<< Chapter < Page Chapter >> Page >
This module is part of the collection, A First Course in Electrical and Computer Engineering . The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.

When we design a filter, we design it for a purpose. For example, a moving average filter is often designed to pass relatively constant data whileaveraging out relatively variable data. In an effort to clarify the behavior of a filter, we typically analyze its response to a standard set of test signals. Wewill call the impulse , the step , and the complex exponential the standard test signals.

Unit Pulse Sequence. The unit pulse sequence is the sequence

u n = δ n = 1 , n = 0 0 , n 0 .

This sequence, illustrated in Figure 1 , consists of all zeros except for a single one at n = 0 . If the unit pulse sequence is passed through a moving average filter (whether finite or not), then the output is called the unit pulse response :

h n = k = 0 w k δ n - k = w n .
This Cartesian graph is relatively simple. The positive portion of the x axis is marked with dashed dots spaced equidistant from each other. The first three are labeled 1, 2, and 3. There are three more dots to the right of those dots. The x axis is labeled n. The y axis only has one dashed point and the it is labeled 1. This Cartesian graph is relatively simple. The positive portion of the x axis is marked with dashed dots spaced equidistant from each other. The first three are labeled 1, 2, and 3. There are three more dots to the right of those dots. The x axis is labeled n. The y axis only has one dashed point and the it is labeled 1.
Unit Pulse Sequence

(Note that δ n - k = 0 unless n = k . ) So the unit pulse sequence may be used to read out the weights of a moving average filter. It is common practice to use w k (the k t h weight) and h k (the k t h impulse response) interchangeably.

Unit Step Sequence. The unit step sequence is the sequence

u n = ξ n = 1 , n 0 0 , n < 0 .

This sequence is illustrated in Figure 2 . When this sequence is applied to a moving average filter, the result is the unit step response

g n = k = 0 n w k = k = 0 n h k .

The unit step response is just the sequence of partial sums of the unit pulse response.

This Cartesian graph has three dashed dots on the negative portion of the x acis. The origin is labeled 0. On the poisitive side of the x axis there are four dashes. The third dash is labeled 3 and the x axis is labeled n. On the positive portion of the y-axis there is a single dashed dot labeled 1. To the right of this dashed dot there are four dots that are directly above the dashes on the positive portion of the x axis. Just above and to the right of the dots is the expression ξ_n vs n. Directly below this this symbol are three more dots. These dots are slightly above and to the right of the dashes on the x axis. This Cartesian graph has three dashed dots on the negative portion of the x acis. The origin is labeled 0. On the poisitive side of the x axis there are four dashes. The third dash is labeled 3 and the x axis is labeled n. On the positive portion of the y-axis there is a single dashed dot labeled 1. To the right of this dashed dot there are four dots that are directly above the dashes on the positive portion of the x axis. Just above and to the right of the dots is the expression ξ_n vs n. Directly below this this symbol are three more dots. These dots are slightly above and to the right of the dashes on the x axis.
Unit Step Sequence

Complex Exponential Sequence. The complex exponential sequence is the sequence

u k = e j k θ , k = 0 , ± 1 , ± 2 , ...

This sequence, illustrated in Figure 3 , is a “discrete-time phasor” that “ratchets” counterclockwise (CCW) as k moves to k + 1 and clockwise (CW) as k moves to k - 1 . Each time the phasor ratchets, it turns out an angle of θ . Why should such a sequence be a useful test sequence? There are two reasons.

This is a circle divided in to many slices. The bottom left is the size of a quarter of the circle. Above this section is a 1/8th sized portion and the rest of the circle is divided into 1/16th sized portions. The right half of the circle has mathematical expression associated with the different slices. The second slice from the top is labeled e^{j2θ} the next slice has an arch capped on either end with horizontal lines and the actual arch is labeled θ. The next few slices proceeding down are labeled e^{jθ}, then e^{j0}, then e^{-jθ}, and finally e^{-j2θ}. This is a circle divided in to many slices. The bottom left is the size of a quarter of the circle. Above this section is a 1/8th sized portion and the rest of the circle is divided into 1/16th sized portions. The right half of the circle has mathematical expression associated with the different slices. The second slice from the top is labeled e^{j2θ} the next slice has an arch capped on either end with horizontal lines and the actual arch is labeled θ. The next few slices proceeding down are labeled e^{jθ}, then e^{j0}, then e^{-jθ}, and finally e^{-j2θ}.
Discrete-Time Phasor

(i) e j k θ represents (or codes) cos k θ . The real part of the sequence e j k θ is the cosinusoidal sequence cos k θ :

Re [ e j k θ ] = cos k θ .

Therefore the discrete-time phasor e j k θ represents (or codes) cos k θ in the same way that the continuous-time phasor e j ω t codes cos ω t . If the moving average filter

x n = k = 0 h k u n - k

has real coefficients, we can get the response to a cosinusoidal sequence by taking the real part of the following sum:

x n = k = 0 h k cos ( n - k ) θ = Re [ k = 0 h k e j ( n - k ) θ ] = Re [ e j n θ k = 0 h k e - j k θ ] .

