1. Convert the following:
Base 2/8/16 Decimal
10111101
2
111001111
2
377
8
756
8
D8
16
3EC
16
Decimal Binary Octal Hexidecimal
199
10
335
10
2. Do the following binary operations (unsigned numbers):
1 0 1 0
+ 0 1 1 0
1 0 1 0
- 0 1 1 0
1 0 1 1 0 0 1 1
AND 0 1 1 1 0 1 0 1
1 0 1 1 0 0 1 1
XOR 0 1 1 1 0 1 0 1
Hex value after rotating A4 left 3 bits?
Hex value after arithmetic shifting
3. Fill in the following ASCII text chart:
B2 right 2 bits?
Text Binary ASCII Octal ASCII Hex ASCII
M+5
4. Convert the following decimal numbers to signed 8-bit numbers:
Decimal Numbers: -68
130 117 122 58 4F 52
-91
Sign/magnitude
1's complement
2's complement

5. What range of values can be represented using a: a) 6-bit unsigned number to __________________ b) 6-bit sign/magnitude number to _________________ c) 6-bit 2's complement number to __________________
6. Perform the indicated operations on these 2's complement signed 6-bit values. Check your answers by converting to decimal. Circle unrepresentable values.
0 0 1 0 1 0
+0 0 1 1 0 1
0 1 1 1 1 0
-0 0 1 0 1 1
1 1 0 0 1 0
+1 0 0 1 0 0
1 1 1 1 1 0
-1 0 1 1 0 0
1 0 1 1 0 1
+0 1 0 1 1 1
1 1 1 0 0 1
-0 1 0 0 1 1
7. Convert the following numbers:
Decimal Number Floating Point Representation
44.1875
10
-319.5625
10
10
0 10000111 11101011000000000000000
10
1 01111101 11000000000000000000000

10111101
2
111001111
2
377
8
756
8
D8
16
3EC
16
189
10
463
10
255
10
494
10
216
10
1004
10
Decimal Binary Octal Hexidecimal
199
10
11000111 307
335
10
101001111 517
2. 1 0 1 0
+ 0 1 1 0
1 0 0 0 0
1 0 1 1 0 0 1 1
AND 0 1 1 1 0 1 0 1
0 0 1 1 0 0 0 1
Hex value after rotating A4 left 3 bits?
A4 = 10100100
Rotate
 00100101
Convert back to hex = 25
1 0 1 0
- 0 1 1 0
0 1 0 0
1 0 1 1 0 0 1 1
XOR 0 1 1 1 0 1 0 1
1 1 0 0 0 1 1 0
Hex value after arithmetic shifting B2 right 2 bits?
B2 = 10110010
ASH
 11101100
Convert back to hex = EC
3.
C7
14F
Text
M+5
XOR
4.
Decimal Numbers:
Sign/magnitude
Binary ASCII Octal ASCII Hex ASCII
1001101 0101011 0110101 115 53 65 4D 2B 35
1011000 1001111 1010010 130 117 122 58 4F 52
-68
11000100
-91
11011011

1's complement
10111011
2's complement
10111100
5. Range for 6-bit unsigned: 0 to 63
Range for 6-bit sign/magnitude: -31 to 31
Range for 6-bit 2's complement: -32 to 31
6.
0 0 1 0 1 0 (+10)
+0 0 1 1 0 1 (+13)
0 1 0 1 1 1 (+23)
0 1 1 1 1 0 (+30)
-0 0 1 0 1 1 (+11)
0 1 0 0 1 1 (+19)
1 1 0 0 1 0 (-14)
+1 0 0 1 0 0 (-28)
(-42)
(-42) unrepresentable
1 1 1 1 1 0 ( -2)
-1 0 1 1 0 0 (-20)
0 1 0 0 1 0 (+18)
7.
44.1875
10
10100100
10100101
1 0 1 1 0 1 (-19)
+0 1 0 1 1 1 (+23)
0 0 0 1 0 0 ( +4)
1 1 1 0 0 1 ( -7)
-0 1 0 0 1 1 (+19)
1 0 0 1 1 0 (-26)
101100.0011 ==> 0 10000100 01100001100000000000000
-319.5625
10
-100111111.1001 ==> 1 10000111 00111111100100000000000
491
10
0 10000111 11101011000000000000000 ==> 1.11101011 x 2
8
-0.4375
10
1 01111101 11000000000000000000000 ==> -1.11 x 2
-2
8.
4.625 + 15.5625 = 20.1875
4.625 = 100.101
15.5625 = 1111.1001
Normalize:
1.00101 x 2 2
1.1111001 x 2 3
Align decimal points:
0.100101 x 2 3
1.1111001 x 2 3
Add:
0.1001010
1.1111001
--------------
10.1000011 x 2 3
De-normalize:
10100.0011
Convert back to decimal:
20.1875

9. Multiply 32.5 x 6.25 = 203.125
32.5 = 100000.1
6.25 = 110.01
Normalize:
1.000001 x 2 5
1.1001 x 2 2
Add exponents:
5 + 2 = 7
Multiply significands:
1.000001 x 1.1001 = 1.1001011001
Combine Results:
1.1001011001 x 2 7
Remove exponent:
11001011.001
Convert to decimal
203.125
10. Booth's Algorthim: 12 x -12
Multiplier = 12 (decimal) = 01100 (5-bit binary)
Multiplicand = -12 (decimal) = 10100 (5-bit binary 2’s complement)
Initial Product = Multiplier w/five leading zeros (for 5-bit operands)
00000 01100
(1a) Always use 0 as your initial previous LSB (pLSB). That means
the LSB & previous LSB above are 00, so there is no arithmetic
operation (no-op).
ASR
00000 00110 0
(1b)
(2a)
(2b)
(3a)
LSB and pLSB above are again 00 - no-op
ASR
00000 00011 0
LSB and previous LSB above are 10 - so subtract the multiplicand from left half:
00000 - 10100 = 01100 (with phantom borrow)
Product is now: 01100 00011
(3b) ASR
00110 00001 1

(4a)
(4b)
(5a)
LSB and previous LSB above are 11 - no-op
ASR
00011 00000 1
LSB and pLSB above are 01 - add the multiplicand to left half
00011 + 10100 = 10111
Product is now: 10111 00000
(5b) ASR
11011 10000
We have completed 5 passes, so answer is 1101110000.
1101110000 is a negative 2’s complement number, equivalent to -144, which is the correct product of 12 x -12.