That is the best and most accurate way I have ever heard this explained. You are absolutely right and its amazing that Brain can be such an amazing teacher without even having feedback from us. while our "professors" can't when we are sitting in front of them begging them to teach us.
By LR did you mean Laplace Transform? The simple explanation is that FT breaks time signals into just sinusoids (or their frequency content). You can't use the FT to solve differential equations because it doesn't cover the exponential part. But you can use them FT for all sorts of frequency related problems like noise, sound, filters, and so on. LT breaks time signals into sinusoids and exponentials (just like the solution to Diffy Q's) so that's the motivation.
I have been following your lectures since 6 months now and I can't thank you enough. I wish I had seen these way earlier. Awesome way of explaining even the most confusing concepts!!
I took differential equations in 1994, and never understood what the s-plane was (honestly, I don't think my professor understood it either). I cannot thank you enough for finally providing a sensible explanation of what in the world the Laplace transform is actually doing. Now I've got to go back and re-read every control theory book I've ever bought, since I can probably make sense of them now.
Man, after so many classes and so many videos I finally understood it! Thanks for the "real world" approach! I was struggling just with correlating with reality! Awesome work, keep up!
imma be real, this video blew my fuckin mind. the part where you went from the 3D s-plane plot to the poles and zeros? holy shit. it's like i just found the secrets to the universe.
understood the whole point of doing Laplace transforms and finding poles and zeroes for the first time. Great work. Thank you very much for posting this videos
Studying for my FE exam after I've taken all my signals classes and control electives; this really helps bring home some of the intuition that they miss. Thanks!
JUST PERFECT! I came back to this after many years and I have to say there are a LOT of insights this video. Perfect for remembering or if you're trying to understand the intuition of what the Laplace transform does.
In computer engineering. Started this class and is the hardest part of the whole degree. Watching this, it took until you drew the circuit until things started to click. Thank you.
Great!! You just gave me an entire new point of view of Laplace transform, if you had given me this explanation long time ago, my engineering career would've been even more fun!!!! Thanks!!
Finally an interesting, intuitive and colourful series on control systems! I'm in my final year in my aerospace engineering program and I'm using your videos as a refresher for control systems. I'm actually learning new perspectives I never thought about and they are helping me to understand topics I didn't quite get. My final year design project is purely based on control systems so this is going to help me immensely. Thank you!
Thanks for the visual approach. At last someone who will draw a few pictures and not just fill a blackboard with greek letters!! I wish people would explain things more this way.
Around 10:50 when you move from A to B to C, I believe the probing signal amplitude should be increasing (decaying slower and then growing faster) rather than decreasing. As sigma decreases, e^-(sigma*t) increases faster due to the negative sign. From my understanding, at points B and B' the probing signal should be increasing at the "opposite" rate that the impulse response is decaying so that the product of the two signals is a constant amplitude sign wave. If I'm wrong please to correct me anyone.
You are correct the Fourier Transform returns a complex number. I think I confused a few people by only drawing one 3D plot (where I also drew the red line). But at 6:50 I explained that there was an imaginary and real component at that point. The graphic was just supposed to show visually how you fill in the S plane with information using the FT. Unfortunately, it didn't accurately represent the real and imaginary response. Does that clear it up a bit?
If you look at w^2/(s^2+w^2) you'll notice that it's a classic 2nd order system with damping, zeta = 0. Which means if you put an impulse into this system an undamped oscillation will be the output, or a sine wave as expected. Remember that with Laplace Transform the only numbers that mean anything are the poles and zeros. In this case the poles are at, S = sqrt(-w^2) which is +- w*i. So there are two poles and both are on the imaginary axis, which again means undamped oscillation.
Also, if you set s = jw in the equation to see the steady response you should get exactly what the Fourier Transform would give us. But you'll notice it doesn't look like the same form, what gives!? Well this is because the Region of Convergence (ROC) of the S-plane doesn't include the imaginary axis for this function (it blows up to infinity). When you take the Fourier transform of sine we have to do some tricks to get it (using dirac delta function). But they do represent the same thing.
The Laplace Transform is defined from -inf to +inf ... but this is really just math notation (hopefully no mathematicians read that and get upset!) The physical world is causal (which means something happens as a consequence of something earlier) so for engineers we don't need to deal with negative time. We set time 0 to be the first action and everything before that is 0. So in that case you can simplify the transform to 0 to +inf. Hope that makes sense.
Hi Shouvik, great suggestion! I've just filled out the form to get my channel reviewed by TH-cam to see if it meets the criteria for their education filter. I don't know how long it'll take but hopefully it'll be available soon. Thanks for the comment.
