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high frequency content in rising and falling edges, data rate and bandwidth, etc.

Discussion in 'Mathematics and Physics' started by PG1995, Aug 31, 2013.

  1. PG1995

    PG1995 Active Member

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    Hi

    Q1: It is said that rising and falling edges, such as those of square wave, imply high frequency content. What does in really mean? For example, periodic square, sawtooth and triangular waves, all three have infinite number of harmonics of the fundamental frequency extending to infinity, so all three have high frequency content. Right? But a square wave has rising and falling edges which are more vertical than those of other two, so this should mean that a square has higher frequency content than the other two. But all three have harmonics extending to infinity. Please help me with this.

    This is the reason I can think of. Although the harmonics extend to infinity but higher harmonics can be ignored without actually affecting the waveform because they have such a small amplitude. Let's take the case of a square and sawtooth waves. Perhaps, in case of a sawtooth wave all the harmonics above 1000 Hz can be ignored but ignoring the harmonics above 1000 Hz for a square wave will greatly affect the shape of the waveform.



    Q2: Please have a look here.



    Q3: In analog communication a medium should be able to support at least that much bandwidth as is required by a signal. For instance, suppose that we want to send the an audio baseband signal (speech and music) by a wire instead of transmitting it using AM then this would require that the wire at least support 5 kHz bandwidth. So, for analog communication the rule is simply that the medium should be able to support the bandwidth required by a baseband signal.

    In case of digital communication the bandwidth is given in terms of data rate (bits per second). For example, if data rate is 500 bps, it means that the medium can 'transport' five hundred bits from one place to another in one second. The Shannon formula gives us relationship between bandwidth (range of frequencies) supported by a medium and data rate supported by it. If you need more data rate (wants to 'transports' more number of bits in one second) then you need to have more bandwidth assuming signal to noise ratio remains constant. What's the underlying reason for this relationship?

    If I had been asked to explain this, I would go like this. (Please first note that my whatever I'm going to say is based on my reasoning given in last paragraph of Q1. If my reasoning is not correct there then whatever I'm going to say isn't also correct.)

    For the sake of starting this discussion assume that a periodic square wave is used to send the bits (periodicity means that bits are sent like this 010101010101010101). When amplitude of the square wave is "1" it means bit "1" and when amplitude is "-1" it stands for bit "0". Now please have a look here. Do you agree with me?

    In almost all practical situations the sequence of bits in non-periodic and it is represented by non-periodic square wave. It is my experience that non-periodic signals almost always require a large bandwidth. In other words, they always have significant higher harmonics. The more non-periodicity or randomness there is, the more bandwidth is required. We kind of say that the amount of non-periodicity or randomness is proportional to the required bandwidth. Do you think what I say is generally true?

    You might find links #3 and #4 helpful for this question.



    Q4: What's the relationship between bandwidth and noise? Does the noise decrease when bandwidth is decreased? In SSB-SC amplitude modulation, it is said that SSB-SC uses less bandwidth compared to DSB-SC therefore it accounts for less noise. I don't understand the reason for this relationship.



    Thank you very much for the help and your time.

    Regards
    PG


    Helpful Links:
    1: http://www.music.mcgill.ca/~gary/307/week5/additive.html
    2: http://www.matrixlab-examples.com/fourier-series.html
    3: http://www.electro-tech-online.com/attachments/cs_56k_single-jpg.75258/
    4: http://imageshack.com/a/img713/383/x4ix.jpg
    5: http://imageshack.com/a/img18/2826/ab0c.jpg
     

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    Last edited: Sep 1, 2013
  2. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Too many questions for one post i think, but that's really up to you.

    For Q1 i think you have it right. Looking at it mathematically (as we should almost always do anyway) we find that the square wave has harmonics that are 1/n where n is the harmonics number 1,3,5,7,9, etc., all the odd guys. The triangle wave however has harmonics that are 1/n^2.
    So except for the first harmonic, the triangle wave always has harmonics that are less than the corresponding harmonics in the square wave. So no matter how many we look at we will see less content in the triangle wave.

