Subtitles section Play video Print subtitles 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