Resistor attenuator circuits

- for use in radio frequency circuits including receivers and transmitters, etc

Attenuator circuits are used in a variety of radio frequency circuit design applications. The attenuators reduce the level of the signal   and this can be used to ensure that the correct radio signal level enters another circuit block such as mixer or amplifier  so that it is not overloaded. As such attenuators are widely used by radio frequency circuit designers. While it is possible to buy ready made attenuators, it is also easy to make attenuators for many applications. Here a simple resistor network can be used to make attenuators that provide levels of attenuation up to figures of 60 dB and at frequency of 1 GHz and more, provided that care is taken with the construction and the choice of components.

One important feature that is required for radio frequencyapplications is that the characteristic impedance should be maintained. In other words the impedance looking into and out of the attenuator should be matched to the characteristic impedance of the system.

T and Pi networks

There are two basic formats that can be used for resistive attenuators. They are the T and pi networks. Often there is little to choose between them and the choice is often down to the preference of the designer.

As the name suggests the "T"section attenuator is in the shape of the letter T with two resistors in the signal line and one in the centre to ground.

T section attenuator

The two resistor values can be calculated very easily knowing the ratio of the input and output voltages, Vin and Vout respectively and the characteristic impedance Ro.

The pi section attenuator is in the form of the Greek letter pi and has one in line resistor and a resistor to ground at the input and the output.

Pi section attenuator

Similarly the values for the pi section attenuator can be calculated

Practical aspects

It is generally good practice not to attempt to achieve any more than a maximum of 20 dB attenuation in any one attenuator section. Even this is possibly a little high. It is therefore common practice to cascade several sections. When this is done the adjoining resistors can be combined. In the case of the  T section attenuator this simply means the two series resistors can be added together. For the pi section attenuators there are parallel resistors.

When making large value attenuators, great care must be taken to prevent  the signal leaking past the attenuator and reaching the output. This can result from capacitive or inductive coupling and poor earth arrangements. To overcome these problems good earth connection and careful layout, keeping the output and input away from one another are required. It may also be necessary to place a screen between the different sections.

Using these attenuators a surprisingly good frequency response can be obtained. Non-inductive resistors are required to ensure the best performance, and using good printed circuit board techniques and surface mount resistors, a good performance at frequencies in excess of 1 GHz are easy to achieve.

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future endeavors with success.
Happy New Year

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ANSWERS For Questions directions-for-questions-from-1-to-2

Answer for question  1 & 2 :- 

Let p be the number of problems on the test. 

Let n be the number of problems solved by the tenth student. 

Then n <= p. 

The number of problems solved, counted with multiplicity, is 7p 

since each student solved seven questions. 

However, we also know that the first nine students solved four questions each and the tenth solved n questions, 

so 

      7p = (4×9) +n. 

Re-arranging gives us 

p = (36 + n)/7 

Recalling that n <= p, we have

n <= p ⇒ 7n <= 7p ⇒ 7n <= 36 + n ⇒ 6n <= 36 ⇒ n <= 6.

Therefore, we are looking for non-negative n such that p = (36 + n)/7 and n <= 6

Therefore n = 6 and so p = 6. That is, the number of problems solved by the tenth 

student was six and there were six questions total on the test. Hence, option (2) is the correct choice

3.)    Given

                 (xy)^z = 64

            

            

since  positive integral 

           

 Whenz = 1, xy = 64. The total number of factors of 64 is 7.
z = 2, xy = 8. The total number of factors of 8 is 4.
z = 3, xy = 4. The total number of factors of 4 is 3.
z = 6, xy = 2. the total number of factors of 2 is 2
Hence, total number of positive integral solutions is 16.

4.)  Diameter of the circle is equal to the side of the smallest

   square is πr² = 77
  ⇒r = 3.5 (sqrt(2))
  ⇒ 2r = 7 sqrt(2) , side of the smallest square.
  Diagonal of the innermost square = 14 units.
  Diagonal of the largest square will be 14+ [(1.5) × 2] × 4
  = 14 + 12 = 26 units
  ⇒ Area of the largest square
  =  0.5 × (diagonal)² = 338 sq units and area of the smallest square =  (7 sqrt(2) )² sq. units.
 Required difference = 338 – 98 = 240 square units.

 5 .)  Total number of ways of dividing 24 boys in two distinct sub-sections is ²⁴C₁₂ = 24!/(12! 12!)
There can be two ways in which two shortest boys lie in different sub-section and correspondingly the remaining 22 boys can be divided into two distinct sub-sections in ²²C₁₁ ways.
∴ Probability = (2 x ²²C₁₁ )/²⁴C₁₂ = 12/23

6.)


