Showing posts with label Digital Electronics. Show all posts
Showing posts with label Digital Electronics. Show all posts

Friday, 9 December 2011

Model question paper of DIGITAL ELECTRONICS of Diploma 1st Sem - E C

            MODEL QUESTION PAPER          
        FIRST SEMESTER          
           SUBJECT :  DIGITAL ELECTRONICS-1          
        Common to E&C ,EI&C,CS&E and IS&E.           
        SUBJECT: DIGITAL ELECTRONICS-I  
        
        TIME :3 HRS
                                                                         MAX.MARKS :100          
                  
        Note:  1)Section A is  compulsory.          
                  2) Answer any two main questions from each of the remaining Sections
          
                  

        SECTION:  A          
                  
1    a.    Fill in the blanks with suitable words                                    5X1      
    (i)    The radix of hexadecimal number system is _______________          
    (ii)    The number of flip-flops required for decade counter is ______________          
    (iii)    Race around condition is eleminated in ________________ flip flop          
    (iv)    An example for unsaturated logic _________ Ics          
    (v)    The SISO shift register with feed back _________ counter          
                  
    b.    Write a note interfacing between TTL and CMOS Ics                         5      
                  
        SECTION:  B          
                  
2    a.    Define the radix of a number system    2      
    b.    Convert the following from Binary to Decimal number system    8      
         (i) 11001001 (ii) 1110.1111          
        (iii) 11001111 -->( )h   (iv) 1AF--> ( ) D          
    c.    Subtract the following using 2's complemnet method          
        (i) 1FE2- 0AC2            (ii) 412.6 - 255.0    5      
                  
3    a.    Convert the following gray code to binary code (i) 110101 (ii) 111110     2      
    b.    Convert the following Excess-3 code to binary code (i) 11011 (ii) 10110   3      
    c.    Differentiate analog and digital signals                                                          5      
    d.     Expalin AND logic with symbol,  logic expression and truth table, .             5      
                  
4    a.    Define Noise Margin and power dissipation with reference to logic families  4      
    b.    State Associative and Distributive laws of Boolean Algebra                       4      
    c.    Simplfy the following expressions and realize the same using gates          
        (i) A+(BC.(D+E)+ B))     (ii) (A+B).(C+D).(AB+CD)                     7      
                  
                  
        SECTION: C          
                  
5    a.    What is a Full adder? Use K- map to derive logic expressions for its output    7      
    b .   Compare serial and parallel adders                                 3      
    c .   Explain the operation of 4-bit parallel adder                      5      
                  
6    a .   What is a priority encoder?                                 2      
    b.    Write the logic circuit and truth table of Decimal to BCD encoder   8      
    c.    Write the logic circuit and truth table of BCD to seven segment decoder for common
         anode display                                                            5                     
7    a.    What is a multiplexer?                                          2      
    b    Explain the 8:1 multiplexer with the help truth table and logic circuit      6      
    c .   What is the use of De-multiplexer? Write the truth table and logic circuit of
           4:1 De-multiplexer                                                     7      
                  
        SECTION: D          
                  
8    a.     Differentiate between combinational and seqential circuits         3      
    b .   Explain the operation of clocked RS-FF using NAND gates         6      
    c.    What is meant by Race around condition? How it is overcome?    7      
                  
9    a .   Explain the logic diagram, truth table and timing diagram,
       the operation of 4-bit ripple counter                              8      
    b.    What do you by modulus of counter?                         2      
    c .   Explain how decoding gates can be used for construction of mod-11 counter   5      
                  
10    a .   Explain with logic diagram, truth table and timing diagram,
              the operation of a 4-bit shift counter  8    
    b.    Write the logic diagram of a 4-bit shift register.
           Write the truth table and timing diagram for a 4-bit SISO register for a data of 1011  ( 7 )

Monday, 10 October 2011

Make your own Burglar Alarm


Let’s begin!

