## Monday Math 105

The Laplace Transform
Part 6: Frequency shifting, time shifting, and scaling properties

Expanding upon last time’s work on the Laplace transform of the exponential, we can prove quickly the “frequency shifting” property of the Laplace transform. Let real function f(t), t≥0, have Laplace transform $F(s)=\mathcal{L}\left[f(t)\right](s)=\int_0^{\infty}e^{-st}f(t)\,dt$. Then, for real number a, we see that
$\begin{eqnarray}\mathcal{L}\left[e^{at}f(t)\right](s)&=&\int_0^{\infty}e^{-st}e^{at}f(t)\,dt\\&=&\int_0^{\infty}e^{-(s-a)t}f(t)\,dt\\\mathcal{L}\left[e^{at}f(t)\right](s)&=&F(s-a)\end{eqnarray}$,
and thus multiplication by an exponential in the time domain equals a translation in the s domain.
This can be combined with our results for sine and cosine to get
$\mathcal{L}\left[e^{at}\cos(\omega{t})\right](s)=\frac{s-a}{(s-a)^2+\omega^2}$
$\mathcal{L}\left[e^{at}\sin(\omega{t})\right](s)=\frac{\omega}{(s-a)^2+\omega^2}$.
And our results for powers of t to get
$\mathcal{L}\left[t^re^{at}\right](s)=\frac{\operatorname{\Gamma}(r+1)}{\left(s-a\right)^{r+1}}$.
Note that for all of these, the shift in the “frequency” domain entails an equal shift in the region of convergence; if $\mathcal{L}\left[f(t)\right](s)=F(s)$ for $\Re(s)\gt{b}$, then $\mathcal{L}\left[e^{at}f(t)\right](s)=F(s-a)$ for $\Re(s)\gt{b+a}$.

Next, let’s consider what happens when we scale our time variable. Again, we have f(t), t≥0, with Laplace transform $F(s)=\mathcal{L}\left[f(t)\right](s)=\int_0^{\infty}e^{-st}f(t)\,dt$. Now, let us find the Laplace transform for f(at), with a>0.
$\mathcal{L}\left[f(at)\right](s)=\int_0^{\infty}e^{-st}f(at)\,dt$.
Making u-substitution u=at, then
$\begin{eqnarray}\mathcal{L}\left[f(at)\right](s)&=&\int_0^{\infty}e^{-s\frac{u}{a}}f(u)\,\frac{du}{a}\\&=&\frac{1}{a}\int_0^{\infty}e^{-\frac{s}{a}u}f(u)\,du\\\mathcal{L}\left[f(at)\right](s)&=&\frac{1}{a}F\left(\frac{s}{a}\right)\end{eqnarray}$.
As with our other result, the region of convergence will be transformed along with the scaling.

We found what happens in the time domain when the transform is shifted in the “frequency” domain, but what happens when we shift in the time domain? Recall that our function f(t) is defined for t≥0; when we shift our function forward, we must set those portions that are at t<0 before our shift to zero, so that our translated function is f(t-τ)H(t-τ), where H(t) is the Heaviside step function, and τ>0 is our time delay.
Integrating as we did for the transform of the Heaviside step function,
$\begin{eqnarray}\mathcal{L}\left[f\left(t-\tau\right)H\left(t-\tau\right)\right](s)&=&\int_0^{\infty}e^{-st}f\left(t-\tau\right)H\left(t-\tau\right)\,dt\\&=&\int_{\tau}^{\infty}e^{-st}f\left(t-\tau\right)\,dt\end{eqnarray}$.
Now, making u-substitution u=t-τ, we find
$\begin{eqnarray}\mathcal{L}\left[f\left(t-\tau\right)H\left(t-\tau\right)\right](s)&=&\int_{\tau}^{\infty}e^{-st}f\left(t-\tau\right)\,dt\\&=&\int_{0}^{\infty}e^{-s(u+\tau)}f(u)\,du\\&=&e^{-s\tau}\int_{0}^{\infty}e^{-su}f(u)\,du\\\mathcal{L}\left[f\left(t-\tau\right)H\left(t-\tau\right)\right](s)&=&e^{-s\tau}F(s)\end{eqnarray}$.
Thus, a time shift corresponds to multiplication by an exponential in the “frequency” domain.

### 5 Responses to “Monday Math 105”

1. Monday Math 108 « Twisted One 151's Weblog Says:

[...] a matter of finding the inverse transform of that last term. Using partial fractions, , and from here, we see , and so . Next, we consider integrals. Let us have, as before, f(t), t>0, with Laplace [...]

2. Monday Math 111 « Twisted One 151's Weblog Says:

[...] know from here that the Laplace transform for cosine is , so Thus, via the frequency shift formula here, . Considering, then, our frequency differentiation formula, . Now consider the “sine [...]

3. Monday Math 114 « Twisted One 151's Weblog Says:

[...] and so we see that , and using from here that the Laplace transform of an exponential is , and from here that the frequency-shifted sine and cosine formulas are and , we see that we have , and we have [...]

4. Monday Math 115 « Twisted One 151's Weblog Says:

[...] we found, using the periodic function rule, that the square wave has Laplace transform , so using the scaling rule, we see that the Laplace transform of our function f(t) is . So, to continue with our solution, , [...]

5. Monday Math 117 « Twisted One 151's Weblog Says:

[...] and proof by induction to show that the general formula is , for integer n≥0. Using the scaling rule and some algebra, this generalizes to [...]