Thanks for Explaining it so Well! I felt really lost in my Signals and Systems subject this semester. But your Videos explain it so well. Love from Germany
thanks a lot Iain...hearty thanks...you made my day...really searching for one like this for about an hour....really underrated video...post more videos like this buddy....
00:02 Laplace transform is related to the Fourier transform 01:28 Laplace transform is a generalization of Fourier transform. 02:48 The Fourier transform of a cos waveform leads to two Delta functions at Omega and -Omega. 04:02 Exploring other functions in Laplace plane 05:01 Laplace transform helps in dealing with waveforms without Fourier transform 06:08 Laplace transform maps exponential increase in time domain to radial increase in frequency domain. 07:12 Region of Convergence defines valid Fourier transforms 08:29 Laplace transform generalizes Fourier transform for signals with growing energy Crafted by Merlin AI.
I was aware of the rules of ROC for right handed and left handed signals. never understood why they worked, though. thank you so much for this explanation.
perfectly explained. thanks really much. i had this in my bachelor but i never understood it. know it makes sense and hopefully i can use it for my masters.
you are amazing sir! Idk why everyone thinks, math should be learned formally with proofs and high level language, instead of exampels and visuals.. Its the same with languages, we learn grammer instead of conversations..
Thanks for this video. It's very illustrative. I understand that, for every sigma, the LT gives you an amplitude and a phase, just like the FT would do. But you only plotted the amplitude, right?
These videos might help: "Why are Poles Negative in Laplace Transform of Real Stable LTI Systems?" ua-cam.com/video/apdh8ZXW3a0/v-deo.html and "How do Poles and Zeros affect the Laplace Transform and the Fourier Transform?" ua-cam.com/video/iP4fckfDNK8/v-deo.html
@@iain_explains Thanks, Professor. However, I'm scratching my head over this definition. Take for instance the signal x(t)=sin(t). The integral of (sin(t))^2 from -Inf to +Inf is unbounded, since the area under the curve is always positive and grows limitless due to the repeating pattern it forms. Yet the FT of x(t) does exist.
Excellent observation! Sinusoids are special cases. Actually in the case of sinusoids, the integral in the Fourier transform equation does not converge. We find the Fourier transform via the inverse Fourier transform of the delta function (which is a mathematical construct with infinite amplitude, infinitesimally small width, but unit area). The integral in the inverse FT formula does converge for a delta function, and the answer is a complex exponential function of time (which is a complex sinusoid in the time domain).
very conceptual and intuitive video lain, but I have a small doubt, let a system having impulse response exp(-t) so the pole would be at s= -1 which implies that system output will converge for all values of s right side of the pole s= -1, assuming causal system now let us give input to the system exp(2t) so the system output would bey(t) = conv(exp(2t),exp(-t)) and thus taking Laplace of y(t) will get( 1/(s-2) * 1/(s+1) ) no taking the partial fraction and solving further we get (0.33/(s-2) - 0.33/(s+1) ) and taking inverse Laplace we y(t)= get 0.33*( exp(2t)- exp(-t)) and now when t tends to infinity output does not seem to converge rather it tends to infinity so why it's not converging despite the point s= 2 coming under the ROC of the system? Hope you have understood my question...thankyou!
Well just because a system has a stable impulse response, it doesn't mean it can stabilise any unstable input. In the example you gave, the input signal has infinite energy!
@@iain_explains ok that's clear lain that the input signal has infinite energy but that's my doubt despite being an infinite energy signal it's still coming under the ROC of the system, any specific reason for it being unstable though it's clearly depicted by the ROC that it should converge that's why its called region of convergence.
No, that's not why it's called the region of convergence. The region of convergence does not relate to the input signals. It relates to the region of the s-plane over which the Laplace Transform exists for the system.
The Laplace transform is complex-valued, just like the Fourier transform is complex-valued. The s-plane is the region of the complex-valued "independent variable", s, which is a generalisation of the real-valued "independent variable", frequency, in the Fourier transform.