    Lets take a quick look at a particular harmonic, number n=9...

    The square wave amplitude is 1/9, while the triangle amplitude is 1/81. That's quite a bit different even though we are only at harmonic number 9. The triangle wave content is much lower.
    In decimals the square wave content is 0.11111111 while the triangle is 0.012345679, so that's a pretty big difference.
    So if we used a sharp low pass filter we would see a bigger difference in the square wave than the triangle wave. This means if we have a bandwidth limited media we should see less effect for the triangle.

    Put another way that is also practical, if we have a square wave that is nearly perfect with very fast rise and fall times, if we want to limit the frequency content we would try to get those rise and fall times to 'slant' up or down, on an angle, so that they take on the appearance more of a triangle ramp than a square wave sharp rise or fall. That helps reduce radiated EMI which helps get past certain required government compliance tests.
     
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  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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  4. dave

    Dave New Member

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  5. steveB

    steveB Well-Known Member Most Helpful Member

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    Q1: Fast edges and sharp corners both imply high frequency content. That's why both square an triangle waves have high frequency content. Infinitely fast edges or perfectly abrupt slope changes imply infinite harmonics.

    Q2a: It is generally true, but I would guess that you can find some interesting exceptions to the general rule.

    Q2b: Is more than generally true. It is always true.

    Q3: This question has so much in it that it almost becomes a loaded question. It's hard to answer without saying something that is not quite right. I'll give more thought on the right way to answer, and I probably have to look up some things I forgot. Damned, if I could just remember all the things I've learned, I'd be a pretty smart guy. :)

    Q4: Noise and bandwidth are generally related because the noise is spread across the frequency spectrum. Hence, typically noise can be reduced by deliberately filtering out frequencies that are not needed because they are outside of the band required by the signal you care about. A good example is Gaussian white noise. It is equally spread across the frequency spectrum. Hence, you pay a high penalty if you open up the bandwidth beyond what the signal needs.
     
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  6. PG1995

    PG1995 Active Member

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    Thank you very much, MrAl, Steve.

    MrAl: Your both posts were very helpful.

    Okay. It makes sense. But don't you think this text is suggesting that the more bandwidth you use, the more reduction there is in noise? I thought I should clear this seemingly contradictory statement. Thanks.

    There is another point I would like to clear. Mostly modulation is taken to be shifting of a frequency spectrum of a baseband signal to some higher value spectrum. Can modulation also be taken to be shifting of a baseband spectrum to some lower value spectrum? Radio waves have frequencies from 300 GHz to as low as 3 kHz. Suppose, the baseband signal exists in the range 250-251 GHz. Can we use 220 GHz carrier in this case? Thank you.

    Regards
    PG
     

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  7. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi again,

    You could actually do the rise and fall times separately and that would show you the difference in harmonic content for just those places in the wave. A ramp with variable slope computed for harmonic content for example. This would show how increasing the ramp slope (and thus getting closer and closer to a perfect square wave rise or fall) increases the high order harmonic content. I might do this myself just for illustration if i get a chance.

    BTW, when we say "Harmonic Content" sometimes we mean the overall content such as "Harmonics from 3 to 103" and we dont specify how great each one is, but usually what we mean when we say "Harmonics from 3 to 103" is that the harmonics are significant in amplitude.
    So the real meaning is a little ambiguous and so a little experience is required sometimes.

    For example, we can say that both the square wave and triangle wave have "all the odd harmonics" but notice we didnt really specify how high in amplitude they were until the question came up as to which one has the greater harmonic content. Then we were forced to reconcile with a comparison of each harmonic in each wave case and we found that whatever harmonic we look at the square wave always had a higher amplitude. So what this boils down to mathematically is at first we seem to imply that they both have something the same about them, and they do (roughly shown as):
    ContentSW=ContentTW=all the odd harmonics

    but then to really compare them we have to state something like this:
    HarmonicAmplitudeSW[n]>HarmonicAmplitudeTW[n], for n>m

    SW=square wave
    TW=triangle wave

    where m might be 0 or 1, we'd have to check the fundamental amplitude for each wave.

