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FM Analysis including Bessel Function

  1. Aug 18, 2013 #1
    1. The problem statement, all variables and given/known data
    An FM broadcast system has the following parameters:
    *Deviation sensitivity 5 kHz/V.
    *Information signal consists of 2 frequency components; 12sin(2π10000t), 10sin(2π15000t).
    *Transmitter antenna impedance is 50Ω.

    a) What are the modulating indexes for the 2 components?

    b) Plot the power distribution of the transmitted signal. (Hint: Consider the
    frequency components in the transmitted signal and the power of each

    c) What percentage of transmitted power spills into the guard band and what
    percentage of power spills to the adjacent channel?

    d) Discuss methods that can be used to avoid transmitted power spilling outside the
    allowed 150 kHz band.
    Use Bessel function to calculate the BW

    3. The attempt at a solution
    I am pretty sure that deviation sensitivity is not the same as deviation ratio but I don't think there is enough info not to treat them the same? - maybe a typo and its meant to be ratio???

    for the first component:
    The frequency is (by inspection) 10000=10kHz

    [tex] m_f=\frac{\delta}{f_i} =\frac{5kHz}{10k} =0.5 [/tex] within the realms of plausibility at least/

    for the second component:
    The frequency is (by inspection) 15000=15kHz
    [tex] m_f=\frac{\delta}{f_i} =\frac{5kHz}{15k} =0.33 [/tex]

    These modulation indexes seem reasonable to me but the 'deviation sensitivity' label is annoying me .
    I am more thinking out loud here and looking for some pointers..... could be way off base
    From the Bessel Function frequency table above,
    for the first component (mf=0.5),J0(carrier)=0.94,J1=0.24 and J2=0.03

    If my understanding is correct, this means that I will get significant components at J0,J1&J2 and that the values in the tables are effectively a multiplier for the power seen at these frequencies? BUT, I don't have the carrier signal so I am stuck here....

    UNLESS this is correct_doubt it but might as well check.
    [tex]p=(0.94*\frac{12}{\sqrt(2)})^2/50=1.27W[/tex] This is for the carrier? J0? seems small
    for J1=82.94mW
    for J2=1.296mW

    If this is correct, again not confident, I need the frequency that these powers occur at in order to plot the power distribution wouldn't I? How would I go about finding this?

    I will need to hit up matlab to get this for the 33% mindex and repeat the above steps for the second frequency component given.

    Getting long for a post anyway, all help/tips/pointers welcome & appreciated.
  2. jcsd
  3. Aug 18, 2013 #2

    rude man

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    Modulation index for FM = (max nominal carrier deviation)/(max. modulation freq.).
    The max. nominal carrier deviation depends on the modulating signal amplitude, not frequency. You have not considered the modulating signal amplitude in determining the numerator of the modulation index.

    "Nominal" because in reality we get frequency deviations of the carrier out to infinity (Bessel functions).

    So if the deviation sensitivity is 5 KHz/V and the modulation amplitude is 12V, what is the numerator in the index The denominator?

    (Hint: this is a case of "wideband" FM where the index is > 1.).
  4. Aug 18, 2013 #3
    Hi again RM, still no feedback on the other questions/

    I did consider this, read about it in lecture notes or somewhere, but gave up because I don't have any carrier information. \Conceptually this makes sense to me as the carrier frequency is modulated in response to the amplitude of the information signal, so the highest carrier frequency will occur at the signals peak 12V?

    so... if 'deviation sensitivity' is actually the amount that the carrier will be modulated in response to the info signal then the biggest 'swing' in frequency (for signal 1) is 5*12=60kHz? * for signal 2 is 5*10=50kHz
    mf=60kHz/10kHz=6 (signal 1)?
    mf=50kHz/15kHz=3.33 (signal 2)?

    If thats correct, am I on the right track with part b of this question?

  5. Aug 18, 2013 #4

    rude man

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    Well, you need the carrier voltage, agreed, but you can deduce it from what is given. Your chart gives the Bessel functions which represent amplitude (including in-phase + vs. out-of-phase -) of each component. So you can compute the 1st sideband with J1, 2nd sideband with J2 etc. And the chart gives you the ratio of J0/J1 which is the amplitude ratio of the carrier to the 1st sideband.