Materials


  1. Insulated Wire
    • Get 4 strands of wire, each 1 foot in length
    • You can find the wires around the house, inside battery-operated appliances that no longer work. Make sure that your parents give you permission to take them apart.
    • You can also buy the wires and other supplies listed from a Radio Shack, a hardware store, or the electrical section of big stores like Walmart.
    • Note: In order to make this alarm, you will need to peel off the rubber/plastic insulation from two of the wires. Try to peel the wires before starting this lesson. If it is too difficult for you, then you can buy non-insulated wire that does not need peeling.
  2. 1.5-volt battery.
  3. Kite string or button-and-carpet thread, 3-5 feet long.
  4. Scissors
  5. Electrical tape
  6. 1.5-volt mini-buzzer
  7. Spring-type wooden clothes pin
  8. A scrap piece of board or plywood 4 inches by 12 inches or bigger.
  9. Super Glue

Principles

The burglar alarm is made of three parts:
  1. The device that makes noise - a buzzer
  2. The battery
  3. The buzzer switch

The Buzzer

We cannot have an alarm without something that makes sounds. That is why we use a buzzer, which is something that makes a loud sound. To make the sound, the buzzer needs to be connected to the battery.

The Battery

The battery is very important to the alarm. For the buzzer to work, it needs energy. This energy comes from the battery. If the buzzer is not connected to the battery, it will not work. You have a switch to control this connection.

The Buzzer Switch

See how the buzzer works by connecting your 1.5-volt battery to the buzzer directly. (The buzzer should come with two different wires. If the wire is insulated, you will need to peel a small length so that you are able to see the metal wire inside.) Get two other pieces of wire and connect them to opposite ends of the battery. Connect one wire to (+) and the other to (-). The battery should have the (+) and (-) labeled. Also, if you look closely, there is a (+) and (-) on the battery that you bought.
You may want to use electrical tape to hold the wires in place on the battery.
Connect the buzzer’s wires and the battery’s wires by twisting them, just like in the diagram 1. The buzzer should be making a sound; if it does not, check the connections at the ends of the wires to the battery.

Question :
  • What happens when you disconnect a wire from the buzzer?
    Now, connect it again as often as you want and see how the buzzer turns on and off. By connecting and disconnecting the wires, what you have made is called a switch.
When you disconnect a wire and the buzzer stops sounding, we say that the switch is open. When it is open, no energy from the battery reaches the buzzer.
When you connect the wire and the buzzer sounds, we say that the switch is closed. When it is closed, the energy from the battery reaches the buzzer.

Make your own Burglar Alarm


Let’s begin!

Materials


  1. Insulated Wire
    • Get 4 strands of wire, each 1 foot in length
    • You can find the wires around the house, inside battery-operated appliances that no longer work. Make sure that your parents give you permission to take them apart.
    • You can also buy the wires and other supplies listed from a Radio Shack, a hardware store, or the electrical section of big stores like Walmart.
    • Note: In order to make this alarm, you will need to peel off the rubber/plastic insulation from two of the wires. Try to peel the wires before starting this lesson. If it is too difficult for you, then you can buy non-insulated wire that does not need peeling.
  2. 1.5-volt battery.
  3. Kite string or button-and-carpet thread, 3-5 feet long.
  4. Scissors
  5. Electrical tape
  6. 1.5-volt mini-buzzer
  7. Spring-type wooden clothes pin
  8. A scrap piece of board or plywood 4 inches by 12 inches or bigger.
  9. Super Glue

Principles

The burglar alarm is made of three parts:
  1. The device that makes noise - a buzzer
  2. The battery
  3. The buzzer switch

The Buzzer

We cannot have an alarm without something that makes sounds. That is why we use a buzzer, which is something that makes a loud sound. To make the sound, the buzzer needs to be connected to the battery.

The Battery

The battery is very important to the alarm. For the buzzer to work, it needs energy. This energy comes from the battery. If the buzzer is not connected to the battery, it will not work. You have a switch to control this connection.

The Buzzer Switch

See how the buzzer works by connecting your 1.5-volt battery to the buzzer directly. (The buzzer should come with two different wires. If the wire is insulated, you will need to peel a small length so that you are able to see the metal wire inside.) Get two other pieces of wire and connect them to opposite ends of the battery. Connect one wire to (+) and the other to (-). The battery should have the (+) and (-) labeled. Also, if you look closely, there is a (+) and (-) on the battery that you bought.
You may want to use electrical tape to hold the wires in place on the battery.
Connect the buzzer’s wires and the battery’s wires by twisting them, just like in the diagram 1. The buzzer should be making a sound; if it does not, check the connections at the ends of the wires to the battery.

Question :
  • What happens when you disconnect a wire from the buzzer?
    Now, connect it again as often as you want and see how the buzzer turns on and off. By connecting and disconnecting the wires, what you have made is called a switch.
When you disconnect a wire and the buzzer stops sounding, we say that the switch is open. When it is open, no energy from the battery reaches the buzzer.
When you connect the wire and the buzzer sounds, we say that the switch is closed. When it is closed, the energy from the battery reaches the buzzer.

Friday, 2 September 2011

Review Questions Of Number Systems

Review Questions
1. What is meant by the radix or base of a number system? Briefly describe why hex representation is
used for the addresses and the contents of the memory locations in the main memory of a computer.
2. What do you understand by the l’s and 2’s complements of a binary number? What will be the
range of decimal numbers that can be represented using a 16-bit 2’s complement format?
3. Briefly describe the salient features of the IEEE-754 standard for representing floating-point
numbers.
4. Why was it considered necessary to carry out a revision of the IEEE-754 standard? What are the
main features of IEEE-754r (the notation for IEEE-754 under revision)?
5. In a number system, what decides (a) the place value or weight of a given digit and (b) the maximum
numbers representable with a given number of digits?
6. In a floating-point representation, what represents (a) the range of representable numbers and (b)
the precision with which a given number can be represented?
7. Why is there a need to have floating-point standards that can take care of decimal data and decimal
arithmetic in addition to binary data and arithmetic?
Problems
1. Do the following conversions:
(a) eight-bit 2’s complement representation of (−23)10;
(b) The decimal equivalent of (00010111)2 represented in 2’s complement form.
(a) 11101001; (b) +23
2. Two possible binary representations of (−1)10 are (10000001)2 and (11111111)2. One of them
belongs to the sign-bit magnitude format and the other to the 2’s complement format. Identify.
(10000001)2
= sign-bit magnitude and (11111111)2
= 2’s complement form
3. Represent the following in the IEEE-754 floating-point standard using the single-precision format:
(a) 32-bit binary number 11110000 11001100 10101010 00001111;
(b) (−118.625)10.
(a) 01001111 01110000 11001100 10101010;
(b) 11000010 11101101 01000000 00000000
4. Give the next three numbers in each of the following hex sequences:
(a) 4A5, 4A6, 4A7, 4A8, ;
(b) B998, B999,
(a) 4A9, 4AA, 4AB; (b) B99A, B99B, B99C
5. Show that:
(a) (13A7)16
= (5031)10;
(b) (3F2)16
= (1111110010)2.
6. Assume a radix-32 arbitrary number system with 0–9 and A–V as its basic digits. Express the mixed
binary number (110101.001)2 in this arbitrary number system.

Range of Numbers and Precision

Range of Numbers and Precision :

          The range of numbers that can be represented in any machine depends upon the number of bits in the
exponent, while the fractional accuracy or precision is ultimately determined by the number of bits
in the mantissa. The higher the number of bits in the exponent, the larger is the range of numbers
that can be represented. For example, the range of numbers possible in a floating-point binary number
format using six bits to represent the magnitude of the exponent would be from 2−64 to 2+64, which
is equivalent to a range of 10−19to 10+19. The precision is determined by the number of bits used to
represent the mantissa. It is usually represented as decimal digits of precision. The concept of precision
as defined with respect to floating-point notation can be explained in simple terms as follows. If the
mantissa is stored in n number of bits, it can represent a decimal number between 0 and 2n−1 as the
mantissa is stored as an unsigned integer. If M is the largest number such that 10M −1 is less than or
equal to 2n−1, then M is the precision expressed as decimal digits of precision. For example, if the
mantissa is expressed in 20 bits, then decimal digits of precision can be found to be about 6, as 220 −1
equals 1 048 575, which is a little over 106−1. We will briefly describe the commonly used formats
for binary floating-point number representation.

Floating Point Number Formats :

           The most commonly used format for representing floating-point numbers is the IEEE-754 standard.
The full title of the standard is IEEE Standard for Binary Floating-point Arithmetic (ANSI/IEEE STD
754-1985). It is also known as Binary Floating-point Arithmetic for Microprocessor Systems, IEC
60559:1989. An ongoing revision to IEEE-754 is IEEE-754r. Another related standard IEEE 854-
1987 generalizes IEEE-754 to cover both binary and decimal arithmetic. A brief description of salient
features of the IEEE-754 standard, along with an introduction to other related standards, is given below.
ANSI/IEEE-754 Format

The IEEE-754 floating point is the most commonly used representation for real numbers on
computers including Intel-based personal computers, Macintoshes and most of the UNIX platforms.
It specifies four formats for representing floating-point numbers. These include single-precision,
double-precision, single-extended precision and double-extended precision formats. Table 1.1 lists
characteristic parameters of the four formats contained in the IEEE-754 standard. Of the four formats
mentioned, the single-precision and double-precision formats are the most commonly used ones. The
single-extended and double-extended precision formats are not common.
Figure 1.1 shows the basic constituent parts of the single- and double-precision formats. As shown in
the figure, the floating-point numbers, as represented using these formats, have three basic components
including the sign, the exponent and the mantissa. A ‘0’ denotes a positive number and a ‘1’ denotes
a negative number. The n-bit exponent field needs to represent both positive and negative exponent
values. To achieve this, a bias equal to 2n−1− 1 is added to the actual exponent in order to obtain the
stored exponent. This equals 127 for an eight-bit exponent of the single-precision format and 1023 for
an 11-bit exponent of the double-precision format. The addition of bias allows the use of an exponent
in the range from −127 to +128, corresponding to a range of 0–255 in the first case, and in the range
from −1023 to +1024, corresponding to a range of 0–2047 in the second case. A negative exponent
is always represented in 2’s complement form. The single-precision format offers a range from 2−127
to 2+127, which is equivalent to 10−38 to 10+38. The figures are 2−1023 to 2+1023, which is equivalent to 10−308 to 10+308 in the case of the double-precision format.

           The extreme exponent values are reserved for representing special values. For example, in the case
of the single-precision format, for an exponent value of −127, the biased exponent value is zero,
represented by an all 0s exponent field. In the case of a biased exponent of zero, if the mantissa is zero
as well, the value of the floating-point number is exactly zero. If the mantissa is nonzero, it represents
a denormalized number that does not have an assumed leading bit of ‘1’. A biased exponent of +255,
corresponding to an actual exponent of +128, is represented by an all 1s exponent field. If the mantissa
is zero, the number represents infinity. The sign bit is used to distinguish between positive and negative
infinity. If the mantissa is nonzero, the number represents a ‘NaN’ (Not a Number). The value NaN is
used to represent a value that does not represent a real number. This means that an eight-bit exponent
can represent exponent values between −126 and +127. Referring to Fig. 1.1(a), the MSB of byte 1
indicates the sign of the mantissa. The remaining seven bits of byte 1 and the MSB of byte 2 represent
an eight-bit exponent. The remaining seven bits of byte 2 and the 16 bits of byte 3 and byte 4 give a
23-bit mantissa. The mantissa m is normalized. The left-hand bit of the normalized mantissa is always


‘1’. This ‘1’ is not included but is always implied. A similar explanation can be given in the case of
the double-precision format shown in Fig. 1.1(b).
Step-by-step transformation of (23)10 into an equivalent floating-point number in single-precision
IEEE format is as follows:
• (23)10
= (10111)2
= 1.0111e + 0100.
• The mantissa = 0111000 00000000 00000000.
• The exponent = 00000100.
• The biased exponent = 00000100 + 01111111 = 10000011.
• The sign of the mantissa = 0.
• (+23)10
= 01000001 10111000 00000000 00000000.
• Also, (–23)10
= 11000001 10111000 00000000 00000000.
IEEE-754r Format
As mentioned earlier, IEEE-754r is an ongoing revision to the IEEE-754 standard. The main objective of
the revision is to extend the standard wherever it has become necessary, the most obvious enhancement
to the standard being the addition of the 128-bit format and decimal format. Extension of the standard
to include decimal floating-point representation has become necessary as most commercial data are
held in decimal form and the binary floating point cannot represent decimal fractions exactly. If the
binary floating point is used to represent decimal data, it is likely that the results will not be the same as
those obtained by using decimal arithmetic.
In the revision process, many of the definitions have been rewritten for clarification and consistency.
In terms of the addition of new formats, a new addition to the existing binary formats is the 128-bit
‘quad-precision’ format. Also, three new decimal formats, matching the lengths of binary formats,
have been described. These include decimal formats with a seven-, 16- and 34-digit mantissa, which
may be normalized or denormalized. In order to achieve maximum range (decided by the number of
exponent bits) and precision (decided by the number of mantissa bits), the formats merge part of the
exponent and mantissa into a combination field and compress the remainder of the mantissa using
densely packed decimal encoding. Detailed description of the revision, however, is beyond the scope
of this book.
IEEE-854 Standard
The main objective of the IEEE-854 standard was to define a standard for floating-point arithmetic
without the radix and word length dependencies of the better-known IEEE-754 standard. That is why
IEEE-854 is called the IEEE standard for radix-independent floating-point arithmetic. Although the
standard specifies only the binary and decimal floating-point arithmetic, it provides sufficient guidelines
for those contemplating the implementation of the floating point using any other radix value such
as 16 of the hexadecimal number system. This standard, too, specifies four formats including single,
single-extended, double and double-extended precision formats.
Example 1.11
Determine the floating-point representation of −14210 using the IEEE single-precision format.
Solution
• As a first step, we will determine the binary equivalent of (142)10. Following the procedure outlined
in an earlier part of the chapter, the binary equivalent can be written as (142)10
= (10001110)2.
• (10001110)2
= 1.000 1110 × 27 = 1.0001110e + 0111.
• The mantissa = 0001110 00000000 00000000.
• The exponent = 00000111.
• The biased exponent = 00000111 + 01111111 = 10000110.
• The sign of the mantissa = 1.
• Therefore, −14210
= 11000011 00001110 00000000 00000000.
Example 1.12
Determine the equivalent decimal numbers for the following floating-point numbers:
(a) 00111111 01000000 00000000 00000000 (IEEE-754 single-precision format);
(b) 11000000 00101001 01100 45 0s (IEEE-754 double-precision format).
Solution
(a) From an examination of the given number:
The sign of the mantissa is positive, as indicated by the ‘0’ bit in the designated position.
The biased exponent = 01111110.
The unbiased exponent = 01111110−01111111 = 11111111.
It is clear from the eight bits of unbiased exponent that the exponent is negative, as the 2’s
complement representation of a number gives ‘1’ in place of MSB.
The magnitude of the exponent is given by the 2’s complement of (11111111)2, which is
(00000001)2= 1.
Number Systems 17
Therefore, the exponent =−1.
The mantissa bits = 11000000 00000000 00000000 (‘1’ in MSB is implied).
The normalized mantissa = 1.1000000 00000000 00000000.
The magnitude of the mantissa can be determined by shifting the mantissa bits one position to the left.
That is, the mantissa = (.11)2
= (0.75)10.
(b) The sign of the mantissa is negative, indicated by the ‘1’ bit in the designated position.
The biased exponent = 10000000010.
The unbiased exponent = 10000000010−01111111111 = 00000000011.
It is clear from the 11 bits of unbiased exponent that the exponent is positive owing to the ‘0’ in
place of MSB. The magnitude of the exponent is 3. Therefore, the exponent = +3.
The mantissa bits = 1100101100 45 0s (‘1’ in MSB is implied).
The normalized mantissa = 1.100101100 45 0s.
The magnitude of the mantissa can be determined by shifting the mantissa bits three positions to
the right.
That is, the mantissa = (1100.101)2
= (12.625)10.
Therefore, the equivalent decimal number =−12625.

Floating Point Numbers

          Floating-point notation can be used conveniently to represent both large as well as small fractional
or mixed numbers. This makes the process of arithmetic operations on these numbers relatively much
easier. Floating-point representation greatly increases the range of numbers, from the smallest to the
largest, that can be represented using a given number of digits. Floating-point numbers are in general
expressed in the form
N = m×be (1.1)
where m is the fractional part, called the significant or mantissa, e is the integer part, called the
exponent, and b is the base of the number system or numeration. Fractional part m is a p-digit number
of the form (±d.dddd dd), with each digit d being an integer between 0 and b – 1 inclusive. If the
leading digit of m is nonzero, then the number is said to be normalized.
Equation (1.1) in the case of decimal, hexadecimal and binary number systems will be written as
follows:
Decimal system
N = m×10e (1.2)
Number Systems 13
Hexadecimal system
N = m×16e (1.3)
Binary system
N = m×2e (1.4)
For example, decimal numbers 0.0003754 and 3754 will be represented in floating-point notation
as 3.754 × 10−4 and 3.754 × 103 respectively. A hex number 257.ABF will be represented as
2.57ABF × 162. In the case of normalized binary numbers, the leading digit, which is the most
significant bit, is always ‘1’ and thus does not need to be stored explicitly.
Also, while expressing a given mixed binary number as a floating-point number, the radix point is
so shifted as to have the most significant bit immediately to the right of the radix point as a ‘1’. Both
the mantissa and the exponent can have a positive or a negative value.
The mixed binary number (110.1011)2 will be represented in floating-point notation as .1101011
× 23 = .1101011e+0011. Here, .1101011 is the mantissa and e+0011 implies that the exponent is
+3. As another example, (0.000111)2 will be written as .111e−0011, with .111 being the mantissa
and e−0011 implying an exponent of −3. Also, (−0.00000101)2 may be written as −.101 × 2−5 =
−.101e−0101, where −.101 is the mantissa and e−0101 indicates an exponent of −5. If we wanted
to represent the mantissas using eight bits, then .1101011 and .111 would be represented as .11010110
and .11100000.

The Four Axioms

                 Conversion of a given number in one number system to its equivalent in another system has been discussed at length in the preceding sections. The methodology has been illustrated with solved examples. The complete methodology can be summarized as four axioms or principles, which, if understood properly, would make it possible to solve any problem related to conversion of a given number in one number system to its equivalent in another number system.
These principles are as follows:
1. Whenever it is desired to find the decimal equivalent of a given number in another number system,
it is given by the sum of all the digits multiplied by their weights or place values. The integer and
fractional parts should be handled separately. Starting from the radix point, the weights of different
digits are r0, r1, r2 for the integer part and r−1, r−2, r−3 for the fractional part, where r is the radix
of the number system whose decimal equivalent needs to be determined.
2. To convert a given mixed decimal number into an equivalent in another number system, the integer
part is progressively divided by r and the remainders noted until the result of division yields a
zero quotient. The remainders written in reverse order constitute the equivalent. r is the radix of
the transformed number system. The fractional part is progressively multiplied by r and the carry
recorded until the result of multiplication yields a zero or when the desired number of bits has been
obtained. The carrys written in forward order constitute the equivalent of the fractional part.
3. The octal–binary conversion and the reverse process are straightforward. For octal–binary
conversion, replace each digit in the octal number with its three-bit binary equivalent. For
hexadecimal–binary conversion, replace each hex digit with its four-bit binary equivalent. For
binary–octal conversion, split the binary number into groups of three bits, starting from the binary
point, and, if needed, complete the outside groups by adding 0s, and then write the octal equivalent
of these three-bit groups. For binary–hex conversion, split the binary number into groups of four
bits, starting from the binary point, and, if needed, complete the outside groups by adding 0s, and
then write the hex equivalent of the four-bit groups.
4. For octal–hexadecimal conversion, we can go from the given octal number to its binary equivalent
and then from the binary equivalent to its hex counterpart. For hexadecimal–octal conversion, we
can go from the hex to its binary equivalent and then from the binary number to its octal equivalent. 
Example 1.9
Assume an arbitrary number system having a radix of 5 and 0, 1, 2, L and M as its independent digits.
Determine:
(a) the decimal equivalent of (12LM.L1);
(b) the total number of possible four-digit combinations in this arbitrary number system.
Solution
(a) The decimal equivalent of (12LM) is given by
M×50 +L×51+2×52 +1×53 = 4×50 +3×51+2×52 +1×53L = 3M= 4
= 4+15+50+125 = 194
The decimal equivalent of (L1) is given by
L×5−1+1×5−2 = 3×5−1+5−2 = 064
Combining the results, (12LM.L1)5
= (194.64)10.
(b) The total number of possible four-digit combinations = 54 = 625.
Example 1.10
The 7’s complement of a certain octal number is 5264. Determine the binary and hexadecimal
equivalents of that octal number.
Solution
• The 7’s complement = 5264.
• Therefore, the octal number = (2513)8.
• The binary equivalent = (010 101 001 011)2
= (10101001011)2.
• Also, (10101001011)2
= (101 0100 1011)2
= (0101 0100 1011)2
= (54B)16.
• Therefore, the hex equivalent of (2513)8
= (54B)16 and the binary equivalent of (2513)8
=(10101001011)2.

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