Regarding the last point about processing, is this related to audio compression and decompression? I can imagine a flattened waveform would be easier to work with?
you gave the example of a microphone recording a voice having infinite energy thus Fourier transform can't be found out , so how we are finding the fourier transform of sine and cosine wave since they are also signal with infinite energy??
Great question. That's where the delta function comes in. The Fourier transform of a sinusoidal wave is only able to be defined with the use of delta functions in the frequency domain.
@@iain_explains ok sir that means any signal which is having infinite energy can be written in the sum of sine and cosine, and hence its Fourier transform can be found in form of impulses. can we not break the microphone recording a voice with positive feedback into the sum of sine and cosine and then we can find its Fourier transform
the value of pole of any system is dependent on the damping provided by the system ?? sir what is the significance of pole of any system in term of laplace?
As-salamu alaykum professor I have a very basic question that why we consider ROC of causal system to be right side of rightmost pole what is the reason/intuition behind it ?
Causal systems have impulse responses that are "right handed" (in time). ie. they equal 0 for all negative time. If the system is unstable, then the impulse response will grow unbounded as time goes to positive infinity. Therefore, in order to be able to take a Fourier transform, they would require a 'damping function' that goes to zero as time goes to positive infinity. This corresponds to values in the s-plane that are in the region to the right of the most positive pole. This video might help: "Why are Poles Negative in Laplace Transform of Real Stable LTI Systems?" ua-cam.com/video/apdh8ZXW3a0/v-deo.html
What would the region of convergence be if there are multiple poles at different exponential values. For example if your input signal was something like x(t) = e^(-t)cos(2*t) + e^(-2t)cos(3*t) I believe you would get poles -1+2j/-1-2j and -2+3j/-2-3j. How would you determine the region of convergence for this multi poled system?
The overall expression only holds when all of the component terms hold. In other words, only for values of "s" that are in the overlapping region of all of the ROC's of the component expressions. Mathematically, this is ROC = ROC1 "intersection" ROC2
Because then you can perform calculations in the "z/Fourier domain", and if you need to, then you can also transform back into the (discrete) time domain.
Are all of the transformed signals going to lie along a vertical axis some distance away from the w-axis? Is it possible to have sloped lines along which the signals lie? Also, are all of the signals going to be orientated orthogonal to these vertical axes or is that just the case here because you used delta functions? Also, I have watched a few other videos and so far yours is the clearest, however other videos make mention of something called the s-plane , what is that?
I explained the overall transform in terms of a collection of "transformed signals", where each different signal only differed by the sigma parameter in the exponential that multiplied the base signal, x(t). And yes, each of these signals corresponds to a vertical line in the sigma-omega plane. But it's important to realise that these signals are all just a function of the base signal x(t). There are no "other" signals. And you ask about the s-plane, well that's exactly what I've drawn - it's the sigma-omega plane. Notice that I've written the equation s=sigma+j omega in the middle of the page.
There are two different definitions people use. One for all time, and the other for only positive time. It doesn't really matter which one you use, as long as you are consistent.
I was showing an example waveform where the magnitude of x(t) grew with time (which means it does not have a FT, and therefore the LT does not exist for sigma=0). The Laplace Transform will exist (only) for values of sigma that are more positive than the rate of the growth in x(t). So, I showed the boundary as being between 0 and sigma1 because the LT for this example does not exist at sigma=0, but I said it does exist at sigma=sigma1, so the boundary must be between the two.
Glad you found it useful. If you haven't already, you might like to check out the other videos on the channel and pdf summary sheets available at iaincollings.com
There are two versions of the LT. I prefer the "bilateral LT", where the integral is over the range -∞ to ∞. The "unilateral LT" is a special case, where the integral is over 0 to ∞. In both definitions, the LT integral is over the range 0 to ∞ when considering impulse responses of systems that are causal - ie. the impulse response, h(t), is zero for t in the range -∞ to 0.
