Yes, that's right. Have you seen my webpage with all the videos categorised. iaincollings.com There are a few videos on the Fourier transform and the basis functions, including these: "Fourier Transform Equation Explained" ua-cam.com/video/8V6Hi-kP9EE/v-deo.html and "Orthogonal Basis Functions in the Fourier Transform" ua-cam.com/video/n2kesLcPY7o/v-deo.html
@@iain_explains I was looking for playlist in your yt channel and you gave me this thankful comment!! I really appreciate your comment and help!! The website would be really helpful for me!! I really appreciate for sharing amazing classes!
Thank you for this awesome explanation . I have few doubts which I would like to clarify . The Fourier transform of a train of impulse function is also a train of impulse function and if you multiply in time domain then it is equivalent to convolution in frequency domain . Going by this fact, the diagram which you drew should have as many replicas of baseband FT as there are delta functions , you drew only two . But there are many time shifted delta functions which you depicted in the plot
Yes, that's right. I only drew frequency domain copies at +/- omega_s because I didn't have any more room on the page. But they also occur at all the integer multiples of +/- omega_s
It shows that if you have a set of discrete samples, then they can represent multiple different continuous-time signals. In the example I showed, it was a cos wave at a frequency w1 and another cos wave at the frequency w1+2pi. I could also have drawn another cos wave at the frequency w1+4pi, and it would also go through all the same points, and so on. And so in the frequency domain plots there will be repetitions every integer multiple of +/-2pi.
Yes. Check you these videos: "Sampling Discrete Time Signals" ua-cam.com/video/mKGp7kus4yo/v-deo.html and "Discrete Time Sampling Example" ua-cam.com/video/vGxvrq5kCzQ/v-deo.html plus other videos at www.iaincollings.com/signals-and-systems#h.2pw12d5n83qf
Yes, that's right. You can confirm it by comparing the equation for the continuous time FT with the equation for the discrete time FT. The integral in the continuous time equation becomes a sum when the signal is a (sampled) sequence of delta functions, and if the elements of that sum are at integer values of t (ie. T=1), then you've got the same equation as the discrete time FT equation.
On moving from Fourier Series to Fourier Transform, the period of the signal increased, so much so that the resolution in F Series spectrum decreased and we obtained a continuous Fourier Transform. Can you please explain this evolution in case of DTFS to DTFT? I understand how DTFT becomes continuous, but I have been having a hard time trying to comprehend as to why the DTFT has to be periodic?
what is the difference between continuous time sampling and discrete-time sampling as we are sampling with the same impulse function only why its like that in case of continious time sampling the signal is zero between the two samples where as its undefined between two samples in case of discrete time sampling but we are using the same delta function in both the cases. suppose we are sampling a signal at rate 1mega hertz than the sample spacing would be 1 micro sec that is not an intiger value means sampling in discrete time is not possible if we are sampling at that high rate ??? where as sampling in continious time is possible as 1 micro sec is not an intiger value .
Well, it's not actually "the same impulse function". The continuous-time basis functions are different from the discrete-time basis functions. In discrete time, nothing exists between the values. There is no sense of "time", there is only the integer indexing of the values. "Discrete time" signals only relate to "time", when you associate them with a "time difference between the values", eg. T. More specifically, the continuous-time impulse function has "infinite height and zero width", whereas the discrete-time impulse function has unit height (and in "discrete time" there is no notion of "width" - since all the values are "discrete"). Also, you need to be careful not to confuse the word "sampling". Strictly speaking, "sampling" is the process of taking discrete samples from a continuous-time signal, and in doing so, generating a discrete-time signal. Sometimes people talk about "upsampling" and "downsampling" the discrete-time signal, but that just involves adding or removing values in the discrete-time signal/vector/sequence (... however you prefer to think of it). It's not the same as "sampling".
