# Z-transform and even/odd signals.

• ace1719
In summary, the conversation discusses the even and odd properties of |X(e^jω)| and ∠X(e^jω) for a real-valued DT signal x[n]. It is determined that X(z) will have a form of \frac{(e^jω-z1)(e^jω-z2)...}{(e^jω-p1)(e^jω-p2)...}, and that even functions are defined by x(n)=x(-n) and odd functions by x(n)=-x(-n). It is also discovered that real-valued signals result in complex-conjugate pairs and that this can be used to show that |X(e^jω)| is even and ∠X(e
ace1719

## Homework Statement

"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)?

None.

## 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.

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.

1 person
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.

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:
Sorry none of the spacing worked... I think you get the idea.

## 1. What is the Z-transform?

The Z-transform is a mathematical tool used in the field of signal processing to convert a discrete-time signal into a function of a complex variable, also known as the Z-domain.

## 2. How is the Z-transform different from the Fourier transform?

While the Fourier transform is used for continuous-time signals, the Z-transform is used for discrete-time signals. The Z-transform also takes into account the entire history of a signal's values, while the Fourier transform only considers the current value of a signal.

## 3. What is the significance of even and odd signals in the Z-transform?

Even and odd signals are important in the Z-transform because they have specific properties that make the calculation of the transform easier. Even signals have symmetry about the y-axis, while odd signals have symmetry about the origin. This allows for simplification in the equations used for the Z-transform.

## 4. How are even and odd signals represented in the Z-transform?

Even signals are represented by real coefficients in the Z-transform, while odd signals are represented by imaginary coefficients. This is due to the symmetry properties mentioned earlier.

## 5. What is the role of even and odd signals in signal processing applications?

Even and odd signals are useful in signal processing applications because they allow for the separation of signal components. For example, in image processing, even signals represent the brightness or intensity of an image, while odd signals represent the edges or texture. This allows for more efficient analysis and manipulation of signals.

• Engineering and Comp Sci Homework Help
Replies
1
Views
2K
• Engineering and Comp Sci Homework Help
Replies
2
Views
2K
• Engineering and Comp Sci Homework Help
Replies
6
Views
4K
• Engineering and Comp Sci Homework Help
Replies
2
Views
2K
• Engineering and Comp Sci Homework Help
Replies
1
Views
1K
• Engineering and Comp Sci Homework Help
Replies
6
Views
2K
• Engineering and Comp Sci Homework Help
Replies
26
Views
3K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Engineering and Comp Sci Homework Help
Replies
1
Views
2K
• General Math
Replies
1
Views
1K