Name: What does the Laplace Transform really tell us? A visual explanation (plus applications)
Uploaded: 2022-11-16T09:52:50.000Z
Duration: 20 min 25 s
Description: VoiceTubeの動画で発音を聞きながら英語表現を覚えよう！学べる英語：

Here we have a mass on a spring. if I pull it back and release, one of four things is going to happen.

First is completely sinusoidal motion, as it'll oscillate back and forth forever,

which will happen if there's no air resistance or damping present.

this happens if we put in some thick or viscous fluid which will cause the mass to exponentially decay to equilibrium,

Yes, this entire video will assume damping force is a multiple of velocity.

Three is a combination of the first two, which happens when the damping isn't as strong, or the spring itself is stronger.

In this case, the spring will oscillate but still decay in the process until it eventually settles.

four, literally anything else can happen if we have some input force or just non-ideal conditions.

We're not directly as concerned with this fourth case in this video, but we will have inputs later. The main emphasis, though,

is these three functions, which is really two, while the third is just a combination.

Now, if I had to summarize what the Laplace transform visually tells us, in just a few seconds,

Well, the Fourier transform tells us which frequencies or sinusoids are present in a function,

the Laplace transform tells us which sinusoids and exponentials are present in a function.

In fact, we're soon going to see that the Fourier transform is just a slice of the Laplace transform.

Now, here's the Fourier transform equation.

It takes in some function of time and outputs a function of Omega, that tells you which sinusoids are present in your signal.

If you put in a pure cosine curve, then out comes the function with one spike, since your original curve was one sinusoidal

Well, for real functions, these are symmetric about the y-axis,

so you actually see two spikes, and the x-coordinates will match the angular frequency of the original signal.

which tells us that takes infinitely many sinusoids to make up this function on the right.

This animation does a great job at showing the original function

being a sum of all those sinusoids, infinitely many,

while the magnitude of the Fourier transform tells us relatively how strong each of those sinusoids is,

One thing to note is that the y-intercept of the Fourier transform is the area under the curve of the original.

Remember: This is the output when Omega equals zero

and when Omega equals 0, then this term goes to 1,

and we have just the integral of the original function, aka "the area under the curve."

Now for a lot of this video, I'll be working with this equation or something similar,

because it has a sinusoid and exponential component.

However, we will assume it's zero for all negative values of 't'

and it essentially turns on at time equals zero, which avoids the function diverging to infinity.

So, when we put it into the Fourier transform, out comes a complex function

I'm not worried about the calculus in this video.

But if you just know integration by parts, you can do this,

Anyways, I'll fill the bottom here to get a new equation,

just with a separated real and imaginary component,

This function only has one input, Omega, so that can just go on a number line.

But out of the function will come a complex number

So, we need two dimensions to represent that.

If ω equals 1, then the output becomes 1 over 1 + 2i.

But that can also be rewritten as 0.2 - 0.4i,

which will be plotted on the real and imaginary axis, respectively, for ω = 1.

Now, this is a distance of roughly 0.45 from the origin, known as the magnitude,

and that we can plot against the input at ω equals 1.

This will be the magnitude of the Fourier transform.

Now for ω = 2 as the input, we get out 1/(-2 +4i),

and that will also go on the output graph.

and that will also go on the magnitude plot for ω = 2.

And lastly, I'll plug in ω = 0, which outputs just 0.5, no imaginary component,

and that will also go on the magnitude plot.

Remember, that 0.5 is just the area under the curve of our original function,

well, accounting for the fact that areas below the x-axis are negative.

If I were to plot all the magnitudes for any input ω,

we'd get this - the magnitude of the Fourier transform.

Keep this plot in mind, because it will show up soon.

But now, let's get to the Laplace transform,

so you can see just how similar it is to what we've done so far.

So here we have the Fourier transform again,

This could also be negative infinity, like above, by the way,

but especially in engineering, we usually deal with signals that, like I said earlier, turn on at time equals zero.

so we can analyze the transient response from there.

The only other difference though is we have an 's' here, instead of an 'iω'.

But s is really α +iω, which I'll substitute in,

because now we can separate that exponent into two terms.

Now look at this: the Fourier transform of something is found by multiplying that function by e^(-iωt), and integrating.

Well, in the Laplace transform, we have the integral of (something)*e^(-iωt)

Meaning, the Laplace transform of some function is just

the Fourier transform of that function times an exponential term.

Doing this for all values of α, all real numbers, gives you the entire Laplace transform.

Let's look at the Laplace transform of the same function as earlier,

But, it's nearly identical to the Fourier transform.

The only difference is: 's' represents a complex number,

which means the input will require two dimensions, one for the real and one for the imaginary component.

The output, just like before, will be complex.

So, we need four dimensions for the Laplace transform in total.

But notice that when α is zero, we get the exact same outputs as the Fourier transform,

as in this alpha-equals-zero-line, or the imaginary axis of the Laplace transform, is the Fourier transform.

Another way to see this is with the integral, because when alpha equals zero,

the exponential goes to 1, and we're left with the original Fourier transform equation.