I think you mentioned that x_s(t) evaluates to zero in between the delta functions, whereas in x_s[t] it is not defined. Then, for the case where T=1, they are different signals, yet they have the same F.T. How can that be? Thank you
You make an interesting observation. Yes, a continuous-time signal composed of regularly spaced delta functions 1 second apart (or even just a single delta function) can have the same Fourier transform as a discrete-time signal (ie. the corresponding discrete-time signal), precisely because the continuous-time delta function is a mathematical concept constructed to capture the same phenomenon as the discrete-time delta function exactly captures - namely "energy at an instant of time". In discrete-time, the "time instances" are well defined (the integers), but in continuous-time, there is no "instant" of time, since there is infinite resolution in time. If you are given a Fourier transform function that repeats every 2pi, and invert it with the formula for discrete-time signals, you will get a discrete time signal. If you invert it with the formula for continuous-time signals, you will get a continuous-time signal (which will consist of continuous-time delta functions at regular spacings 1 sec apart with areas that match the heights of the discrete-time signal).
@@iain_explains I see. Thank you very much! It looks like it's better to think of a continuous-time signal as something like a probability distribution function. So the delta functions are just a mathematical concept. It might be the case that real signals can never be a delta function, but it turns out that as you add the FT components, you get a delta function in the time domain. Like a limit of a function: as you approach to infinity, you happen to approach to a certain value. It follows that if you have a (theoretical) signal composed of delta functions separated at a fixed interval, you can think of it as a discrete-time signal.
That's exactly right. It's interesting to think about. In our lives (which are happening in continuous time) nothing ever happens at an instant. Everything happens over a time period (even if that time period is extremely small - eg computer chips that clock data at Gbps rates). Every possible change is limited by the speed of light.
Sir to get back our original xt we will face aliasing but if I only want to get back xn then is there any role of aliasing because I can do that with formula or xn ?
Using trignometric relation you proved that cos[(w+2*pi)n] = cos(wn)-sin(wn). how is this equal to cos(wn) ? I am little confused here..can you please clarify?
Just to clarify: I didn't prove that cos[(w+2*pi)n] = cos(wn)-sin(wn). I actually just expanded out cos(A+B) using the standard expansion cos(A)cos(B)-sin(A)sin(B). Then I mentioned that sin(2 pi n) = 0 , since n is an integer, the sin function equals 0 for any integer multiple of 2pi. This means that the second term in the expansion is 0 (ie. the term sin(A)sin(B) ). And also cos(2 pi n) = 1 (since cos = 1 for any integer multiple of 2pi), which means that the overall expansion equals cos(wn).
No, you're not correct. If the time domain delta functions all had the same value, then you would be right - the Fourier transform would be a series of delta functions (ua-cam.com/video/t0NLutwAi3c/v-deo.html). But in this case, the time domain delta functions each have different values (given by the function x(t) values at those times). See this video for more details: "Sampling Signals" ua-cam.com/video/AcuQnIXiZ2A/v-deo.html
I got Time domain function multiply the pure delta functions equal to convolution of both two functions in frequency domain. It would be series of bell shape.
It depends on the type of discretisation you're talking about. Sometime discretisation refers to generating a discrete-time approximation to a continuous-time system. In this case, yes, "sampling" is an important step in that process. In other cases, discretisation refers to quantising the amplitude levels of a signal. For example, both types of discretisation are performed when your voice signal is "digitised" in a mobile phone (before it is transmitted). The continuous-time waveform is sampled at 8 thousand samples per second (ie. time domain sampling), and the sampler (ie. analog-to-digital converter) records/stores each sample value by mapping it to one of 256 possible amplitude levels (ie. amplitude domain quantisation). These 256 levels are indexed by 8 digital bits (2^8 = 256). Therefore the overall digital signal requires 64 kbps.
Thank you very much for the videos! Super helpful. Would you mind clarifying why is e^{jw} used for discrete time, and {jw} used for continuous time? I understand w means angular frequency, but what does {jw} mean as an entity? Thanks a lot!
It's a good question. Perhaps it's best to think in the time domain for this question. Periodic signals have a fundamental period (of course), and this means that all the sinusoidal signal components that make up that periodic signal, must have periods that are an integer fraction of that fundamental period. In other words, the overall signal is only periodic if all the component sinusoids undergo an integer multiple number of cycles over the fundamental period time (and then start again exactly, in the next fundamental period time).
