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  • This video was sponsored by brilliant

  • 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.

  • Two is exponential decay;

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

  • without overshooting.

  • 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.

  • And

  • 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,

  • it'd be this:

  • 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.

  • If you put in a non periodic function,

  • you'll get out something more complex,

  • 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,

  • where the sum is seen in yellow;

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

  • for any given frequency.

  • 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

  • that I won't graph quite yet.

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

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

  • just treat 'i' as a constant.

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

  • just with a separated real and imaginary component,

  • and this is what we'll work with.

  • 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

  • with the real and imaginary component.

  • So, we need two dimensions to represent that.

  • So, let's put in some values now.

  • 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),

  • which simplifies to -0.1 -0.2i,

  • and that will also go on the output graph.

  • The magnitude for this is roughly 0.224,

  • 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,

  • and this is the Laplace transform.

  • Nearly identical.

  • 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.

  • To see what I mean by this,

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

  • which, again, I'm not gonna derive.

  • 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.

  • So, if I plug in something on that axis,

  • We should get the same magnitudes as before.

  • Like, if we put in 0 for α and 0 for ω, which goes here on the input,

  • then out comes 0.5, that area under the curve, which goes here on the output.

  • And, since I don't really have another dimension, I'll just write that magnitude above the corresponding input.

  • If I plug in 1 for ω and 0 for α, which is also on the same axis,

  • then out comes the 0.2 -0.4i from before, which on the output is a magnitude of roughly 0.45.

  • And I won't show it, but if I did plug in this point where ω equals 2,

  • the output would have a magnitude of about 0.224

  • If I was using a third dimension to plot those magnitudes,

  • then we'd see the exact same function as before, above that imaginary axis.

  • We could also make a contour plot though.

  • Like just imagine looking down on this graph, where colors are assigned to different y-values,

  • because then those colors can represent the magnitudes.

  • Then, if I move the green line over to, let's say, alpha equals -0.5,

  • the outputs will be the Fourier transform of something slightly different.

  • Let me just put the original Laplace equation back up top,

  • and you'll note that, since there's already a negative sign here,

  • when I plug in -0.5 for alpha, this will just calculate the Fourier transform of our original function times e^0.5t.

  • That Fourier transform will look like this. And again, I could write the magnitudes to show the outputs,

  • or I could use colors.

  • If α is swept through the plane, we get the entire Laplace transform plot.

  • And if we actually use a third dimension for those magnitudes, it would look like this.

  • This plane here shown in green, which I'll fade the plot a little so you can see, is all the inputs 's'.

  • X represents α and y is really iω.

  • The 3D plot itself is the magnitude of the complex outputs.

  • This is the Laplace transform of the original function e^(-t)*sin(t).

  • The thing is, this doesn't tell us much just by looking at it.

  • So what I'm going to do is graph the equation alpha or x equals 0 that'll give us a plane as shown

  • Because remember from before the alpha equals zero line actually yields the Fourier transform of the original function

  • Which we can see with the intersecting curve

  • This is what I meant by the Fourier transform is a slice of the Laplace transform

  • But now I'm going to decrease alpha while also plotting the original function times e to the minus alpha T

  • Right now alpha is 0 so that

  • Exponential is just 1 but as I sweep alpha we start to see both plots change and what you're seeing with that

  • Intersection on the right is the Fourier transform of the plot on the left at any given time

  • Like here

  • I'll pause it alpha equals negative 0.5 because this is what we just saw a minute ago where the original curve times e to the

  • Positive 0.5 T because that double negative has a Fourier transform with those two small Peaks

  • Then I'll extend the range of Z because as we get closer to alpha equals negative 1 we get an intersection with too much higher

  • And narrower peaks. We're rendering also isn't looking so good. But anyways, this happens because the left plot is approaching just a regular sinusoidal

  • ZAR about to cancel once we get to alpha equals negative 1 and we saw before that a pure sinusoid as a Fourier

  • Transform of 2 infinite spikes, which we can also see on the 3d plot

  • So the quick summary to what we've seen already is that to construct a Laplace transform take whatever function you want to work with

  • Multiplied by e to the minus alpha T for some alpha

  • Let's just change it something random like negative 0.93 and graph the 2d Fourier transform of that function

  • Then keep doing that as you sweep through all values of alpha stacking

  • two-dimensional 48 plot side by side until you get your 3d Laplace transform

  • You will likely not be shown this in a classroom setting and that's because most of it pretty much doesn't matter

  • All we care about are these two peaks, which do go to infinity known as the poles

  • Now if we go back to the 2d plot those poles are located at negative 1 plus and minus. I

  • Which we can represent with an X

  • Poles and also zeros, which doesn't apply to our function are pretty much all you'll ever see for these plots

  • The reason the poles are there though is because if we plug in negative 1 plus or minus I into the Laplace equation

  • We get 1 over 0 which I'll just write as infinity because on the plots those are represented with an X

  • And one more thing you'll notice when I was sweeping the Alpha plane

  • I stopped at those poles or alpha equals negative 1 and never went behind that

  • That's because once alpha goes below negative 1 the function. We were plotting diverges which means the Fourier transform of this does as well

  • So the Laplace transform doesn't actually exist in that region

  • Whereas this is the good region, which we give a name to the region of convergence

  • Just think of this region is all the Alpha is such that this part of the Laplace transform eventually converges to zero

  • This means everything I said earlier with taking the Fourier transform of this function and at being a slice of our plot is true

  • If that slice is in the region