1. Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
    Dismiss Notice

Fourier Transform of QAM

Discussion in 'Mathematics and Physics' started by derick007, Oct 19, 2014.

  1. derick007

    derick007 Member

    Joined:
    Feb 5, 2013
    Messages:
    89
    Likes:
    1
    Location:
    NORTHERN IRELAND, UK
    I would like to find the FT of a QAM signal, please refer to the attached pdf.

    As you can see I would like to do this using convolution.

    First of all I am not sure if the FT of the phase shifted carrier is correct ?

    Secondly, I don't know how to convolve the baseband modulating signal with the phase shifted carrier ?

    Any help would be greatly appreciated.
     

    Attached Files:

  2. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,297
    Likes:
    629
    I didn't check the transform of c(t) carefully, but it looks right, and certainly the structure looks correct. I can check it more carefully later, but the structure is what matters for the method of calculating the answer. You have a sum of two impulse functions in the frequency domain. I think the frequency domain typically is in terms of omega (2pi f) rather than f, but maybe that is just the way you prefer to do it, in terms of f.

    Next you need to find the transform of the a(t), which is a square wave. This will be an infinite sum of impulse functions in the frequency domain. Once, you have both A(w) and C(w) correct, you will be ready to find the convolution of the two.

    Convolution of impulse functions are relatively easy because the convolution integral is sliding one function past the other, and only when the impulses overlap is there a non-zero value. Since one function only has a sum of 2 impulses, break the problem into a sum of separate convolution integrals, each with one impulse function only from C(w). Then you will see that each of the two integrals is easy to calculate as one impulse slides through the infinite sum of impulses in A(w). Then add the result of the two integrals together to get A(w) * C(w). Transforming back to the time domain will give the sum of two sine waves, and I think you will basically discover a trig identity in the time domain. This will allow you to check your answer. Using the trig identity on c(t) a(t) should give you the same answer.

    The other way to solve the answer for A(w) * C(w) is to use the trig identity first on a(t) c(t) and then transform that back to the frequency domain.
     
    Last edited: Oct 20, 2014
  3. MrAl

    MrAl Well-Known Member Most Helpful Member

    Joined:
    Sep 7, 2008
    Messages:
    11,026
    Likes:
    951
    Location:
    NJ
    Hello there,

    I have a quick question myself:
    Isn't the phase shift supposed to be constant, such as 90 degrees, which would make the cos(wt+ph) equal to -sin(wt) ?
    It's been a long while since i looked at Q anything not just QAM, but the Q usually stands for quadrature which implies two signals 90 degrees out of phase. No problem if it's not like that, but just thought i would ask as it is always good to start with the simplest possible form.

    Also, and again i could be incorrect here, but if you intend to amplitude modulate a sinusoidal with another signal then it's not a convolution in time it's a multiplication in time, and a multiplication in time is not a multiplication in frequency. A convolution in time is a multiplication in frequency, and probably a multiplication in time is a convolution in frequency, or at least something like that. But with a square wave amplitude modulator, i would think it would just be some sort of 'on/off' type of operation because 1 times anything is just that thing, and 0 times anything is just zero.
    Again, QAM is not something i use every day :)

    Example from regular AM:
    Y(t)=(1+sin(2*pi*fm*t))*cos(2*pi*Fc*t)
    That's multiplication in time (not convolution in time) which is not multiplication in frequency but is probably a convolution in frequency or something like that.
     
    Last edited: Oct 21, 2014
  4. dave

    Dave New Member

    Joined:
    Jan 12, 1997
    Messages:
    -
    Likes:
    0


     
  5. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,297
    Likes:
    629

    Mr Al,

    You are correct about the phase factor, but look more carefully at what he wrote. There are two impulse functions, so the changing phase factor with frequency only matters at the frequencies (there are two of them) where the impulse functions are. You will notice that you get +θ or -θ at each of those impulse functions.

    Also, you are correct about multiplication in the time domain is convolution in the frequency domain, but it's not clear why you point this out. This appears to be what the OP did.
     
