Information Technology Reference
In-Depth Information
The remainder is thus 101, so
R
(
x
) is
x
2
+ 1. The transmitted polynomial will be
T
(
x
)=
x
3
M
(
x
)+
R
(
x
)=
x
8
+
x
7
+
x
5
+
x
3
+
x
2
+ 1 (110101101)
To check this, use modulo-2 division to give
x
5
+1
x
3
+
x
2
+
1
x
8
+
x
7
+
x
5
+
x
3
x
8
+
x
7
+
x
5
x
3
x
3
+
x
2
+
1
x
2
+
1
Remainder
This gives the same answer as the state table, i.e.
x
2
+1.
Prove that the transmitted message does not generate a remainder when divided by
P
(
x
).
The transmitted polynomial,
T
(
x
), is
x
8
+
x
7
+
x
5
+
x
3
+
x
2
+ 1 (110101101) and the generator
polynomial,
G
(
x
), is 1+
x
2
+
x
3
. Thus
x
5
+1
x
3
+
x
2
+
1
x
8
+
x
7
+
x
5
+
x
3
+
x
2
+
1
x
8
+
x
7
+
x
5
x
3
+
x
2
+
1
x
3
+
x
2
+
1
0
Remainder
As there is a zero remainder, there is no error.
D.2 Longitudinal/vertical redundancy checks (LRC/VRC)
RS-232 uses vertical redundancy checks (VRC) when it adds a parity bit to the transmitted
character. Longitudinal (or horizontal) redundancy checks (LRC) adds a parity bit for all bits
in the message at the same bit position. Vertical coding operates on a single character and is
known as character error coding. Horizontal checks operate on groups of characters and de-
scribed as message coding. LRC always uses even parity and the parity bit for the LRC char-
acter has the parity of the VRC code.
In the example given next, the character sent for LRC is thus
10101000
(28h) or a '
(
'.
The message sent is '
F
', '
r
', '
e
', '
d
', '
d
', '
y
' and '
(
'.
Without VR checking, LR checking detects most errors but does not detect errors where
an even number of characters have an error in the same bit position. In the previous example
if bit 2 of the '
F
' and '
r
' were in error then LRC would be valid.