    So the way we were able to find this was to compare each harmonic amplitude of the two cases. If we had a more complex wave however then we'd have to resort to what is called the "Total Harmonic Distortion" of the wave, and compare that, and make some concessions about what we were accepting and not accepting according to the application's demands.

    Also, when we say "It increases the high order harmonic content" we dont necessarily mean that it adds MORE harmonics, it might (and usually does) mean that those harmonics where already there but now they are even higher in amplitude than before. Thus, if we say that a particular action increased the harmonic content that might mean that the 11th through 21st harmonics went up from what they were before that action took effect.
    So there could be times when we mean that more harmonics were actually introduced ie the 11th harmonic was 0 and now it is 0.3, but usually it means that say the 11th harmonic was 0.02 or something and now it is 0.3 or so. We would probably be prone to explain in more detail why a totally new harmonic was introduced though if it was never present before.
     
    Last edited: Sep 7, 2013
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  8. steveB

    steveB Well-Known Member Most Helpful Member

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    Ah, ok. I didn't realize this is what you are talking about. Honestly, this is an advance method, and I'm rusty on the theory of this. i haven't studied this for a couple of decades. This is similar to the old spread spectrum techniques. The idea is to spread the signal across a wide band, much larger than what is needed for the signal. Theory then says that you can sacrifice your frequency band for better SNR. Rather than trying to brush up on this and restate it, I'll just refer you to this link.

    http://www.ausairpower.net/AC-0900.html
     
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  9. PG1995

    PG1995 Active Member

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    Thanks a lot, MrAl, Steve.

    MrAl: Thank you for getting back to that topic with more detail. At the moment I can't pursue that topic in more detail. But after reading your reply my understanding has improved. As a matter of fact, your two posts at the start had cleared my confusion quite enough.

    Q1:
    Steve: So, don't you think that these are two contradictory statements? In statement #1, it is said that decreasing the bandwidth decreases noise, and in statement #2 it is said that increasing the bandwidth decreases the noise. Please note that I found these statements in two different books under two different topics; "statement #1" under the topic of AM SSB and "statement #2" under the topic of Advantages of Modulation.

    Q2:
    If possible, please also help me with the query below.

    Mostly modulation is taken to be shifting of a frequency spectrum of a baseband signal to some higher value spectrum. Can modulation also be taken to be shifting of a baseband spectrum to some lower value spectrum? Radio waves have frequencies from 300 GHz to as low as 3 kHz. Suppose, the baseband signal exists in the range 250-251 GHz. Can we use 220 GHz carrier in this case? Thank you.

    Regards
    PG
     
    Last edited: Sep 8, 2013
  10. steveB

    steveB Well-Known Member Most Helpful Member

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    Q1: I wouldn't say those ideas are contradictory, but I would say that the advanced wideband modulation schemes based on Shannon's Information Theory is somewhat counter-intuitive at first. I remember studying spread-spectrum techniques using coding 25 years ago in a communications class. When you study the details, it actually makes perfect sense. But again, the details have faded from my mind. I can't say all wideband technique are as intuitive because I haven't studied them.

    Q2: Generally, you don't want to do this. The resulting spectrum can in principle still have the baseband information in it, but I'm not sure how you would demodulate it. Still, I can't say for sure that nobody hasn't played around with this idea. People try all kinds of crazy things.
     
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  11. PG1995

    PG1995 Active Member

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    Hi

    Q1:
    My query is an extension to Q1 from post #1 above. MrAl told me that my thinking was correct.

    It was said that rising and falling edges, such as those of square wave, imply high frequency content. Is there an intuitive explanation possible for this? Why would rising and falling edges imply high frequency content? Thanks.

    Q2:
    I had been under the impression that FT of a function, either time-limited or not, will contain frequencies extending to infinity assuming we don't ignore higher harmonics because of their small amplitudes. This is the case with FS too where every periodic function has frequencies extending to infinity.

    I understand that coming up with an exact answer isn't possible in this discussion so I want to know your general opinion.

