Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

FM SpecAN question

  1. Jul 10, 2007 #1
    I'm looking for a "physical" explanation (or as close to one as I can get) for the displayed spectrum of an FM signal on a spectrum analyzer.

    I'm using a sinusoidal carrier of 200 MHz, a sinusoidal modulating tone, and a deviation of 5 KHz for my question.

    Lets say I use a very slow rate for the modulating tone, like 1 Hz, the spectrum analyzer will display a waveform that moves across the screen side to side at a rate of 1 Hz.

    If I continue to increase the rate for the modulating tone until I get to, say 4 KHz, I will see a classic FM spectrum with sidebands every 4 KHz. What I am wondering about is why I see such a spectrum.

    The actual time domain signal is moving through various frequencies between the two extremes of carrier-5KHz and carrier+5KHz at a 4 KHz rate, yet the spectrum analyzer appears to show little power at certain frequencies.

    What I am looking for is a good "physical" way to connect what is happening in the time domain and what the spectrum analyzer displays.

    Its possible I need to think more about what happens inside the spectrum analyzer when it receives such a frequency varying signal, then about the FM signal itself.

    Ive looked at the equations and I understand them mathematically, but I'm having trouble "seeing" such a spectrum physically.

    Anyone have a good way to think about this?
  2. jcsd
  3. Jul 10, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    At 1 Hz you are just changing the carrier frequency.
    While there are sidebands as a practical matter this will be far below the device resolution.

    At 4khz the device can resolve the +/- difference sidebands resulting from modulation.
  4. Jul 10, 2007 #3
    Hi there NoTime....

    Well you are doing that at 4 KHz as well, just much faster...

    I suspect that your comment about device resolution is likely true...

    Any thoughts on the general question?
  5. Jul 11, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    General question?
    Are you asking "Why do we have sidebands"?
  6. Jul 12, 2007 #5
    Not "why do we have sidebands?", but why does the spectrum physically take the shape it does. There are certain frequencies that have very little displayed power on the SPECAN.

    The general question was...

    That was the question I was asking.

    The spectrum makes sense if you look at the transform, but sometimes a mathematical answer is very unsatisfying and seems shallow. Im looking for as much of a "physical" answer as I can get as to why there are certain frequencies that seem to have little to no power even though the time domain signal does indeed have these frequencies present.
  7. Jul 12, 2007 #6
    You are not always going to see complex waveforms correctly on a swept spectrum analyzer because remember the analyzer takes some finite time to make a full sweep. For better readings increase the sweep rate, but at the same time the spectrum analyzer is giving up the resolution bandwidths.

    So it's a comprise, either have low sweep rate with high resolution, or fast sweep rate with low resolution. FM is affected by this alot because the modulation goes back and forth, and gets caught up in the spectrum analyzer's sweep.
    Last edited: Jul 12, 2007
  8. Jul 12, 2007 #7
    Fair enough...

    So there is some interaction between the shifting frequency of the FM signal and the sweep rate of the SPEC AN....makes sense.

    But thats not the source of the spectrum we see, is it?

    Mathematically, for the signal I described, we would expect to see the sidebands occur at multiples of 4 KHz (the FM rate) from the carrier frequency. The math seems to imply that we would expect to see a spectrum that has equally spaced (with the FM rate as the spacing) signals that decrease in amplitude.

    I'm wondering why in the receiver (in this case a SPEC AN) we see such a spectrum that has this "FM rate spaced" spectrum with the in between frequencies having little power.

    How does the FM time domain signal create such a spectrum, since the FM signal does pass through those "in between" frequencies with a non zero voltage level? How are we getting such low power in the frequencies not spaced at multiples of 4 KHz?

    I hope my question is making sense.....
  9. Jul 12, 2007 #8


    User Avatar
    Science Advisor
    Gold Member

    Not sure I completely agree with waht on this. Yes, waht has the operation of a spectrum analyzer correct. Sweep vs resolution bandwidth and all that jazz. BUT, it is not the fault of the instrument itself that there are pockets in the spectrum that have little to no power. The spectrum is truly like steadele describes. I've never been able to wrap my mind completely around that either steadele. I appologize if I misunderstand you waht.
  10. Jul 13, 2007 #9
    Hopefully someone will be able to shed some light on the subject...