    With two modulating sinusoids the Fourier transform of f(t) + g(t) is F(jw) + G(jw) so your spectrum is just the addition of the two sinusoidal modulations taken one at a time.

    I have appended an excellent tutorial below. Be careful with your Ac terms.

    Attached Files:

  6. Aug 18, 2013 #5
    Thanks, that is a good article but

    I'm still confused, the sideband amplitudes for the first signal (mf=6) are:
    -3.36,-2.88,1.32,4.32,4.32,3,1.56,0.72&0.24 This is the 9 (J1-J9) from the table multiplied by the amplitude 12. This is then converted to powers simply (^2/Antenna Impedance) which will gove me some sideband powers but what freqeuncy corresponds to each power? Given that these are the amplitudes and as established the amplitudes correspond to the frequency modulation I figure that --- (maybe easier to explain my thoughts with an example)

    for J1, P=3.36^2/(2*50)=112.9mW
    and this 'spike' would occur at a frequency of f(carrier)+16.8Hz...................16.8=3.36*5

    repeat for the other 8 sidebands and for the other info signal.

    Problem 2,

    I am missing a large power component at the carrier fundamental. As I understand it, this is one component for both information signals as the same carrier is modulated by both signals? You mentioned a ratio of J0/J1 in the chart but I can't see where that is? Or is it just a property of the table that J0/J1 = carrier/first sideband which would give me 0.15/-0.28=carrier/3.36..........................carrier = -1.8V???

    This doesn't seem right to me, why would this ratio not hold true for the other sidebands? J2-J10?
  7. Aug 18, 2013 #6

    rude man

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    The spikes occur + and - n times the max modulating frequency, where n = 1, 2, .... which in this instance is n times 10KHz. So for the 1st modulation you get spikes at fc - 90 KHz, fc - 80 KHz, fc - 70 kHz, ... fc - 10KHz, fc, fc + 10 KHz, fc + 20 KHz, ... all the way to fc + 90 KHz. You were not given the actual carrier frequency fc so on your graph just indicate fc as fc.

    The amplitudes of each component is given by the table and so power of each component is V2/R as you say. Except that power is for both the 10 KHz offsets, both the 20 KHz offsets, etc.

    BTW I believe there is an error in the pdf file, in the example they give they say Ac2/2 = 1 I think it should be Ac/2 = 1.

    If I said J(0)/J(1) it was a big goof. I meant J0(β)/J1(β). β = modulation index = 6 here.

    The carrier for a modulation index of 6 is J0(6)/J1(6) = 0.15/-0.28 times the 10 KHz offset amplitude. That is what your chart says. It also says the ratio of carrier voltage to the 2nd offset frequency (20KHz) is J0(6)/J2(6). Consistent all the way across the board. So if you know the 10 KHz offset voltage you also know the carrier voltage. Note that in this case the fc component voltage is 180 deg out of phase with the 10 KHz offset component.

    The offset at fc - 10 KHz and the offset at fc + 10 KHz each get one-half the total voltage indicated in the chart.

    This is not an easy field to wade thru & I suggest you study the link pdf file carefully. It nicelly relates the time and frequency functions of a sinusoidally modulated carrier.

    In relation to that link it might help to point out that given v(t) = Acos(2πf0t), the Fourier transform is (A/2)[δ(f - f0) + δ(f + f0)] and the power is (A2/2).
  8. Aug 18, 2013 #7
    Thanks RM, this is enough to compliment my materials I hope - spend the next few hours going back over everything and try and get a response to part B and C before I post with the problems that will inevitably arise.
  9. Aug 19, 2013 #8
    First progress first, Matlab to generate the Bessel first kind table is incredibly easy! So I have those values for the 3.33 mf.

    Secondly, after rereading your post a dozen times I don't think I was that far off the mark with my powers - although it was more a fluke than a result of understanding. My new & (hopefully) improved understanding of things...

    For the fist signal, mf=6, I will have a carrier 15% of its unmodulated self and sidebands of amplitude (-)28%,(-)24%,11%,36%,36%,25%,13%,6% and 2% of the original carrier amplitude. And, these (as you stated explicitly) occur at +-10% ->+-90% of the unmodulated carrier frequency.

