Modulation and coding course- lecture 1

... exercises and formulae  Old exams Lecture 1 8 Schedule  13 lectures:  from week 5 to week 8  10 tutorials:  week 5 to week 8  1 mandatory laboratory work:  Week 9  Final written exam on 12 th ... March 2007 kl 9.00 -14 .00. Lecture 1 9 Staff  Course responsible and lecturer and giving tutorials:  Catharina Logothetis  Office: Hus 7 (våning 6), Ångström  Tel.: 01...

Ngày tải lên : 08/11/2013, 18:15

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Modulation and coding course- lecture 10

... 0/00 0 /11 0 /10 0/ 01 1 /11 1/ 01 1/00 0/00 0 /11 0 /10 0/ 01 1 /11 1/ 01 1/00 0/00 0 /11 0 /10 0/ 01 1 /11 1/ 01 1/00 0/00 0 /11 0 /10 0/ 01 1 /11 1/ 01 1/00 0/00 6 t 1 t 2 t 3 t 4 t 5 t 1 0 1 0 0 11 10 00 10 11 Input ...  Example: 11 0 01 010 10 11 100 11 1 011 :sequenceOutput 0 01 :sequenceInput 21 uu Branch word Register contents 11 100 010 11 111 011 1 000...

Ngày tải lên : 25/10/2013, 06:15

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Modulation and coding course- lecture 9

... of subspace a is ) }11 11( ) ,10 10(), 010 1(),0000{( 4 V n V n V ) }11 11( ) ,11 01( ) ,11 00() ,10 11( ) ,10 10() ,10 01( ) ,10 00( ), 011 1(), 010 1(), 010 0(),0 011 (),0 010 (),00 01( ),0000{( 4 =V Lecture 9 14 Some definitions… ... set.  Example: {} .for basis a is )00 01( ),0 010 (), 010 0() ,10 00( 4 V {} . spans )10 01( ),0 011 () ,11 00(), 011 0() ,10 00( 4 V { } n G vvv ,,, 21...

Ngày tải lên : 25/10/2013, 06:15

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Modulation and coding course- lecture 12

... Viterbi decoding )10 1( ˆ =m )11 100 010 11( ˆ =U )10 1(=m )11 100 010 11( =U )1, 3 2 ,1, 3 2 ,1, 3 2 , 3 2 , 3 2 , 3 2 ,1( − − −− =Z 5/3 -5/3 4/3 0 0 1/ 3 1/ 3 -1/ 3 -1/ 3 5/3 -5/3 1/ 3 1/ 3 -1/ 3 6 t 1 t 2 t ... )10 1(=m )11 100 010 11( =U ) 011 011 1 011 (=Z 0 2 0 1 2 1 0 1 1 0 1 2 2 1 0 2 1 1 1 6 t 1 t 2 t 3 t 4 t 5 t 1 0 2 3 0 1 2 3 2 3 20 2 30 ( ) ii ttS ),(Γ B...

Ngày tải lên : 29/10/2013, 13:15

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Modulation and coding course- lecture 11

... 0/00 0 /11 0 /10 0/ 01 1 /11 1/ 01 1/00 0/00 0 /11 0 /10 0/ 01 1 /11 1/ 01 1/00 0/00 0 /11 0 /10 0/ 01 1 /11 1/ 01 1/00 0/00 0 /11 0 /10 0/ 01 1 /11 1/ 01 1/00 0/00 6 t 1 t 2 t 3 t 4 t 5 t 1 0 1 0 0 11 10 00 10 11 Input ... Tail bits Lecture 11 10 Trellis – cont’d 1/ 11 0/00 0 /10 1/ 11 1/ 01 0/00 0 /11 0 /10 0/ 01 1 /11 1/ 01 1/00 0/00 0 /11 0 /10 0/ 01 0/0...

Ngày tải lên : 29/10/2013, 13:15

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Modulation and coding course- lecture 2

... 1. 3657 10 0 0.4552 011 -0.4552 010 -1. 3657 0 01 -2.2762 000 -3 .18 67 PCM codeword 11 0 11 0 11 1 11 0 10 0 010 011 10 0 10 0 011 PCM sequence amplitude x(t) Lecture 2 12 Quantization error  Quantizing ... and Lecture 2 11 Quantization example t Ts: sampling time x(nTs): sampled values xq(nTs): quantized values boundaries Quant. levels 11 1 3 .18 67 11 0 2.2762 10 1 1...

Ngày tải lên : 08/11/2013, 18:15

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Tài liệu Modulation and coding course- lecture 3 pptx

... values boundaries Quant. levels 11 1 3 .18 67 11 0 2.2762 10 1 1. 3657 10 0 0.4552 011 -0.4552 010 -1. 3657 0 01 -2.2762 000 -3 .18 67 PCM codeword 11 0 11 0 11 1 11 0 10 0 010 011 10 0 10 0 011 PCM sequence amplitude x(t) Lecture 3 ... pulse) 11 ’ 22 10 BA = Lecture 3 6 Example of M-ary PAM … 0 Tb 2Tb 3Tb 4Tb 5Tb 6Tb 0 Ts 2Ts 0 T 2T 3T 2.2762 V 1. 3657 V 1 1 0 1...

Ngày tải lên : 27/01/2014, 08:20

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Tài liệu Modulation and coding course- lecture 4 pdf

... space ),()()()( ),()()()( ),()()()( 32 313 23 213 13 22 212 22 212 12 12 111 212 111 1 aatatats aatatats aatatats =⇔+= =⇔+= = ⇔ + = s s s ψψ ψψ ψ ψ )( 1 t ψ )( 2 t ψ ),( 12 111 aa = s ),( 22 212 aa = s ),( 32 313 aa=s Transmitted ... procedure: Tt )( 1 ts Tt )( 2 ts )( )( )()( )()( 21 12 11 AA tAts tAts −== −= = ss ψ ψ )( 1 t ψ -A A 0 1 s 2 s T A T A− 0 0 Tt )( 1 t ψ T 1...

Ngày tải lên : 27/01/2014, 08:20

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Tài liệu Modulation and coding course- lecture 5 doc

... ) ˆ (Pr)( 1 i M i E mmMP ≠= ∑ = )|( 1 1 )( 1 1)( 1 )( 1 11 ∑ ∫ ∑∑ = == −= −== M i Z i M i ic M i ieE i dmp M mP M mP M MP zz z Lecture 5 15 Example for binary PAM )( 1 t ψ b E b E− 0 1 s 2 s )|( 1 mp z z )|( 2 mp ... ss= i s k s ∑ ≠ = ≤ M ik k ikie PmP 1 2 ),()( ss ∑∑ = ≠ = ≤ M i M ik k ikE P M MP 11 2 ),( 1 )( ss Lecture 5 17 Union bound: Example of union boun...

Ngày tải lên : 27/01/2014, 08:20

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Digital Communication I: Modulation and Coding Course-Lecture 2 potx

... 1. 3657 10 0 0.4552 011 -0.4552 010 -1. 3657 0 01 -2.2762 000 -3 .18 67 PCM codeword 11 0 11 0 11 1 11 0 10 0 010 011 10 0 10 0 011 PCM sequence amplitude x(t) Lecture 2 12 Quantization error  Quantizing ... and Lecture 2 11 Quantization example t Ts: sampling time x(nTs): sampled values xq(nTs): quantized values boundaries Quant. levels 11 1 3 .18 67 11 0 2.2762 10...

Ngày tải lên : 23/03/2014, 10:21

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