Professor De Boer's list of
TEXTBOOK ERRATA
(last update 12/05/2019)
Mano, Kime, and Martin, Logic and Computer Design Fundamentals,
Fifth Edition, Pearson, 2016.
ISBN-13: 978-0-13-376063-7 or ISBN-10: 0-13-376063-4
(link to errata for the 4th edition)
(link to errata for the 3td edition)
If you are considering purchasing this textbook and worrying that it is a poor choice due to the length of this list of errata, please reconsider. Competing textbooks have about as many errata, but perhaps no list like this. "Better the devil you know than the devil you don't." Professor De Boer likes this book enough to find it worthwhile to publish this list of errata.
List of Errata:
Page 34, Problem 1-10.
The answers given online at the publisher's web site have errors of precision. Correct answers are:
a.) 16612.346 8
b.) 792.41D16
c.) 1010 1111.0010 1100 112
In part (a) the original number, 7562.45 base ten, has significance down to the 1/100’s place. It would be desirable to convert the number maintaining at least that much accuracy, but not signifying much more. In octal one can round to the 1/8’s place, 1/64’s place, 1/512 place, 1/4096 place etc. Rounding to the 1/512’s place seems an appropriate choice compared to the original 1/100’s place accuracy. That will preserve all the given significant digits and not signify too much additional precision. Since 512 = 83 this means I should round to three digits to the right of the radix point. The textbook’s answer rounds to the nearest 1/4096’s place. The original number does not have that much place significance. Thus 16612.346 octal is a more reasonable answer.Page 35, Problem 1-17
The rationale for parts (b) and (c) is the same, only the bases are changed. (Posted 6/10/2019, updated 8/29/2019)
Change "1480" to "1460"
(Posted 6/08/2018)
Page 36, Problem 1-24, second line
In the sixth group of digits, change "110001" to "1100001"
(Posted 6/08/2018)
Page 104, Problem 2-12.
The answer given online at the publisher's web site has an error in it. The error is in the answer for the p.o.s. form in part (b). The correct answer is
(the same as the s.o.p. form.)(Posted 6/10/2019)
Page 106, Problem 2-25.
The answer for part (c) given online at the publisher's web site uses non-standard notation. Basically, it is wrong. Part (c) has two possible correct answers, only one of which needs to be presented.
One possible correct answer is
The other possible correct answer is
(Posted 9/16/2019)
Pages 127-128, Example 3-5
In the definition of variables and again in Figure 3.11 change IG to IS.
(Posted 6/10/2019)
Page 190, Problem 3-31
Change "4-to-6 line decoder" to "3-to-6 line decoder"
(Posted 6/08/2018)
Page 193, Problem 3-51.
In the problem statement, change the word "unsigned" to "signed." Note that when you "obtain" the complement you negate the number.
In the solution posted on the web for this problem, again change the word "unsigned" to "signed."
(The problem statement as given is ambiguous. "Obtain" could be construed to mean converting unsigned numbers into a signed representation by adding a leading zero as a positive sign. The other way to construe the problem statement is to negate the given number by finding the complement as described in the text on pages 157 and 158. This was the author's intent. For students this is an unexpected maneuver. They wonder how one can negate an unsigned number, or why one would do this—although there are valid reasons. Simply stating that the given numbers are already signed clears up the confusion.)Page 243, Last three lines on the page.
(Posted 6/10/2019)
On the third line from the bottom of the page,
change the first instance of "Fill_2" to "Fill_2(t + 1)"
On the next line, change the first instance of "Fill_3" to "Fill_3(t + 1)"
On the next line, change "Mix" to "Mix(t + 1)"
(Posted 6/10/2019)
Page 286, Problem 4-18, part (c)
Change "for the odd parity generator" to "for the serial 2s complementer"
(Posted 6/08/2018)
Page 319, Problem 5-5
Change "32x8" to "64x8"
(Posted 6/08/2018)
Page 319, Problem 5-7.