    Here is where my understanding gets a bit dodgy;

    Voltage of fc=J0(6)/J1(6)*fi = 0.15/-0.28*12V....Hence V(fc)=6.45V.....This should be a single power point on my chart for both signals? Which I can confirm does not work out mathematically, for mf=3.333, Vfc=0.36/0.2*10=18V not even close.......

    ...Maybe this is a fault with my understanding though, I can't see how this would ever result in two identical carrier voltages from two different information signals...... But the question states that we are dealing with two components of an information signal - so how can I possibly have two distinct carrier amplitudes. The different sideband 'amplitude' I understand to not really be different voltagea of the signal that is fed to the antenna but a theoretical way to represent there impact on the instantaneous power of the modulated signal? This is FM and not AM which is why I'm having a brain explosion over the different amplitudes were working with......

    Onto part C.
    What percentage of transmitted power spills into the guard band and what
    percentage of power spills to the adjacent channel?

    The gaurd band is an ACMA (or in your case FCC) mandated channel separation typically 25kHz - spillage out of the 200kHz (Again ACMA& FCC) allotted BW past the guard band is into the next channel.

    Since I want a percentage i can probably just work with the frequency... although the square in the power eq may screw me - soon find out.

    The biggest frequency 'spread' is obviously from the larger modulation index, 6, although, any spillage from the 3.33 mod index will contribute to power spill percentage, damn! & yes i am thinking about this as I type, hope your not averse to some friendly rambling.

    for mf=6 I have 9 non-negligable sidebands so BW=(2*9+1)10k=190kHz which is within the allowable BW. There is no spillage which is obviously wrong???/

    Thanks, it's late now- hopefully there will be some wisdom waiting for me on this forum when I wake up/
    Last edited: Aug 19, 2013
  10. Aug 19, 2013 #9

    rude man

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    Hi Fisher- I wll be delayed today but will try to get back in 4 hrs or so. I am working on a diagram model of the fm process.

    My time is GMT-7. Where are you?

    EDIT: I will be using terminology of the linked file I sent. Let's forget about the Fourier transform part, it's not really needed. You need to understand how the final xfm(t) was derived.
    I could not possibly improve on that derivation.

    A key idea is that phase is the fundamental parameter, not frequency. Do you know what a VCO is? It's called a voltage-to-frequency converter but really it's a "time-integral of voltage - to - phase" converter. Do you have PowerPoint? I was thinking of appending a diagram of the general fm process via a PPT slide.

    I'm not a communications expert but my professional background fortuitously includes a very heavy exposure to Bessel functions.

    See you in a few hrs.
    Last edited: Aug 19, 2013
  11. Aug 19, 2013 #10

    rude man

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    The equation I want to deal with is from the link file:

    xFM(t) = Ac ∑ Jn(β)cos[2π(fc + nfm)t]

    with the summation taken from n = -∞ to ∞.

    If you look at this carefully you will see that this formula gives you all the sidebands for a sinusoidal modulation. The sidebands start at +/- fm away from the carrier frequency fc and theoretically extend to +/- infinity, but your chart indicates the significant number of 2n sidebands. For example, if β = 6, n goes from -9 to 9 incl. zero. Of course, n=0 is the carrier.

    Important: J-n(β) = (-1)nJn(β). This formula plus the signs in your table indicate whether each sideband is in or out of phase with the modulation source.

    Once you got this far, playing devil's advocate with yourself, you might ask: what if nfm > fc? The answer lies in cox(-x) = cos(x).

    Of course this is not commonly encountered. Usually fc >> nfm for all n. The link file has a good dicussion of limits placed on β and fm in the commercial 88 - 108 MHz band. It shows that for β = 5 and fm = 15 KHz there is actually (per your table) a small amount of leakage into the adjacent (200 KHz separation) spots.
  12. Aug 19, 2013 #11
    RM. GMT+10, close to the other side of the world. VCO - voltage controlled oscillator, remember that fun from analogue electronics last year.

