Yes, it's a continuous function. Yes, the "height" of the delta function is infinite. Of course, we don't have a piece of paper that's infinity tall, so instead we draw an arrow, to indicate that the actual "height" is infinite. However, also, the area under the delta function is finite! So, often we draw the arrows to be as tall as their areas, just to give an intuitive visual representation, to indicate the delta functions with different areas. Here's a video with more details: "Delta Function Explained" th-cam.com/video/lyraqtMWtGk/w-d-xo.html
What happens if we convolute a general signal (e.g. d(t) ) to this summation in frequency domain? What would the result be? I assume I would get the summation from n=-infinity to +infinity.D(n/T) but the answer states it's the summation from n=-infinity to +infinity.D(n/T).delta(f--n/T). Would appreciate it if you could explain how the dirac function still remains after the multiplication.
I'm not totally clear about what you're asking, but I think you're talking about sampling. Check out this video which might help: th-cam.com/video/AcuQnIXiZ2A/w-d-xo.html Basically, the multiplication in the time domain is the same as a convolution in the frequency domain. And since the Fourier transform of a sum of delta functions is also a sum of delta functions (in the frequency domain) you get copies of the transform of d(t) appearing at each of the multiples of the sampling rate (1/T) in the frequency domain, since convolving with a delta function moves a function to the location of the delta function (see th-cam.com/video/TIcfY19dk0c/w-d-xo.html )
@@iain_explains It's tough writing my question here as a comment, but yes, it's kind of like sampling. In your video: th-cam.com/video/AcuQnIXiZ2A/w-d-xo.html , you had a product of x(t)p(t) on the left side of the paper, where p(t) would be the summation of delta functions. My question was instead of x(t)p(t) in the time domain, I had (summation from k=-∞ to k=+∞)Σ 𝛿(f-k/T).X(f) in the frequency domain - a multiplication and not a convolution. Trying to solve that product I expected to get just (summation from k=-∞ to k=+∞)Σ X(k/T), but the correct solution provided was (summation from k=-∞ to k=+∞)Σ 𝛿(f-k/T).X(k/T). I'm assuming the summation had something to do with the difference in our answers. Why is this so? I hope this was clearer
@@jeremykua8525 Ah, I see. This is a common thing that people get confused with. It is important to realise that X(k/T) is just a number. It's not a function. It is the value of the function X(f) at the frequency f=k/T. In contrast, 𝛿(f-k/T).X(f) is a function. It is zero for all values of f, except at f=k/T where there is an impulse (zero width, infinite height, but with area = X(k/T) ). Hope this helps.
Sorry about that. I made this video a long time ago - before I knew that I was able to upload bigger file sizes. I've also been using a better microphone on my more recent videos.
Can we find the trigonometric fourier series of it ?
And how
A very good video, it really helped visualize the sum function
but isn't this a continuous function? wouldn't the values of dirac be infinity? And wouldn't we plot it at infinite ys?
Yes, it's a continuous function. Yes, the "height" of the delta function is infinite. Of course, we don't have a piece of paper that's infinity tall, so instead we draw an arrow, to indicate that the actual "height" is infinite. However, also, the area under the delta function is finite! So, often we draw the arrows to be as tall as their areas, just to give an intuitive visual representation, to indicate the delta functions with different areas. Here's a video with more details: "Delta Function Explained" th-cam.com/video/lyraqtMWtGk/w-d-xo.html
What happens if we convolute a general signal (e.g. d(t) ) to this summation in frequency domain? What would the result be? I assume I would get the summation from n=-infinity to +infinity.D(n/T) but the answer states it's the summation from n=-infinity to +infinity.D(n/T).delta(f--n/T). Would appreciate it if you could explain how the dirac function still remains after the multiplication.
I'm not totally clear about what you're asking, but I think you're talking about sampling. Check out this video which might help: th-cam.com/video/AcuQnIXiZ2A/w-d-xo.html Basically, the multiplication in the time domain is the same as a convolution in the frequency domain. And since the Fourier transform of a sum of delta functions is also a sum of delta functions (in the frequency domain) you get copies of the transform of d(t) appearing at each of the multiples of the sampling rate (1/T) in the frequency domain, since convolving with a delta function moves a function to the location of the delta function (see th-cam.com/video/TIcfY19dk0c/w-d-xo.html )
@@iain_explains It's tough writing my question here as a comment, but yes, it's kind of like sampling. In your video: th-cam.com/video/AcuQnIXiZ2A/w-d-xo.html , you had a product of x(t)p(t) on the left side of the paper, where p(t) would be the summation of delta functions. My question was instead of x(t)p(t) in the time domain, I had (summation from k=-∞ to k=+∞)Σ 𝛿(f-k/T).X(f) in the frequency domain - a multiplication and not a convolution. Trying to solve that product I expected to get just (summation from k=-∞ to k=+∞)Σ X(k/T), but the correct solution provided was (summation from k=-∞ to k=+∞)Σ 𝛿(f-k/T).X(k/T). I'm assuming the summation had something to do with the difference in our answers. Why is this so? I hope this was clearer
@@jeremykua8525 Ah, I see. This is a common thing that people get confused with. It is important to realise that X(k/T) is just a number. It's not a function. It is the value of the function X(f) at the frequency f=k/T. In contrast, 𝛿(f-k/T).X(f) is a function. It is zero for all values of f, except at f=k/T where there is an impulse (zero width, infinite height, but with area = X(k/T) ). Hope this helps.
Thank you sir
You're welcome
what will be sumation of del [ n-2p ] . please tell as fast as you can
you just solved a question about my final exam. 😁😁
Great to hear!
very poor sound amd vid quality
Sorry about that. I made this video a long time ago - before I knew that I was able to upload bigger file sizes. I've also been using a better microphone on my more recent videos.