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Calling T.Engineer

  1. Aug 1, 2007 #1


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    As per PF posting guidelines:

    if your problem is related to undergrad material, please post it under the relevant homework section,

    if your problem can be considered graduate or post-graduate level, then:
    -- please identify the problem as completely as possible as best as you can.
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    -- please post specific questions. Describe your problem as narrowly as possible.

    Thank you.
  2. jcsd
  3. Aug 2, 2007 #2
    I am going to study the properties of Hermite polynomials mathematically and by simulation.
    Hermite Polynomial has the following formula:
    Hn(t) = (-1)^n * e^(t^2) * d^n/dt^n e-^(t^2) … (1)

    Hermite polynomials had been modulated and modified to be as follows

    Hn(t) = (-1)^n * cos(2π*fc*t)e^(t^2) * d^n/dt^n e-^(t^2)…. (2)

    Where fc= 6.85GHz, and n= 1,2,3,…,N

    The modulated and modified Hermite polynomials are orthogonal to each other for different numbers of n.

    In order to transmit a signal which represented by eq(2) with k users active in the system, multiple access interference is a factor limiting for large number of users.
    Let each user use different value of n. For example: user1 uses eq(2) with n=1, user2 uses eq(2) with n=2,…., etc.

    The medium access to the system is achieved through the assignment of a unique Time Hoping code sequence per user, to reduce the multi-User interference.
    Also, the transmitted signal is modulated by pulse position modulation.

    So, the transmitted signal will be represented by:

    S(t)= [tex]\sum^{\infty}_{j=-\infty} Hn(t- jTf - cj Tc - d^kj)[/tex]

    My question is:
    I’d like to find an expression for the probability density function, the mean, and variance using eq(2).

    My results:
    1. The mean and variance depend on pulse shape which is represented by
    eq(2) .
    2. From research I had found the following equations that I couldn’t
    understand what it refer to exactly.
    For example:
    [tex]\int^{\infty}_{-\infty}e^-(t^2) Hn(t)Hm(t) dt [/tex] =
    0 if n \neq m
    2^n*n! [tex]\sqrt{π}[/tex] if n =m

    And I have a paper and it contain all the related equation of Hermite polynomial but I couldnt understand it very well .
    if you want I can send it to you to explain it to me.
    Thanks alot!
  4. Aug 2, 2007 #3


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    1. What is each user doing? Is each user transmitting a signal (e.g. at random times)? Or are they listening to signals?
    2. Where does the signal originate from? One of the users? Or not?
    3. Who receives the signal? All users? Some of them?
    4. What is the objective? Can you state it in non-technical terms? (E.g., "to send a signal from user A to user B, but other users C to Z create a noise.")
  5. Aug 3, 2007 #4
    If we will have k users, where every user has a transmitted signal represented by eq(2) with different values of n.
    The time of transmitting a signal for each user will be specified according to time hoping techniques which suggest:
    For each user it specified a pseudorandom cj taking integer range 0<cj<Nh.
    Where Nh is the number of hops.
    Let Tf is the time duration of a frame. Tc is the hop width in Time hopping systems satisfying Tf = Tc*Nh
    d is the delay associated with PPM.
    as the following equation
    S(t)= [tex]\sum^{\infty}_{j=-\infty} Hn(t- jTf - cj Tc - d^kj)[/tex] ....(3)

    If you mean how many users will be activated in a specific time, If this is right, let say 4 users in a specific time.

    Let assume we will have a single receiver. The receiver will behave just like a differentiated device.
    If perfect synchronization between transmitter and receiver is assumed, given the orthogonality property of the waveforms, the transmit code word can be detected in a
    symbol-by-symbol maximum likely hood fashion by correlating the received signal with each of the N possible waveforms.

    We suggest a system with 4 users (where k=4).
    User1 will transmit the following sequence (0110), where at t=t0, bit0 is transmitted.
    User2 will transmit the following sequence (1010), where at t=t0, bit1 is transmitted.
    User3 will transmit the following sequence (1110), where at t=t0, bit1 is transmitted.
    User4 will transmit the following sequence (0010), where at t=t0, bit0 is transmitted.

    So, at time t=to , the receiver should detect the following (0110).

    The transmitted bit from any user will be considered as a noise for the others.
  6. Aug 3, 2007 #5


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    Since cj is random, # = t - jTf - cjTc - dkj is random.

    Suppose cj is distributed (discrete) uniformly over the interval [0,Nh] for each user. If nothing else is random in the expression # = t - jTf - cjTc - dkj, then # is distributed (discrete) uniformly over interval [A,B] where A = t - jTf - dkj and B = t - jTf - NhTc - dkj.

    See http://en.wikipedia.org/wiki/Uniform_distribution_(discrete)

    For given k and n, you can find the mean and the variance of Hn(#) either analytically or by simulation.

    I understand k indexes the user. What does j index? Does it index bit sequence (first bit, second bit)?