    Phase is the fundamental parameter not frequency? That makes sense from an instantaneous point of view since phase is what the electronics would actually be altering. -> Obviously phase and frequency are closely related, cant really change one without the other.

    Yes i have ppt, look forward to seeing your FM process diagram & Ill go through the document again and see if I can't get some more understanding out of it.
  13. Aug 19, 2013 #12
    The problem is with my modulation indexes, there is no leakage outside of the 200kHz BW, I have 6 &3.33 or 9 & 6 non-negligible sidebands.


    This again leads me back to my basic problem (misunderstanding), I have two components of ONE information signal which modulates a SINGLE carrier?

    *How can I have two bandwidths as is the result of my current procedure?
    *Same for the carrier amplitude derived from the Bessel first kind table?

    I think this is where my understanding is lacking - I get (mostly) what the bessel table is giving me and I understand the alotted BW & Gaurd band & adjacent channels.

    Have to jump in a car for 4hrs now - Thanks & really appreciate your time on this.
  14. Aug 19, 2013 #13

    rude man

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    Here's my diagram, it's a pdf file, you don't need ppt.

    Attached Files:

  15. Aug 19, 2013 #14

    rude man

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    This is a rule-of-thumb, I would use the table. If you get spillage beyond 200 KHz that's the way it is. But for the large n sidebands the voltages are very small & negligible (or ignored) in practice.
    That's whence your rule-of-thumb derives. There's a name for it, I forget what. The table is authoritative.
    Yes, you get two spectra about the common carrier.

    You have to pick the larger of the two bandwidths to accommodate the two modulation signals together.

    The carrier is not affected by the presence of two or even an infinite number of modulation signals. The carrier amplitude is always Ac. Again, look at the pdf tutorial.
  16. Aug 20, 2013 #15

    rude man

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    Fisher, I may have jumped the gunn on having two sinusoidal modulations applied. I thought one could simply consider one modulation at a time, then add the results (superposition principle). But now I have grave doubts about that thought, since the two modulations are actually subjected to a nonlinear process, to wit, a cosine function. This may be a trickier business than seemed at first blush. Unfortunately, it's my bedtime so I have to hold the matter in abeyance until tomorrow.

    What is needed is to equate

    xfm(t) = Accos[2πfc1sin(2πfm1t) + β2sin(2πfm2t)]

    to the corresponding series expansion in the various Jn terms. There will obviously be Jn1) and Jn2) terms, corresponding to the two modulation frequencies fm1 and fm2, and the two modulation indices β1 adn β2. Stay tuned, if you'll pardon the pun ...
  17. Aug 20, 2013 #16
    Thinking about part c for hours on end has ironically given rise to some new ideas for part b.
    *I think I may have been confusing you with my queries about the carrier amplitude, or maybe I'm tripping now?

    In a nutshell I have decided that there is no way I can know the carrier amplitude with the information given, the carrier amplitude is in no-way dependent on the information signal which is the only 'real' information I have. I DID NOT for some reason realise this immediately and so was stuck trying to calculate it from the ratio you were talking about (something I still do not get?). -As I write this, it seems blatantly obvious - hope I'm not mistaking now.

    If I assume a 1V rms carrier amplitude in order to plot the distribution I get the carrier power J0 as
    for mf = 3.33......P=(-0.36*1)^2/50=2.592mW at fc
    mf =6 ...........P=(0.15*1)^2/50=0.45mW at fc
    ...This is a bit of a problem, would this be just be 2.592-0.45 since the minus sign indicates they are not in phase?

    When I do the same procedure for the sidebands I get some on the same frequency, so would these be sum if they are in phase and difference if they are out of phase? i.e if one is - and 1+ then take the difference for the power?

  18. Aug 20, 2013 #17

    rude man

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    The total fm-modulated signal is xfm(t) and this total signal, a sum of sinusoidal frequency components, has constant amplitude. That is not the same thing as the carrier amplitude of the modulated carrier. The carrier is one of the components of xfm(t) and is given by your table as a fraction of the unmodulated carrier amplitude. The carrier amplitude varies with the modulation index which is why your table requires you to specify the index before giving you the relative amplitude of the carrier and all the sidebands. The actual amplitude of the carrier and all the sidebands is given in that table but you have to multiply each number by the unmodulated carrier amplitude. The unmodulated carrier amplitude is Ac and yes you were not given that. You were also not given the carrier frequency. So when you plot the voltage spectrum, the relative amplitudes of each frequency component is all you can plot. Relative to Ac = the unmodulated xfm(t).