    Last edited: Oct 21, 2014
  6. MrAl

    MrAl Well-Known Member Most Helpful Member

    Joined:
    Sep 7, 2008
    Messages:
    11,026
    Likes:
    951
    Location:
    NJ
    Hi there Steve,

    Thanks for the reply as i am currently trying to remember some of this stuff myself. Since i use Laplace most of the time and almost never Fourier anymore i am trying to remember some of the theory about Fourier Transforms. Please keep this in mind :)

    Not really sure what you are trying to show here. Wouldnt a constant phase angle simplify things a bit?

    Actually it appears that he did not do this yet, and is multiplying in frequency, or intends to? Maybe not as i do see the little asterisk there with the note. Maybe i read that wrong.
    So you are saying he actually intends to do a convolution in frequency rather than A(f) times C(f) ?

    Also, i dont remember what we do with the imaginary part, if any, that could come up in the transform such as F(-sin(wt)).
    Dang, it's been a while. Laplace is so much cleaner.
     
  7. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,297
    Likes:
    629
    Each impulse funtion has its own phase factor, or complex coefficient. If phi=0, then the two constants are equal. If phi=90 degrees, then they are equal in magnitude, and oposite in sign. So, there is really no frequency dependence to the phase factor, but the f/Fc basically encodes a plus and minus sign, if we think of sine and cosine only. I agree it would be simpler to put the negative sign with the left side impluse function, in those cases, or whenever possible. But, the notation shown is compact.

    The * is a standard symbol for convolution. This is a it confusing because * is also used for multipication in computer programing languages. But, the math convention preceeded the programing language convention, I think.

    The imaginary part is important, but we always get complex conjugate pairs when the time domain signal is real.
     
  8. derick007

    derick007 Member

    Joined:
    Feb 5, 2013
    Messages:
    89
    Likes:
    1
    Location:
    NORTHERN IRELAND, UK
    steveB

    Many thanks for replying. I havent had a chance to go through it yet, but hopefully I will very soon. I will let you know how I get on.

    Regards derick007
     
  9. derick007

    derick007 Member

    Joined:
    Feb 5, 2013
    Messages:
    89
    Likes:
    1
    Location:
    NORTHERN IRELAND, UK
    steveB, MrAl

    Just to confirm, I use the asterisk * to denote convolution in the frequency domain.

    derick007
     
  10. MrAl

    MrAl Well-Known Member Most Helpful Member

    Joined:
    Sep 7, 2008
    Messages:
    11,026
    Likes:
    951
    Location:
    NJ
    Hello again

    derick:
    Ok thanks i see that now :)
    BTW, do you intend to use 90 degrees for ph or let it vary?
    Also, do you have another sinusoidal signal in the system that is not at the same phase as cos(wt+ph) ?

    Steve:
    What i meant was the transform for cos(wt) is:
    pi*Impulse(w,w0)

    while the transform for sin(wt) is
    pi/j*Impulse(w,w0)

    where Impulse(w,w0) is
    curlydelta*((w-w0)+(w+w0))

    curlydelta is the symbol usually used to indicate an impulse.

    One is real while the other is imaginary.
     
  11. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,297
    Likes:
    629
    Yes, but that's not quite right. The transform for sin(wt) has a negative sign on one of the impulses. We can see this using Euler's relations.

    2 cos(x) = exp(jx) + exp(-jx)

    2j sin(x) = exp(jx) - exp(-jx)

    So, what I was saying is that the coefficients of the impulses are always complex conjugates of one another. One is exp(-j phi) and the other is exp(+j phi). When phi=0, then the coefficients are equal (to 1), and when phi=pi/2, then the coefficients are +j and -j. For other angles, they are two complex numbers that are complex conjugates of each other.

    Getting back to your original point, I think what is bothering you is the "f/Fc" that shows up in the formula. All this does is encode the plus and minus sign so that the coefficients are complex conjugates of one another. It is probably less confusing to just put the exp(j phi) and exp(-j phi) in front of each impulse function, rather than trying to have one overall exp(f phi / Fc) that applies to both impulse functions.
     