Change "8 x 3" to "8 x 4".
(Posted 6/10/2019)
Page 391, Problem 6-3.
This problem is poorly worded. The problem statement below is a better expression of the authors' intent. Work this version of the problem instead.
6-3. Given any 16-bit operand, what operation must be performed
with what mask word to. . .
(a) clear all odd bit positions to logic-0 but leave all even
bit positions the same as in the given operand? (Assume
bit positions are 15 down to 0 from left to right.)
(b) set the rightmost 4 bits to logic-1 but leaving all other bit
positions as in the given operand?
(c) complement the most significant 8 bits but leaving all other
bit positions as in the given operand?
For each case above, demonstrate how the operation and mask
word you specify would achieve the goal using the
operand word 10101100 010100112.
(Posted 11/14/2019, updated 11/21/2019)
Page 432, Problem 7-7.
Add this sentence to the problem statement after the first sentence:
"A single refresh operation takes 60 ns on this DRAM."
Page 432, Problem 7-8.
Add a phrase to the start of part (b) so that it reads:
"(b) Given that the memory system is organized into 16 bit words, how many address lines. . . "
Page 481, Problem 8-15
In the table that accompanies part (a) the header row includes "BA" twice, once in column 4 and again in column 5. Change the instance of "BA" that is in column 5 to "MB." Also add one more colunm, which will be column 12, and enter "BC" into the header row of that new column. (Posted 12/05/2018)
Page 483, Problem 8-25
Append the following to the end of the problem statement:
"Find the state machine diagram." (Posted 6/08/2018)
Page 484, Problem 8-27
Add the word, "in" to the last line of the problem statement so it reads:
". . . store the minimum value in address 2 of the data memory."
(Posted 6/08/2018)
Page 514, Table 9-7.
No errata, however Table 9-7 has little content.
Here is a supplement to Table 9-7.
(posted 6/10/2019)
Pages 587-589, "Hard Drive" section.
No errata, however hard drives with rotating parts and "cylinder, head, sector" (CHS) addressing are obsolete for new products.
The coverage here is also very simplified (which is appropriate for this context). For more detailed information about hard drives with rotating parts one may read about zoned recording and logical block addressing. Hard drives without rotating parts are now common. You can read about solid-state hard drives. (posted 6/10/2019)
Page 616, Problem 11-1.
The unit prefixes in the solution posted on the Web do not follow industry practice. The hard disk drive industry usually publishes capacities using decimal-based definitions of kilo, mega, giga, tera, etc. (More on binary-based prefixes here and here.) In the solutions below prefixes "k," "M," or "G," are decimal-based and "Ki," "Mi," or "Gi" are binary based.
Solutions illustrating various formats
a.) (1023 cylinders)(1 head)(63 sectors/track)(512 bytes/sector)
= 32 997 888 bytes (exactly)
= 33 MB (approximately)
The line above is the number typically advertised and the preferred answer.
= 32 224.5 KiB (exactly)
= 31.5 MiB (approximately)
b.) (8191 cylinders)(4 heads)(63 sectors/track)(512 bytes/sector)
= 1 056 835 584 bytes (exactly)
= 1 GB (approximately)
The line above is the number typically advertised and the preferred answer.
= 1 032 066 KiB (exactly)
= 1008 MiB (approximately)
= 0.984 GiB (approximately)
c.) (16 383 cylinders)(16 heads)(63 sectors/track)(512 bytes/sector)
= 8 455 200 768 bytes (exactly)
= 8.45 GB (approximately)
The line above is the number typically advertised and the preferred answer.
= 8 257 032 KiB (exactly)
= 8 063 MiB (exactly)
= 7.87 GiB (exactly)
(posted 6/10/2019)
Disclaimer: This list of errata is provided by Professor De Boer for the use of his students in his courses. This list is offered as is, with no guarantee of any kind. It is likely to be incomplete at the least.