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Exercise Solutions for ELEC 2200 Section 2 Spring 2013 - Conversions, Assignments of Electronics

AMET UniversityElectronics

The solutions to exercise #1 of elec 2200 section 2 spring 2013, which involves performing various binary conversions. The conversions include converting binary numbers to different bases, such as binary to decimal and decimal to binary.

Typology: Assignments

2019/2020

Uploaded on 12/27/2020

anh-khoa
anh-khoa 🇬🇧

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ELEC 2200 Section 2 Spring 2013
Exercise #1
Perform the following conversions:
1. 4310 = ?2
2. 104310 = ?2
3. 110110102 = ?10
4. 011011012 = ?10
5. 110110102 = ?8
6. 01111101012 = ?8
7. 11111010110011102 = ?16
8. 1001000112 = ?16
9. 1238 = ?2
10. ECE16 = ?8
pf2

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Download Exercise Solutions for ELEC 2200 Section 2 Spring 2013 - Conversions and more Assignments Electronics in PDF only on Docsity!

Perform the following conversions:

Exercise #1 Solutions

  • ELEC 2200 Section 2 Spring
    • Exercise #
    1. 43 10 =?
    1. 1043 10 =?
    1. 11011010 2 =?
    1. 01101101 2 =?
    1. 11011010 2 =?
    1. 0111110101 2 =?
    1. 1111101011001110 2 =?
    1. 100100011 2 =?
    1. 123 8 =?
    1. ECE 16 =? - ELEC 2200 Section 2 Spring
    1. 43 10 =? Perform the following conversions:
    • 21/2 = 10 remainder = 43/2 = 21 remainder = 1 (LSB)
    • 10/2 = 5 remainder =
    • 5/2 = 2 remainder =
    • 2/2 = 1 remainder =
    • 1/2 = 0 remainder = 1 (MSB) therefore, 43 10 =
    1. 1043 10 =?
    • 521/2 = 260 remainder = 1043/2 = 521 remainder = 1 (LSB)
    • 260/2 = 130 remainder =
    • 130/2 = 65 remainder =
    • 65/2 = 32 remainder =
    • 32/2 = 16 remainder =
    • 16/2 = 8 remainder =
    • 8/2 = 4 remainder =
    • 4/2 = 2 remainder =
    • 2/2 = 1 remainder =
    • 1/2 = 0 remainder = 1 (MSB) therefore, 1043 10 =
    1. 11011010 2 =?
    • 1×2 7 +1×2 6 +0×2 5 +1×2 4 +1×2 3 +0×2 2 +1×2 1 +0×2 0 = 128 + 64 + 16 + 8 + 2 =
    1. 01101101 2 =?
    • 1×2 6 + 1×2 5 + 1×2 3 + 1×2 2 + 1×2 0 = 64 + 32 + 8 + 4 + 1 =
    1. 11011010 2 =?
      • 3 3 2 8 therefore, 11011010 2 =
    1. 0111110101 2 =? - 7 6 5 8 therefore, 0111110101 2 =
    1. 1111101011001110 2 =? - F A C E 16 therefore, 111110101100 2 = FACE
    1. 100100011 2 =?
      • 1 2 3 16 therefore, 100100011 2 =
    1. 123 8 =?
      • 1 2 3
    • 001 010 011 2 therefore, 123 8 =
    1. ECE 16 =? - E C E
      • 7 3 1 6 8 therefore, ECE 16 = 111 011 001 110 2 (rearranging into groups of 3)