7 ) here are 18 possible values of x and 9 possible values of y.
Since, BC is parallel to the X – axis, therefore AB will be parallel to the Y – axis.
The X – coordinate of points B and A will be same and the Y – coordinate of points B and C will be same.
For, every chosen co – ordinate of point B, the Y – coordinate of point A can be chosen in
9 – 1 = 8 ways and the X – coordinate of point A can be chosen in 18 – 1 = 17 ways.
Number of ways in which co – ordinate of point B can be chosen = 18 x 9
Total number of distinct triangles = 18 x 9 x 17 x 8 = 22032

8 .) Question

 The maximum possible area (in square units) of such a triangle ABC is ?

For maximum possible area of such a triangle ABC, the co – ordinates of point B, A and C should be


(- 8, 9), (- 8, 17) and (17, 9) respectively.


Maximum Area = (1/2) x 25 x 8 = 100 square units.


9 .)



10 )  Let, the average age of the family when the first child was born be 'x' years.
Therefore, the total age of the family when the first child was born is 3x.

Let, after a years, the second child was born:
Therefore, (3x + 3a) = 18 x 4 = 72

Let, after another b years, the third child was born
Therefore, 72 + 4b = 16 x 5
b = 2,
Let, after another c years, the fourth child was born
Therefore, 80 + 5c = 20 x 6 = 120
c = 8

Let, after another d years, the fifth child was born
Therefore, 120 + 6d = 24 x 7 = 168
d = 8.

Let, after another 'e' years the average age of the family is 28 years.
Also, 168 + 7e = 28 x 8 = 224
7e = 56
e = 8

Also, a + b + c + d + e = Present age of the eldest child = 27
a = 27 – 26 = 1 year.

3x + 3 = 72
x = 23. 

 I AM  EXTREMELY SORRY FOR DELAY IN ANSWERS  

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Points to remember while preparing for GATE 2011

Systematic preparation is only the way to qualify the GATE 2011 and all public sector exams. Following these suggestion for success in GATE and public sector exams.

1. Bring previous year GATE exam question papers (gate exam questions are available very easily in the market).
2. Start back study (i.e. pick up GATE questions and find the solution) I personally believe if you practice to solve all the gate questions from the previous asked (at least 6-8 year) you can definitely qualify the gate exam.
3. Another important fact is: do not answer wrong during the gate examination even your total score is only 35 to 45 % (mostly people could not qualify gate exam because they attempt to answer wrong question).
4. Keep in mind, most of the time negative marking is only responsible to the failure.
5. Join some Gate exam discussion and online sharing forum and discuss your doubts and get touch with Gate Aspirants.
I hope if you follow all these suggestion you will definitely qualify GATE exam.

Preparing for GATE 2011 can be made easy in 10 steps

Preparing for GATE can be as easy as preparing for your college examinations. Just take little cautions while studying any topic and do remember that GATE paper focus on your in depth knowledge of subject, your basics, presence of mind during examination etc.

1. Always follow standard books for GATE. Try to cover complete syllabus. If not possible expertise in what ever portion of syllabus you practice.

2. Try preparing notes after reading every chapter topic. This may initially take some time but will help you while revising before paper.

3. While reading any chapter topic does ask your self following questions What, How, and why and see improvement.

4. Best way to prepare is to follow cycle Learn, Test, Analyze, Improve, Learn, Test, Analyze, and Improve.

5. Do remember that GATE is completely objective question based test. Most of time solving objective questions is tricky. Learn tips to solve GATE objective questions.

7. In case of doubts do ask some expert or use forums to discuss questions as provided by httpwww.indiastudychannel.com or ask me.

8. Don’t worry if your percentage in university exams is low as GATE admissions do not consider them. Just maintain minimum percentage required by many colleges including IIT’s. Look at GATE Cutoff and eligibility section to know eligibility and cut off of various colleges.

9. Group study is one of the best ways of preparation. Divide sections topics between you and your partner and have a brief session on topic from your friend before you actually start topic. This will save your time and efforts and will improve your and your partner’s understanding on the topic.

10. Normally coaching GATE is not required but if you are not able to concentrate much then this is a good option.

GATE – 2011 Preparation Tips

GATE – 2011 Preparation Tips

1. Material Collection

• Syllabus
• All the relevant books based on the subject(Divide the books in two groups - (1)Fundamental and basic concepts (2)Problem oriented
• Some books helpful for pre-requisite knowledge on the subject
• Some good guide books for GATE
• Previous questions papers

2. Keep contact with some expert and GATE experienced persons

3. Study - Syllabus and Previous questions papers
4. Start from the first chapter

• Read at least 5 books, it will widen your knowledge(if necessary consult with the books for pre-requisite knowledge or with some expert)
• Note down the probable concepts (definitions, unit, dimension etc.)
• Note down necessary theories, formulae etc
• Solve problems as maximum as possible(from text books, Guide books etc)
• Think about various tricks in solving problems(if necessary, note it)
• Go for series of self tests based on this chapter(take other's help to conduct tests)
• Continue the self tests until getting a very good score

5. Solve more and more problems, discover more and more new tricks…

6. Follow the same procedure for the rest chapters
7. Finally, go for self tests based on whole syllabus(take other's help to conduct these tests)
8. On the exam day…you will be at the Pick, who can stop you? 