    In case of FT I think I had it wrong all along because I had solved at least one problem where a time-unlimited sinc function had FT as a frequency-limited rectangular function; the frequencies didn't extend to infinity. What is your general statement about time-unlimited functions and the frequencies contained in their FT? Don't you think most of the time their frequencies extend to infinity?

    Can a time-limited function have a FT with limited frequencies? Thank you.

    Regards
    PG
     
    Last edited: Sep 29, 2013
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  12. steveB

    steveB Well-Known Member Most Helpful Member

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    Q1: I'm not sure of a good intuitive explanation if the idea isn't already intuitive to you. For me, it is so intuitive, that I never thought about it too much. In some sense, you might say that a sharp edge is the only possible with high frequency sinusoidal waves because low frequency ones have no hope of creating a fast transition. But, I dont' know if that will help you at all.

    Q2: It sounds like you discovered on your own that it is possible for a time unlimited signal to have a frequency limited FS or FT. The sinc function is a good example of a non0periodic signal and a simple sine wave is an example of a periodic one. A time limited signal should always require infinite frequency, unless it's a trivial case (f(t)=0 for example) or some type of exotic signal that a mathematician might look at but has no practical value (I'm not aware of any of these, but I just mention it because those crazy math guys invent all kinds of things that make no sense to me).
     
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  13. PG1995

    PG1995 Active Member

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    I think a periodic signal (which also means it's going to be time unlimited) will always have frequency limited spectrum.

    But is this always true for a non-periodic signal to have frequency limited spectrum? I don't think it's true that every time-unlimited non-periodic signal (sinc function is a non-periodic time unlimited signal) is always going to have a frequency limited spectrum? In my opinion, majority of time-unlimited non-periodic signals will have unlimited frequency spectrum. Thanks.

    Regards
    PG
     
    Last edited: Jan 9, 2014
  14. steveB

    steveB Well-Known Member Most Helpful Member

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    What about a square wave, triangle wave or sawtooth wave?

    It shouldn't be a matter of opinion. You are making a statement than can be easily proved by examples. Please provide some example of this.
     
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  15. PG1995

    PG1995 Active Member

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    Thank you.

    Thanks a lot. Now I will never forget it because now we have documented it here!


    Yes, I agree with you. Actually I wasn't able to come up with any example. So, in a way, I was requesting you to let me know of any simple example, if possible. But now I will review some of the stuff later. Thanks.

    Regards
    PG
     
  16. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    All you have to do really is remember that when you know the harmonics and amplitudes of the wave to recreate that wave in the time domain you must add the sinusoidal components together to get that wave again in the time domain. This constant awareness of what is happening in the frequency domain and what is happening in the time domain is the key to understanding this.

    If we look at the first harmonic of say a 1Hz square wave with amplitude 1, we find that it is just a 1Hz sine wave with amplitude a bit higher than 1. If we plot these two with the square wave on the top graph and the sine wave on the bottom graph we see that they both go through zero at the same points in time and when one goes positive the other goes positive too, and when one goes negative the other goes negative also. So they follow each other in step more or less even though the shapes are different.

    But what we notice when we compare them this way is that the sine wave of the same frequency as the square wave takes a LOT longer to rise to the peak, while the square wave rises immediately to the peak. Now if we had two or more 1Hz sine waves of different phases and we added them together, we could NEVER produce a steep rising front like the square wave has. That is because of the basic way that sine waves add together.

    When we add a higher non infinite frequency harmonic to the fundamental however (or even just look at a higher harmonic by itself) we see right away that the wave has the innate ability to RISE FASTER than the fundamental. There is still a limit because it's a sine wave of non infinite frequency, but now the charge in the wave can occur much faster.

    It's because we can get a faster change in the time domain with a sine wave that is of a higher harmonic that allows us to force a change in the final wave such as the square wave. By adding the higher harmonic, we get a fast rising wave front and that is why we end up with high harmonic content in the square wave. If we limit this to some finite values like say 1000Hz, then that means that we limit the rise time of the square wave to some value that is slower than a true square wave. If we triple that limit to 3000Hz, then we (probably) decrease the rise time by 1/3 of what it was before (although i did not compute this it should follow some rule like that or similar). But since we need a rise time that is zero, we would need a frequency that is very very high to allow us to force that fast of a change in the time domain. In fact for a pure square wave it would have to be infinite, and that only happens in the theoretical realm. In practice, we find that there is no real square wave only a squarish wave that is almost a true square wave and the fronts rise very quickly but not in zero time.

