Typical Exam Question on Sampling

[@electro.engineering]

The kind of professor I wish I had for Signals and Systems haha. Prof. Ian, thank you so much for dedicating some of your time for this KZbin Channel. You have no idea how much it has helped me. I found you because of your videos on continuous time convolution and I have been a fan of yours since then. Stay safe! Greetings from Germany.

[@iain_explains]

That's great to hear. I'm so glad you have found my videos helpful. Continuous time convolution was the topic that got me started making these videos. It's something that so many students find difficult, and I think it's poorly explained by many teachers, so I'm really happy to hear that you liked my videos on that topic.

[@luckyboy-ih5hd]

Sir, I am a student in college. I am doing research on narrowband in IoT. Can you do some topic about it ? I would love to hear your analysis. Thank you

[@Sid_music]

MAY YOU PROSPER. MAY NATURE BLESS YOU WITH ABUNDANCE. You teach amazing. KUDOS!! Love from India

[@iain_explains]

I'm so glad you like the video.

[@imk820]

Hi, professor. I have a question. It is actually quite an open question. Let's say, we got a signal in the frequency domain from our measurent instrument and assume the signal has been processed through ADC and DFT. We don't know any information about the siganl. There seems to be many harmonics (although we don't know if it's a real signal or just created from aliasing) and we want to find out what is the actual signal and what are from noise, interference, distortion, aliasing, nonlinearity, and so on. I think this is an real-life set-up when we want to analyze signals on screen. What strategies can we make to find out what's what? I can only come up with 2 idea which (1) try another signal into the system and see what happens and (2) try a better ADC to avoid nonlinearity and errors. But they are all just abstract ideas and not clear.I want to know many more strategies to solve this problem. What do you think, professor? Thank you.

[@iain_explains]

It depends on what ADC chip you're using. ADC chips typically have an analog pre-filter before the digitisation, to eliminate the frequency components of the analog signal that would cause aliasing in the sampled output.

[@aidanabregov1412]

Yes!! Please do more of these! What was the Nyquist solution you mentioned when you made the sampling bounds +/-6pi for 3Hz? (3 * 2pi = 6pi ?)

[@iain_explains]

According to the Nyquist sampling theorem, any signal that has a Fourier transform that equals zero for all frequencies above 3 Hz (ie. is band limited) can be fully reconstructed from samples taken at 6 samples per second. The corresponding ideal low pass reconstruction filter, for a 6 sample per second sampler, has a frequency response that is constant between 0-3 Hz and equals zero above 3 Hz. And f = 3 Hz equals w = 6pi radians per second.

[@muhammadahmedtariq2357]

However, a pure sine wave sampled at 2B results in zero samples. Therefore reconstruction is impossible. It is a must to sample such signals at rate greater than Nyquist rate.

[@iain_explains]

Yes, you're right, strictly speaking I should have said greater than 6 samples per second. But it only needs to be epsilon greater, where epsilon can be infinitesimally small.

[@muhammadahmedtariq2357]

Can epsilon be smaller than one sample ? Please elaborate.

[@iain_explains]

A pure sine wave only has three parameters: amplitude, phase, frequency. So, in principle, you only need three equations to solve for the three unknowns (ie. you only need three samples in total!). However, no matter what you do, as soon as you sample, there will always be unavoidable multiple solutions (or ambiguities) from all the multiples of frequency which will also satisfy the equations, no matter how fast you sample, or how close your samples are). Another class of ambiguity comes if your three samples are exactly a wavelength apart. Another is if they are exactly half a wavelength apart. Of course you can avoid these last two classes by always sampling faster than twice the highest frequency component (but it only needs to be epsilon higher, so that the spacing of samples is not _exactly_ half the wavelength).

[@gemacabero6482]

Thanks for the videos. I didn't understand one thing quite well. How do you know that the cutoff frequency for the filter is w_c = 6pi. How should I figure this out? And also you can't recover the original signal right? Because the max.frequency component of the signal 's FT is 16 pi right? And so the sampling frequency of 12 pi is not at least twice the max. frequency component. Thanks !

[@iain_explains]

Since the sampling rate is 6 samples per second, therefore the maximum frequency that can be exactly reconstructed is 3 Hz, which equals 3 x 2pi radians per second (angular frequency).

[@irrationalpie3143]

@@iain_explains I think the example given here is that of undersampling. So if the sin(12*pi*t) carrier was modulated by a 1Hz signal, the fact that we placed the carrier at 3Hz through undersampling would not be a problem.

[@AsadullahSarkar-m3y]

what will be amplitude after sampling?

[@ravindratomar9916]

Dear professor what will be the sampling rate for the base band signal which is nonzero from -30kHz to +20kHz. One of my friend asked this question. Sir, is 50kHz is correct??. Pls reply sir.

[@iain_explains]

Well, first of all, this would be a complex valued signal, since it is not symmetric around f=0. So then you'd need to decompose it into its real and imaginary components, and then sample those separately. See: "Sampling Bandlimited Signals: Why are the Samples "Complex"?" kzbin.info/www/bejne/gJjPg3qInt-kfa8

[@deanvanzelst5011]

my formula sheet gives the fourier transform of sin(16 pi t) different from what you have written. π/ j [δ(ω − ω0) − δ(ω + ω0)] is what i get. so now i am confused.

[@iain_explains]

Don't forget that j^2 = -1 . If you multiply your expression by j/j (ie. j on the top, and j on the bottom) then you'll get j^2 terms on the bottom ... which equal -1 ... and the "j" will have moved from the denominator to the numerator (which is what I showed in my expression in the video).

[@deanvanzelst5011]

@@iain_explains tnx for the clear explanation

[@ghaidaaal7281]

Hi l have a difficult question can u answer it pleaaaaaase 😢😢😢

[@hohunghau]

Thank you.

[@iain_explains]

You're welcome!

[@gamingandmusic9217]

First viewew ,👍