Fourier Transform magnitude

Discussion in 'Math' started by jag1972, Jul 25, 2012.

  1. jag1972

    Thread Starter Active Member

    Feb 25, 2010
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    Hello,
    I am attempting and a bit stuck question on page 113 from the book: Fourier series by W.Bolton. Determine the magnitude for the following Fourier transform:

    \frac{1}{1+j\varpi} Equation 1

    To determine the magnitude the normal process is to seperate the real and complex and the apply pythagoras thereom.

    [Fω] = \sqrt{(\frac{1}{1+\varpi^{2}})^{2}+(\frac{-\varpi}{1+\varpi^{2}})^{2}} Equation 2

    I dont quite know how it equals this :mad::

    \frac{1}{\sqrt{1+\varpi^{2}}} Equation 3

    I can not get from equation 2 to equation 3. The best I can do is this, even this has not got the square root applied:

    \frac{1+\varpi^{2}}{1+2\varpi^{2}+\varpi^{4}}

    I just want a hint, Is it partial fractions I have to look at or is there another way.

    Thank you in advance :)

    PS: I dont know where the omega sign is, the one I have has a line above it
     
  2. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    Hi,
    What you have got is right ! The denominator is Square of the numerator ,if you just cancel, and apply Square Root your equation gets the form of equation 3.
     
  3. jag1972

    Thread Starter Active Member

    Feb 25, 2010
    60
    0
    Thank you very much for your help.
     
  4. jag1972

    Thread Starter Active Member

    Feb 25, 2010
    60
    0
    I have a Fourier Transform: \frac{2}{2-cos \varpi + jsin \varpi}

    To obtain the magnitude I have done exactly the same as my original post:

    \frac{2(2-cos \varpi - jsin \varpi)}{(2-cos \varpi + jsin \varpi)((2-cos \varpi - jsin \varpi)}





    The denominator is:

    \frac{4- 2cos \varpi - j2 sin \varpi}{2^{2} - 2cos \varpi - j2sin \varpi -2 cos\varpi + cos ^{2}\varpi +j sin\varpi cos \varpi + j2 sin \varpi - j sin \varpi cos \varpi -j^{2} sin^{2}\varpi

    j^{2}= -1

    sin^{2} \varpi+cos^{2} \varpi = 1

    -j2 sinω + j2 sinω = 0

    + jsinω cosω - j sinω cosω =0

    This makes:

    \frac{4- 2cos \varpi - j2 sin \varpi}{4- 2 cos\varpi -2cos \varpi + 1}

    \frac{4- 2cos \varpi - j2 sin \varpi}{5- 4 cos\varpi}

    Split the real and the imaginary:

    (\frac{4- 2cos \varpi}{5- 4 cos\varpi})^{2}+ (\frac{-j sin \varpi}{5- 4 cos\varpi})^{2}

    I have:

    \frac{25-40 cos\varpi +16}{16 - 16 cos\varpi + 4 cos^{2} \varpi}

    The actual answer is supposed to be:
    \frac{2}{\sqrt{5-4 cos\varpi}}

    Could someone please let me know where I have gone wrong. Thank you in advance.
     
  5. Wendy

    Moderator

    Mar 24, 2008
    20,764
    2,535
    I have moved the discussion of using special math symbols over here...

    Showing special math symbols.

    This is to avoid distracting from the OPs thread. It does look like a discussion that is needed on an offline basis.
     
  6. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    You should not include 'j' while calculating the magnitude as you have done here.
     
  7. jag1972

    Thread Starter Active Member

    Feb 25, 2010
    60
    0
    vvkannan, I will remember for next time, however I cannot see how this solves my problem, can you give me a bigger hint please :)
     
  8. vvkannan

    Active Member

    Aug 9, 2008
    138
    11
    You missed a 2 before the sin term,and remove the j.
    The common denominator is (5-4cosw)^2.

    Simplifying the numerators

    (4-2cosw)^2 + -2sinw ^ 2 =16-16cosw+4=4(5-4cosw)

    This gets cancelled with denominator.

    So we have 4/(5-4cosw). Taking the square root we have the required answer.
    While doing these type of calculations do make sure you have not missed any signs or numbers from the previous step and you are through :)
     
  9. jag1972

    Thread Starter Active Member

    Feb 25, 2010
    60
    0
    Thank you very much for your help:). I will try and not be clumsy (missing numbers or signs) with my equations in the future.
     
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