# Z-transform and even/odd signals.

1. Mar 25, 2014

### ace1719

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, $\frac{(e^jω-z1)(e^jω-z2)...}{(e^jω-p1)(e^jω-p2)...}$, 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. Mar 25, 2014

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.

3. Mar 25, 2014

### ace1719

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.

4. Mar 25, 2014

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
5. Mar 25, 2014