  12. MrAl

    MrAl Well-Known Member Most Helpful Member

    Joined:
    Sep 7, 2008
    Messages:
    11,026
    Likes:
    951
    Location:
    NJ
    Hi,

    I probably should not have defined that Impulse function because it looks like i am stating that there is only one impulse when really there are two where one is negative as you say. Let me write it out trying to use the symbols on this site:
    π*(δ(w-w0)+δ(w+w0)), for cos(wt)
    (π/j)*(δ(w-w0)+δ(w+w0)), for sin(wt)


    which makes it more clear i was stating that there are two impulses.

    So the Impulse function probably should have been called "Impulses":
    Impulses(w,w0)=δ(w-w0)+δ(w+w0)

    δ="curlydelta"

    and this function creates two impulses one at w0 and one at -w0.

    So you see i had understood that there are two impulses, one negative and one positive, but yet one of the transforms has a "j" in the denominator, even though it also has two impulses.

    So to put it another way, if you were to make a table of transforms of common functions, how would you define the transform of cos(wt) and sin(wt) ?
    cos(wt)==>
    sin(wt)==>

    More to the point of this thread would be -sin(wt) but no big deal for now.

    My first thought is that the imaginary part comes into play at the end when we need to render the output into real world terms.

    The confusion about the asterisk comes in because sometimes it is written for example:
    Aw*Bw

    where "*" is stated as being "convolution" when really it is just multiplication. True convolution requires the integral even in the frequency domain, so i think what it is is sometimes the notation is overstated.

    ie:
    Integral(xt*h(t-T))dT=X*H
    where "*" denotes convolution.

    The above is not correct because "*" must be multiplication, but sometimes we say "convolve" anyway:
    "Let us convolve the two functions X and H: X*H"
    Really when we convolve it should be x*h=X.H
    where the dot means multiply.
    If we write X*H where * really means convolve, then we mean:
    Integral(X(w).H(w-W))dw
    where again the dot means actual multiplication.
     
    Last edited: Oct 22, 2014
  13. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,297
    Likes:
    629
    Hi Mr. Al,

    Yes, your meaning was clear to me. The notation you used originally had the issues you mentioned, but still your meaning was clear in the context of the thread.

    So, when I was commenting on "one of the impulses being negative", for the case of the sine function. I was not referring to the left or right impluses. Clearly there are two impulses for any phase shifted cosine wave. When I say, positive or negative impulse, I just mean the amplitude is positive or negative. But even that is a bit ambiguous because we are dealing with complex coefficients. But, the difference between the sine and cosine is that if we want to factor out one common phase factor, we need 2 positive impulses for cosine and we need 1 positive and 1 negative impulse for sine, as follows.

    cos(wt) --> 0.5 ( + δ(w-wo) + δ(w+wo) )

    sin(wt) --> 0.5 j ( - δ(w-wo) + δ(w+wo) )

    This is basically what I was trying to say. We can see that we are basically dealing with the Euler relations for sine and cosine. We don't see the exp(wt) and exp(-wt) parts explicitly because the Fourier transform implies that these sinusoids are there, and the delta functions identify the frequency locations.

    The other point I wanted to make is that the form is much less confusing if we dont try to have a common factor, but instead explicitly write the correct coefficient in front of each impulse, as follows.

    2 cos(wt+phi) --> exp(j phi) δ(w-wo) + exp(-j phi) δ(w+wo)

    In this way we see that the coefficients are always complex conjugates of one another (complex conjugate is replacing j with -j). This ensures that when you reconstruct the time domain signal, you will get a real function.
     
    Last edited: Oct 22, 2014
  14. derick007

    derick007 Member

    Joined:
    Feb 5, 2013
    Messages:
    89
    Likes:
    1
    Location:
    NORTHERN IRELAND, UK
    steveB

    The attached pdf demonstrates the answer, however I am not entirely sure how the convolution works and what the resulting phase angle would be.