In this formula, the sum

k = 0 h k e - j k θ

is called the complex frequency response of the filter and is given the symbol

H ( e j θ ) = k = 0 h k e - j k θ .

This complex frequency response is just a complex number, with a magnitude | H ( e j θ ) | and a phase arg H ( e j θ ) . Therefore the output of the moving average filter is

x n = Re [ e j n θ H ( e j θ ) ] = Re [ e j n θ | H ( e j θ ) | e j arg H ( e j θ ) ] = | H ( e j θ ) | c o s [ n θ + arg H ( e j θ ) ] .

This remarkable result says that the output is also cosinusoidal, but its amplitude is | H ( e j θ ) | rather than 1, and its phase is a r g H ( e j θ ) rather than 0. In the examples to follow, we will show that the complex “gain” H ( e j θ ) can be highly selective in θ , meaning that cosines of some angular frequencies are passed with little attenuation while cosines of other frequencies are dramatically attenuated. By choosing the filter coefficients, we can design the frequency selectivity we would like to have.

(ii) e j k θ is a sampled data version of e j ω t . The discrete-time phasor e j k θ can be produced physically by sampling the continuous-time phasor e j ω t at the periodic sampling instants t k = k T :

e j k θ = e j ω t | t = k T = e j ω k T θ = ω T .

The dimensions of θ are radians, the dimensions of ω are radians/second, and the dimensions of T are seconds. We call T the sampling interval and 1 T the sampling rate or sampling frequency. If the original angular frequency of thephasor e j ω t is increased to ω + m ( 2 π T ) , then the discrete-time phasor remains e j k θ :

e j [ ω + m ( 2 π / T ) ] t | t = k T = e j ( ω k T + k m 2 π ) = e j k θ .

This means that all continuous-time phasors of the form e j [ ω + m ( 2 π / T ) ] t “hide under the same alias” when viewed through the sampling operation. Thatis, the sampled-data phasor cannot distinguish the frequency ω from the frequency ω + m 2 π T . In your subsequent courses you will study aliasing in more detail and study the Nyquist rule for sampling:

T 2 π Ω ; 1 T Ω 2 π

This rule says that you must sample signals at a rate ( 1 T ) that exceeds the bandwidth Ω 2 π of the signal.

Let's pass the cosinusoidal sequence u k = cos k θ through the finite moving average filter

x n = k = 0 N - 1 h k u n - k h k = 1 N k = 0 , 1 , ... , N - 1 .

We know from our previous result that the output is

x n = | H ( e j θ ) | cos [ n θ + arg H ( e j θ ) ] .

The complex frequency response for this example is

H ( e j θ ) = k = 0 N - 1 1 N e - j k θ = 1 N 1 - e - j N θ 1 - e - j θ

(Do you see your old friend, the finite sum formula, at work?) Let's try to manipulate the result into a more elegant form:

H ( e j θ ) = 1 N e - j ( N / 2 ) θ [ e j ( N / 2 ) θ - e - j ( N / 2 ) θ ] e - j ( θ / 2 ) [ e j ( θ / 2 ) - e - j ( θ / 2 ) ] = 1 N e - j [ ( N - 1 ) / 2 ] θ sin ( N 2 θ ) sin ( 1 2 θ ) .

The magnitude of the function H ( e j θ ) is

| H ( e j θ ) | = 1 N , | sin ( N 2 θ ) sin ( 1 2 θ ) | .

At θ = 0 , corresponding to a “DC phasor,” H ( e j θ ) equals 1; at θ = 2 π N | H ( e j θ ) = 0 | . The magnitude of the complex frequency response is plotted in Figure 4 .

This Cartesian graph contains a series of arches progressing along the x axis. The first arch starts at the far left end of the negative portion of the x axis and returns to the x axis about halfway to the origin. This point is labeled -2π/N. The next arch begins at this same point and returns to the x axis the same distance away from the origin as it left  but now on the positive side of the x axis. This point is labeled 2π/N. At about the peak of this arch on the negative side of the x axis is the mark 1. The third arch begins at point 2π/N and ends at a point labeled 2(2π/N). There is one final arch that starts at the last point and an unlabeled point. The x axis is labeled θ. This Cartesian graph contains a series of arches progressing along the x axis. The first arch starts at the far left end of the negative portion of the x axis and returns to the x axis about halfway to the origin. This point is labeled -2π/N. The next arch begins at this same point and returns to the x axis the same distance away from the origin as it left  but now on the positive side of the x axis. This point is labeled 2π/N. At about the peak of this arch on the negative side of the x axis is the mark 1. The third arch begins at point 2π/N and ends at a point labeled 2(2π/N). There is one final arch that starts at the last point and an unlabeled point. The x axis is labeled θ.
Frequency Selectivity of a Moving Average Filter

This result shows that the moving average filter is frequency selective, passing low frequencies with gain near 1 and high frequencies with gain near 0.

Questions & Answers

so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!
QuizOver.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, A first course in electrical and computer engineering. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10685/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'A first course in electrical and computer engineering' conversation and receive update notifications?

Ask