Why is the Laplace transform of $1-\cos(4\omega t)$ equal to $\frac{1}{s} -\frac{s}{s^2+16\omega^2}$?

Question

Why is Laplace transform of $1-\cos(4\omega t)$ equal to $\frac{1}{s} -\frac{s}{s^2+16\omega^2}$?

I can't get this. How do I calculate this?

Answer 1

Use the definition:

$$\text{Y}\left(\text{s}\right)=\mathcal{L}_t\left[y\left(t\right)\right]_{\left(\text{s}\right)}=\int_0^\infty y\left(t\right)e^{-\text{s}t}\space\text{d}t$$

So, for your problem:

$$\mathcal{L}_t\left[1-\cos\left(4\omega t\right)\right]_{\left(\text{s}\right)}=\int_0^\infty\left(1-\cos\left(4\omega t\right)\right)e^{-\text{s}t}\space\text{d}t=\int_0^\infty e^{-\text{s}t}\space\text{d}t-\int_0^\infty\cos\left(4\omega t\right)e^{-\text{s}t}\space\text{d}t$$

Now, use:

1. Substitute $u=-\text{s}t$ and $\text{d}u=-\text{s}\space\text{d}t$: $$\int_0^\infty e^{-\text{s}t}\space\text{d}t=\lim_{\text{n}\to\infty}\int_0^{-\text{s}\text{n}}-\frac{e^u}{\text{s}}\space\text{d}u=-\frac{1}{\text{s}}\lim_{\text{n}\to\infty}\int_0^{-\text{s}\text{n}}e^u\space\text{d}u=-\frac{1}{\text{s}}\lim_{\text{n}\to\infty}\left(e^{-\text{s}\text{n}}-1\right)$$
2. Use integration by parts, twice: $$\mathcal{I}=\int_0^\infty\cos\left(4\omega t\right)e^{-\text{s}t}\space\text{d}t=\lim_{\text{n}\to\infty}\left[-\frac{\cos\left(4\omega t\right)e^{-\text{s}t}}{\text{s}}\right]_{t=0}^\text{n}-\frac{4\omega}{\text{s}}\int_0^\infty\sin\left(4\omega t\right)e^{-\text{s}t}\space\text{d}t=$$ $$\lim_{\text{n}\to\infty}\left[-\frac{\cos\left(4\omega t\right)e^{-\text{s}t}}{\text{s}}\right]_{t=0}^\text{n}+\lim_{\text{m}\to\infty}\left[\frac{4\omega e^{-\text{s}t}\sin\left(4\omega t\right)}{\text{s}^2}\right]_{t=0}^\text{m}-\frac{16\omega^2}{\text{s}^2}\int_0^\infty\cos\left(4\omega t\right)e^{-\text{s}t}\space\text{d}t$$

So, we get:

$$\mathcal{I}=\lim_{\text{n}\to\infty}\left[-\frac{\cos\left(4\omega t\right)e^{-\text{s}t}}{\text{s}}\right]_{t=0}^\text{n}+\lim_{\text{m}\to\infty}\left[\frac{4\omega e^{-\text{s}t}\sin\left(4\omega t\right)}{\text{s}^2}\right]_{t=0}^\text{m}-\frac{16\omega^2}{\text{s}^2}\cdot\mathcal{I}$$

When you solve this for $\mathcal{I}$:

$$\mathcal{I}=\frac{\lim_{\text{n}\to\infty}\left[-\frac{\cos\left(4\omega t\right)e^{-\text{s}t}}{\text{s}}\right]_{t=0}^\text{n}+\lim_{\text{m}\to\infty}\left[\frac{4\omega e^{-\text{s}t}\sin\left(4\omega t\right)}{\text{s}^2}\right]_{t=0}^\text{m}}{1+\frac{16\omega^2}{\text{s}^2}}$$

When:

$$1+\frac{16\omega^2}{\text{s}^2}\ne0$$

Now, solving the limits:

• When $\Re\left[\text{s}\right]>0$: $$-\frac{1}{\text{s}}\lim_{\text{n}\to\infty}\left(e^{-\text{s}\text{n}}-1\right)=\frac{1}{s}$$
• When $\Re\left[\text{s}\right]>0$ and $\omega\in\mathbb{R}$: $$\lim_{\text{n}\to\infty}\left[-\frac{\cos\left(4\omega t\right)e^{-\text{s}t}}{\text{s}}\right]_{t=0}^\text{n}=-\frac{1}{\text{s}}$$
• When $\Re\left[\text{s}\right]>0$ and $\omega\in\mathbb{R}$: $$\lim_{\text{m}\to\infty}\left[\frac{4\omega e^{-\text{s}t}\sin\left(4\omega t\right)}{\text{s}^2}\right]_{t=0}^\text{m}=0$$

Answer 2

Hint. This comes from the standard fact that $$ \int_0^\infty e^{-(a+ib)t}\:dt=\frac1{a+ib},\qquad a>0, \,b \in \mathbb{R} \tag1 $$ then one may just write, for $s>0$, $$ \int_0^\infty\left(1-\cos\left(4\omega t\right)\right)e^{-\text{s}t}\space\text{d}t=\int_0^\infty e^{-\text{s}t}\space\text{d}t- \frac12\int_0^\infty e^{-(s+4i\omega)t}\space\text{d}t-\frac12\int_0^\infty e^{-(s-4i\omega)t}\space\text{d}t \tag2 $$ and conclude using $(1)$.

Answer 3

$Hint$: $\mathcal{L}[f(t]=\int_0^\infty e^{-st}f(t)dt$ where $Re(s)>0$

Plugging in $f(t)=e^{at}$ in the above definition, you will easily obtain $\mathcal{L}[e^{at}]=\frac{1}{s-a}$, $s-a>0$.

Now, $\mathcal{L}[1]=\mathcal{L}[e^{0t}]=\frac{1}{s}$.

and $\mathcal{L}[cos wt]=$$\mathcal{L}[\frac{e^{iwt}+e^{-iwt}}{2}]$$=\frac{1}{2}\left(\frac{1}{s-iw}+\frac{1}{s+iw}\right)$

$=\frac{s}{s^2+w^2}$.

Answer 4

We have $$\mathcal L\{1\}=\frac1s \;\;\;\text{ and } \;\;\; \mathcal L\{\cos(at)\}=\frac{s}{s^2+a^2}$$ when integrals converge. ($s>0$ in Real case).

Using the linearity, we have $$\mathcal L\{1-\cos(4\omega t)\}=\mathcal L\{1\}-\mathcal L\{\cos(4\omega t)\}.$$