Directions for Questions from 1 to 2: A class of ten students took a math exam. 

Each problem was solved by exactly seven of the ten students. The first nine students in the class 

solved exactly four problems each.

Question 1.

(linear equations)

What is the number of problems solved by the tenth student? Correct Answer: 2 6 4 7 3 Skip this question

Question 2.

(linear equations)

What is the number of problems on the test? Correct Answer: 2 5 6 4 7 3 Skip this question

Question 3.

Find the number of positive integral solutions of the equation (xy)z = 64. Correct Answer: 3 14 15 16 17 18 Skip this question

Question 4.

5 concentric squares each having different area are drawn on a sheet of paper.  If the area of the circle inscribed in the smallest square is 77 square units and the distance between  the corresponding vertices of consecutive squares is 1.5 units, then find the difference between  the areas of the largest and the smallest square in square units. (Take pie = 22/7) Correct Answer: 1 240 264 224 216 256 Skip this question

Question 5.

A class of 24 boys was divided into two sub-sections containing 12 boys each.  What is the probability that the two shortest boys of the class are in different sub-sections,  if all boys have different heights? Correct Answer: 3 7/24 6/23 12/23 14/23 None of these Skip this question

Question 6.

Two wheels of diameters 7 cm and 14 cm start rolling simultaneously towards each other from  two points which are 1990.5 cm apart. Both wheels make the same number of revolutions per second.  If both the wheels meet (touch externally) after 10 seconds, then what is the speed of the smaller wheel? Correct Answer: 2 132 cm/sec 66 cm/sec 44 cm/sec 22 cm/sec 67 cm/sec Skip this question

Directions for Questions from 7 to 8:

 ABC is a right – angled triangle right angled at B. BC is parallel to the X – axis. 
Given that the co-ordinates (x, y) of A, B and C are integers such - 9 < x < 10 and 8 < y < 18. 

Question 7.

(permutaion and combination)

How many distinct triangles ABC are possible? Correct Answer: 2 24000 22032 18724 27564 Infinitely Many Skip this question

Question 8.

(permutaion and combination)

The maximum possible area (in square units) of such a triangle ABC is  Correct Answer: 2 84 100 104 120 92 Skip this question

Question 9.

‘S’ denotes the sum to infinity and ‘Sn’ denotes the sum to ‘n’ terms of the series  1 + 4/5 + 16/25 + ...+ If S - Sn < 5/2. then the least value of ‘n’ is Correct Answer: 3 2 3 4 5 6 Skip this question

Directions for Questions from 10 to 10:

 A family consists of the husband, wife and six children. 
The age of the eldest child and the average age of the entire family at present is 27 years and 28 years
 respectively. The average age of the entire family after the fifth, fourth, third and second child
 was born is 24, 20, 16 and 18 respectively.

Question 10.

(linear equations)

What is the difference between the age of the first and the second child Correct Answer: 3 3 years 2 years 1 year 5 years Cannot be determined. Skip this question

For the students: online test

For the students: online test: "check your basic knowledge on electronics : ATTENTION: A browser version of 4.0 or higher  or Explorer must be used otherwise th..."

check once

check your basic knowledge on electronics :
R U Electronics students ? then Y waiting come and check out yours stuff .....
visit: http://ece-08.blogspot.com/p/attention-browser-version-of-4.html

academics

ACADEMIC CALENDAR FOR  THE COMMENCEMENT OF CLASS WORK  

FOR B.TECH/B.PHARM II, III & IV YEAR-II SEMESTER

            The Proposed Academic Calendar for B.Tech/B.Pharm II, III & IV Year-II Semester  Regular During the year 2010-11 is detailed below.

B.TECH/B.PHARM II, III & IV YEAR-II SEMESTER
Description	From	To	Weeks
Commencement of Class Work	29-11-2010	-	-
I Unit of Instructions	29-11-2010	22-01-2011	8W
I Mid Examinations	24-01-2011	29-01-2011	1W
II Unit of Instruction	31-01-2011	26-03-2011	8W
II Mid Examination	28-03-2011	2-04-2011	1W
Preparation &  Practicals	4-04-2011	16-04-2011	2W
End Examinations	18-04-2011	30-04-2011	2W

NEXT ACADEMIC YEAR CLASS WORK:13-06-2011

Mobile Number Portability (MNP) India launched: SMS format & Rules

Mobile Number Portability (MNP) India launched: SMS format & Rules

The much awaited Mobile Number Portability (MNP) service will be launched today in Rohtak, Haryana. The Union Minister of Communications & IT Shri Kapil Sibal with Shri Bhupinder Singh Hooda, Chief Minister of Haryana as the Chief Guest will inaugurate the service at a function in Rohtak at 3 p.m. on 25.11.2010.

