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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 Ï‰ 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 Ï‰ = 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 Ï‰ equals 1.

• This will be the magnitude of the Fourier transform.

• Now for Ï‰ = 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 Ï‰ = 2.

• And lastly, I'll plug in Ï‰ = 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 Ï‰,

• 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 'iÏ‰'.

• But s is really Î± +iÏ‰, 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^(-iÏ‰t), and integrating.

• Well, in the Laplace transform, we have the integral of (something)*e^(-iÏ‰t)

• 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 Î±, 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 Î± 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 Î± and 0 for Ï‰, 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 Ï‰ and 0 for Î±, 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 Ï‰ 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 Î± 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 Î± and y is really iÏ‰.

• 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