2:40 Only true for linear differential equations with constant coefficients. A couple examples of exceptions are the Hermite Equation (defines solutions to the quantum harmonic oscillator), the Associated Legendre Equation (which defines spherical harmonics), and the Bessel Equation (unsurprisingly defines the Bessel functions, which are defined as a power series). In addition, sinusoids are also exponentials, just complex exponentials ( cos(x) = (exp(ix) + exp(-ix))/2, or exp(ix) = cos(x) + i*sin(x) ). So really the solutions to linear differential equations with constant coefficients are just various combinations of exponentials with complex valued coefficients in the exponent.
Hi Andy, for some reason your comment has been marked as spam. The power series is a very specific transform, it turns a time history into a series of sines and cosines (1D). The Laplace transform is more general in that it is a superset of the Fourier series (2D). The Laplace transform contains the Fourier series in it plus all of the exponential information as well. Therefore, you can think of the Laplace transform as a generalized form of the Fourier Series.
great, got a preview of what ill be using this for next year...one thing that confused me about LT was that i had no idea what it was doing graphically other than the simple bland exponential behavior example they always give at the beginning...This makes more sense for practical purposes.
I just wanted to say thank you so much for this video it has really helped me to understand laplace transforms in a way that I never did before. Also thank you for making these available to us all, I really appreciate what you do.
At 2m40s, the claim is way too broad. Exponentials are the only solutions to *homogeneous linear constant-coefficient* differential equations, or in physical terms, they are the only possible *transient* responses of *linear time-invariant* systems. For example, the linear time-invariant system y'' + y' + y = x^2 has a non-exponential (particular) solution y = x^2 - 2*x just due to its inhomogeneity. It'd be bad for people to believe an (incorrect) statement like "the solution to every differential equation is an exponential." That'd be an extremely powerful game-changer if it were true.
Can you think of a function or a signal (other than exp or any sinusoidal func for that matter) that if you take it's derivatives (1st, 2nd, etc) and if you add them all up will get a zero? With the mathematics that we know now, there isn't any. I'm not sure if there is a proof for that but for now it's a (very) valid statement.
I think you misunderstood what the original comment was saying. The video states that the solution to ALL differential equations are exponentials, sinusoids, or combinations of the two. This is just not true. It may be true for most physical differential equations such as Laplace's equation or the heat equation, but it is not true for ALL differential equations. Hell, most physically described systems are described by Legendre polynomials while are neither exponentials nor sinusoids. You can put sinusoids as the argument for Legendre polynomials, and most of the time you want to because of symmetries, but they are not inherently exponentials NOR sinusoids.
Thank god, I thought I was the only one who got super confused by the statement the video made. The first differential equation we learned in ordinary differential equations were ones where you could use simple integration to find solutions. However, I see how it could be a valid statement that every solution to a differential equation is *comprised* of sinusoidals and exponentials, as this is true of all signals.
Absolutely clear. Brilliant. I like this kind of video because it shows that we can explain some abstracts concepts with precise words and illustrations.
Wow! Just exactly what kind of a mind does it take to be able to just trip this stuff off the tongue like Brian does? I was mesmerized by this video and understood practically none of it . . . .!!
I think this might be quite a good idea! Having an image to quickly scan over to refresh my mind at the end of each of these videos would be quite useful. Thanks for the videos Brian.
Wow. This is a really excellent explanation. Well paced too and clearly drawn. I like the fact I don't have to wait for you to write / draw things. That can get a bit tedious on Khan Academy
I thought that "solution to differential equations must be either ..." was misleading. He's specifically talking about linear differential equations with constant coefficients.
You go, girl! (I'm at a loss to say anything probative.) Is math a conspiracy of smart people over the rest of us? I mean, i'm not dumb (stop sniggering), but this could be total baloney and I have no way to discern. For example the quote "...just below infinity..." I don't believe in shaming myself, but, huh?
Hugo, No, you're not shaming yourself. This guy is a wonderful example of David Hilbert's wise remark "You get all sorts of nonsenses when you bring in infinity." What he says about the declining case of a sinusoidal signal being "unfathomably large but not infinite," for instance, is a hoot. How be you try "limitless," baby?
You can even get terms of type t*exp(at), t^2*exp(at), ..., t^k*exp(at) if the characteristic polynomial of the linear diff. eq. has a root of multiplicity larger than one. These terms arise from taking the exponential of the Jordan form of the associated linear system.
Your video is unique. It answers my questions perfectly.Thank you so much Brian, I regret so much that I pay a ton to university, hoping to learn things step by step, but the only things I get are complications.
Just to be sure the laplace transform is only defined from 0 to infinity? Cause at 5:15 the laplace transform is written with the bounds of the fourier transform (i.e.. -infinity to infinity)
Yes that is correct. Really what it represents though is a 2nd order system that will oscillate forever ( as a sinusoidal wave) after it has been subjected to an impulse input. Basically once you get the system started it'll continue without growing or shrinking in amplitude. Imagine a ball rolling inside a completely frictionless cup. It'll just rock back and forth forever since no energy is ever lost or gained.
