Laplace Transform - asking for a tip

Question

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.

Answer 1

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

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

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

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

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

Again:

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

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

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

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

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

Finally:

$$\mathcal{L}(f)=I\ |_0^\infty=I(\infty)-I(0)={a\cos b +s\sin b\over s^2 + a^2}$$

In your case:

$$f(t) = \cos(2t-\frac{\pi}{4})=\sin(\frac\pi2 - 2t+\frac{\pi}{4})=\sin(-2t+\frac{3\pi}{4})\implies a=-2, \ b=\frac{3\pi}{4}$$

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

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

Answer 2

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 $cos(t)$:

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

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.

Answer 3

$\require{cancel}$ 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