1. ## [SOLVED]Laplace transform problem

Compute the Laplace transform of $\displaystyle cos(at-b)$, where $\displaystyle a,b$ are real numbers.

I've tried this several times and I am ending up with an answer that I know is wrong. I seem to be making the same mistake over and over.

I set up the problem like this:

$\displaystyle \int{e^{-st}cos(at-b)}$

and then use integration by parts.

Judging by WolframAlpha my answer should be

$\displaystyle \frac{s}{s^2+1}$

but I keep getting answers similar to this:

$\displaystyle \frac{-s}{s^2-a^2}e^{-st}[cos(at-b)+\frac{a}{s}sin(at-b)]$

Help would be greatly appreciated.

2. Two things
1. the deniminator of your antiderivative should be s^2 + a^2

2. You nedd to evaluate this between 0 and infinity

See attachment

3. Alternatively

cos(at-b) = cos(at)cos(b) + sin(at)sin(b)

We can use the basic formulas for cos(at) and sin(at)

L{cos(at-b)} = L{cos(at)cos(b)} + L {sin(at)sin(b)}

= cos(b)s/(s^2+a^2) + asin(b)/(s^2+a^2)

4. Thanks so much for the help!

When evaluating from 0 to infinity:
$\displaystyle$
$\displaystyle \frac {-scos(at-b)e^{-st} + asin(b-at)e^{-st}}{a^2+s^2}$

1. Why does the first part of the evaluation (with infinity) simply get discarded?

2. When evaluating 0 with
$\displaystyle scos(at-b)$, why does it become $\displaystyle -scos(b)$ and not $\displaystyle scos(-b)$

I attatched a picture to help explain where I am confused.

5. neglecting the constants lim e^(-t) cos(t) or lim e^(-t) sin(t) are 0

This comes from squeezing theorem - e^(-t) < e^(-t) cos(t) < e^(-t)

lim -e^(-t) = lime^(-t) = 0

Remember your under damped harmonic oscillator

2. you have - s cos(at-b)

a. because you are evaluating at the lower limit you subtract

you have scos(-b)

b cos(-b) = cos(b) you have scos(b)

6. Originally Posted by pkrogel
Compute the Laplace transform of $\displaystyle cos(at-b)$, where $\displaystyle a,b$ are real numbers.

I've tried this several times and I am ending up with an answer that I know is wrong. I seem to be making the same mistake over and over.

I set up the problem like this:

$\displaystyle \int{e^{-st}cos(at-b)}$

and then use integration by parts.

Judging by WolframAlpha my answer should be

$\displaystyle \frac{s}{s^2+1}$

but I keep getting answers similar to this:

$\displaystyle \frac{-s}{s^2-a^2}e^{-st}[cos(at-b)+\frac{a}{s}sin(at-b)]$

Help would be greatly appreciated.
Well, that can't be right because the Laplace transform is a function of s only, not t. Did you not evaluate at s= 0 and $\displaystyle \infty$?

To integrate by parts, let $\displaystyle u= e^{-st}$ and $\displaystyle dv= cos(at- b)dt$. Then $\displaystyle du= -se^{-st}dt$ and $\displaystyle v= \frac{1}{a}sin(at- b)$. $\displaystyle \int_0^\infty e^{-st}cos(at-b)dt= \frac{s}{a}e^{-st}\right|_0^\infty+ \frac{s}{a}\int_0^\infty e^{-st}sin(at-b)dt$$\displaystyle = -\frac{s}{a}sin(b)+ \frac{s}{a}\int_0^\infty e^{-st}sin(at-b)dt. To do [/b]that[/b] integral, let \displaystyle u= e^{-st} again and let \displaystyle dv= sin(at-b)dt. Then \displaystyle du= -se^{-st}dt and [tex]v= -\frac{1}{a}cos(at- b). \displaystyle \int_0^\infty e^{-st}sin(at-b)dt=frac{1}{a}e^{-st}cos(at-b)\right|_0^\infty- \frac{s}{a}\int_0^\infty e^{-st}cos(at-b)dt$$\displaystyle = \frac{1}{a}cos(b)- \frac{s}{a}\int_0^\infty e^{-st}cos(at-b)dt$.

Putting that into the first integral, $\displaystyle \int_0^\infty e^{-st}cos(at-b)= -\frac{s}{a}sin(b))- \frac{s}{a}\int_0^\infty e^{-st}cos(at-b)dt$.
Adding that integral on the right to the left side gives $\displaystyle (1+ \frac{s}{a})\int_0^\infty e^{-st}cos(at-b)dt= \frac{a+s}{a}\int_0^\infty e^{-st}cos(at-b)dt= -\frac{s}{a}sin(b))$ so $\displaystyle \int_0^\infty e^{-st}cos(at-b)dt= -\frac{a}{a+s}\frac{s}{a}sin(b)= \frac{s}{s+a}sin(b)$.

That's the basic idea- better check for errors.

7. Alternatively

cos(at-b) = cos(at)cos(b) + sin(at)sin(b)

We can use the basic formulas for cos(at) and sin(at)

L{cos(at-b)} = L{cos(at)cos(b)} + L {sin(at)sin(b)}

= cos(b)s/(s^2+a^2) + asin(b)/(s^2+a^2)

Gives the correct Laplace transform -- the attachment of the post
previous to this one does so using the definition of laplace transform

8. For the details of the integration see the attachment

9. thanks so much guys I finally got it!