Amplitude Modulation AM Radio Signal Transmission Explained

[@janismikits]

This is probably the best explanation I've found on youtube so far. Made my homework a WHOLE lot easier and more intuitive, thank you.

[@iain_explains]

Glad it helped!

[@RoenbergMusic]

Just had an exam in DSP, and your videos have been a great help in addition to my lectures, so thank you so much. I attribute part of my success in said exam to this channel. Currently preparing for my upcoming telecommunication classes watching more of your videos. Keep up the great work!

[@iain_explains]

That's wonderful to hear. It's always great to know that people are finding the videos helpful across a number of classes/subjects.

[@jerryhsu4497]

Your lessons are so amazing, useful and easy to understand. Many thanks to you for your dear contribution!

[@iain_explains]

You're welcome. Glad you like them!

[@kareemabdellah3647]

thank you i desperately needed that explanation

[@iain_explains]

I'm so glad you found it useful.

[@jasonshum888]

I think its good to add some equations on the note. But so far 8s easy to understand . Thanks for your video!

[@PawanSingh-yk8uy]

Sir, your videos are just boon to engineering students, which help us to understand the topic in easiest way on edge.

[@iain_explains]

Thanks for your nice comment. I'm glad you like the videos.

[@Lolipop5370]

Great explanation! You have helped me so much in my study of electrical engineering. Maybe you could make another for SSB-modulation :)

[@iain_explains]

Thanks, I'm glad the videos have been helpful. I'll add SSB to my "to do" list.

[@katze582]

Amazing explanations! Your videos help me a lot! - Student of Information Engineering

[@iain_explains]

Glad you like them!

[@bernardm3066]

Hi. Great video. Are you planning to do one for FM too?

[@iain_explains]

Thanks for the suggestion, I'll add it to my "to do" list.

[@kneeko1039]

great video! really helpful for undergrad electrical engineering communications paper

[@iain_explains]

Glad to hear that!

[@mertgoksel8586]

Beautiful video

[@iain_explains]

Glad you liked it.

[@willyturing1803]

Hi Lain,great video as always,on the timeline 9:11,i don't understand why they will lie on the center rather than on the side.What do i mean is that when we trying to do convolution of two functions,we are actually first flip one function in 180 degrees regarding to some value t,this way when time t passed,our flipped function will move,and eventually across the delta function and remain the convolution result shape just the same as the original function,except the origin(start) of the function now is at the time where the delta function at. say if original function is start at time 0,and delta function is at time 4,so the convolution of the function will start at time 4 with the same function shape as the original. What am i missing here?

[@iain_explains]

Convolving a function with a delta function has the result of moving the original function to be centred at the location of the delta function. In other words, the "zero point" of the original function (where t=0) would move to the location of the delta function. So, looking at the 9:11 point of this video, imagine there was only one delta function in the third graph on the right hand side of the page at 702 kHz (instead of the two delta functions). Then the second graph would be shifted so that the "zero point" of the second graph is located at 702 kHz. This would mean that one of the "lumps" would be centred at 0 kHz and the other would be centred at 2x702 kHz. Now think about doing the same for the delta function at -702 kHz, and then adding the two results together (since convolution is a linear operation).

[@willyturing1803]

@@iain_explains Hi lain,i understand it now,i was wrong about the shifting and re-locating part of convolution,i thought if you the delta function is at 702kHz,then the "lump" should appeared exactly at this location to indicate that it is exactly when this pulse happens,there's the function.But i neglected that the function we are shifting is actually f(t-t')(t' is tao) instead of f(t),so if there are some shift in the original function f(t) where t starts at say 5,then if we consider f(t-t'),there are "extra shift" in time t we need to take into account when we shifting the function f(t-t') to overlap with the delta function. Thanks for your explanation,i put my understanding for the question i had after your clarification here just so everyone who might wondering the same could get a little perspective,correct me if i'm wrong,thanks again,cheers mate.

[@akme-lrnworld810]

Thanks a lot, dear sir. Exceptionally well. Exactly what I'm looking for.

[@iain_explains]

Glad it was helpful!

[@RajenderSinghCharan]

Great video, very aptly explained. I wanted to know why do we need to use BPF before convolution with the carrier frequency at receiver, if we skip that step, still after LPF we would only have our main signal only. Could you please explain this ?