Thanks for Explaining it so Well! I felt really lost in my Signals and Systems subject this semester. But your Videos explain it so well. Love from Germany
Glad the videos have helped!
taking a signals and systems subject this semester so thank you for this!
Glad you found it useful. Have you checked out the other videos on the channel too?
That was THE best explanation I've seen for Laplace ROC. Thanks.
Thanks. I'm glad you found it helpful.
thanks a lot Iain...hearty thanks...you made my day...really searching for one like this for about an hour....really underrated video...post more videos like this buddy....
Glad you liked it!
00:02 Laplace transform is related to the Fourier transform
01:28 Laplace transform is a generalization of Fourier transform.
02:48 The Fourier transform of a cos waveform leads to two Delta functions at Omega and -Omega.
04:02 Exploring other functions in Laplace plane
05:01 Laplace transform helps in dealing with waveforms without Fourier transform
06:08 Laplace transform maps exponential increase in time domain to radial increase in frequency domain.
07:12 Region of Convergence defines valid Fourier transforms
08:29 Laplace transform generalizes Fourier transform for signals with growing energy
Crafted by Merlin AI.
It's so easy to understand ROC now. Moreover I can understand it's properties even better since the origin is understood.Thank you Iain sir
Glad it was helpful!
This was such a clear and easy explanation! Thank you
Glad it was helpful!
I was aware of the rules of ROC for right handed and left handed signals. never understood why they worked, though. thank you so much for this explanation.
I'm so glad the video helped you to understand some of the more fundamental aspects.
such a clear explanation. Was looking at other clips but they were difficult to understand and convoluted. Like!
I'm glad you found it helpful.
@@iain_explains Very helpful. Exam was a piece of cake after studying your videos and making exercises!
A beautiful crystal clear explanation. Many thanks Iain.
Thanks Navod.
This explanation was so clear, Thank you Iain!
Glad it was helpful!
damn, i didn't realize how far i was from understanding Laplace untill watching this. thank you.
I'm so glad it helped.
Actual mind blowing explanation, thanks a lot!
Glad you liked it!
perfectly explained. thanks really much. i had this in my bachelor but i never understood it. know it makes sense and hopefully i can use it for my masters.
Glad it was helpful!
you are amazing sir!
Idk why everyone thinks, math should be learned formally with proofs and high level language, instead of exampels and visuals..
Its the same with languages, we learn grammer instead of conversations..
I'm glad you like the approach I've taken in my videos.
Perfectly explained. Thank you!
Glad it was helpful!
Thanks for this video. It's very illustrative.
I understand that, for every sigma, the LT gives you an amplitude and a phase, just like the FT would do. But you only plotted the amplitude, right?
Yes, that's right. Phase is important, of course, but it's the amplitude that more clearly provides intuition.
Really nice and easy to understand explanation! Thank you for this video..
Thanks. Glad you found it useful.
Amazing explanation. Was mindblown after understanding what e^-(sigma)t actually does !!! Thank you soo much for this video.
Glad it was helpful!
Incredible explanation! Cheers from Oklahoma!
Glad you liked it!
Great explanation ❤
Glad you liked it
Thank you for the video! Why is it that the region of convergence is defined by the poles of a transfer function?
These videos might help: "Why are Poles Negative in Laplace Transform of Real Stable LTI Systems?" ua-cam.com/video/apdh8ZXW3a0/v-deo.html and "How do Poles and Zeros affect the Laplace Transform and the Fourier Transform?" ua-cam.com/video/iP4fckfDNK8/v-deo.html
Can you explain what a finite-energy signal means, more precisely? Thank you
The integral of the square of the function, over all time, is finite.
@@iain_explains Thanks, Professor. However, I'm scratching my head over this definition. Take for instance the signal x(t)=sin(t). The integral of (sin(t))^2 from -Inf to +Inf is unbounded, since the area under the curve is always positive and grows limitless due to the repeating pattern it forms. Yet the FT of x(t) does exist.
