2 1 2 1 cos cos sin sin dt t n m dt t n m dt t n t m T Let us first assume that

# 2 1 2 1 cos cos sin sin dt t n m dt t n m dt t n t m

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2 1 . 0 2 1 0 ) ) cos(( ) ) cos(( ) sin( ) sin( dt t n m dt t n m dt t n t m T Let us first assume that m=kn n n t n k n k t n k n k dt t n k dt t n k / 1 0 / 1 0 . 0 2 1 . 0 2 1 ) ) 1 sin(( ) 1 ( 2 1 ) ) 1 sin(( ) 1 ( 2 1 ) ) 1 cos(( ) ) 1 cos(( 0 0 0 ) ) 1 sin(( ) 1 ( 2 1 ) ) 1 sin(( ) 1 ( 2 1 k n k k n k (for all integer k >1) There is no guarantee that the integral will be zero for all other k ’s. So we conclude that ) sin( t n and ) sin( t m are orthogonal if the second one is a harmonic of the first one. Similar arguments shall be made for all other possible pairings. 72. 02.10.2010 1 st Midterm Show that single bit errors can be corrected in a schema of ( n,k )=(63,57) (with a good selection of generator matrix, of course). Solution The requirement for single bit error correction is that all single bit errors are individually addressed by the syndrome vector which happens to be the same size with parity vector. In our case the number of parity bits is 63-57=6, so is the number of syndrome bits. With 6 bits, we can address 2 6 =64 different error positions. Since our block size is 63, one is reserved for no-error case.
151227621 DIGITAL COMMUNICATIONS 44 73. 14.12.2010 2 nd Midterm A convolutional encoder is defined with the output equations of 2 1 2 i i i x x O and 2 1 1 2 i i i i x x x O . Assuming all initial states are zero, what would the code sequence transmitted be when the information bit sequence is 1011101 ? Solution x i x i-1 x i-2 O 2i O 2i+1 1 0 0 0 1 0 1 0 1 1 1 0 1 1 0 1 1 0 1 0 1 1 1 0 1 0 1 1 0 0 1 0 1 1 0 0 1 0 1 1 0 0 1 1 1 Output stream = 01111010010010 1111 (last 4 bits are optional, just to empty the registers) 74.14.12.2010 2ndMidterm Input to the following trellis is “11000001011100”. What is the decoded sequence? States and the state transition graph are given on the rightmost section. 00 01 10 11 00 11 11 10 01 01 10 00
151227621 DIGITAL COMMUNICATIONS 45 Solution Decoded sequence : 1011000 (solid line=1, dashed=0) 75. 14.12.2010 2nd Midterm What is the main advantage of differential signaling against single ended signaling? Solution Differential signaling alleviates the negative effects that possible ground potential difference between the transmitter and the receiver may cause. Single ended connections would cause jitter under unsteady potential difference. 76. 14.12.2010 2nd Midterm What is the purpose of the 8B10B encoding? Solution Purpose of 8B10B encoding is to keep average signal voltage about zero, and consequently preventing charge buildup on the possible capacitors along the signal path. Another important contribution of 8B10B is to help bit synchronization by continuously changing the signal. 77. 14.12.2010 2nd Midterm What do interleavers do and why? 00 01 10 11 00 11 11 10 01 01 10 00 11 00 00 01 01 11 2 0 0 1 1 2 0 0 2 3 1 1 0 2 1 1 2 2 1 1 1 1 0 0 1 1 1 2 2 1 1 2 1 1 0 2 0 1 1 2 0 1 0 0 1 1 1 6 3 4 4 5 4 5 5 1 3 3 4 4 4 4 4 3 4 3 5 1 3 3 3 3 3 3 3 2 1 4 2 3 4 1 5 2 5 2 2 4 1 1 2 0
151227621 DIGITAL COMMUNICATIONS 46 Solution Interleavers distribute possible multi-bit errors over multiple codewords as single bit errors, so that they are correctable by single bit ECC’s. It is done by constructing new codewords with bits collected from multiple codewords.

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• Fall '18
• Mr. Bhullar
• Hamming Code, Error detection and correction, Parity bit

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