[@iain_explains]

There are two reasons for the BPF. I should have mentioned them in the video. The first is that in real wireless channels there are other AM radio stations at other neighbouring carrier frequencies (plus all the other wireless signals such as mobile/cellular, WiFi, etc) and you want to filter them all out before down converting to recover just your own signal. The second reason is that even if there were no other wireless signals at other frequencies, there is still noise at all the other frequencies, since noise is broadband, and you want to filter that out. For more details on that, see: "What is Noise Power in Communication Systems?" kzbin.info/www/bejne/laLRZYWwgteLeMU

[@RajenderSinghCharan]

@@iain_explains Thanks for replying Iain

[@gym5959]

Hello professor, i have a question , I have an assignment problem to modulate an audio signal that is sampled at 44.1kHz with a carrier with 101MHz. We are required to then extract the complex envelope of the transmitted signal and also demodulate the original audio signal. My question is that the sampling frequency of the audio signal is lesser than the carrier frequency, if I were to use the sampling frequency of 44kHz to sample the career signal and modulate the audio signal, wont there be problems like aliasing ? And if I wanted to see the spectrum of the modulated signal what fs should I use , as the frequency of spectrum would be limited to (-fs/2 ,fs/2) but the spectrum will extend to fc+fs

[@iain_explains]

I think you're referring to sub-sampling or under-sampling techniques. Yes, if the passband is conjugate symmetric then you can sample at a low rate, chosen such that there are aliased copies centred on f=0, and also chosen to ensure that the other aliased copies do not overlap. This is a way to avoid multiplying by the cos and sin. If the passband is not conjugate symmetric, then the sampling frequency needs to be twice as high as the conjugate symmetric case (which is equivalent to needing twice as many samples, which is equivalent to needing the same number of complex valued samples - nothing comes for free). Here's a video with more details: "Sampling Bandlimited Signals: Why are the Samples Complex?" kzbin.info/www/bejne/gJjPg3qInt-kfa8

[@Make_Boxing_Great_Again]

So for half decent sound we need around 5,000 variations of frequency per second for pitch, where does variations in sound volume come into it then?

[@aravindc9232]

Pure Class. Just what you need .

[@iain_explains]

Thanks. Glad it was helpful.

[@AS-nx9fu]

How did u get the bottom wave of the signal in the time domain after the multiplication

[@iain_explains]

There is no "bottom wave". The sin wave, r(t), has an amplitude envelope of 1. After the multiplication, it has an amplitude envelope given by x(t).

[@akme-lrnworld810]

Sir, I didn't get the summary sheet for this video. Is there a way you can provide it to me? As well, I checked your personal website and could not find it.

[@iain_explains]

Yes, you're right, it's a bit hard to find, but since it's not "digital" communications, it's on the Signals and Systems page, under the heading Applications of the Fourier Transform. ... and also here: drive.google.com/file/d/1y8VxQzEqoQIONUIgxsDXyH2ht3rvVoGC/view

[@nanjiang4158]

Do we use this method in 5G system? I suppose yes. Is that mean we always need to know the carrier waveform in the receiver? So, the question is whether the carrier waveform is fixed thus the receiver always knows it. Or, the carrier waveform is not fixed, does the transmitter requires some control signalling to tell it to the receiver?

[@iain_explains]

The transmitter and receiver need to agree on which carrier frequency to use, and they need to "lock" their frequency generators (crystal oscillators) to the same frequency. This is done via control channels and training resource blocks. This video might help from a "systems" perspective: "How is Data Received? An Overview of Digital Communications" kzbin.info/www/bejne/nquZq5uXeMadpKs

[@joaquinortiz279]

Great video!

[@iain_explains]

Thanks, glad you found it useful.

[@vincentvanvegan2776]

Awesome videos! :) You always talk in your other videos about pulses, that go into a transmit filter and do some pulse shaping with a sinc or square root raised cosine. But here you say now "[...] amplitude of a signal you are going to send is modulated by a carrier at the frequency you are going to send [...]" (or something like this). Can I also just describe "digital amplitude modulation" as a series of delta impulses with symbol time T, that is then put into a square root raised cosine filter and shifted by cos multiplication to the carrier frequency? I guess then the waveform would not look like you draw, however, the sampling points at kT would exactly match your drawing. Why are we no longer talking about "sending pulses"? What is the difference between amp/freq/phase modulation and sending pulses?

[@iain_explains]

This video is about modulating an analog signal. Analog signals don't have pulses, or symbol times.

