78 14122010 2nd Midterm For 8 PSK determine I Q values Solution Bin I Q Bin I Q

# 78 14122010 2nd midterm for 8 psk determine i q

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78. 14.12.2010 2nd Midterm For 8-PSK determine I-Q values. Solution Bin I Q Bin I Q 000 1 0 100 -1 0 001 0.707 0.707 101 -0.707 -0.707 010 0 1 110 0 -1 011 -0.707 0.707 111 0.707 -0.707 79. 13.01.2011 Final Exam Design an m-sequence generator with [3,1] taps. What is the output of this generator? Assume initial state of FF’s is 111. Solution The m-sequence is then Q 3 =1110100. 80.13.01.2011 Final Exam Describe FHSS in 3-5 sentences. Solution In Frequency Hopping Spread Spectrum RF communication technique, the transmitter frequency is continuously shifted to another pseudo-randomly predetermined value at hop-rates lower than but comparable to bit-rates. A receiver that is aware of this pseudo- Q 1 Q 2 Q 3 Q 1 Q 2 1 1 1 0 0 1 1 1 1 0 1 0 0 1 0 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 1 0
151227621 DIGITAL COMMUNICATIONS 47 random hop sequence can follow the transmitter; otherwise the frequency spectrum of entire transmission seems spread over the available bandwidth. This helps resist interference and fading effects. It also allows existence of multiple transmitters simultaneously in the same band. 81. 13.01.2011 Final Exam Determine the generator matrix for a (7,4) systematic-cyclic code whose generator polynomial is given as 1 ) ( 3 p p p g . Solution Parity is the reminder of the division ) ( / ) ( p g p X p k n for systematic codes. So, 3 0 4 1 5 2 6 3 0 1 2 2 3 3 3 4 7 ) ( ) ( p x p x p x p x x p x p x p x p p X p . Dividing by ) ( p g using long division 1 ) ( ) ( ) ( ) ( ) ( 1 3 3 2 0 2 1 0 2 3 2 1 3 2 0 3 1 2 2 3 3 3 3 0 4 1 5 2 6 3 p p x x x p x x x p x x x x x x p x x p x p x p p p x p x p x p x we see that the reminder consists of 3 2 1 2 x x x p , 2 1 0 1 x x x p and 3 2 0 0 x x x p terms, and they are the parity equations. Since P I G and 0 1 2 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 p p p G , the parity section of the generator matrix are to be calculated by taking 1000, 0100, 0010 and 0001 vectors as inputs to the parity equations. We find 1 1 0 0 1 1 1 1 1 1 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 G . The generator matrix can also be found using a method described as follows. 1. A n k matrix consisting of cyclic shifts of binary string representing the generator polynomial is constructed as 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 Other elements are set to zero, so 1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 1 1 0 1 . 2. Using linear row operations in modulo arithmetic, left k k sub-matrix is made unity matrix. For example if you add bottom two rows onto the first row, first row becomes 1000101 which is the same as the first row of the generator matrix we have already calculated. Continuing such linear row operations, the resulting matrix is the generator matrix itself.
151227621 DIGITAL COMMUNICATIONS 48 82. 13.01.2011 Final Exam The following linear-systematic-cyclic code is given with some codes missing. Determine the missing codes. Indicate how you calculated those codes.

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

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