100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
The z-plane $16.99   Add to cart

Exam (elaborations)

The z-plane

 1 view  0 purchase
  • Course
  • Institution

Exam of 30 pages for the course psycologie at psycologie (The z-plane)

Preview 3 out of 30  pages

  • March 13, 2024
  • 30
  • 2023/2024
  • Exam (elaborations)
  • Questions & answers
avatar-seller
3 The z-plane


3.1 CHAPTER PREVIEW
We start this chapter with a review of the s-plane, a concept with which you are
probably familiar. We will then move on to its discrete equivalent- the z-plane.
The z-plane has an extremely important role to play in the design and analysis of
discrete systems. The significance of z-plane pole-zero diagrams will be discussed,
both in terms of signal shapes and system frequency response. The link between
the s-plane and z-plane will also be examined.



3.2 POLES, ZEROS AND THE s-PLANE
It is very likely that you have already encountered the 's-plane'. If so, you will
appreciate how useful this concept is when it comes to the analysis and design of
analogue systems. In a similar way, the z-plane is an invaluable tool when it
comes to the analysis and design of discrete systems. As it's much easier to
explain the significance and the use of the z-plane via the s-plane, we will begin
with a very brief review of the s-plane. If you are completely at ease with the
Laplace transform, the s-plane, and the idea of poles and zeros, then you can
probably afford to skip the next few pages. However, if you know little about
these, or are very rusty, then you will need to do some extra reading, as it is
unlikely that the following review will be sufficient for you. You will find
comprehensive explanations in most textbooks dealing with circuit analysis and
control systems in particular. Just three of many suitable sources are Bolton
(1998), Powell (1995), and Dorf and Bishop (1995).
The s-plane is simply an Argand diagram, used to display the values of s.
Remember that s represents the complex frequency, i.e. s = ~r + jo), and so s
values can be naturally shown on an Argand diagram. As is usual for Argand
diagrams, the real part of s, i.e. or, is plotted along the horizontal axis and the
imaginary part, jco, along the vertical axis.
As an example, consider a continuous system with a transfer function given
by:

s+4
T(s) =
(s + 2)(s + 3)
This transfer function can be represented on an s-plane diagram by displaying its
poles and zeros (Fig. 3.1).
System zeros are the s-values that cause the numerator, and so T(s) itself, to
become zero, while poles are defined as the values of s which result in T(s)

,42 The z-plane

Imaginary
je
3

System
zero
1
Real
o x x I I I
-4 -3 ~(~'-2 -1 1 2 c
-1
System - -2
poles
-3



Figure 3.1

becoming infinite, i.e. its denominator becoming zero. It follows that this particular
transfer function has a single zero at s = - 4 and two poles at s = - 2 a n d - 3 . On
pole-zero diagrams, poles are traditionally represented by crosses and zeros by
circles.
In this simple example all values are real and so appear on the real (or) axis -
but what if the transfer function had been more complicated? For example,

S2 +2S+5
T(s) -
(s + 3)(s 2 + 2s + 2)

We will begin by finding the zeros for this transfer function and so must first
equate the numerator to zero, i.e. s 2 + 2s + 5 - 0. As this quadratic expression
factorizes to (s + 1 +j2)(s + 1 - j 2 ) , then the system must have complex conjugate
zeros at s = - 1 + j2.
To find the poles we need to examine the denominator. Clearly, there must be
one pole at s = -3 (this is obviously a real pole). Also, as s 2 + 2s + 2 factorizes
to (s + 1 + j)(s + 1 - j ) , there must be two complex conjugate poles at s = - 1 + j .
The pole-zero diagram for this system is shown in Fig. 3.2.
Pole-zero diagrams of system transfer functions contain a surprising amount
of information about the system and are extremely useful when it comes to the
design and analysis of signal processors. For example, it can be shown that a
system with poles in the right-hand half of the s-plane is an unstable system. We
will examine system pole-zero diagrams in much more detail later in the chapter.


3.3 POLE-ZERO DIAGRAMS FOR
CONTINUOUS SIGNALS
As signals can also be transformed into the s-domain, i.e. expressed as functions
of s, then they too can be represented by pole-zero diagrams. These 'p-z' diagrams
contain information about the signal shapes and so are very useful.
To illustrate this important link between the signal p - z diagram and the signal
shape consider a signal, y(t), having a Laplace transform, Y(s), given by:

, 3.3 Pole-zero diagrams for continuous signals 43

Imaginary
j(o
3

0 2

X 1
Real
I )< I I I I
-4 -3 -2 -1 1 2 c~
X -1

0 -2

-3




Figure 3.2

3s 2 + 8s + 14
r(s) -
(s + 2)(s 2 + 2s + 10)

We will first find the time variation of the signal by using the traditional 'inverse
Laplace transform' method. However, before the Laplace transform tables can be
consulted the expression needs to be broken down into its partial fractions, i.e."

Y(s)- 1 + 2 ( s + 1)
s+2 s 2 + 2 s + 10

(You are strongly advised to derive this expression for yourself.)
By inspection of the Laplace transform tables (Appendix A), you should find
that the 1/(s + 2) part corresponds to a signal in the time domain of e - 2 t .
As the other component can be expressed, very conveniently, as 2(s + 1)/[(s +
1) 2 + 32], it 'inverse Laplace transforms' to 2e -t cos 3t. ( A g a i n - it's important
that you check this for yourself.)
The time variation of the output is therefore given by:
y(t) - e -2t + 2e -t cos 3t (3.1)

i.e. an exponentially decaying d.c. signal plus an exponentially decaying sinusoidal
(strictly 'cosinusoidal') oscillation.
We now need to look f o r some link between the signal shape and the pole-zero
diagram f o r Y(s).
As Y(s) factorizes to:

3(s + 1.33 + j l . 7 ) ( s + 1 . 3 3 - j l . 7 )
r(s) =
(s + 2)(s + 1 + j 3 ) ( s + 1 - j 3 )
then Y(s) has zeros a t - 1 . 3 3 + j l . 7 and poles a t - 2 a n d - 1 _+j3 (Fig. 3.3).
So we have a pole a t - 2 and a ' - 2 ' also appears in the 'e -2t' term of equation
(3.1). We also have two complex poles at -1 + j3. If we look at the 2e -t cos 3t
component of the signal, then a '1' occurs in the '2e -t' term, i.e. we can think of
this as 2e -it, and obviously a '3' appears in the 'cos 3t' term. In other words, it's
possible that the pole a t - 2 corresponds to the exponentially decaying d.c. signal

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying these notes from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller samsonronald256. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy these notes for $16.99. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

67096 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy study notes for 14 years now

Start selling
$16.99
  • (0)
  Add to cart