Ytukyg

I'm trying to solve this function using the Laplace Transform theorem.

$f(t) = cos2(t-frac{1}{8}pi)$

Sure I could just use the table, which would give me the answer right away, but pretend that I do not have that kind of luxury, I'd like to know how to approach these types of functions, more specifically, $f(t)=cos(at+b)$ or $f(t)=sin(at+b)$.

Thank you in advance.

To use the Laplace Transform, I recommend you use the fact that $cos(t) = frac{e^{it}+e^{-it}}{2}$ (this can be easily proven using Euler's identity). Start by taking Laplace transform of $cos(t)$:

$I =int_0^{infty}e^{-st}cos(t)dt = int_0^{infty} e^{-st}frac{e^{it}+e^{-it}}{2}dt = int_0^{infty}frac{e^{t(i-s)}+e^{-t(s+i)}}{2}dt = int_0^inftyfrac{e^{t(i-s)}}{2}dt +int_0^inftyfrac{e^{-t(i+s)}}{2}dt$

Solving the integral, we get that:

$I = frac{e^{-t(-i+s)}}{2(i-s)}Big|_0^infty + frac{e^{-t(s+i)}}{-2(s+i)}Big|_0^infty = frac{1}{2}Big[frac{-1}{i-s}+frac{1}{s+i}Big] = frac{1}{2}Big[frac{-(s+i)+(i-s)}{(i-s)(s+i)}Big]= frac{1}{2}Big[frac{-2s}{i^2-s^2}Big] = frac{1}{2}Big[frac{2s}{-i^2+s^2}Big]=frac{1}{2}Big[frac{2s}{s^2+1}Big]= frac{s}{s^2+1}$

This corresponds to the item on the table.

To get the Laplace transform of $cos(at+b)$,just use the fact that $cos(at+b)= cos(b)cos(at)-sin(b)sin(at)$, and pull out constants & apply linearity, essentially repeating the calculation above.

$require{cancel}$
Write

begin{eqnarray}
f(t) &=& cos2(t - pi/8) = cos(2t - pi/4) = cos(2t)cos(pi/4) - sin(2t)sin(pi/4) \
&=& frac{sqrt{2}}{2}cos(2t) - frac{sqrt{2}}{{2}}sin(2t)
end{eqnarray}

We need to calculate the transform of $cos(2t)$ and $sin(2t)$, will do the first one and leave the second one for you to practice. In general

$$
mathcal{L}[f] = int_0^{+infty}{rm d}s ~e^{-st} f(t)
$$

So we have

$$
mathcal{L}[cos (2t)] = int_0^{+infty}{rm d}t~e^{-st}cos(2t)
$$

Now do integration by parts $u = e^{-st}$, ${rm d}v = cos(2t)$, so that

begin{eqnarray}
mathcal{L}[cos (2t)] &=& cancelto{0}{left.frac{1}{2}e^{-st}sin(2t)right|_0^{+infty}} - int_0^{+infty}{rm d}t~left[-s e^{-st}right]left[frac{1}{2}sin(2t)right] \
&=& frac{s}{2}int_0^{+infty}{rm d}t~ e^{-st}sin(2t)
end{eqnarray}

Let's do yet another integration by parts $u = e^{-st}$, ${rm d}v = sin(2t)$

begin{eqnarray}
mathcal{L}[cos (2t)] &=& frac{s}{2}left{left. -frac{1}{2}e^{-st}cos(2t) right|_{0}^{+infty} - int_0^{+infty} {rm d}t~ left[-s e^{-st}right]left[ -frac{1}{2}cos(2t)right] right} \
&=& frac{s}{2}left{ frac{1}{2} - frac{s}{2}int_0^{+infty}{rm d}t~e^{-st}cos(2t)right} \
&=& frac{s}{2} - frac{s^2}{4}mathcal{L}[cos (2t)] \
left(1 + frac{s^2}{4} right) mathcal{L}[cos (2t)]&=& frac{s}{4} \
mathcal{L}[cos (2t)] &=& frac{s/4}{1 + s^2/4} = frac{s}{4 + s^2}
end{eqnarray}

The other one is pretty similar to this one

$$f(t)=sin(at+b)$$

$$mathcal{L}(f) = int_0 ^infty e^{-st} f(t) dt=int_0 ^infty e^{-st} sin(at+b) dt$$

You can solve $$I=int e^{-st} sin(at+b) dt$$ ...by using integration by parts:

$$u=e^{-st},space du=-se^{-st},space dv=sin(at+b), space v=-{cos(at+b) over a}$$

$$I=-{e^{-st}cos(at+b) over a}-frac saint e^{-st} cos(at+b) dt$$

Again:

$$u=e^{-st},space du=-se^{-st},space dv=cos(at+b), space v={sin(at+b) over a}$$

$$I=-{e^{-st}cos(at+b) over a}-frac saleft( {e^{-st}sin(at+b) over a} +frac sa int e^{-st} sin(at+b) dt right)$$

$$I=-{e^{-st}cos(at+b) over a}-frac saleft( {e^{-st}sin(at+b) over a} +frac sa I right)$$

$$(1+{s^2over a^2}) I=-{ae^{-st}cos(at+b) +se^{-st}sin(at+b)over a^2}$$

$$I=-{ae^{-st}cos(at+b) +se^{-st}sin(at+b)over s^2 + a^2}$$

Finally:

$$mathcal{L}(f)=I |_0^infty=I(infty)-I(0)={acos b +ssin bover s^2 + a^2}$$

In your case:

$$f(t) = cos(2t-frac{pi}{4})=sin(fracpi2 - 2t+frac{pi}{4})=sin(-2t+frac{3pi}{4})implies a=-2, b=frac{3pi}{4}$$

$$mathcal{L}(f)={-2cos frac{3pi}{4} +ssin frac{3pi}{4}over s^2 + 4}=frac{sqrt{2}+sfrac{sqrt{2}}{2}}{s^2 + 4}={sqrt{2}over2}frac{s+2}{s^2+4}$$

you can check the final result on WolframAlpha: https://www.wolframalpha.com/input/?i=laplace+transform+cos(2t-pi%2F4)