At 5:46 I didn't get how did he plot the response as rising and falling exponential over sigma=0 line? Can anyone help explain it? Any help will be appreciated. TIA
The Fourier Transform transforms x(t) in the time domain into the frequency domain which is a 2D space (the frequency w and the amplitude of that frequency). However, here in the s-domain it becomes a 3D space (sigma, w and the amplitude). Note that what he said is using the Laplace transform when sigma=0 thus in the s-domain it corrsponds to a plot in the sigma=0 plane. Hope that helps.
I only get half of this because I haven't gotten much of the mathematics yet (was just looking for Laplace transform because we vaguely saw it) but this is epic. I'll so dig into this after my exams.
However, this is a bit misleading I think, because we could also add a third dimension and transform time signals in sinusoids, exponentials, and square waves. There's nothing stopping us mathematically from doing this. But there is no physical reason to make that transformation. So transforms just move from one domain to another. The frequency domain is good for some analysis, and the S-domain is good for differentials. They are related, but typically thought of as separate.
Brian Douglas Well, I used the "Amara" system to incorporate subtitles to your videos. I have the .sub file with the transcription, give me a email address to send you the file. In the link you can see the result: www.amara.org/es/videos/dnjeirof00BL/info/the-laplace-transform-a-graphical-approach/?tab=video
your lectures are very good to understand as your way of teaching by clolourful depction of example as well as less than half hour lecture which can not bored. please uplod more lecture regarding control system components. Thank and reagrds
awesome information. at first the maths class look boring but after know what i'm doing. it get pretty interesting. don't fully understand but i think i will get there
Hi Brian, That's the best explanation of Laplace Transform I have seen, which is touching on originating conceptual ideas. All your videos have this charecteristics. Sure, you have extraordinary talent and expertise. But, I would like to learn where this culture of conceptual understanding of you comes from. Is it related to university you graduated from or is there conceptual formal textbooks/resources that you can strongly suggest in this manner?
yes. many profs had no practical experience in the real world. Many never had a job and had to do actual stress calculations or real force solutions where they quickly found out that analytical methods are not sufficient and one had to do numerical simulations or just build a prototype and measure the stuff
Hi, nice explantiona! correction:? in time = 6:01, when \sigma == -1, the graph should be a +ve exponetial wrt to time, as well as when \sigma == -1 at time 6.15, the exponetial curve should be a +ve going function.
Thank you for your brilliant explanation, I always hate when teachers "parachute" methods and equations without explaining the Why, well you did just the opposite and thank you for that :D
In the 6:10 - 6:20 region, are the plots for \sigma=-1 and \sigma=1 swapped? The Fourier Transform part does require -j in the exponent but before that we are multiplying x(t) by exp(-\sigma), not exp(\sigma), correct? So, for \sigma = -1, the modulating function will be a growing exponential. Correct?
2:30 interesting, solutions to differential equations representing physical phenomena results in exponentials or sinusoids, nice, it clears a lot of things.
Hi, great work in general; but, I think their is an error in your explanation. The Laplace Transform has the exponential term (e^(-st) = e^(-(sigma + (j)omega)t). So, when you separate out the exponent's real term (sigma), you forgot to include the negative sign. i.e., e^(-(sigma + (j)omega)t) = e^-(sigma)t * e^-(j)omega)t) So, in the LHP, the exponential terms ( e^-(sigma)t ) are increasing with time - and, in the RHP, the exponential terms ( e^-(sigma)t ) are decreasing with time. Later, around 11:00, you are looking at sets of points in the LHP and describing them as decreasing exponentials - the are increasing exponentials. For instance,the ( e^-(sigma)t ) term @ A-A' is exponentially increasing; but the impulse response is exponentially decreasing at a faster rate - so that when you multiply and integrate these two terms you get a finite value. The ( e^-(sigma)t ) term @ B-B' is exponentially increasing just as fast as the impulse response is exponentially decreasing - so that when you multiply and integrate these two terms you get a a barely infinite value. This is a 'pole'. For instance,the ( e^-(sigma)t ) term @ C-C' is exponentially increasing much faster than the impulse response is exponentially decreasing - so that when you multiply and integrate these two terms you get a value that rapidly goes to infinity.
+awkinga It is exactly what I am thinking when I try to find some examples to visualize the A-A', B-B' and C-C'. exp(-s) at 2nd and 3rd quadrant (when sigma < 0) should be exponentially increasing.
I don't know I'm late to the party but I don't understand the comment at 3:50. I thought that to move from time domain to frequency domain we would divide the time domain signal by a sinusoid to determine the "amount of that sinusoid" in the time domain. He goes over this in Fourier Transform (part 2) at minute 6. So why are we multiplying the signals now?
Hmm, I think this is a tricky question for me to answer. The thing is that we really don't perform the Laplace Transform and its inverse the way I explain in this video. The intent of the video was to give you a better understanding of what the LT is doing behind the scenes when you solve that integral ... or when you look it up in a table. By reverse I was just saying that you usually don't take the LP of the impulse response, you take it on the Diffy Q and solve for the impulse with it.