A plan for phased migration of networks in all the remaining 22 (excluding Haryana) Licensed Service Areas in the country for working in MNP scenario is being chalked out in consultation with service providers. The detailed program for the entire country will be announced separately.


From tomorrow, mobile phone customers in Haryana Licensed Service Area (LSA) will have the choice of selecting their telecom service provider (operator) with out changing their number, provided a minimum period of 90 days has elapsed after subscription to the mobile service of the current service provider. For change of service provider i.e. porting, a subscriber has to send an SMS (PORTMobile Number) from the number he wishes to be ported, to number 1900 whereby a Unique Porting Code (UPC) will be received on SMS from his current service provider. The subscriber will need to apply in the prescribedapplication form to the chosen new service provider quoting the UPC which will act as a reference while filling up the application form with new service provider.

The new service provider will then take action to get the required processes completed to enable the subscriber to get connected to his network. Porting has to be completed within 7 working days. Telecom Regulatory Authority of India (TRAI) has put a ceiling of Rs. 19/- on porting charges which the new service provider may collect from the subscriber. Post-paid subscribers before making the porting request, have to make sure that their last bill has been paid failing which the request for change to new service provider shall be rejected. In the case of pre-paid subscriber any balance amount left will not be carried forward when the number is transferred to the new service provider.

quiz

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Hi friends here i am glad to propose a new page in blog named "Puzzles And Quizzes" to develop the analytical skills and aptitude skills...


Here i would like to give weekly puzzles and quiz questions. Answer to the puzzles are sent the blogs email id : blogece08@gmail.com
correct answer with appropriate solutions will be sent back.....

Books for ECE gate

It is time to prepare for gate who are appearing for gate 2011 and 2012. it is well known to all that February 13th(second sunday) is exam. some important text books for ECE
****    The reference books for GATE ECE students are below.

Linear Algebra, Numerical Methods,Transform Theory
Higher Engineering Mathematics by B.S.Grewal

Calculus, Differential equations, Complex variables:
Intermediate Mathematics, S.chand publications by B.V.Sastry , K.Venkateswarlu ( if i remember ) both the volumes.

Probability and Statistics: 
Probability , statistics and queuing theory by S.C.Gupta & V.K.Kapoor

Networks: Network graphs:
Network Analysis by Van Valkenburg
Network Theory by Alexander Sadiku, UA Bakshi
Network Theory by Shaum’s outline, Robbins and Miller

Electronic Devices
Electronic Devices and Circuits by Boylestead
Solid State Electronic Devices by Benjamin G Streetman
Integrated electronics by Milman Halkias
Electronic Devices and Circuits by David A Bell
Integrated Circuits: Sedra Smith, K.R. Botkar
Electronic Principals by Malvino

Analog and Digital Circuits 
Engineering Circuit Analysis Hayt & Kemmerly
Electric Circuits by Joseph A. Edminister
Fundamentals of Electric Circuits – Sadiku
Electronic Circuit Analysis by Donald Neamen
Electronic Devices and Circuits by Boylestead
Micro Electronic Circuits by Sedra & Smith
Digital Circuits Anand Kumar, Morris Mano

Signals and Systems:
Signals and System by Oppenham and Schiener
Signals and System: by Schaum Series, Willsky & Nacob
Signals and System by Sanjay Sharma

Control Systems:
Control Systems by Ogatta,Kuo
Linear Control Systems by B.S Manke.

Communications:
Communication Systems by Simon Haykin
Principle of Communication System by B. P. Lathi
Principle of Communication System by Taub & Schilling
Communication Systems by Carlson
Communication Systems by Sanjay Sharma

Electromagnetics
Electromagnetic Waves & Radiating Systems by Hayt, JD Kraus
Elements of Electromagneticsby Sadiku
Electromagnetics by J.D.Kraus
Engineering Applications of Electromagnetic Theory by Liao

Few other useful books for GATE ECE preparation.
Gate 2010 by R.S. Kanodia (for questions only)
Previous GATE Question papers(Made Easy Publishers,GK publishers etc)
Previous Engg Services question papers(Made Easy Publishers)
Aptitude Test-D R Choudhary
GATE Physics Self Study Book by Surekha Tomar
Multiple Choice Questions with Explainatory Answers by UPKAR Publication , Agra

gate 2011 application status

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