    I've asked several of my co workers, PhD's included, and so far I have come up empty. :frown:

    I actually took 5 minutes yesterday in the lab and created the very signal I spoke of...and looked at it again. I tried averaging the signal and doing a max hold as well, but those damn pockets in the spectrum just mocked me...
  11. Jul 13, 2007 #10
    No problem, I agree with you now, it's not the SA in question. If I recall correctly, the FM spectrum is governed my the Bessel functions. Zeros of that function are when the sidebands cancel eachother out to an empty pocket.

    Bessel functions come from that fact that the FM carrier might take the form

    [tex] x(t) = A cos(Bt + sin(Ct)) [/tex]

    Flip it to frequency domain and there is the theoretical model. The answer should lie in here.

    [tex] J_n(x)= \frac{1}{\pi} \int_0^\pi\;cos(n\theta - x sin(\theta))d\theta [/tex]

    The way to derive this is to rewrite the trigs under an integral as a function of e such that if [tex] \zeta = e^{i\theta}[/tex]

    The integrand will take the form [tex] e^{\frac{x}{2}(\zeta-\frac{1}{\zeta})}[/tex] which can now be easily expressed as an infinite sum.

    What is interesting to note, that the term [tex] \phi=x sin(\theta)[/tex] can act also as a variable phase shift to the carrier's main frequency. Where the phase does the canceling you get the dead spot. You can think of the Bessel functions as a sort of interference pattern.
  12. Jul 13, 2007 #11
    Yep...the spectrum is described by bessel functions. The amplitude of each sideband is related to the bessel function of nth order according to the modulation index.

    How do they cancel each other out though? The time domain signal sweeps through different frequencies..as far as I know they aren't really adding together in time like several sinusoids added together would.

    This is what I'm not seeing...how are the empty pockets created if the frequency is just shifting instead of seperate frequencies adding together?

    Yes..mathematically it seems to fit together, but when I think about what the FM signal is really doing...which is just sweeping in frequency it doesn't seem to make sense.

    At time t(0) the frequency is f1 and then at time t(1) the frequency has shifted to frequency f2...I don't see a situation where we would have two different frequencies combining to give us the empty pockets.

    Yet it seems something like this must be happening physically, but I can't seem to work it out.

    I like the idea of thinking about the Bessel functions as an interference pattern, but I am wondering how such a pattern is created in an FM signal.

    What exactly is interfering?

    Different frequencies? Arent those different frequencies occuring at different times? How then would they interfere if they don't occur at the same time?

    I'm not trying to be difficult here....just trying to understand what is going on. The Bessel functions describe what is happening but they don't explain why (at least I can't see it).

    Any thoughts?
    Last edited: Jul 13, 2007
  13. Jul 14, 2007 #12
    This is not very intuitive, but I think the confusion arises when you think the carrier is shifting its frequency linearly (like a sawtooth). But actually the shift in frequency is non-linear, following sin(x). Therefore there are going to be places on the spectrum where a change in frequency in some places will be faster than others. Since the spectrum analyzer is sweeping linearly at a constant rate, in the places where the frequency shift is the fastest, vertical spikes should appear more attenuated, than in places where the rate in frequency deviation is the smallest. This alone will not account for the spectrum observed, but should give a basic idea of the frequency density distribution.

    When there is non-linearly involved, you can expect harmonics to be produced, and it's sums and differences, just like in a mixer. Not only that, but all the junk produced is also going to have different phase.

    I suspect, there also has to be a reflected wave whether from an open or closed circuit, that with every cycle of the modulation signal, the reflected wave will be 180 degrees out of phase canceling out the next cycle, and constructively interfering at sums and differences. So in a sense, it would be interfering with itself.

    That's the only explanation I can come up.
  14. Jul 15, 2007 #13
    Agreed....a sine waves rate of change is not linear and so neither would the frequency shifting since the frequency shifts follow the voltage swings.

    I don't understand why they would appear more attenuated, unless you are trying to look at the spectrum analyzers filter response.

    If that is true, then wouldn't we expect the carrier frequency to be more attenuated, since that frequency is at the zero crossings of the modulated signal? The zero crossings are where the maximum rate of voltage change occur so this should correspond to the maximum rate of frequency change in the FM signal.

    Yes but wouldn't the harmonics, in this case, be at multiples of "200 MHz + whatever the frequency deviation is at that moment"? I'm not sure if the harmonics or phase differences really affect the general shape of the spectrum we see falling out of the Bessel functions.....

    This would seem to be additional information...like we see the entire spectrum complete with the "empty pockets" repeated at multiples of 200 MHz.

    Interesting thought there...