Starting at 5:30, you say you are multiplying by a real exponential e^rho. When rho equals zeros, you're essentially applying a unit gain to the original function of time. However, once rho starts becoming non zero, it start damping the fourier transform. Why is this? Why does multiplying a time signal by an exponential function dampen the fourier transform? I can't visualize what multiplying a time signal by an exponential function does to the time signal.
Brian thanks for this informative video and recommending this book " Steven W. Smith - The Scientist and Engineer's Guide to Digital Signal Processing "..... awesome book .
Awesome! Love the intuitive graphical approach. The Laplace transform is a pretty magical mystery transform! Always wanted to understand how it worked. Thanks!!!
Point of contention... Newton's second law as you wrote it initially where sum(F) = d(mv)/dt is not only valid when mass is constant. The simplified version sum(F) = ma is the version where mass must be constant. But the rate of change of momentum obviously can include a time-dependent mass, and is the most general form of Newton's second law.
Yes, thank you. So is it correct to say that, even though the transform returns complex numbers, you can still plot this graph by considering only the real or the imaginary part or the magnitude of the complex number?
Hi Brian great video. I've found that explaining the transform as projection over Orthogonal basis vectors is also a good interpretation which high school students grasp very easily. Dollar example becomes a scalar example which may not put the true perspective. Something I would like to add your points is that dimension of the fourier transform is not really magnitude but rather magnitude per hertz(density). Surprising why no text mentions about this particular thing.
I am pretty sure that the Fourier Transform returns a function with complex numbers. Infact, for each frequency it returns a complex number whose magnitude is the magnitude of the sine wave at that frequency and whose argument (angle) is the phase of the sine wave.
Hi MultiNova100 I just got like 20 emails from you! :) The "it" I was referring to was the impulse response. Now I'll go see about your other questions ... standby
What is this vertical axis starting at 5:42? I know that poles going to infinity vertically at a location (sigma, omega) are indicative of the behavior of the system, and that these poles are the roots of the characteristic equation of the system. I also know the characteristic equation is generally A*x=0, and if x is non zero (non-trivial), then A has to be zero. Why does setting A = 0 tell you anything about x?
He has given a very explanation on poles and zeros the positive and negative part cancelling each other as zerp but is not so by a vertical shift of increment except how a root locus diagram is scribbbled to give some explanation between damness as exponential changed for frequency transfer.
2:16 must admit I've built a career on control system but had no idea that differential equations led to an expectation of a sinusoidal element on its own. Obviously it does in control theory but I didn't think it did in pure maths. Any opinions from mathematicians on this?
You're a "teacher". My 'professors' at uni, they're "tellers". Nice lecture, from a nice lecturer. Thank you!
My Controls Teacher at U of Florida was Dr. Bullock a walking incompetent idiot. I hope you get this message!
I'm relearning controls.
That is the best and most accurate way I have ever heard this explained. You are absolutely right and its amazing that Brain can be such an amazing teacher without even having feedback from us. while our "professors" can't when we are sitting in front of them begging them to teach us.
Exactly. IMO teacher holds a higher status than professor.
Teachers actually "teach", while professors merely "profess"
@@vaughnmonkeynot begging fakin paying for them to teach us
It's hard to put into words how good these videos actually are. What an amazing piece of work. I'll make sure to watch and like every one of them.
By LR did you mean Laplace Transform? The simple explanation is that FT breaks time signals into just sinusoids (or their frequency content). You can't use the FT to solve differential equations because it doesn't cover the exponential part. But you can use them FT for all sorts of frequency related problems like noise, sound, filters, and so on. LT breaks time signals into sinusoids and exponentials (just like the solution to Diffy Q's) so that's the motivation.
is it possible to know exact value of magnitude and phase for arbitrary frequency from continuous frequency spectrum?
I have been following your lectures since 6 months now and I can't thank you enough. I wish I had seen these way earlier. Awesome way of explaining even the most confusing concepts!!
I took differential equations in 1994, and never understood what the s-plane was (honestly, I don't think my professor understood it either). I cannot thank you enough for finally providing a sensible explanation of what in the world the Laplace transform is actually doing. Now I've got to go back and re-read every control theory book I've ever bought, since I can probably make sense of them now.
Man, after so many classes and so many videos I finally understood it! Thanks for the "real world" approach! I was struggling just with correlating with reality!
Awesome work, keep up!
imma be real, this video blew my fuckin mind. the part where you went from the 3D s-plane plot to the poles and zeros? holy shit. it's like i just found the secrets to the universe.
THANKS A LOT ! first time someone explains it in a way that I can actually grasp the idea behind the Laplace transform
I feel like this is where my class left off, and you picked back up perfectly
First video on youtube, where one "thumbs up" is not enough. Amazing video, after so many years it is not magic for me anymore!
understood the whole point of doing Laplace transforms and finding poles and zeroes for the first time. Great work. Thank you very much for posting this videos
Studying for my FE exam after I've taken all my signals classes and control electives; this really helps bring home some of the intuition that they miss. Thanks!