Excellent observation! Sinusoids are special cases. Actually in the case of sinusoids, the integral in the Fourier transform equation does not converge. We find the Fourier transform via the inverse Fourier transform of the delta function (which is a mathematical construct with infinite amplitude, infinitesimally small width, but unit area). The integral in the inverse FT formula does converge for a delta function, and the answer is a complex exponential function of time (which is a complex sinusoid in the time domain).
very conceptual and intuitive video lain, but I have a small doubt,
let a system having impulse response exp(-t) so the pole would be at s= -1 which implies that system output will converge for all values of s right side of the pole s= -1, assuming causal system now let us give input to the system exp(2t) so the system output would bey(t) = conv(exp(2t),exp(-t)) and thus taking Laplace of y(t) will get( 1/(s-2) * 1/(s+1) ) no taking the partial fraction and solving further we get (0.33/(s-2) - 0.33/(s+1) ) and taking inverse Laplace we
y(t)= get 0.33*( exp(2t)- exp(-t)) and now when t tends to infinity output does not seem to converge rather it tends to infinity so why it's not converging despite the point s= 2 coming under the ROC of the system? Hope you have understood my question...thankyou!
Well just because a system has a stable impulse response, it doesn't mean it can stabilise any unstable input. In the example you gave, the input signal has infinite energy!
@@iain_explains ok that's clear lain that the input signal has infinite energy but that's my doubt despite being an infinite energy signal it's still coming under the ROC of the system, any specific reason for it being unstable though it's clearly depicted by the ROC that it should converge that's why its called region of convergence.
No, that's not why it's called the region of convergence. The region of convergence does not relate to the input signals. It relates to the region of the s-plane over which the Laplace Transform exists for the system.
Very clear explanation!
Glad you think so!
So where does the imaginary part of the FT go? Or how does it translate into this? Or are quaternions required to see that?
The Laplace transform is complex-valued, just like the Fourier transform is complex-valued. The s-plane is the region of the complex-valued "independent variable", s, which is a generalisation of the real-valued "independent variable", frequency, in the Fourier transform.
Regarding the last point about processing, is this related to audio compression and decompression? I can imagine a flattened waveform would be easier to work with?
Sorry, I'm not sure what you're referring to here.
thank you so much! very clear and precise!
Glad it was helpful!
you gave the example of a microphone recording a voice having infinite energy thus Fourier transform can't be found out , so how we are finding the fourier transform of sine and cosine wave since they are also signal with infinite energy??
Great question. That's where the delta function comes in. The Fourier transform of a sinusoidal wave is only able to be defined with the use of delta functions in the frequency domain.
@@iain_explains ok sir that means any signal which is having infinite energy can be written in the sum of sine and cosine, and hence its Fourier transform can be found in form of impulses.
can we not break the microphone recording a voice with positive feedback into the sum of sine and cosine and then we can find its Fourier transform
the value of pole of any system is dependent on the damping provided by the system ??
sir what is the significance of pole of any system in term of laplace?
You might like to watch: "How do Poles and Zeros affect the Laplace Transform and the Fourier Transform?" ua-cam.com/video/iP4fckfDNK8/v-deo.html
Awesome explanation, thank you sir!!
Glad it was helpful!
I suppose that there will still be signals that don't have a LT. I'm thinking about x(t)=e^t^t and things like that.
Yes, that's right. Here's another one x(t) = e^(-2t)u(-t) + e^(-t)u(t)
As-salamu alaykum professor
I have a very basic question that why we consider ROC of causal system to be right side of rightmost pole
what is the reason/intuition behind it ?