    At t0, does the receiver detect 0110, or just 0?
    Last edited: Aug 3, 2007
  7. Aug 3, 2007 #6
    how? where should I applied my function? I couldnt understand?
    please, can you show me an example?
    Thanks alot!

    I think it should detect just 0.
  8. Aug 3, 2007 #7
    j is the index for pseudorandom time hopping sequence.
  9. Aug 3, 2007 #8
    ok I will try to do somthing just give me little time?
    I think I find it "I hope so".
    Thanks alot!
  10. Aug 4, 2007 #9
  11. Aug 4, 2007 #10


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    E(Hn(#)) = [itex]\sum_{q=1}^{Nh}[/itex] P(cjq) Hn(#(cjq))

    where P(cjq) = 1/Nh for q = 1, ..., Nh; cjq = q and #(cjq) is the value of Hn's generic argument # evaluated at cjq, that is # = t - jTf - cjqTc - dkj = t - jTf - q Tc - dkj.

    This mean is conditional on the (given) values of all parameters other than cj. For example, t, jTf, Tc, dkj and n are all assumed given (constant).
    Last edited: Aug 5, 2007
  12. Aug 20, 2007 #11
    Actually they find the autocorrelation function for just Hn(t)

    [tex]\int^{\infty}_{-\infty} Hn(t)Hm(t) dt [/tex]

    where m not equal to n.
    also Hm(t) is the first derivative function of Hn(t)
    And then they try to find the mean and variance.
    So, I believe we should get the autocorrelation function and then use the result to find the mean and variance
  13. Aug 20, 2007 #12
    can you for example explain to me what it means each equation in the attached file, please.
    Thank you!

    Attached Files:

  14. Aug 20, 2007 #13


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    The text says (15) is the orthogonality condition.

    (16) implies [itex]1/N^2 = I/\delta[/itex] where I is the integral in Equation (15).

    (17) is a re-statement of (15) with [itex]\psi=H/N[/itex].

    (18) is a linear approximation to [itex]\psi_n(t-\tau)[/itex] with the c coefficients to be determined.

    (19) derives the c coefficients from (17) and (18). Each c coefficient is a correlation function between [itex]\psi_n[/itex] and [itex]\psi_m[/itex].
  15. Aug 21, 2007 #14
    Why it is important to find the Normalization coefficient Nn which is represented by equation (16)
  16. Aug 21, 2007 #15


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    My guess is, in order to simplify things. To use (18) as a linear approximation to H/N, and to derive (19) as a correlation function between [itex]H_n/N_n[/itex] and [itex]H_m/N_m[/itex].
  17. Aug 22, 2007 #16
    this is if :
    Hn(t) = (-1)^n * e^(t^2)* d^n/dt^n * e^(-t^2)

    what about if :
    Hn(t)= (-1)^n * e^(t^2)* d^n/dt^n * e^(-t^2) * cos (2[tex]\pi[/tex] fc t)

    where fc is a constant.
    will we get the same result?
  18. Aug 22, 2007 #17


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    This will change both sides of (15), so either of two things will have to change:

    EITHER the definition of N will change,

    OR the definition of [itex]\psi[/itex] ("psi") will change. Specifically, new psi = old psi * cos(...). I don't know whether (19) will still be valid.
    Last edited: Aug 22, 2007
  19. Aug 23, 2007 #18
    Accroding to what, they implies [itex]1/N^2 = I/\delta[/itex]
    If I will apply it to my new function with cos (...)
    then how would I work.
    May be you can help me to make it more cleare to me,because till now I dont know how they get the result in eq(16) and according to what?
    Thanks alot!
  20. Aug 23, 2007 #19


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    If you take (15) and divide both sides with [itex]\delta[/itex] you will get 1/N^2.

    (16) is not a result; it is a definition.

    Let's say that you introduce the cos(...) term but N remains the same. In that case, new psi = old psi * cos(...).
  21. Aug 24, 2007 #20
    and what about equations 17, 18, and 19.
    will they still the same.
    when the new psi= old psi* cos (2 pi fc t)
    will eqations 17, 18 , and 19 give the same results?
  22. Aug 24, 2007 #21


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    (17) will apply.

    (18) is a definition conditional upon the c's, so my guess is it will apply.

    (19) relies on (18) and (17), so my guess is (19) will still apply.

    But you should verify these.
  23. Aug 24, 2007 #22
    but I dont know how?
  24. Aug 24, 2007 #23


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    Study your textbook and seek help from your teachers.

    Do you know how to integrate?

    Can you find antiderivatives?
    Last edited: Aug 24, 2007
  25. Aug 24, 2007 #24
    I dont think it will be so easy?
    May be I should use a formula that it will help me to find the result of the integrals, or may be I should use a mathimatical program to find it out.
  26. Aug 24, 2007 #25


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    Do you know how to integrate? Yes or no?

    What is the integral of "t dt" from 0 to 1?
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