第五章作业
1.which type of memory is volatile?
A.ROM
B. E2PROM
C. RAM
D. flash memory
2.which type of memory has 6-transistor structure?
A. DRAM
B. SRAM
C. ROM
D. EPROM
ing hamming code, its purpose is of one-bit error.
A. detecting and correcting
B. detecting
C. correcting
D. none of all
4.Flash memory is .
A. read-only memory
B. read-mostly memory
C. read-write memory
D. volatile
5.Which answer about internal memory is not true?
A. RAM can be accessed at any time, but data would be lost when power down..
B. When accessing RAM, access time is non-relation with storage location.
C. In internal memory, data can’t be modified.
D. Each addressable location has a unique address.
Page161 Problems: 5.4 5.5 5.6 5.7 5.8
参考答案：CBABC
5.4. word stored as 0111
word fetched as 0011 “异或”
syndrome word：0100 =>syndrome word contains one and only one “1”.
5.5. Answer: (1) 2k-1≥m+k =>k=4-bit
1
1
1
C1(1,2,4,5,7)=0⊕1⊕0⊕0⊕1=0
C2(1,3,4,6,7)=0⊕0⊕0⊕0⊕1=1
C4(2,3,4,8) =1⊕0⊕0⊕1 =0
C8(5,6,7,8) =0⊕0⊕1⊕1 = 0=> old check bits：C8C4C2C1=0010
5.6. Answer:
(1) 2k-1≥m+k =>k=4-bit
(2)Set up a table (同上题)
(3)Syndrome word: 0111⊕1101=1010
=>D6 is wrong, so the data from memory is 00011001.
5.7. 2k-1≥m+k , m=1024-bit =>k=11-bit
5.8.Answer: (1) 2k-1≥m+k , m=16-bit =>k=5-bit
1
1
1
1
1
1
(3)Calculates old check bits:
C1(1,2,4,5,7,9,11,12,14,16) =1
C2(1,3,4,6,7,10,11,13,14)=0
C4(2,3,4,8,9,10,11,15,16) =0
C8(5,6,7,8,9,10,11) =0
C16(12,13,14,15,16) =0
=> old check bits：C16C8C4C2C1=00001
(4) Suppose data bit D5 is changed from 1 to 0 in storage. 0101000000101001.When fetches the data , Re-calculates new check bit :
C1’(1,2,4,5,7,9,11,12,14,16) =0
C2’(1,3,4,6,7,10,11,13,14)=0
C4’(2,3,4,8,9,10,11,15,16) =0
C8’(5,6,7,8,9,10,11) =1
C16’(12,13,14,15,16) =0 => old check bits：C16’C8’C4’C2’C1’=01000
(5)Makes syndrome:
C16C8C4C2C1=00001old
C16’C8’C4’C2’C1’ =01000 new
=> syndrome word =01001
It means position 9 is error, D5 is wrong, data bit D5 is changed from 0 to 1.。