[@gamingwithdad5774]

Forgive me. But I am playing around with concepts that are way over my head. Is the impulse response going back to negative infinity the same as a frequency going back to negative infinity? If so, can that frequency be fine tuned?

[@iain_explains]

The impulse response is the response of a system when an impulse gets applied at t=0. For all realisable systems (causal systems) the impulse response is (naturally, by definition) equal to zero for all t

[@gamingwithdad5774]

@@iain_explains what I mean is to isolate a frequency at a specific negative hz.

[@gamingwithdad5774]

@@iain_explains also, if this is the case, how does relativity account for a wave traveling towards negative infinity? Does that not violate relativity?

[@iain_explains]

Perhaps you missed it in the video called "What is negative frequency?", but I made the point that a "negative frequency" is actually a positive frequency, but with the phase changing in a negative way (in other words, it is a complex number that is rotating in the negative direction. ie. clockwise). All real signals are made up of a "negative" frequency component, and a "positive" frequency component in equal measure (so that the complex component of the signal cancels out, and leaves only the "real" component). Have you watched my video on how complex numbers relate to real signals? That might give more insights: kzbin.info/www/bejne/in26dmZuba-KfdU

[@deraan4388]

You are absolutely amazing!

[@iain_explains]

Thanks for your nice comment. I'm glad you're finding the videos useful.

[@prasanthr3875]

In the amplitude modulation when the messege signal m(t) is multiplied with carrier signal c(t) how we get the upper sideband c(t) +m(t). I get a visualisation of how the lower sideband c(t)-m(t) is created when watched the w2ea youtube videos.but don't able to get how upper sideband is created, as we said that in amplitude modulation the carrier frequency doesn't change the w2ea youtube video link kzbin.info/www/bejne/bZ_QnpVmpdaHi5I

[@iain_explains]

I think you might be getting the frequency domain and the time domain mixed up. The upper sideband is not c(t)+m(t) and the lower sideband is not c(t)-m(t). As I say in the video at time 2:20, m(t) and c(t) are multiplied together, and this generates a signal that is centred at the carrier frequency, and therefore has a component above the carrier frequency (upper sideband) and a component below the carrier frequency (lower sideband), as shown at 3:20.

[@prasanthr3875]

If 5kHz message signal of a single tone(For eg) when modulated with 15 kHz carrier frequency. The lower sideband is 10kHz(15_5)khz and that I can able to visualize .w2ea youtube video shows it correctly in oscilloscope. But the upper sideband 20khz(15+5)kHz is not abel to visualize. Can you do any video regarding this through any vector model representation

[@iain_explains]

I'm not sure you are understanding exactly what is shown on the oscilloscope in that video. It shows the overall signal c(t)m(t) (ie. they are multiplied together, not added!) He shows the differential component signals - but that is differential with respect to the constant bias voltage. Not with respect to the carrier signal. I'll try to make a video to explain this better.

[@prasanthr3875]

@@iain_explains Thanks for spending this much time for me. Expecting a detail video I am thinking as the lower side band is generated the time period of lower band will be T (lower)=1/ f(Lower).It is clear from that video ,the oscilloscope showing lower sideband is not a single tone(frequency) but an envelope. Like wise the upper and time period will be T(upper)= 1/f(upper).

[@prasanthr3875]

If the massage singal taken as as a single of 1Hz frequency. ie 1 positive and 1 negative cycle. And if the carrier frequency is 1000 Hz. When they are mixed After the 1st cycle of the messege signal the phase of the carrier is shifted by 180 degree and after the messege signal complete the 2nd cycle again the carrier frequency is shifted by 180 degree. So in result the carrier complete 1001 (fc+fm) due the phase shit rather than 1000 cycles. I s it correct

[@wssz112]

i see why my filter did not work =. = i keep convolving my signal in Frequency domain with my filter.....

[@iain_explains]

I'm glad the video helped you to understand that important point. I think you've just given me an idea for another video on why convolution in one domain is the same as multiplication in the other domain.

[@TheGmr140]

very careful here, what you are explaining is DSB-SC modulation, AM broadcast use DSB-SC plus carrier, or often called AM-LC so the AM broadcast equation is: [1 + p * m(t) ] * cos(wc * t) , where p is modulation percentage, and m(t) is message with amplitude range 1 to -1

[@iain_explains]

Yes indeed. This is intended to be an introductory video on Amplitude Modulation, not to go into all the different versions. Maybe that can be the topic for a future video.