Thank you so much for these videos. Question, if I had a continuous time signal 5*Cos(wt). Will the amplitude on the plot Xs(jw) be 5*pi*(1/Ts) ? Or simply pi. What about the Xs(e^jw)? Will the amplitude be 5*pi ?
Yes, it's the second option you suggested: 5*pi*(1/Ts), and this is the same amplitude for the Fourier Transform of the discrete time version of the signal, Xs(e^jw), too.
Watched over 10 videos but this is the best! You made it easy by providing a good overview to motivate people to learn.
I'm so glad you found the video helpful. It's great to know that you think it's the best one you've seen on the topic.
You are incredible in explaining these Fourier stuff
I'm so glad you like my explanations. Thanks for your nice comment.
Thank you so much for this video, only 10 minutes to watch and it has helped my understanding hugely! Fantastic work
Glad it helped!
3:33 is the reason you choose cos(w_0n) as example is we can make any function by cos... is this right??!
Yes, that's right. Have you seen my webpage with all the videos categorised. iaincollings.com There are a few videos on the Fourier transform and the basis functions, including these: "Fourier Transform Equation Explained" ua-cam.com/video/8V6Hi-kP9EE/v-deo.html and "Orthogonal Basis Functions in the Fourier Transform" ua-cam.com/video/n2kesLcPY7o/v-deo.html
@@iain_explains I was looking for playlist in your yt channel and you gave me this thankful comment!!
I really appreciate your comment and help!! The website would be really helpful for me!! I really appreciate for sharing amazing classes!
Thank you so much for teaching these in such an intuitive way. Really helpful.
I'm so glad you like them!
Thank you for this awesome explanation . I have few doubts which I would like to clarify . The Fourier transform of a train of impulse function is also a train of impulse function and if you multiply in time domain then it is equivalent to convolution in frequency domain . Going by this fact, the diagram which you drew should have as many replicas of baseband FT as there are delta functions , you drew only two . But there are many time shifted delta functions which you depicted in the plot
Yes, that's right. I only drew frequency domain copies at +/- omega_s because I didn't have any more room on the page. But they also occur at all the integer multiples of +/- omega_s
@ 5:24 what is the significance of drawing two waves of different frequencies with same number of the samples
It shows that if you have a set of discrete samples, then they can represent multiple different continuous-time signals. In the example I showed, it was a cos wave at a frequency w1 and another cos wave at the frequency w1+2pi. I could also have drawn another cos wave at the frequency w1+4pi, and it would also go through all the same points, and so on. And so in the frequency domain plots there will be repetitions every integer multiple of +/-2pi.
Excellent explanation!
Glad it was helpful!
i've watched 10+vids and 3+ books including oppenheim;s but this video is killing it..! thank you preofessor!!
Glad you found the video helpful. And thanks for your nice comment.
I have a doubt, is there such thing as sampling of a discret time signal?
Yes. Check you these videos: "Sampling Discrete Time Signals" ua-cam.com/video/mKGp7kus4yo/v-deo.html and "Discrete Time Sampling Example" ua-cam.com/video/vGxvrq5kCzQ/v-deo.html plus other videos at www.iaincollings.com/signals-and-systems#h.2pw12d5n83qf
@@iain_explains tysm, ur videos are of great help!!
sir your video is mind-blowing crystal clear simplistic explanation
Glad you found it useful. For a full listing of videos categorised into topic sections, see my webpage: iaincollings.com
Thanks Iain. IS it safe to say that when T=1, the F.Ts of both the continuous sampled signal and Discrete signal will be the same?
Yes, that's right. You can confirm it by comparing the equation for the continuous time FT with the equation for the discrete time FT. The integral in the continuous time equation becomes a sum when the signal is a (sampled) sequence of delta functions, and if the elements of that sum are at integer values of t (ie. T=1), then you've got the same equation as the discrete time FT equation.