    The above applies for one-tone modulation. The difficulty is solving for the two-tone-modulated carrier. I have looked thru the Web and did not find an analysis giving the voltage spectrum (the Fourier series) of a two-tone-modulated fm system. I was also disappointed in wolfram alpha's inability to come up with this. We need the Fourier series expansion of cos[2πfct + β1sin(2πfm1t) + β2sin(2πfm2t)] whereas all we have is the series for one or the other modulation but not both. Hello, any math geniuses out there? Everything is a constant except for t. :confused:

    (Actually, I know how to approach this problem but it would be a lot of dog-work. If you want I can give you the approach).

    As I said, you can't just assume superposition because the two modulations are included in the argument of the cosine function which is a nonlinear process. You're doing this: determine the spectrum for the 1st modulation by itself, determine the spectrum for the second modulation by itself, then just superpose the two spectra. I don't think that works.

    Problem is, I don't think you can just use superposition which is what you're doing.

    If you really did have two voltages at the same frequency you'd have to take the algebraic sum before squaring. So if you had two voltages of equal magnitude but different sign the power would be zero for that frequency component.

    Once you have the true voltage spectrum (the elusive Fourier series for two-tone modulation), power is just squaring each frequency component voltage and dividing by (in this case) 50 ohms. Again, since you were not given the carrier amplitude you cannot plot the absolute power spectrum, just the spectrum relative to the carrier amplitude squared. This would be squaring the table entries as you did.

    Consider a simulation like with Simulink? BTW I erred in the diagram I appended & append a corrected one herewith. The integrator 1/s does not belong. The VCO already time-integrates the input voltage if the output is considered phase instead of frequency. You could try to use the model in your simulation.

    Attached Files:

  19. Aug 20, 2013 #18
    RM, Talked to my lecturer about this... He intends us to effectively treat this as two single tones for the purposes of this question.

    So effectively, while it is probably not technically correct, what I was doing for part b is what is expected.
    Apparently, there is a trick to the question where I only need the % power so it is fine to just assume 1V to get the percentage distribution. This kind of makes sense but should it add up to roughly 100%?

    If the Same frequency occurs I need to take the sum of the amplitudes before squaring.... I'll get this chart done now and post a pic when im done - if you wouldn't mind checking it looks right.

    I am also pretty much certain that there should be some spillage into the sidbands and next channel but I can't get the BW greater than 200kHz even using the Bessel table? ---Also, the approximation is called Carson's rule I believe.
    Last edited: Aug 20, 2013
  20. Aug 20, 2013 #19

    rude man

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    I was beginning to suspect as much. That is, forgive me, an absurd assumption in this case. Now, if β < 0.2 or so it may be OK since this is considered "narrow-band" fm. Anyway, I did get a chuckle out of it.

    Yes. Fact of math: Ʃ J2n(β) from n = -∞ to ∞ = 1 for any β.
    Cramer's rule solves determinants. The rule you're thinking of is called Carson's rule. It's quite good, works well for small as well as large betas. The worst error is when beta = 1.

    Looks to me like no spillage either. Why do you think there should be any? Say you're sitting at fc = 100 MHz then the farthest-away sideband for beta = 6 is at 9 x 10 KHz = 90 KHz & you're allowed 100 KHz (the adjacent station may also have beta = 6 and fm = 10 KHz & it's 200 KHz away carrier-wise.

    For beta = 3.3 it's of course even better.

  21. Aug 20, 2013 #20
    The way the question is worded is what makes my think there should be some, although it never says that it's a 200kHz BW which is an assumption... might need to talk to lecturer about this too. -I also read somewhere that radio stations effectively need to keep there mf <5 which means that the 3.33 will not spill and that the mf=6 is only on the verge... maybe the answer is simply that it doesn't spill - - - - everything we have done thus far makes sense to me - no real reason to think it's wrong.
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