Q.8. (a): For (6,3) systematic linear block code, the three parity-check bits C4, C5 and C6 are formed from the following equation C4 = d1 ⊕ d3 ; C5 = d1 ⊕ d2 ⊕ d3; C6 = d1 ⊕ d2
(i) Write the generator matrix G.
(ii) Construct all possible code words.
(iii) If the received work is 010111 find the location of the error and the transmitted data bits.
Solution:
(i) The Generator matrix G as defined below -
| 1 0 0 1 1 1 |
G = | 0 1 0 0 1 1 |
| 0 0 1 1 1 0 |
Also,
| 1 1 1 |
| 0 1 1 |
H' = | 1 1 0 |
| 1 0 0 |
| 0 1 0 |
| 0 0 1 |
(ii) Now, all possible code-words are -
C = dG
here,
d G C
| 1 1 1 | | 1 1 1 0 1 0 |
| 1 1 0 | | 1 1 0 1 0 0 |
| 1 0 1 | | 1 0 0 1 1 1 | | 1 0 1 0 0 1 |
| 1 0 0 | | 0 1 0 0 1 1 | = | 1 0 0 1 1 1 |
| 0 1 1 | | 0 0 1 1 1 0 | | 0 1 1 1 0 1 |
| 0 1 0 | | 0 1 0 0 1 1 |
| 0 0 1 | | 0 0 1 1 1 0 |
| 0 0 0 | | 0 0 0 0 0 0 |
(iii) to find the location of error, we need to calculate decoding table as S = eH'
S e
| 1 1 1 | | 1 0 0 0 0 0 |
| 1 1 0 | | 0 1 0 0 0 0 |
| 1 0 1 | | 0 0 1 0 0 0 |
| 1 0 0 | | 0 0 0 1 0 0 |
| 0 1 1 | | 0 0 0 0 1 0 |
| 0 1 0 | | 0 0 0 0 0 1 |
| 0 0 0 0 1 1 |
here, the received work is 010111, hence to decode this code, we will use s = rH'
| 1 1 1 |
| 1 1 0 |
s = rH' = | 0 1 0 1 1 1| | 1 0 1 | = | 1 0 0 |
| 1 0 0 |
| 0 1 1 |
| 0 1 0 |
we can tabulate all the above codes in following fashion -
r | s | e | C | d
0 1 0 1 1 1 | 1 0 0 | 0 0 0 1 0 0 | 0 1 0 0 1 1 | 0 1 0
Q.8. (b): Calculate the ratio of circular waveguide cross-sectional area to the rectangular waveguide cross-sectional area assuming that both these waveguides have equal cutoff frequency for the dominant mode, if P11 = 1.841
Solution: For the dominant (TE11) mode in the circular waveguide, we have
The area of a circle with radius r is given by -
In the rectangular waveguide, for the TE10 mode -
If the two cut off wavelengths are to be the same, then
The area of a standard rectangular waveguide is -
Hence, the ratio of circular waveguide cross-sectional area to the rectangular waveguide cross-sectional area will be -
(i) Write the generator matrix G.
(ii) Construct all possible code words.
(iii) If the received work is 010111 find the location of the error and the transmitted data bits.
Solution:
(i) The Generator matrix G as defined below -
| 1 0 0 1 1 1 |
G = | 0 1 0 0 1 1 |
| 0 0 1 1 1 0 |
Also,
| 1 1 1 |
| 0 1 1 |
H' = | 1 1 0 |
| 1 0 0 |
| 0 1 0 |
| 0 0 1 |
(ii) Now, all possible code-words are -
C = dG
here,
d G C
| 1 1 1 | | 1 1 1 0 1 0 |
| 1 1 0 | | 1 1 0 1 0 0 |
| 1 0 1 | | 1 0 0 1 1 1 | | 1 0 1 0 0 1 |
| 1 0 0 | | 0 1 0 0 1 1 | = | 1 0 0 1 1 1 |
| 0 1 1 | | 0 0 1 1 1 0 | | 0 1 1 1 0 1 |
| 0 1 0 | | 0 1 0 0 1 1 |
| 0 0 1 | | 0 0 1 1 1 0 |
| 0 0 0 | | 0 0 0 0 0 0 |
(iii) to find the location of error, we need to calculate decoding table as S = eH'
S e
| 1 1 1 | | 1 0 0 0 0 0 |
| 1 1 0 | | 0 1 0 0 0 0 |
| 1 0 1 | | 0 0 1 0 0 0 |
| 1 0 0 | | 0 0 0 1 0 0 |
| 0 1 1 | | 0 0 0 0 1 0 |
| 0 1 0 | | 0 0 0 0 0 1 |
| 0 0 0 0 1 1 |
here, the received work is 010111, hence to decode this code, we will use s = rH'
| 1 1 1 |
| 1 1 0 |
s = rH' = | 0 1 0 1 1 1| | 1 0 1 | = | 1 0 0 |
| 1 0 0 |
| 0 1 1 |
| 0 1 0 |
we can tabulate all the above codes in following fashion -
r | s | e | C | d
0 1 0 1 1 1 | 1 0 0 | 0 0 0 1 0 0 | 0 1 0 0 1 1 | 0 1 0
Q.8. (b): Calculate the ratio of circular waveguide cross-sectional area to the rectangular waveguide cross-sectional area assuming that both these waveguides have equal cutoff frequency for the dominant mode, if P11 = 1.841
Solution: For the dominant (TE11) mode in the circular waveguide, we have
λo = 2πr / (Kr) = 2πr / (P11) = 2πr /1.841 = 3.41r
The area of a circle with radius r is given by -
Ac = πr.r
In the rectangular waveguide, for the TE10 mode -
λo = 2a
If the two cut off wavelengths are to be the same, then
2a = 3.41 r
=> a = 1.705 r
The area of a standard rectangular waveguide is -
Ar = ab = a.a/2 = squr (1.705 r) / 2 = 1.45.r.r
Hence, the ratio of circular waveguide cross-sectional area to the rectangular waveguide cross-sectional area will be -
Ac/Ar = πr.r/1.45.r.r = 2.17
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