JUST PERFECT! I came back to this after many years and I have to say there are a LOT of insights this video. Perfect for remembering or if you're trying to understand the intuition of what the Laplace transform does.
Brian, I love the way your videos are built up and edited.
You have really put a lot of thought in it.
Brilliant!
I must say you have excellent hand writing. Makes following much easier.
Awesome! THis video will help thousands to understand laplace transform.
MOST IMPORTANT VIDEO IN THE ENGINEERING UNIVERSE EVER!
It's a great idea you came up with instead of simply writing while talking, wasting time in the process.
Good work!
In computer engineering. Started this class and is the hardest part of the whole degree. Watching this, it took until you drew the circuit until things started to click. Thank you.
Great!! You just gave me an entire new point of view of Laplace transform, if you had given me this explanation long time ago, my engineering career would've been even more fun!!!! Thanks!!
Man your tutorials are awesome. Its a lot better to watch your tutorial than going to college. Applause !!
Finally an interesting, intuitive and colourful series on control systems! I'm in my final year in my aerospace engineering program and I'm using your videos as a refresher for control systems. I'm actually learning new perspectives I never thought about and they are helping me to understand topics I didn't quite get. My final year design project is purely based on control systems so this is going to help me immensely. Thank you!
Thanks for the visual approach. At last someone who will draw a few pictures and not just fill a blackboard with greek letters!! I wish people would explain things more this way.
Around 10:50 when you move from A to B to C, I believe the probing signal amplitude should be increasing (decaying slower and then growing faster) rather than decreasing. As sigma decreases, e^-(sigma*t) increases faster due to the negative sign.
From my understanding, at points B and B' the probing signal should be increasing at the "opposite" rate that the impulse response is decaying so that the product of the two signals is a constant amplitude sign wave. If I'm wrong please to correct me anyone.
You are correct the Fourier Transform returns a complex number. I think I confused a few people by only drawing one 3D plot (where I also drew the red line). But at 6:50 I explained that there was an imaginary and real component at that point. The graphic was just supposed to show visually how you fill in the S plane with information using the FT. Unfortunately, it didn't accurately represent the real and imaginary response. Does that clear it up a bit?
The clarity and detail into each topic is amazing, it is so clear and easy to understand. Thank you so much!
AMAZING! best part was the 3d part going to 2d to show the poles and zero, best explanation ever
If you look at w^2/(s^2+w^2) you'll notice that it's a classic 2nd order system with damping, zeta = 0. Which means if you put an impulse into this system an undamped oscillation will be the output, or a sine wave as expected. Remember that with Laplace Transform the only numbers that mean anything are the poles and zeros. In this case the poles are at, S = sqrt(-w^2) which is +- w*i. So there are two poles and both are on the imaginary axis, which again means undamped oscillation.
This was really good for actual understanding and imaginative approach. Now we can really get what the plot is.
Also, if you set s = jw in the equation to see the steady response you should get exactly what the Fourier Transform would give us. But you'll notice it doesn't look like the same form, what gives!? Well this is because the Region of Convergence (ROC) of the S-plane doesn't include the imaginary axis for this function (it blows up to infinity). When you take the Fourier transform of sine we have to do some tricks to get it (using dirac delta function). But they do represent the same thing.
High quality visuals keeping pace with your lecture was fantastic. Excellent job with this.
The Laplace Transform is defined from -inf to +inf ... but this is really just math notation (hopefully no mathematicians read that and get upset!) The physical world is causal (which means something happens as a consequence of something earlier) so for engineers we don't need to deal with negative time. We set time 0 to be the first action and everything before that is 0. So in that case you can simplify the transform to 0 to +inf. Hope that makes sense.
Hi Shouvik, great suggestion! I've just filled out the form to get my channel reviewed by TH-cam to see if it meets the criteria for their education filter. I don't know how long it'll take but hopefully it'll be available soon. Thanks for the comment.
2:40 Only true for linear differential equations with constant coefficients. A couple examples of exceptions are the Hermite Equation (defines solutions to the quantum harmonic oscillator), the Associated Legendre Equation (which defines spherical harmonics), and the Bessel Equation (unsurprisingly defines the Bessel functions, which are defined as a power series). In addition, sinusoids are also exponentials, just complex exponentials ( cos(x) = (exp(ix) + exp(-ix))/2, or exp(ix) = cos(x) + i*sin(x) ). So really the solutions to linear differential equations with constant coefficients are just various combinations of exponentials with complex valued coefficients in the exponent.
The 3d plot explanation was amazing.....cleared a lot of things......thanks a loooottttt !!!
This video is insanely good.
Direct and clear in explanation! Great lecture.
