1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    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:


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted