Modulation and coding course- lecture 2
... Quantization noise variance: 2 Sat 2 Lin 22 2 )()(})]({[ σσσ +==−= ∫ ∞ ∞− dxxpxexqx q E ll L l l qxp q )( 12 2 12/ 0 2 2 Lin ∑ − = = σ Uniform q. 12 2 2 Lin q = σ Lecture 2 14 Uniform and non-uniform quant. ... Digital Communications I: Modulation and Coding Course Period 3 – 20 0/ Catharina Logothetis Lecture 2 Lecture 2 2 Last time, we talked about: Importa...
Ngày tải lên: 08/11/2013, 18:15
... AGC Quantization noise variance: 2 Sat 2 Lin 22 2 )()(})]({[ σσσ +==−= ∫ ∞ ∞− dxxpxexqx q E ll L l l qxp q )( 12 2 12/ 0 2 2 Lin ∑ − = = σ Uniform q. 12 2 2 Lin q = σ Lecture 2 14 Uniform and non-uniform quant. Uniform ... Communications I: Modulation and Coding Course Term 3 – 20 08 Catharina Logothetis Lecture 2 Lecture 2 20 Spectra of PCM waveforms...
Ngày tải lên: 23/03/2014, 10:21
Modulation and coding course- lecture 10
... sequence is found as follows: 22 )2( 2 )2( 1 )2( 02 22) 1( 2 )1( 1 )1( 01 1..)( 1..)( XXgXggX XXXgXggX +=++= ++=++= g g )()( with interlaced )()()( 21 XXXXX gmgmU = Lecture 10 14 Encoder representation ... 0 1 t 1 u 2 u 11 21 uu 0 1 0 2 t 1 u 2 u 01 21 uu 1 0 1 3 t 1 u 2 u 00 21 uu 0 1 0 4 t 1 u 2 u 01 21 uu )101(=m Time Output OutputTime Message sequence: (Branch word)...
Ngày tải lên: 25/10/2013, 06:15
Modulation and coding course- lecture 9
... I: Modulation and Coding Course Period 3 - 20 07 Catharina Logothetis Lecture 9 Lecture 9 2 Last time we talked about: Evaluating the average probability of symbol error for different bandpass ... error detection and correction codes Lecture 9 7 Why using error correction coding? Error performance vs. bandwidth Power vs. bandwidth Data rate vs. bandwidth Capacity v...
Ngày tải lên: 25/10/2013, 06:15
Modulation and coding course- lecture 12
... )0110111011(=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 Branch metric Partial metric Lecture 12 9 Example of soft-decision Viterbi decoding ... denoted by ffree dd or Lecture 12 12 Free distance … 2 0 1 2 1 0 2 1 1 2 1 0 0 2 1 1 0 2 0 6 t 1 t 2 t 3 t 4 t 5 t Hamming weight of the branch Al...
Ngày tải lên: 29/10/2013, 13:15
Modulation and coding course- lecture 11
... Modulation, Demodulation and Coding Course Period 3 - 20 07 Catharina Logothetis Lecture 11 Lecture 11 2 Last time, we talked about: Another class ... Decoding based on soft-bits, is called the “soft-decision decoding”. On AWGN channels, 2 dB and on fading channels 6 dB gain are obtained by using soft-decoding over hard-decoding. Lecture ... sequence and is one of th...
Ngày tải lên: 29/10/2013, 13:15
Modulation and coding course- lecture 1
... Digital Communication I: Modulation and Coding Course Period 3 - 20 07 Catharina Logothetis Lecture 1 Lecture 1 2 Course information Scope of the course Digital ... Hall, 20 02, 2 nd edition, ISBN: 0-13- 095007-6 “Introduction to digital communications”, by Michael B. Pursley, Pearson, Prentice Hall, 20 05, International edition, ISBN: 0-13- 123 3 92- 0 ”Digital ... (...
Ngày tải lên: 08/11/2013, 18:15
... matched filter Tt Tt Tt 02T )()()( thtsty opti ∗ = 2 A )(ts i )(th opt Tt Tt Tt 02T )()()( thtsty opti ∗ = 2 A )(ts i )(th opt T /2 3T /2 T /2 TT /2 2 2 A − T A T A T A T A− T A− T A Lecture 3 18 Properties ... M-ary PAM … 0 Tb 2Tb 3Tb 4Tb 5Tb 6Tb 0 Ts 2Ts 0 T 2T 3T 2. 27 62 V 1.3657 V 1 1 0 1 0 1 0 T 2T 3T 4T 5T 6T Rb=1/Tb=3/Ts R=1/T=1/Tb=3/Ts Rb=1/Tb=3/Ts R=1/T=1/2Tb=3/2Ts=...
Ngày tải lên: 27/01/2014, 08:20
Tài liệu Modulation and coding course- lecture 4 pdf
... Vector to waveform conversion Lecture 4 17 Example of projecting signals to an orthonormal signal space ),()()()( ),()()()( ),()()()( 323 1 323 21313 22 2 122 221 2 12 121 1 121 21111 aatatats aatatats aatatats =⇔+= =⇔+= = ⇔ + = s s s ψψ ψψ ψ ψ )( 1 t ψ )( 2 t ψ ),( 121 11 aa = s ),( 22 2 12 aa = s ),( 323 13 aa=s Transmitted ... ==><= ∫ ∞ ∞− 2 )()(),()( )()( txatax = )()(...
Ngày tải lên: 27/01/2014, 08:20
Tài liệu Modulation and coding course- lecture 5 doc
... bound 1 ψ 1 s 4 s 2 s 3 s 1 Z 4 Z 3 Z 2 Z r 2 ψ 1 ψ 1 s 4 s 2 s 3 s 2 A r 1 ψ 1 s 4 s 2 s 3 s 3 A r 1 ψ 1 s 4 s 2 s 3 s 4 A r 2 ψ 2 ψ 2 ψ ∫ ∪∪ = 4 32 )|()( 11 ZZZ e dmpmP rr r ∫ = 2 )|(),( 1 122 A dmpP rrss r ∫ = 3 )|(),( 11 32 A dmpP ... )|( 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 PA...
Ngày tải lên: 27/01/2014, 08:20