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Z-transform and even/odd signals.

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data
    "Let x[n] be a real-valued DT signal. Using the Pole-Zero plot of X(z), argue as to why:
    a. |X(e^jω)| is even in ω.
    b. ∠X(e^jω) is odd in ω.

    Hint: if x[n] is real, what can we say about the poles and zeros of X(z)?


    2. Relevant equations

    None.

    3. The attempt at a solution
    I figure that X(z) will probably of the form, [itex]\frac{(e^jω-z1)(e^jω-z2)...}{(e^jω-p1)(e^jω-p2)...}[/itex], and I know even functions are defined by x(n)=x(-n) and odd functions by x(n)=-x(-n), but beyond that, not much.
     
  2. jcsd
  3. Mar 25, 2014 #2
    If x(n) is real valued then:

    x(n)=x*(n)

    DTFT x(n) <-------> X(e^jw)
    DTFT x*(n) <-------> X*(e^-jw)

    X(e^jw)=Re{X(e^jw)} + jIm{X(e^jw)}
    X*(e^-jw)=Re{X(e^-jw)} - jIm{X(e^-jw)}

    since x(n)=x*(n), X(e^jw)=X*(e^-jw)
    thus Re{X(e^jw)} + jIm{X(e^jw)}= Re{X(e^-jw)} - jIm{X(e^-jw)} so
    |Re{X(e^jw)} + jIm{X(e^jw)}| = |Re{X(e^-jw)} - jIm{X(e^-jw)}|
    Even symmetry.

    sorry that looks sloppy...

    You can use the same logic to show the phase is odd.


    Hint: real-valued signals result in complex-conjugate pairs.
     
  4. Mar 25, 2014 #3
    I came up with an elegant solution to this problem, albeit it didn't use poles and zeros.

    Re(x[n]) -> 0.5(X(z) + X*(z*))
    Im(x[n]) -> -0.5j(X(z) - X*(z*))

    Im(x[n])=0, therefore we can conclude that X(z) = X*(z*). (1)
    I read in a book that, for real signals X(-z)=X*(z). (2)

    |X(z)| = sqrt(X(z*)X*(z*)) (from eq. (1)). (3)

    X*(z)=X(-z) and X(z)=X*(-z) (from eq. (2)). Substituting this into eq. (3) gives you:

    |X(z)| = sqrt(X(z*)X*(z*)) = sqrt(X(-z*)X*(-z*)), which is the definition of an even function.


    As for angle,

    ∠X(z) = arctan(Im(X(z)/Re(X(z)). Arctan is odd, therefore ∠X(z) is also odd.

    If you see any mistakes, please respond.
     
  5. Mar 25, 2014 #4
    That is all correct.

    To use poles/zeros you just need to realize that real values create complex-conjugate pairs, so one real input will give you |a|e^jb and |-a|e^-jb making a z-plane plot look something like this:
    . |Im
    . | x
    . |
    ------------------------------------
    . | Re
    . | x
    . |

    So you can see that magnitude is the same (even) and the angle (pi/4 vs -pi/4 for example) is odd.
     
    Last edited: Mar 25, 2014
  6. Mar 25, 2014 #5
    Sorry none of the spacing worked.... I think you get the idea.
     
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