Thank you, it is clear and useful
Glad it was helpful!
On moving from Fourier Series to Fourier Transform, the period of the signal increased, so much so that the resolution in F Series spectrum decreased and we obtained a continuous Fourier Transform.
Can you please explain this evolution in case of DTFS to DTFT?
I understand how DTFT becomes continuous, but I have been having a hard time trying to comprehend as to why the DTFT has to be periodic?
I know you've found the answer already, but for others who might have the same question, please check out: ua-cam.com/video/P7q2YMQiat8/v-deo.html
Great explanation! Thank you so much for these videos.
Thanks, I'm glad you've found the videos useful.
what is the difference between continuous time sampling and discrete-time sampling as we are sampling with the same impulse function only
why its like that in case of continious time sampling the signal is zero between the two samples where as its undefined between two samples in case of discrete time sampling but we are using the same delta function in both the cases.
suppose we are sampling a signal at rate 1mega hertz than the sample spacing would be 1 micro sec that is not an intiger value means sampling in discrete time is not possible if we are sampling at that high rate ??? where as sampling in continious time is possible as
1 micro sec is not an intiger value .
Well, it's not actually "the same impulse function". The continuous-time basis functions are different from the discrete-time basis functions. In discrete time, nothing exists between the values. There is no sense of "time", there is only the integer indexing of the values. "Discrete time" signals only relate to "time", when you associate them with a "time difference between the values", eg. T. More specifically, the continuous-time impulse function has "infinite height and zero width", whereas the discrete-time impulse function has unit height (and in "discrete time" there is no notion of "width" - since all the values are "discrete").
Also, you need to be careful not to confuse the word "sampling". Strictly speaking, "sampling" is the process of taking discrete samples from a continuous-time signal, and in doing so, generating a discrete-time signal. Sometimes people talk about "upsampling" and "downsampling" the discrete-time signal, but that just involves adding or removing values in the discrete-time signal/vector/sequence (... however you prefer to think of it). It's not the same as "sampling".
I think you mentioned that x_s(t) evaluates to zero in between the delta functions, whereas in x_s[t] it is not defined. Then, for the case where T=1, they are different signals, yet they have the same F.T. How can that be? Thank you
You make an interesting observation. Yes, a continuous-time signal composed of regularly spaced delta functions 1 second apart (or even just a single delta function) can have the same Fourier transform as a discrete-time signal (ie. the corresponding discrete-time signal), precisely because the continuous-time delta function is a mathematical concept constructed to capture the same phenomenon as the discrete-time delta function exactly captures - namely "energy at an instant of time". In discrete-time, the "time instances" are well defined (the integers), but in continuous-time, there is no "instant" of time, since there is infinite resolution in time. If you are given a Fourier transform function that repeats every 2pi, and invert it with the formula for discrete-time signals, you will get a discrete time signal. If you invert it with the formula for continuous-time signals, you will get a continuous-time signal (which will consist of continuous-time delta functions at regular spacings 1 sec apart with areas that match the heights of the discrete-time signal).
@@iain_explains I see. Thank you very much!
It looks like it's better to think of a continuous-time signal as something like a probability distribution function.
So the delta functions are just a mathematical concept. It might be the case that real signals can never be a delta function, but it turns out that as you add the FT components, you get a delta function in the time domain. Like a limit of a function: as you approach to infinity, you happen to approach to a certain value.
It follows that if you have a (theoretical) signal composed of delta functions separated at a fixed interval, you can think of it as a discrete-time signal.
That's exactly right. It's interesting to think about. In our lives (which are happening in continuous time) nothing ever happens at an instant. Everything happens over a time period (even if that time period is extremely small - eg computer chips that clock data at Gbps rates). Every possible change is limited by the speed of light.
sir is this same thing happen in the case of not band limited signal
Yes, but when the signals are not band limited there will be aliasing. For more on that topic, check out: ua-cam.com/video/AcuQnIXiZ2A/v-deo.html
Sir to get back our original xt we will face aliasing but if I only want to get back xn then is there any role of aliasing because I can do that with formula or xn ?