    Doesn't this imply that we could change the spectrum we are seeing by varying the cable lengths between the signal generator and spectrum analyzer?

    Also, from the derivation of the spectrum there isn't any consideration made with respect to transmission line reflections so the spectrum's general shape appears to be independent of this, unless I am missing something.

    While I don't think these things you have mentioned completely explain the empty pockets, you have brought up valid points that we need to consider when looking at a signal on a spectrum analyzer.

    Thanks for making the effort to tackle this somewhat strange question.

    Any other thoughts on how the empty pockets get created?
  15. Jul 15, 2007 #14


    User Avatar
    Science Advisor
    Gold Member

    Steadele: The carrier frequency (center) certainly does get attenuated at times. Sometimes completely nulled out. Where this happens is dependent upon the relationship of the modulating frequency and the deviation. Bessel functions tell us what this relationship is. The procedure of setting the deviation by modulating with an appropriate frequency and adjusting deviation until the carrier nulls is known as (naturally) a bessel null. However, I don't know what the relationship is off the top of my head. When a person thinks about it in the time domain, one would think that all frequencies in the spectrum between the center carrier frequency and the farthest reach of the deviation would have energy in them. However, this is not true. In order to fully wrap your mind around it from a time domain point of view I believe you have to think about that fact that as the carrier frequency is continually varied you actually change the shape of the carrier sine wave itself. You cannot continually change the frequency of the carrier without distorting it. We (at least I) don't think about it this way normally. When we watch a slowly deviated carrier on an oscilloscope we think of it as a perfect sine wave just increasing in frequency. BUT, anytime you change the carrier frequency in ANY way, you no longer have a perfect sine wave. The carrier has to be constantly changing during each cycle so each one is distorted. Now if we could magically keep each cycle symetrical and pure and only change the frequency between each cycle at zero crossing I'm quite certain the spectrum would look much differently.
    Last edited: Jul 15, 2007
  16. Jul 15, 2007 #15
    Of course it does...but based on the modulation index, which is why I chose 200 MHz for the carrier, 5KHz for the deviation, and 4KHz for the rate. I chose a specific example because my question involved the nulls in the spectrum that lie between the carrier and the first sideband.

    I'm deliberately avoiding equations and derivations unless some physical insight can be pulled out of them.

    Yes this is all true...in fact we in the EW community sometimes use this to check and make sure we are getting the desired deviation from a adjustable signal source. You can null the carrier and then use the predicted modulation index from the Bessel function chart to check your deviation.

    What I was referring to when talking to waht was that if we have a situation where the higher rate of change in frequency is producing some kind of null, because of the spectrum analyzers response, then I would expect thh carrier to be nulled out since the maximum rate of voltage change for the modulated signal occurs at the zero crossings.

    IOW, I don't think this explanation is correct.

    Hmmm...true...interesting thought...

    I don't normally think about it that way either...

    You might be on to something here...

    Oh yes.....I think that would definately produce a very different spectrum...

    I'm not sure you have answered the question completely, but I think this might have something to do with the existence of the nulls...

    I'll keep thinking about it.

    Averagesupernova and waht: thanks for the replies and please try to bear with me here as I try and understand this thing.

    If you guys have any additional thoughts please let me know...most people here are probably smarter than me anyway, which is why I ask you guys these questions.
  17. Jul 15, 2007 #16


    User Avatar
    Science Advisor
    Gold Member

    As long as we are talking about a steady sine wave deviating at a fixed deviation we should be able to throw all spectrum analyzer responses out. You can average, peak-hold, set the video bandwidth or resolution bandwidth to whatever you want, you will always see these nulls. Unless the scan rate of the spectrum analyzer is synchronized with the modulating frequency it will display a fairly accurate representation of what is actually in the spectrum especially if the spec-an is sweeping quite slowly compared to the modulating frequency.
  18. Jul 15, 2007 #17
    Yeah...this is exactly what I have observed in the lab...none of these settings remove the nulls so it must be a property of the FM signal.

    Its just very difficult for me to explain what I'm seeing from the time domain perspective, and it bugs the hell out of me.