8:43 What is happening and what is the meaning of probing the impulse response ?
For example how this probing is achieved, is it by changing v(t) ?
Hi Andy, for some reason your comment has been marked as spam. The power series is a very specific transform, it turns a time history into a series of sines and cosines (1D). The Laplace transform is more general in that it is a superset of the Fourier series (2D). The Laplace transform contains the Fourier series in it plus all of the exponential information as well. Therefore, you can think of the Laplace transform as a generalized form of the Fourier Series.
great, got a preview of what ill be using this for next year...one thing that confused me about LT was that i had no idea what it was doing graphically other than the simple bland exponential behavior example they always give at the beginning...This makes more sense for practical purposes.
I just wanted to say thank you so much for this video it has really helped me to understand laplace transforms in a way that I never did before. Also thank you for making these available to us all, I really appreciate what you do.
At 2m40s, the claim is way too broad. Exponentials are the only solutions to *homogeneous linear constant-coefficient* differential equations, or in physical terms, they are the only possible *transient* responses of *linear time-invariant* systems. For example, the linear time-invariant system y'' + y' + y = x^2 has a non-exponential (particular) solution y = x^2 - 2*x just due to its inhomogeneity. It'd be bad for people to believe an (incorrect) statement like "the solution to every differential equation is an exponential." That'd be an extremely powerful game-changer if it were true.
Can you think of a function or a signal (other than exp or any sinusoidal func for that matter) that if you take it's derivatives (1st, 2nd, etc) and if you add them all up will get a zero? With the mathematics that we know now, there isn't any. I'm not sure if there is a proof for that but for now it's a (very) valid statement.
I think you misunderstood what the original comment was saying. The video states that the solution to ALL differential equations are exponentials, sinusoids, or combinations of the two. This is just not true. It may be true for most physical differential equations such as Laplace's equation or the heat equation, but it is not true for ALL differential equations. Hell, most physically described systems are described by Legendre polynomials while are neither exponentials nor sinusoids. You can put sinusoids as the argument for Legendre polynomials, and most of the time you want to because of symmetries, but they are not inherently exponentials NOR sinusoids.
Thank god, I thought I was the only one who got super confused by the statement the video made. The first differential equation we learned in ordinary differential equations were ones where you could use simple integration to find solutions.
However, I see how it could be a valid statement that every solution to a differential equation is *comprised* of sinusoidals and exponentials, as this is true of all signals.
Finally I understand what the laplace transform is for, thanks.
Absolutely clear. Brilliant. I like this kind of video because it shows that we can explain some abstracts concepts with precise words and illustrations.
Wow! Just exactly what kind of a mind does it take to be able to just trip this stuff off the tongue like Brian does? I was mesmerized by this video and understood practically none of it . . . .!!
I think this might be quite a good idea! Having an image to quickly scan over to refresh my mind at the end of each of these videos would be quite useful. Thanks for the videos Brian.
How I can get a good report about (Laplace transform and fourier series )
Wow. This is a really excellent explanation. Well paced too and clearly drawn. I like the fact I don't have to wait for you to write / draw things. That can get a bit tedious on Khan Academy
I thought that "solution to differential equations must be either ..." was misleading. He's specifically talking about linear differential equations with constant coefficients.
yeah exact differential equations wouldn't behave that way for example
Athul Prakash no you can find non sinusoidal and non exponential from simply some separable equations
You go, girl! (I'm at a loss to say anything probative.) Is math a conspiracy of smart people over the rest of us? I mean, i'm not dumb (stop sniggering), but this could be total baloney and I have no way to discern. For example the quote "...just below infinity..." I don't believe in shaming myself, but, huh?
Hugo,
No, you're not shaming yourself. This guy is a wonderful example of David Hilbert's wise remark "You get all sorts of nonsenses when you bring in infinity."
What he says about the declining case of a sinusoidal signal being "unfathomably large but not infinite," for instance, is a hoot. How be you try "limitless," baby?
You can even get terms of type t*exp(at), t^2*exp(at), ..., t^k*exp(at) if the characteristic polynomial of the linear diff. eq. has a root of multiplicity larger than one. These terms arise from taking the exponential of the Jordan form of the associated linear system.
VERY well explained! Thank you, the contour map of the laplace transform plane was really helpful to visualize whats actually going on.
Kinda blew my mind at the end :D
Thanks so much for this video!
Your video is unique. It answers my questions perfectly.Thank you so much Brian, I regret so much that I pay a ton to university, hoping to learn things step by step, but the only things I get are complications.
It's this video that made me finally click. Can't thank you enough, I'm buying your book
Just to be sure the laplace transform is only defined from 0 to infinity? Cause at 5:15 the laplace transform is written with the bounds of the fourier transform (i.e.. -infinity to infinity)
Yes that is correct. Really what it represents though is a 2nd order system that will oscillate forever ( as a sinusoidal wave) after it has been subjected to an impulse input. Basically once you get the system started it'll continue without growing or shrinking in amplitude. Imagine a ball rolling inside a completely frictionless cup. It'll just rock back and forth forever since no energy is ever lost or gained.
