Therefore the filter response in time should be 2009 1 st Midterm A

Therefore the filter response in time should be 2009

This preview shows page 37 - 40 out of 62 pages.

Therefore, the filter response in time should be like 63. 02.11.2009 1 st Midterm A periodic signal ) sin( 2 ) 6 cos( ) ( t t t x is at the input of the ideal band-pass filter whose magnitude response is given below. What would be the output power spectral density? Solution Since the power spectrum of the signal is given as
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151227621 DIGITAL COMMUNICATIONS 38 and 2 ) ( ) ( ) ( f H f G f G X Y the component at f =0.5 is filtered out. Only the component at f =3 remains and it is amplified. 64. 10.12.2009 2 nd Midterm Generator polynomial of a (7,4) systematic cyclic encoder is given as 1 ) ( 3 p p p g . The information word that is coded and transmitted is 1010. Since the syndrome vector is found to be 110, what is the received codeword r ? Solution The systematic codeword polynomial is ) ( ) ( ) ( p rm p X p p c k n where ) ( p rm is the reminder of the division ) ( / ) ( p g p X p k n . Thus, ) 1 /( ) ( Re ) ( 3 3 3 p p p p p m p rm and 1 ) ( p p rm . The codeword polynomial is then 1 ) ( 4 6 p p p p c corresponding to 1010011. Since we have a nonzero syndrome vector, we know that the received word has an error (assumed single bit error). As we do not know which syndrome vector corresponds to which single bit error yet, we need to try out for each possible single bit error until ) ( p s is found. ) 101 ( 1 ) ( / Re 2 6 p p g p m ) 111 ( 1 ) ( / Re 2 5 p p p g p m ) 110 ( ) ( / Re 2 4 p p p g p m this is the syndrome vector we have. Therefore, we know that the fourth bit is in error, which makes the received word (1000011). We can verify this by ) 110 ( ) ( / ) 1 ( Re 2 4 6 p p p g p p p m verified… 65. 10.12.2009 2 nd Midterm A hexadecimal sequence with LRC and parity checking (even parity) is given as {88, 93, 9E, 93, A9, 81, 99, B6} where LSB of each byte are parity bit for the 7-bit word and the last byte is BCC. What would be the erroneous bytes in the sequence, if there were any? Solution After marking the parity bits that are incorrect, we note that they are cause by one data and one BCC bits as marked on the figure. That is, 9E was 8E and B6 was B7 before the errors happened. It is also possible that the third parity bit and fourth bit of the BCC were erroneous, causing the same parity check errors.
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151227621 DIGITAL COMMUNICATIONS 39 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 1 1 0 0 1 1 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 1 0 1 1 1 0 0 0 1 0 1 0 1 1 1 1 0 66. 10.12.2009 2 nd Midterm A convolutional encoder is defined with the output equations of 2 2 i i i x x O and 2 1 1 2 i i i i x x x O . What is the code sequence transmitted when the information bit sequence is 111010 ? Solution Assuming the previous state of the encoder was 00, the encoder outputs would be O 2i = 110100<10> and O 2i+1 = 101001<10>, where the last two bits of each sequence occur if the encoder is fed by 00 after transmitting the input sequence. The output stream would then be O = 111001100001<1100> 67.
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  • Fall '18
  • Mr. Bhullar
  • Hamming Code, Error detection and correction, Parity bit

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