Question 1 – Number Systems
a.
student number is s3482454,then
245410 = 1001 1001 01102
245410 = 46268
245410 = 99616
b.
245410 = 116A16
c.
first name is “CHUN HEI” and surname is “LI” ,then
CHU26 = (2) 10,(7) 10 , (20) 10 ,
LII26 = (11) 10, (8) 10 ,(8)10
add up CHU26 and LII26 ,sum in base 26 is
CHU26 + LII26 = (13) 10, (15) 10 ,(28)10 = NQD26
Question 2 – Binary Addition and SubTraction
student number is s3482454,then
A = 4 , B = 5
a.
A = 01002
B = 01012
–A = 1100
–B = 1011
b.
A +Ｂ＝01002 + 01012 ＝10012
the answer is valid to 4-bit arithmetic
c.
-A complement representation is 1111 11002
-B complement representation is 1111 10112
-A-B= (-A) + (-B) = 1111 11002 + 1111 10112 = 111101112 = (–9) 10
Question 3 – Bitwise Operations
a.
A OR 010000012
b.
( A AND 001111002 ) OR A
Question 4 – Logic Circcuits and Truth Tables
a.
(a)
(b)
b.
(a)
|
A |
B |
C |
Y |
|
0 |
0 |
0 |
0 |
|
0 |
0 |
1 |
0 |
|
0 |
1 |
0 |
0 |
|
0 |
1 |
1 |
0 |
|
1 |
0 |
0 |
0 |
|
1 |
0 |
1 |
0 |
|
1 |
1 |
0 |
1 |
|
1 |
1 |
1 |
1 |
(b)
|
A |
B |
C |
Y |
|
0 |
0 |
0 |
0 |
|
0 |
0 |
1 |
0 |
|
0 |
1 |
0 |
0 |
|
0 |
1 |
1 |
0 |
|
1 |
0 |
0 |
0 |
|
1 |
0 |
1 |
0 |
|
1 |
1 |
0 |
1 |
|
1 |
1 |
1 |
0 |
Question 5 – SECDED code
a.
there an error in transmission. Because of the check code error.
b.
|
bit position |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
info bit |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
|
check code |
R1 |
R2 |
R3 |
R1 = BIT1 ⊕ BIT3 ⊕ BIT5 ⊕ BIT7 = 1
R2 = BIT2 ⊕ BIT3 ⊕ BIT6 = 1
R3 = BIT4 ⊕ BIT5 ⊕ BIT6 ⊕ BIT7 = 0
R1R2R3 = 1102 = 610 ,the BIT6 is error , The correct ASCII is B
Question 6
a.
CSMA/CD working principle :
Before send data, listening channel is idle or not. If it’s idle,sending data immediately.If the channel is busy,wait for a period of time until the transmission of information in the channel is completed and the send the data. If two or more nodes send a request at the same time after the last piece of information is sent, it is determined as a conflict. If it detects a collision, it immediately stops sending data, waits for a random period of time, and tries again.
b.
When a data collision occurs, the delay value is mainly determined by the backoff algorithm.
Binary exponential backoff can achieve better decomposition results. In the case of a shared channel, each node conducts a random delay t when the collision occurs(0<t<T),t follows the exponential distribution with a base of two (0~T). Backoff algorithms include: non-persistence, 1-adherence, P-persistence.
c.
Some adjustments have been made to CSMA/CD in 802.11, using the new protocol CSMA/CA (Carrier Sense Multiple Access with Collision Avoidance) or DCF (Distributed Coordination Function). The CSMA/CA uses the ACK signal to avoid collisions, that is, only when the client receives the ACK signal returned on the network, it confirms that the sent data has correctly reached the destination address.
CSMA/CD:
Carriers with collision detection listen for multiple accesses and can detect collisions but cannot "avoid"
CSMA/CA:
Carriers with collision avoidance monitor multiple accesses, and when packets are sent, they cannot detect whether there is collision on the channel. They can only try to avoid them.
Major difference:
1. The transmission medium of the two is different, CSMA/CD is used for bus Ethernet, and CSMA/CA is used for wireless LAN 802.11a/b/g/n, etc.;
2. The detection method is different. CSMA/CD detects the change of voltage in the cable. When the data collide, the voltage in the cable will change. The CSMA/CA uses energy detection (ED) and carrier detection (CS). And energy carriers are mixed to detect the idleness of the three detection channels;
3. In a WLAN, for a node, the signal strength it has just sent out is much higher than the signal strength from other nodes, which means that its own signal will cover other signals.
4. Conflicts at this node do not imply a conflict at the receiving node.
Question 7
a.
|
Description |
Sign |
Exponent |
Mantissa |
|
Bit position |
15 |
14-11 |
10-0 |
Data storage is divided into three parts, which are: sign、exponent and mantissa
Sign: 0 represents a positive number, 1 represents a negative number
Exponent: Used to store index data in scientific notation
Mantissa: Use shift store mantissa part
b.
1 = 1 * 20
|
Description |
Sign |
Exponent |
Mantissa |
|
Bit position |
1 |
0000 |
00000000000 |
1 = 1000000000000000
10 = 10102 = 1.010 * 23
|
Description |
Sign |
Exponent |
Mantissa |
|
Bit position |
1 |
0011 |
01000000000 |
10 = 1001101000000000
1/3 = 0.01012 = 1.01 * 2-2
|
Description |
Sign |
Exponent |
Mantissa |
|
Bit position |
1 |
1010 |
01000000000 |
1/3 = 1101001000000000
c.
student number is s3482454,then
A = 4 , B = 5
x = A + B/10 = 4 + 5/10 = 4.5
y = B + A/10 = 5 + 4/10 = 5.4
( i )
x = 0100.01012 = 1.000101 * 22
|
Description |
Sign |
Exponent |
Mantissa |
|
Bit position |
1 |
0010 |
00010100000 |
x = 1001000010100000
y = 0101.01002 = 1.010100 * 22
|
Description |
Sign |
Exponent |
Mantissa |
|
Bit position |
1 |
0010 |
01010000000 |
y = 1001001010000000
( ii )
x + y = 1001000010100000 + 1001001010000000
= 1001001100100000
= 10. 011001 * 22
= 1001.10012
= 1*23 + 1*20 + 1*2-1 + 1*2-4
d.
For a certain set of data, if we use different combinations,The accumulated sum is not the same, and the deviation is not the same. Through analysis, we can draw.The deviation of adding a floating number with a number closer to its exponent is smaller, and the deviation when it is added to a floating point whose index is farther. For a floating-point number, in the computer, we store it in three parts. When we add two floating-point numbers, It will go through five steps, zero-operand checking, ranking, summation, normalization, and rounding. Both rounding and normalization have round-off deviation. For two floating-point numbers, it is certain that the number of roundings with a smaller difference in order is less, and the deviation is relatively small.
Therefore, if you want to reduce the deviation, you should make the index levels almost as much as possible and sort the data according to the size of the index.
Eg:
a = 2101(0.11010101) , b = 2111(0.10100010) , c = 2001(0.10011010), a+b+c=?
(1) ((a+b)+c)
= （ 2101(0.11010101) + 2111(0.10100010) ）+ 2001(0.10011010)
= 108.510
(2) ((a+c)+b)
= （ 2101(0.11010101) + 2001(0.10011010) ） + 2111(0.10100010)
= 10910
The correct value is: 108.828125
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