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The calculus of FM super heterodyne receiver

Discussion in 'Electronics Homework Help' started by dietermoreno, Apr 9, 2013.

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  1. dietermoreno


    Dec 30, 2012
    The calculus of FM super heterodyne receiver.

    I found an interesting problem in my calculus text book. The instructor skipped this section in the text book because it is using the graphing calculator to help with the calculus and my instructor is opposed to doing that.

    Well so I'm looking at an example the book gave. I would like a conceptual understanding of this book's example to help me understand how FM radio receiver works. The calculus text book Single Variable Calculus Early Transcendentals is by James Stewart. The edition is 7th edition. The chapter I am in is chapter 4. The chapter section I am in is 4.6

    So the example problem starts with a function and uses the first derivative test and the second derivative test with the graphing calculator to get the graph of the function as well as the graphs of the composition of functions that create the FM modulated function.

    The function is of the family of functions f(x) = sin ( x + sin(cx) ), where c = 2.

    So the function we want to graph is f(x) = sin ( x + sin(2x) )

    So when we first graph it on the graphing calculator we get an inaccurate graph that looks something like this: [​IMG]

    So the first derivative is f ' (x) = cos ( x + sin2x) * ( 1 + 2cos2x), using the chain rule.

    So then the book says that using zooming in on the graphing calculator with the first derivative test, we find the following approxiamate values for f(x):
    Intervals of increase: (0, 0.6), (1.0, 1.6), (2.1, 2.5)
    Intervals of decrease: (0.6, 1.0), (1.6, 2.1), (2.5, pi)
    Local maximum values: f(0.6) =~ 1, f(1.6) =~1, f(2.5) =~1
    Local minimum values: f(1.0) =~0.94, f(2.1) =~0.94

    So the second derivative is f " (x) = - sin ( x + sin2x) * (1 + 2cos2x) + cos (x + sin2x) * (-2sin2x * 2), using the chain rule and the product rule
    = f " (x) = -sin ( x + sin2x) * (1 + 2cos2x) + cos (x + sin2x) * (-4sin2x), simplifying
    = f " (x) = -sin ( x + sin2x) * (1 + 2cos2x) - 4sin2x * cos (x + sin2x)
    = f " (x) = -( (1 + 2cos2x)) * (sin ( x + sin2x) - 4sin2x * cos (x + sin2x)

    So the book says that graphing the second derivative, we find the following approxiamate values for f(x) :
    Concave upward on: (0.8, 1.3), (1.8, 2.3)
    Concave downward on: (0, 0.8), (1.3, 1.8), (2.3, pi)
    Inflection points: (0, 0), (0.8, 0.97), (1.3, 0.97), (1.8, 0.97), (2.3, 0.97)

    So now we can sketch the graph by hand of f(x) for a more accurate graph and it looks something like this (Wikipedia used different values of C, such as 0.1, 10, and even 100):

    Zoomed in it looks something like this: [​IMG]

    I think the wave will look different and will look something like this if a different c value is used: [​IMG]

    So what c value is used in the FM broadcast band from 88MHZ to 107.9MHZ? Google failed to answer my question. Wikipedia says that narrow band FM radio uses a lower order modulation index of c < 1 resulting in a frequency change of 2.5 khz above and 2.5 khz below the center frequency for a bandwidth to carry human speech of 3.5 khz bandwidth. Wikipedia says that wide band FM radio uses a higher order modulation index of c>1 resulting in a frequency change of 75 khz above and 75 khz below the center frequency for a bandwidth to carry music of up to 20khz bandwidth. So is the challenge of finding a good modulation index to use for FM, is the challenge to replicate the full frequency response of the human ear (20khz bandwidth) at a minimum bandwidth to optimize bandwidth so that neighbooring stations do not interfer and more stations can transmit in the same city? Is this why FM uses VHF instead of MW for bandwidth optimization so that nearby stations do not interfer since the shorter wavelength waves don't travel as far so a higher order modulation index can be used with out worrying about interference? From the Wikipedia article, it appears that to calculate the effective bandwidth of a modulation index for FM you have to solve a differential equation with integrals in it, and I am only in calc I so I can't solve these equations, so I let Wikipedia solve it for me.