    The best way to comprehend this is to do a few simple experiments with a graphing calculator or on the computer in a graphing program. Start with a fundamental like 1H and add different harmonics and see how the initial rise time decreases. Note that eventually you reach the limit of the graphics to display a ramp and it starts to look like it has a zero rise time. To truly understand this you just have to take the time to do stuff like this.
     
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  17. steveB

    steveB Well-Known Member Most Helpful Member

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    Ah OK, I didn't realize you were asking for examples. So, your original statement was the following.

    Essentially, I agree with your statement. Saying "majority" is questionable because you are dealing with an infinite number of each type. But, let's just say "the majority of practical idealized time-unlimited non-periodic signals have unlimited frequency spectrums".

    Now, we need to qualify that, in practice, we can approximate these band-unlimited signals by band-limited versions that have limited frequency spectra. Hence, this is just a discussion of principles.

    Simplest examples are, impulse, pulse and step.

    Now, let's use some math tricks to say something powerful that can bolster your statement even more.

    Take any signal, whether band limited or band unlimited. At some point in time T where the signal is nonzero, place a step function there to turn that signal off before that point in time and make the signal zero for t<T and unmodified for t>T. Such a signal will be band unlimited. This is important because in practice we always need to turn signals on and off abruptly. So, mathematically how can I prove what I just said. That's simple. The signal can be represented as the step signal u(t-T) multiplied by the original signal x(t). Well, multiplication in the time domain transforms as convolution in the frequency domain. Hence, it follows that since u(t-T) is band unlimited, then the multiplication of u(t-T) x(t) is also band unlimited because the convolution of an infinite frequency range in the frequency domain with any other spectrum (whether band limited or band unlimited) will give a band unlimited spectrum.
     
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  18. PG1995

    PG1995 Active Member

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  19. steveB

    steveB Well-Known Member Most Helpful Member

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    There are a lot of exceptions. For every infinite bandwidth signal, you can make an infinite number of finite bandwidth approximations to that signal, by chopping all frequencies above a certain value. You can also make an infinite number of infinite bandwidth approximations to those signals by applying a first order low pass filter at a particular cutoff frequency. In that case the upper frequencies are attenuated heavily, but they are not zero in theory. You could also apply a second order low pass filter, or a third order, etc. and still you would have in infinite bandwidth signal in each case.

    This is why the word 'majority' is a bit imprecise, but I'm sure you get the idea now.
     
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  20. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    It sounds to me like you are trying to categorize every signal under the sun and then examine the result to see if there is some trend with respect to the bandwidth of these signals.

    If you are going to do that then the first thing you have to do is create a big division between real signals and theoretical signals.

    Theoretical signals can contain almost any frequency component including what we might call infinite, but real signals always appear bandlimited because the higher frequency components do not pass through anything without at least some frequencies being attenuated even though these may be the very highest ones.

    Aside from that though it is very hard to do this i think because no matter what signal we are using who says that the next guy that comes along wants to use a test signal that is entirely different than the ones we had been using all along. In specific applications there may be a tendency to use the same signals but i think it is hard to specify something like this that has to cover every possible application.

    There are applications that use simple sine waves for test signals and applications that use square waves for test signals. A linear circuit will pass this information while a non linear circuit could very well introduce higher harmonics. The theoretical non linear circuit could produce an infinite number of harmonics, while the real life circuit could produce a large number of harmonics but not an infinite number.

    ADDED:
    We can say that a signal that originates as a theoretical signal and travels through a theoretical linear medium gets attenuated according to the medium's filter function and so the higher harmonics will still remain just at a reduced amplitude, but if that same signal is put through a real medium i am not sure we can still say this because the real medium will probably attenuate some of the signals out to a true zero level or else the noise will mask these lowest levels.
     
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