Causal systems have impulse responses that are "right handed" (in time). ie. they equal 0 for all negative time. If the system is unstable, then the impulse response will grow unbounded as time goes to positive infinity. Therefore, in order to be able to take a Fourier transform, they would require a 'damping function' that goes to zero as time goes to positive infinity. This corresponds to values in the s-plane that are in the region to the right of the most positive pole. This video might help: "Why are Poles Negative in Laplace Transform of Real Stable LTI Systems?" ua-cam.com/video/apdh8ZXW3a0/v-deo.html
thanks professor it was really helpful👍
What would the region of convergence be if there are multiple poles at different exponential values. For example if your input signal was something like x(t) = e^(-t)cos(2*t) + e^(-2t)cos(3*t) I believe you would get poles -1+2j/-1-2j and -2+3j/-2-3j. How would you determine the region of convergence for this multi poled system?
The overall expression only holds when all of the component terms hold. In other words, only for values of "s" that are in the overlapping region of all of the ROC's of the component expressions. Mathematically, this is ROC = ROC1 "intersection" ROC2
Great explanation, Thank you
Glad it was helpful!
Laplace Transform converges (gives finite value) in ROC. How is this information (the finite value of LT) help us anyhow?
Because then you can perform calculations in the "z/Fourier domain", and if you need to, then you can also transform back into the (discrete) time domain.
Are all of the transformed signals going to lie along a vertical axis some distance away from the w-axis? Is it possible to have sloped lines along which the signals lie? Also, are all of the signals going to be orientated orthogonal to these vertical axes or is that just the case here because you used delta functions? Also, I have watched a few other videos and so far yours is the clearest, however other videos make mention of something called the s-plane , what is that?
I explained the overall transform in terms of a collection of "transformed signals", where each different signal only differed by the sigma parameter in the exponential that multiplied the base signal, x(t). And yes, each of these signals corresponds to a vertical line in the sigma-omega plane. But it's important to realise that these signals are all just a function of the base signal x(t). There are no "other" signals. And you ask about the s-plane, well that's exactly what I've drawn - it's the sigma-omega plane. Notice that I've written the equation s=sigma+j omega in the middle of the page.
Great explanation! Thanks a ton.
Glad you found it useful.
Thanks! Very understandable
Glad it helped!
Thank you❤
You're welcome 😊
Great explanation, thank you very much.
Glad it was helpful!
this is amazing
I'm so glad you think so.
Isn't the laplace transform defined for values t greater zero?
There are two different definitions people use. One for all time, and the other for only positive time. It doesn't really matter which one you use, as long as you are consistent.
10/10 explanation
I'm so glad you think so.
Thank you for such a clear explanation!!!!🙏👍..
Glad it was helpful!
thanks a lot sir i understood completely
Glad to hear that
sir but positive plane is full of decay exponential right, but why did you take sigma in between 0 to sigma1
I was showing an example waveform where the magnitude of x(t) grew with time (which means it does not have a FT, and therefore the LT does not exist for sigma=0). The Laplace Transform will exist (only) for values of sigma that are more positive than the rate of the growth in x(t). So, I showed the boundary as being between 0 and sigma1 because the LT for this example does not exist at sigma=0, but I said it does exist at sigma=sigma1, so the boundary must be between the two.
@@iain_explains
May I know how did you evaluate the boundary I mean mathematicaly what should be it's condition ?
Thank you sir! Great explanation. I suddenly understood it after watching your video.
Glad you found it useful. If you haven't already, you might like to check out the other videos on the channel and pdf summary sheets available at iaincollings.com
Thank you sir
Welcome
thank you so much!
You're welcome!
Nice explanation sir
Thanks for liking
Thank you so much sir 🙏💕
Most welcome
Sir, why the range of integration is 0 to ∞ in Laplace
But FT is -∞ to ∞
There are two versions of the LT. I prefer the "bilateral LT", where the integral is over the range -∞ to ∞. The "unilateral LT" is a special case, where the integral is over 0 to ∞. In both definitions, the LT integral is over the range 0 to ∞ when considering impulse responses of systems that are causal - ie. the impulse response, h(t), is zero for t in the range -∞ to 0.
You showed us to big picture, reasons and causes, I hope you will see the biggest picture.
ur the best
Thanks for your nice comment. I'm glad you like the videos.
Brilliant
Thanks. I'm glad you liked it.
I wish to study under you