    Wouldn't an FM receiver only receive stations that use the same modulation index c that the receiver understands?

    Which would have less distortion from EMI, lower value of modulation index c or higher value of modulation index c?

    So FM radio reciever is taking derivatives of frequency harmonics generated by the feedback of tubes or transistors (sort of) and that is why the mixer it uses is called a "differentiator"?

    So is this why FM radio only works with a super heterodyne technique employed in the detector and does not work with a regenerative technique or crystal technique employed in the detector?

    So in the AM super heterodyne receiver, the super heterodyne technique employed in the detector is to greatly increase the gain without an audible feedback squel, improve the frequency selectibility, and reduce the interference of nearby stations, correct? Compared to in the AM regenerative receiver, the regenerative technique employed in the detector greatly increases the gain but the gain is limited to below over drive when in AM mode instead of CW mode or else an audible feedback squel is heard (which is turned on and off to generate CW Morse code beeping), correct?

    I have heard of a "super regenerative receiver" on the internet that can receive FM. So does this mean that the super heterodyne technique is used to convert the FM into AM that the regenerative detector can receive?

    So can an FM radio receiver use any detector, even crystal detector, to demodulate AM to audio, as long as the FM has already been converted to AM by the super heterodyne technique?

    Wikipedia says that FM is used in VHS. So does that make the video quality of VHS superior to that of analog TV (since analog TV video signal uses AM)? Does that also make the audio quality of VHS comparable to the audio quality of CDs if a higher order modulation index capable of replicating the full frequency range of human hearing was used and superior to vinly because vinly uses unmodulated audio AM which AM is prone to noise?

    Edit: I found in the Wikipedia article when I read it again that the ratio of original frequency to synthesized frequency in wideband stereo FM broadcast is 10,000 , compared to the ratio of original frequency to synthesized frequency in VHS is only 2. So I presume that VHS audio can not replicate the full frequency response of human hearing and VHS audio is mono, although it certainly has less hiss than vinly.

    Edit 2: I found in the Wikipedia article about VHS that the audio is umodulated (base band) and is very prone to hiss, which is why Dolby encoding is used, but unfortunately all Dolby does is attempt to restore dynamic range to make the audio sound punchier resulting in the tape hiss also being increased in dynamic range, much the same as crystallizer algorithms used today to attempt to improve the quality of MP3 for MP3 encoded with a noise floor to decrease dynamic range in file storage to decrease file size, so really the quality is not improved, just most people can not tell the difference and ignore the amplified hiss and think it sounds punchier, resulting in most people can not tell the difference between Dolby encoding or MP3 encoding used with a crystalizer algorithm compared to uncompressed PCM audio.

    So to receive the wideband stereo FM broadcast band, would a reciever have to be able to recieve at modulation index of 10,000 and would not recieve at a modulation index higher or lower?

    Ratio of original frequency to synthesized frequency is the same as modulation index, right?

    Okay just tell me if I'm in over my head in the calculus and nothing about FM will make sense since I'm not an engineering student who has taken Calc III and differential equations. Then I will just give up.
    Last edited: Apr 9, 2013
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    A lot of questions here.

    And for a pleasant change, they seem to be well informed.

    Consider what the change in frequency represents.

    The absolute difference between the transmitters nominal frequency and its instantaneous frequency is determined by the instantaneous amplitude of the modulating signal.

    This means that for larger values of c, an equivalent signal will have a greater deviation. or (and this is useful) a lower level signal will have the SAME deviation.

    Any receiver can pick up the signal, however the peaks may be clipped if the bandwidth of the receiver is insufficient or if the "detector" can't cope with the deviation.

    Very difficult to say since EMI can affect many things.

    This is a non-sequitor.