At 5:46 I didn't get how did he plot the response as rising and falling exponential over sigma=0 line? Can anyone help explain it? Any help will be appreciated. TIA
Can someone please give explanation of above question? I also have same doubt. (@Brain Douglas, or @All Electronics, or anybody else.)
The Fourier Transform transforms x(t) in the time domain into the frequency domain which is a 2D space (the frequency w and the amplitude of that frequency). However, here in the s-domain it becomes a 3D space (sigma, w and the amplitude). Note that what he said is using the Laplace transform when sigma=0 thus in the s-domain it corrsponds to a plot in the sigma=0 plane. Hope that helps.
As a controls 2 student, reviewing your videos from Fourier transforms too classical controls theory I am very impressed with your videos! Keep it up!
These talks are stunning!!!
I only get half of this because I haven't gotten much of the mathematics yet (was just looking for Laplace transform because we vaguely saw it) but this is epic. I'll so dig into this after my exams.
However, this is a bit misleading I think, because we could also add a third dimension and transform time signals in sinusoids, exponentials, and square waves. There's nothing stopping us mathematically from doing this. But there is no physical reason to make that transformation. So transforms just move from one domain to another. The frequency domain is good for some analysis, and the S-domain is good for differentials. They are related, but typically thought of as separate.
Excellent video, may I to send you a transcription in spanish? it will be great if this information was available in many languages
Absolutely, I can add the transcript as subtitles maybe on the video. Or did you envision something else? Thanks for volunteering for this! Cheers.
Brian Douglas Well, I used the "Amara" system to incorporate subtitles to your videos. I have the .sub file with the transcription, give me a email address to send you the file.
In the link you can see the result:
www.amara.org/es/videos/dnjeirof00BL/info/the-laplace-transform-a-graphical-approach/?tab=video
Good translation to spanish. Thanks. Muchas gracias.
your lectures are very good to understand as your way of teaching by clolourful depction of example as well as less than half hour lecture which can not bored. please uplod more lecture regarding control system components.
Thank and reagrds
Thanks for doing this in reverse, made so much more sense this way.
I'm working on the root locus videos now! The first in the series will be out next week.
I am finally beginning to connect all the stuff Ive been learning as a electrical engineering student...wow.
I really like your videos. You know your stuff 99.9%. please keep adding more vids on ME Controls. Thanks
awesome information. at first the maths class look boring but after know what i'm doing. it get pretty interesting. don't fully understand but i think i will get there
Hi Brian, That's the best explanation of Laplace Transform I have seen, which is touching on originating conceptual ideas. All your videos have this charecteristics. Sure, you have extraordinary talent and expertise. But, I would like to learn where this culture of conceptual understanding of you comes from. Is it related to university you graduated from or is there conceptual formal textbooks/resources that you can strongly suggest in this manner?
I keep thinking how much I wish my profs provided this much intuition to what,s going on. Was it because they didn't have it to give?
yes. many profs had no practical experience in the real world. Many never had a job and had to do actual stress calculations or real force solutions where they quickly found out that analytical methods are not sufficient and one had to do numerical simulations or just build a prototype and measure the stuff
Hi, nice explantiona!
correction:? in time = 6:01, when \sigma == -1, the graph should be a +ve exponetial wrt to time, as well as when \sigma == -1 at time 6.15, the exponetial curve should be a +ve going function.
Thank you for your brilliant explanation, I always hate when teachers "parachute" methods and equations without explaining the Why, well you did just the opposite and thank you for that :D
MY IB LIFE SAVER!! THANK U SO MUCH
This is the closest I've come to understanding Laplace. I still don't fully get it, but I have glimmers of it. Thank you so much.
In the 6:10 - 6:20 region, are the plots for \sigma=-1 and \sigma=1 swapped? The Fourier Transform part does require -j in the exponent but before that we are multiplying x(t) by exp(-\sigma), not exp(\sigma), correct? So, for \sigma = -1, the modulating function will be a growing exponential. Correct?
This video blew my mind
What is the target of probing the response, started at 8:45 ? is it finding the poles and zeros? or something else
2:30 interesting, solutions to differential equations representing physical phenomena results in exponentials or sinusoids, nice, it clears a lot of things.
Great visualization of the Laplace Transformation! Made my day.
Hi, great work in general; but, I think their is an error in your explanation. The Laplace Transform has the exponential term (e^(-st) = e^(-(sigma + (j)omega)t). So, when you separate out the exponent's real term (sigma), you forgot to include the negative sign. i.e., e^(-(sigma + (j)omega)t) = e^-(sigma)t * e^-(j)omega)t) So, in the LHP, the exponential terms ( e^-(sigma)t ) are increasing with time - and, in the RHP, the exponential terms ( e^-(sigma)t ) are decreasing with time.