    Regards, Derek
     

    Attached Files:

  15. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,297
    Likes:
    629
    Hi Derek,

    I think what you showed is not correct. You did a transform of a single pulse, but your original question said that you have a modulating signal which is a square wave. There is a world of difference between the two. A single pulse has a continuous spectrum for the Fourier Transform (as you showed), while the square wave has (as I mentioned previously) an infinite number of impulses, where the impulse amplitudes correspond with the Fourier coefficients.
     
  16. MrAl

    MrAl Well-Known Member Most Helpful Member

    Joined:
    Sep 7, 2008
    Messages:
    11,026
    Likes:
    951
    Location:
    NJ
    Hi,

    Yeah that does not look right at least for the pulse train. For that you would use the form where you integrate from -T/2 to T/2 to get the cn, then 2*pi*sum(n from -inf to +int) of cn*curlydelta(w-n*w0)

    F(jw)=2π Ʃ[n,-∞ to +∞] cn*δ(ω-nω0)

    (trying out the symbols here)

    If you can integrate from -T/4 to +T/4 instead of 0 to T/2 i think you get a cleaner transform, if those phase shifted pulses are ok for your application.
     
  17. derick007

    derick007 Member

    Joined:
    Feb 5, 2013
    Messages:
    89
    Likes:
    1
    Location:
    NORTHERN IRELAND, UK
    Hi steveB

    As I am learning along the way, I thought I should first try a single rectangular pulse before considering a periodic square wave. Once I am content I have understood as much as I can re. single pulse, I will look at the periodic square wave.

    I still think what I have produced for the single rectangular pulse is correct, although I am still unsure about how the phases are dealt with - I assume simple addition.

    I think I can understand the result of the convolution in the frequency domain if I consider the Fourier Transform (FT) of the carrier, which is a phasor (amplitude 1/2 and phase phi, at + / - Fc) similar to a DIRAC DELTA function in the time domain i.e. unit impulse.

    As we know, if a unit impulse is shifted by To in the time domain and then multiplied by a function f(t) and the result integrated (area under curve), the answer is simply f(To). I understand it is simply f(To), because this is equal to the area under the curve at the point To i.e. the limits of the integral are virtually the same, To ?

    Therefore I would suggest treating the phasor in the frequency domain (which is shifted by + / - Fc ) in a similar fashion. This will yield the sinc function shifted to + / - Fc and scaled in amplitude by 1/2.

    Regards, Derek
     
  18. derick007

    derick007 Member

    Joined:
    Feb 5, 2013
    Messages:
    89
    Likes:
    1
    Location:
    NORTHERN IRELAND, UK
    steveB, MrAl

    Have a look at the attachment. Interesting to note the difference. I presume this can be explained by one of the rules/properties which state a time shift in the time domain is the equivalent of multiplying by an complex exponential in the frequency domain ?

    Must admit I am feeling proud of myself as I think I have it cracked.
     

    Attached Files:

  19. MrAl

    MrAl Well-Known Member Most Helpful Member

    Joined:
    Sep 7, 2008
    Messages:
    11,026
    Likes:
    951
    Location:
    NJ
    Hi,

    Yes that is right, a time shift is a multiplication by a complex exponential like e^-jwT where T is positive to shift to the right.
    And yes, integrating on both sides of t=0 makes a cleaner transform.
    Well done Grasshopper :)
     
  20. steveB

    steveB Well-Known Member Most Helpful Member

    Joined:
    Jan 16, 2009
    Messages:
    1,297
    Likes:
    629
    Yes, I agree, that's correct. I think it is a good idea to do as many of these as you can.
     
  21. derick007

    derick007 Member

    Joined:
    Feb 5, 2013
    Messages:
    89
    Likes:
    1
    Location:
    NORTHERN IRELAND, UK
    SteveB, MrAl

    Thanks again for all your time and help.

    I will try the periodic square wave now - stay tooned

    Regards, Derek
     

Share This Page