    No. The superior bandwidth is probably more the reason.

    if it's recorded on tape, then it is certainly used to modulate a signal. You need to see how tape recording works.

    You also need to look up what Dolby does. It is very useful in reducing noise where the noise spectrum is frequency dependant.

    Not as I understand it. It is the ratio of the deviation to the nominal frequency.

    Don't give up yet. It seems like you're trying to learn this stuff and not just make it up as you go along.
  3. Laplace


    Apr 4, 2010
    The problem here is that your math textbook has constructed an example of FM modulation with no identified modulation index. (Here the implied modulation index is 1.) So it provides no basis for examining the important details of FM modulation. Plotting the modulated carrier wave, while a possibly interesting mathematical exercise, is a fairly useless activity for a communications engineer. And taking the derivative of the modulated wave lends no insight to the important aspects of frequency modulation.

    Introductory communications textbooks will examine the Fourier spectrum of the frequency modulated signal, although this is too complicated to derive mathematically. For the simplest case of pure sinusoidal modulation one can perform a Jacobi expansion of the modulated signal with Bessel function coefficients which leads to a line spectrum having an infinite number of frequency lines. For more complex modulating signals one just applies the Carson Bandwidth Rule.

    Don't give up. But don't look for practical signal analysis answers in a calculus textbook.
  4. davenn

    davenn Moderator

    Sep 5, 2009

    And NO ... Analog TV uses FM for voice and AM for video

    also, there is a better SNR with FM

  5. davenn

    davenn Moderator

    Sep 5, 2009
    higher mod index = wider bandwidth, which always has a better SNR
    this is the huge advantage of spread spectrum communications
    Also FM is a substantial improvement over AM. It has a much better SNR, this is because most EMI is an amplitude modulated single or multiple freqs rather than an FM'ed single freq

    Steve is correct and No, synthesised freq doesnt even enter into it

    Last edited: Apr 10, 2013
  6. Merlin3189


    Aug 4, 2011
    Wow! A very impressive bit of research. Having looked at the WikiP article I'm not surprised you're a bit confused. I'm supposed to understand this stuff (or at least I was once) and I struggled with that article.
    I don't think I can deal with all your questions - you cover a lot of ground. So I'll just try a couple of points that I haven't seen addressed in the other replies. (E&OE!)

    Superregenerative receiver is not superheterodyne and regenerative, it is supersonic regenerative receiver - a regenerative receiver which is allowed to reach self-oscillation, then quenched at supersonic frequency.

    The superheterodyne (or supersonic heterodyne) is a receiver which converts incoming signals to an intermediate frequency which is supersonic, by mixing with a different frequency (hetrodyne-ing).

    In both cases the "super" bit is saying that something happens at a frequency above the audible range (supersonic), but what happens and why, is very different.
    In both cases there is plenty of info on the web if you are interested. (IMO the best explanations are the ones that don't go into the calculus though!)

    No. There are many techniques for demodulating/ detecting FM.
    Regenerative receivers did detect FM, by "slope detection." If a circuit is sharply tuned to a frequency, then it is less sensitive to frequencies away from its centre frequency and gets less sensitive the further away they are. Therefore if a circuit is tuned a bit to one side of an FM signal, as the frequency varies, so does the strength of the signal in the circuit. So the frequency variation (FM) is converted to an amplitude variation (AM).

    I think you may mean "discriminator", a circuit which discriminates between frequencies. Try looking up Foster Seeley Discriminator, which is used for FM detection.

    This is where my IMO comes in. The mod.index is a bit of maths to help us engineer efficient radio systems. I'm pretty confident the guys who thought up FM (or most bits of radio) didn't start out worrying about this. You make an FM signal anyhow you like, then pretty much any FM detector is going to extract some audio from it. Just as any AM radio can pick up any AM signal (and unfortunately most circuits with sparks or switches in generate very AM signals!) If your system is not exactly matched to the transmitting system, then you just get distortion (such as loss of high or low frequencies or peak amplitudes) and extra noise.
    Last edited by a moderator: Apr 15, 2013
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