    Of course this is one of many, many questions I might just never have answered or ever understand. So many things in physics, math, and engineering I just can't seem to ever completely understand. The most interesting things in life are often the most frustrating and humbling things in life.
  19. Jul 15, 2007 #18


    User Avatar
    Science Advisor
    Gold Member

    Something has just ocurred to me that is worth noting. You DO understand how a spectrum analyzer works don't you? I simplified description follows: A local oscillator is swept linearly, usually considerably higher in frequency than the band of interest. A ramp waveform is generated and technically, frequency modulates the local oscillator in order to sweep it through the range needed. Then, it is mixed with the incoming signal. The product is fed through various stages of IF filtering and amplifiers. It is detected and fed through a log amp and then to the display. The display CRT is swept synchronously with the local oscillator. Now think about the local oscillator signal and why it is being frequency modulated but obviously covers ALL parts of its spectrum. Interesting huh?
  20. Jul 15, 2007 #19
    I have a basic understanding...yes

    I never thought about the ramp generator that way, but I guess it really does technically frequency modulate the LO.

    I follow you so far....

    That is interesting....

    What do you make of it? If you are seeing something insightful then please share it because I'm not catching it.....

    I'm not always the brightest bulb on the string :smile:
  21. Jul 16, 2007 #20


    User Avatar
    Science Advisor
    Gold Member

    I posted that in quite a hurry. I really haven't stopped thinking about it since. There are some more points that I should make concerning this. When I have a bit more time later today I will post more.
  22. Jul 16, 2007 #21
    I will be looking forward to it....
  23. Jul 16, 2007 #22
    The sawtooth driving the first local oscillator of a spectrum analyzer is linear. Not only that, but blanking gives more time to the LO to return freely to the starting position. The LO is not tuned backwards like in FM. Therefore a nice linear sweep can cover all the band without any gaps, the other garbage that is produced is removed by the image filter.

    Modern spectrum analyzers have a very advanced synthesized LO's and the IF is a DSP stage capable of advanced signal processing magic.
  24. Jul 16, 2007 #23
    Bah...good catch waht.

    The point about the LO not being tuned backwards does seem to negate the FM viewpoint of the LO....

    Could you elaborate on how blanking gives the LO more time to return to the starting position?

  25. Jul 16, 2007 #24
    I meant the LO does not tune backwards like when the modulating signal is a sine wave. It is still FM by definition, also by fourier analysis the sawtooth is the sum of odd harmonics sine waves. When the tuning begins, the ramp voltage increase linearly driving the VCO until the final frequency has been reached. Then the blanking signal will do two things. It will blank the screen, and give enough time for the VCO to freely return to the starting frequency. It will not tune linearly backwards from final to initial frequency. Once the VCO is at the starting frequency, the blanking signal unblanks the screen and new sweep will begin. The purpose of blanking is to not show the response on the the cathode ray tube when the VCO returns to starting frequency.

    The blanking thing was done in old school spectrum analyzers, todays SA's are synthesized.
  26. Jul 16, 2007 #25


    User Avatar
    Science Advisor
    Gold Member

    Ok. I'm going to cover some things in both of Waht posts. The fact that the LO in the spec-an is swept 'linearly' does not mean that all frequencies are covered. Technically the LO DOES tune backwards as in a normal frequency modulated signal. It just does it very quickly. If we examine the sawtooth signal that drives the LO in a spec-an, we will find that it contains ALL harmonics of the fundamental. Suppose we have spec-an that sweeps at a rate of about 10 hertz. The fundamental sine wave would be 10 hertz and all harmonics of 10 hertz would be included in this complex waveform. If we were able to look at this sweeping signal in the frequency domain wouldn't we find sidebands every 10 hertz away from an imaginary center frequency? Intuitively it seems otherwise, but I do believe that we would. The average spectrum analyzer is not going to resolve low enough to see it unless the sweep speed is slowed down. Haven't you ever noticed that the tighter you set the resolution bandwidth the slower the spec-an sweeps? My guess is that if you took a spec-an and probed its LO output with another spec-an you would see sidebands at the sweep rate. Naturally the measuring spec-an would need to have a peak-hold turned on to sample all of the sidebands because of the different scan rates of the 2 machines. Now that I think about it, the company I used to work for tested RF oscillators in this way. The oscillator was swept with a sawtooth wave and the spectrum scanned at a much slower or faster rate. The peak hold was turned on. The oscillator had to be flat within a certain spec and had to cover a spec'd part of the band. Some oscillators were swept fast while the spec-an scanned slowly and other oscillators were tested the opposite way. I can't say for sure, but it seems like I recall seeing sidebands at the rate of the sawtooth frequency. Just because the sawtooth driving the LO is linear does not mean that all frequencies are covered. It still comes down to the fundamental frequency of modulating signal.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook