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Express the following decimal numbers in binary form and in binary-coded decimal form: a.* \(9.75\); b. \(6.5\); c. \(11.75\); d. \(63.03125\); e. \(67.375\).

Short Answer

Expert verified
9.75 is (1001.11)_2 and (1001 0111.0101)_{BCD}. 6.5 is (110.1)_2 and (0110.0101)_{BCD}. 11.75 is (1011.11)_2 and (0001 0001.0111 0101)_{BCD}. 63.03125 is (111111.00001)_2 and (0110 0011.0000 0011 0001 0010 0101)_{BCD}. 67.375 is (1000011.011)_2 and (0110 0111.0011 0111 0101)_{BCD}.

Step by step solution

01

Converting 9.75 to Binary

To convert 9.75 to binary, separate the integer (9) and fractional parts (0.75). First, convert 9 to binary: - 9 ÷ 2 = 4 remainder 1 - 4 ÷ 2 = 2 remainder 0 - 2 ÷ 2 = 1 remainder 0 - 1 ÷ 2 = 0 remainder 1 Reading remainders from bottom to top, 9 converted to binary is (1001)_2. Next, convert 0.75: - 0.75 × 2 = 1.5 (integer part = 1) - 0.5 × 2 = 1.0 (integer part = 1) Combining both results, 9.75 in binary is (1001.11)_2.
02

Binary-Coded Decimal Form of 9.75

Binary-coded decimal (BCD) represents each decimal digit individually using 4 bits: - 9 = (1001) - 7 = (0111) - 5 = (0101) Therefore, 9.75 in BCD is (1001 0111.0101)_{BCD}.
03

Converting 6.5 to Binary

Separate 6 (integer) and 0.5 (fractional). Convert 6: - 6 ÷ 2 = 3 remainder 0 - 3 ÷ 2 = 1 remainder 1 - 1 ÷ 2 = 0 remainder 1 Reading remainders from bottom to top, 6 in binary is (110)_2. Convert 0.5: - 0.5 × 2 = 1.0 (integer part = 1) Thus, 6.5 in binary is (110.1)_2.
04

Binary-Coded Decimal Form of 6.5

In BCD, each decimal digit is individually converted: - 6 = (0110) - 5 = (0101) Therefore, 6.5 in BCD is (0110.0101)_{BCD}.
05

Converting 11.75 to Binary

Separate 11 (integer) and 0.75 (fractional). Convert 11: - 11 ÷ 2 = 5 remainder 1 - 5 ÷ 2 = 2 remainder 1 - 2 ÷ 2 = 1 remainder 0 - 1 ÷ 2 = 0 remainder 1 Thus, 11 is (1011)_2. Convert 0.75 exactly the same as in Step 1, resulting in (0.11)_2. Combining both parts, 11.75 in binary is (1011.11)_2.
06

Binary-Coded Decimal Form of 11.75

In BCD, convert each decimal digit: - 1 = (0001) - 1 = (0001) - 7 = (0111) - 5 = (0101) Therefore, 11.75 in BCD is (0001 0001.0111 0101)_{BCD}.
07

Converting 63.03125 to Binary

Separate 63 and 0.03125. Convert 63 to binary: - 63 ÷ 2 = 31 remainder 1 - 31 ÷ 2 = 15 remainder 1 - 15 ÷ 2 = 7 remainder 1 - 7 ÷ 2 = 3 remainder 1 - 3 ÷ 2 = 1 remainder 1 - 1 ÷ 2 = 0 remainder 1 So, the binary of 63 is (111111)_2. Convert 0.03125: - 0.03125 × 2 = 0.0625 - 0.0625 × 2 = 0.125 - 0.125 × 2 = 0.25 - 0.25 × 2 = 0.5 - 0.5 × 2 = 1.0 (stops) Combine to get 0.03125 as (0.00001)_2. Thus, 63.03125 in binary is (111111.00001)_2.
08

Binary-Coded Decimal Form of 63.03125

Convert each digit in BCD: - 6 = (0110) - 3 = (0011) - 0 = (0000) - 3 = (0011) - 1 = (0001) - 2 = (0010) - 5 = (0101) Thus, 63.03125 in BCD is (0110 0011.0000 0011 0001 0010 0101)_{BCD}.
09

Converting 67.375 to Binary

Separate 67 and 0.375. Convert 67: - 67 ÷ 2 = 33 remainder 1 - 33 ÷ 2 = 16 remainder 1 - 16 ÷ 2 = 8 remainder 0 - 8 ÷ 2 = 4 remainder 0 - 4 ÷ 2 = 2 remainder 0 - 2 ÷ 2 = 1 remainder 0 - 1 ÷ 2 = 0 remainder 1 Thus, the binary of 67 is (1000011)_2. For 0.375, similar conversion gives (0.011)_2 from known steps of multipliers. Hence, 67.375 in binary is (1000011.011)_2.
10

Binary-Coded Decimal Form of 67.375

Convert each digit to BCD: - 6 = (0110) - 7 = (0111) - 3 = (0011) - 7 = (0111) - 5 = (0101) Thus, 67.375 in BCD is (0110 0111.0011 0111 0101)_{BCD}.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binary-Coded Decimal (BCD)
Binary-Coded Decimal (BCD) is a way to represent decimal digits using binary codes but with a twist. Every decimal digit is turned into its own four-bit binary equivalent. This method serves a useful purpose in various electronic devices such as digital clocks, calculators, and other electronic gadgets that display numeric information.

The key advantage of BCD is its simplicity when dealing with decimal numbers. This simplicity can reduce errors in conversion and is often easier for humans to comprehend when reading output directly. However, BCD isn't as condensed or efficient as pure binary for data storage. For instance, BCD for the decimal number 9.75 would be represented as:
  • The number "9" becomes 1001
  • The number "7" becomes 0111
  • The number "5" becomes 0101
Overall, BCD can be viewed more like a bridge between the decimal and binary worlds, making life easier when shifts between base systems are needed.
Decimal to Binary Conversion
Converting a decimal number into its binary equivalent is a fundamental skill in binary arithmetic and digital electronics. The process might seem tricky at first, but once you understand the steps, it becomes fairly straightforward.

Let's break it down:
  • Start by dividing the integer part of the decimal number by 2.
  • Record the remainder and divide the quotient obtained from the previous step by 2 again.
  • Repeat this process until the quotient reaches zero, and then read the binary digits from bottom to top to construct the binary result.
If you're dealing with fractions, there's a slightly different method:
  • Multiply the fraction by 2.
  • Note the integer part obtained and proceed further with the new fractional part.
  • Continue until the fractional part becomes zero (or as close to zero as needed).
For example, converting the decimal 9.75 involves these basic steps where the whole part, 9, becomes 1001 in binary, and the fractional part, 0.75, gives you 0.11 when converted to binary. Combine these two results, and the complete binary conversion of 9.75 is 1001.11.
This two-fold approach of dealing with integers and fractions separately simplifies the conversion process, ensuring that even complex decimal values can be accurately represented as binary.
Educational Problem Solving in Electrical Engineering
Problem-solving in electrical engineering education often hinges on understanding concepts like binary conversion. While it sounds technical, grasping these small segments of knowledge makes larger concepts much less daunting.

It’s not just about finding the answer but understanding the process:
  • Identifying the problem and knowing what you’re trying to convert (e.g., from decimal to binary).
  • Breaking the problem into manageable steps, such as handling the integer and fractional parts separately.
  • Practicing these conversions regularly to strengthen understanding.
This stepwise, methodical approach is embedded deeply in engineering education. It aids in cultivating analytical thinking and logical reasoning in students.

By actively engaging with these types of exercises, students develop a robust toolkit for tackling not only mathematical problems but also more complex engineering challenges. This iterative style of learning, with an emphasis on clear and practical solutions, is foundational to problem-solving across all areas of engineering, encouraging students to see each challenge as an opportunity to grow in their field.

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Most popular questions from this chapter

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Design a logic circuit to control electrical power to the engine ignition of a speed boat. Logic output \(I\) is to become high if ignition power is to be applied and is to remain low otherwise. Gasoline fumes in the engine compartment present a serious hazard of explosion. A sensor provides a logic input \(F\) that is high if fumes are present. Ignition power should not be applied if fumes are present. To help prevent accidents, ignition power should not be applied while the outdrive is in gear. Logic signal \(G\) is high if the outdrive is in gear and is low otherwise. A blower is provided to clear fumes from the engine compartment and is to be operated for five minutes before applying ignition power. Logic signal \(B\) becomes high after the blower has been in operation for five minis provided so that the operator can choose to apply ignition power even if the blower has not operated for five minutes and if the outdrive is in gear, but not if gasoline fumes are present. a. Prepare a truth table listing all combinations of the input signals \(B, E, F\), and \(G .\) Also, show the desired value of \(I\) for each row in the table. b. Using the sum-of-products approach, write a Boolean expression for \(I\). c. Using the product-of-sums approach, write a Boolean expression for \(I\). d. Try to manipulate the expressions of parts (b) and (c) to obtain a logic circuit having the least number of gates and inverters. Use AND gates, OR gates, and inverters.

Verify the following Boolean expressions with the help of truth tables: a. \({ }^{*} E=A \bar{B}+A B C+C \bar{D}\) b. \(D=\overline{A C}+A \bar{B} C\) c. \(Z=\bar{W} Y+\overline{(\bar{X}+Y+W)}\) d. \(D=\overline{A C} B+\bar{B}\) e. \(D=\overline{(\bar{C}+A B)}\)

Describe how to test whether overflow or underflow has occurred in adding signed two's-complement numbers.

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