Later, around 11:00, you are looking at sets of points in the LHP and describing them as decreasing exponentials - the are increasing exponentials.
For instance,the ( e^-(sigma)t ) term @ A-A' is exponentially increasing; but the impulse response is exponentially decreasing at a faster rate - so that when you multiply and integrate these two terms you get a finite value.
The ( e^-(sigma)t ) term @ B-B' is exponentially increasing just as fast as the impulse response is exponentially decreasing - so that when you multiply and integrate these two terms you get a a barely infinite value. This is a 'pole'.
For instance,the ( e^-(sigma)t ) term @ C-C' is exponentially increasing much faster than the impulse response is exponentially decreasing - so that when you multiply and integrate these two terms you get a value that rapidly goes to infinity.
+awkinga It is exactly what I am thinking when I try to find some examples to visualize the A-A', B-B' and C-C'. exp(-s) at 2nd and 3rd quadrant (when sigma < 0) should be exponentially increasing.
Very important remark
Nobody likes this one? This is the comment that should be pinned! Remarkable!
This is a very important remark!
You are really good at explaining this material!
The black magic of math
I don't know I'm late to the party but I don't understand the comment at 3:50. I thought that to move from time domain to frequency domain we would divide the time domain signal by a sinusoid to determine the "amount of that sinusoid" in the time domain. He goes over this in Fourier Transform (part 2) at minute 6. So why are we multiplying the signals now?
Hmm, I think this is a tricky question for me to answer. The thing is that we really don't perform the Laplace Transform and its inverse the way I explain in this video. The intent of the video was to give you a better understanding of what the LT is doing behind the scenes when you solve that integral ... or when you look it up in a table. By reverse I was just saying that you usually don't take the LP of the impulse response, you take it on the Diffy Q and solve for the impulse with it.
genuine and digestable. thankyou sir!
Always a good refresher. Thanks!
Starting at 5:30, you say you are multiplying by a real exponential e^rho. When rho equals zeros, you're essentially applying a unit gain to the original function of time. However, once rho starts becoming non zero, it start damping the fourier transform. Why is this? Why does multiplying a time signal by an exponential function dampen the fourier transform? I can't visualize what multiplying a time signal by an exponential function does to the time signal.
Bump
I also cannot understand how those bumps to infinity were created? Where did they come from?
this is helping me so much understand the motivation of what i have to do thank you
Brian thanks for this informative video and recommending this book " Steven W. Smith - The Scientist and Engineer's Guide to Digital Signal Processing "..... awesome book .
Awesome! Love the intuitive graphical approach. The Laplace transform is a pretty magical mystery transform! Always wanted to understand how it worked. Thanks!!!
At 6:05 shouldn't you premultiply for the exponential of the opposite of sigma? The exponential inside the integral has a minus to its exponent
Point of contention... Newton's second law as you wrote it initially where sum(F) = d(mv)/dt is not only valid when mass is constant. The simplified version sum(F) = ma is the version where mass must be constant. But the rate of change of momentum obviously can include a time-dependent mass, and is the most general form of Newton's second law.
Ohh man this is great! I wish there were more videos of graphic understanding in mathmatics as well!
Yes, thank you. So is it correct to say that, even though the transform returns complex numbers, you can still plot this graph by considering only the real or the imaginary part or the magnitude of the complex number?
Hi Brian great video. I've found that explaining the transform as projection over Orthogonal basis vectors is also a good interpretation which high school students grasp very easily. Dollar example becomes a scalar example which may not put the true perspective. Something I would like to add your points is that dimension of the fourier transform is not really magnitude but rather magnitude per hertz(density). Surprising why no text mentions about this particular thing.
I am pretty sure that the Fourier Transform returns a function with complex numbers. Infact, for each frequency it returns a complex number whose magnitude is the magnitude of the sine wave at that frequency and whose argument (angle) is the phase of the sine wave.
Hi MultiNova100 I just got like 20 emails from you! :) The "it" I was referring to was the impulse response. Now I'll go see about your other questions ... standby
What is this vertical axis starting at 5:42? I know that poles going to infinity vertically at a location (sigma, omega) are indicative of the behavior of the system, and that these poles are the roots of the characteristic equation of the system. I also know the characteristic equation is generally A*x=0, and if x is non zero (non-trivial), then A has to be zero. Why does setting A = 0 tell you anything about x?
I am reading your ebook. Thanks a lot for you kindly sharing.
He has given a very explanation on poles and zeros the positive and negative part cancelling each other as zerp but is not so by a vertical shift of increment except how a root locus diagram is scribbbled to give some explanation between damness as exponential changed for frequency transfer.
2:16 must admit I've built a career on control system but had no idea that differential equations led to an expectation of a sinusoidal element on its own. Obviously it does in control theory but I didn't think it did in pure maths. Any opinions from mathematicians on this?
You just blew my mind.