So does the basis function(FT of discrete time) repeat at 4π, 6π, and it's negatives or only at 2π and -2π?
Yes, exactly. For more details see: "Discrete Time Basis Functions" ua-cam.com/video/P7q2YMQiat8/v-deo.html
thank you
You're welcome
Using trignometric relation you proved that cos[(w+2*pi)n] = cos(wn)-sin(wn). how is this equal to cos(wn) ? I am little confused here..can you please clarify?
Just to clarify: I didn't prove that cos[(w+2*pi)n] = cos(wn)-sin(wn). I actually just expanded out cos(A+B) using the standard expansion cos(A)cos(B)-sin(A)sin(B). Then I mentioned that sin(2 pi n) = 0 , since n is an integer, the sin function equals 0 for any integer multiple of 2pi. This means that the second term in the expansion is 0 (ie. the term sin(A)sin(B) ). And also cos(2 pi n) = 1 (since cos = 1 for any integer multiple of 2pi), which means that the overall expansion equals cos(wn).
The 2nd set,( the FT of delta functions ), right hand side graphics seems incorrect. It should be still delta functions,am I right?
No, you're not correct. If the time domain delta functions all had the same value, then you would be right - the Fourier transform would be a series of delta functions (ua-cam.com/video/t0NLutwAi3c/v-deo.html). But in this case, the time domain delta functions each have different values (given by the function x(t) values at those times). See this video for more details: "Sampling Signals" ua-cam.com/video/AcuQnIXiZ2A/v-deo.html
I got
Time domain function multiply the pure delta functions equal to convolution of both two functions in frequency domain.
It would be series of bell shape.
Hi professor, I have a very naive question: would sampling in the signal processing be a similar idea to “discretization” in the numerical analysis?
It depends on the type of discretisation you're talking about. Sometime discretisation refers to generating a discrete-time approximation to a continuous-time system. In this case, yes, "sampling" is an important step in that process. In other cases, discretisation refers to quantising the amplitude levels of a signal. For example, both types of discretisation are performed when your voice signal is "digitised" in a mobile phone (before it is transmitted). The continuous-time waveform is sampled at 8 thousand samples per second (ie. time domain sampling), and the sampler (ie. analog-to-digital converter) records/stores each sample value by mapping it to one of 256 possible amplitude levels (ie. amplitude domain quantisation). These 256 levels are indexed by 8 digital bits (2^8 = 256). Therefore the overall digital signal requires 64 kbps.
Thank you very much for the videos! Super helpful.
Would you mind clarifying why is e^{jw} used for discrete time, and {jw} used for continuous time? I understand w means angular frequency, but what does {jw} mean as an entity?
Thanks a lot!
Yes, this is a common confusion. This video explains it: "Transform Notation" ua-cam.com/video/y81O9z5qnBE/v-deo.html
@@iain_explains Thank you very much!
Glad it helped.
Hi Sir, I have a doubt. Why does the periodic signal in time always give a discrete frequency spectrum? Could you please clarify my doubt?
It's a good question. Perhaps it's best to think in the time domain for this question. Periodic signals have a fundamental period (of course), and this means that all the sinusoidal signal components that make up that periodic signal, must have periods that are an integer fraction of that fundamental period. In other words, the overall signal is only periodic if all the component sinusoids undergo an integer multiple number of cycles over the fundamental period time (and then start again exactly, in the next fundamental period time).
Great
Thank you so much for these videos. Question, if I had a continuous time signal 5*Cos(wt). Will the amplitude on the plot Xs(jw) be 5*pi*(1/Ts) ? Or simply pi. What about the Xs(e^jw)? Will the amplitude be 5*pi ?
Yes, it's the second option you suggested: 5*pi*(1/Ts), and this is the same amplitude for the Fourier Transform of the discrete time version of the signal, Xs(e^jw), too.
don't we call this repetition of frequency components as aliasing?
It's only "aliasing" when the "frequency domain copies" overlap with each other. ie. when w_c > w_s - w_c