frac {-j3} {8(cos(-4t) + jsin(-4t))} + \frac {j3} {8(cos(4t) + jsin(4t))})$$

Much better :) And it still looks like a challenge to get rid of all the 'j's (is that even possible?). What did you mean use "Euler"?

I have been waiting for the OP to get done with this method to suggest that. But note that I had a typo in post #16 which I can't edit. It should have said complete the square on$$\frac 1 {s^2+6s+25}$$and use that shifting formula.

You said take the square (s² + 6s + 9 + 16) and do this?

L^-1F(S + a) = e^-atL^-1F(s)

So is s = (s+3)² and a = 16?

What do you have for L^-1F(s) then?

Don't go anywhere Rude Man!

 

[rude man]

I'd be able to make it look better here on the forums if only I knew the way Kurtz posted his fractions (like on post 43 on here)...

$$\frac {1} {e^{-3t}} ( \frac {-j3} {8(cos(-4t) + jsin(-4t))} + \frac {j3} {8(cos(4t) + jsin(4t))})$$

Much better :) And it still looks like a challenge to get rid of all the 'j's (is that even possible?). What did you mean use "Euler"?

This is "much better"? It's not. Not at all. It's a disaster!
Euler: e^jx = cos(x) + jsin(x)
Please go back to post 41, last part.
I'm going to let LCKurtz guide you thru his approach.

 

[LCKurtz]

You said take the square (s² + 6s + 9 + 16) and do this?

L^-1F(S + a) = e^-atL^-1F(s)

So is s = (s+3)² and a = 16?

What do you have for L^-1F(s) then?

Don't go anywhere Rude Man!

You have $\mathcal L^{-1}\frac 1 {(s+3)^2 + 4^2}$ so that fraction is $F(s+3)$. So what does the equation I highlighted in red tell you for this problem?

 

[Color_of_Cyan]

This is "much better"? It's not. Not at all. It's a disaster!
Euler: e^jx = cos(x) + jsin(x)
Please go back to post 41, last part.
I'm going to let LCKurtz guide you thru his approach.

The code works fine for me in Chrome, and I simplified it the way you said before, didn't I? Or was I supposed to simplify the 1/e^{3 +- 4j} out again?

You have $\mathcal L^{-1}\frac 1 {(s+3)^2 + 4^2}$ so that fraction is $F(s+3)$. So what does the equation I highlighted in red tell you for this problem?

Ah, so you use the table as much as possible for wherever it's L^-1F(s)?

Because looking back at it, F(s) with b/(s² + a²) = sin bt

so sin(4t) = L^-1 of that ?

Then it's sin(4t)e^-4t?

 

[LCKurtz]

Ah, so you use the table as much as possible for wherever it's L^-1F(s)?

Because looking back at it, F(s) with b/(s² + a²) = sin bt

Those aren't equal. You have $s$ on the left and $t$ on the right.

so sin(4t) = L^-1 of that ?

Then it's sin(4t)e^-4t?

It's close but you are being very careless. What is $a$? What is $b$? You have to match up the table entries exactly with your problem.

 

[NascentOxygen]

CoC, it's often helpful when you can check your working as you proceed to learn from your mistakes, and wolfram alpha is of incalculable value here. A couple of examples apposite to the mathematics you have been using:-

e.g., http://m.wolframalpha.com/input/?i=partial fractions -3/((s+3+i4)(s+3-i4))&x=0&y=0

e.g., http://m.wolframalpha.com/input/?i=inverse Laplace Transform 1/((s+3)^2+4^2)&x=0&y=0
maybe scroll down to alternate forms (for some reason if I use a lower-case "t" in spelling "transform" it doesn't list the explicit "sin" solution.)

Good luck with your studies!