... Digital communications I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 6 Lecture 6 10 The ... += |)(||)(| fHfH RC = 0=r 5. 0 =r 1=r 1=r 5. 0=r 0=r )()( thth RC = T2 1 T4 3 T 1 T4 3− T2 1− T 1− 1 0 .5 0 1 0 .5 0 T T2 T3 T− T2− T3− s RrW )1( Passband DSB += Lecture 6 12 Pulse shaping and equalization ... with ISI: Binary-PAM, SRRQ pulse Non-ideal c...
Ngày tải lên: 18/10/2013, 06:15
... iEE sii s s ski Ed ki Ed 2 2 min , = ≠ = Digital Communications I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 5 Lecture 5 20 Eb/No figure of merit in digital communications SNR ... symbol. b bb R W N S WN ST N E == / 0 b R W : Bit rate : Bandwidth Lecture 5 21 Example of Symbol error prob. For PAM signals )( 1 t ψ 0 1 s 2 s b E b E− Binary...
Ngày tải lên: 27/01/2014, 08:20
Modulation and coding course- lecture 10
... Digital Communications I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 10 Lecture 10 10 Effective code rate Initialize the memory before encoding the ... vector for each modulo-2 adder). The i:th element in each vector, is “1” if the i:th stage in the shift register is connected to the corresponding modulo- 2 adder, and “0” otherwise. Example: m...
Ngày tải lên: 25/10/2013, 06:15
Modulation and coding course- lecture 9
... 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 vs. bandwidth (dB) ... jnj n tj M pp j n P − += − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ≤ ∑ )1( 1 jnj n tj B pp j n j n P − += − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ≈ ∑ )1( 1 1 p Digital Communications I: Modulation and Coding Course Period 3 -...
Ngày tải lên: 25/10/2013, 06:15
Modulation and coding course- lecture 12
... decoding )101( ˆ =m )1110001011( ˆ =U )101(=m )1110001011(=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 3 t 4 t 5 t -5/ 3 0 -5/ 3 -5/ 3 10/3 1/3 14/3 2 8/3 10/3 13/33 1/3 5/ 35/ 3 ( ) ii ttS ),(Γ Branch metric Partial ... from the received sequence Digital Communications I: Modulation and...
Ngày tải lên: 29/10/2013, 13:15
Modulation and coding course- lecture 11
... decoding )101( ˆ =m )1110001011( ˆ =U )101(=m )1110001011(=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 3 t 4 t 5 t -5/ 3 0 -5/ 3 -5/ 3 10/3 1/3 14/3 2 8/3 10/3 13/33 1/3 5/ 35/ 3 ( ) ii ttS ),(Γ Branch metric Partial ... decoding-cont’d i=4 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...
Ngày tải lên: 29/10/2013, 13:15
Modulation and coding course- lecture 1
... issues in designing a digital communication system (DCS): Utilized techniques Formatting and source coding Modulation (Baseband and bandpass signaling) Channel coding Equalization Synchronization ... bandwidth! Baseband signal Bandpass signal Local oscillator Lecture 1 25 Bandwidth of signal … Different definition of bandwidth: a) Half-power bandwidth b) Nois...
Ngày tải lên: 08/11/2013, 18:15
Modulation and coding course- lecture 2
... Digital Communications I: Modulation and Coding Course Period 3 – 200/ Catharina Logothetis Lecture 2 Lecture 2 10 Encoding (PCM) A uniform linear quantizer is called Pulse Code Modulation ... levels and Lecture 2 11 Quantization example t Ts: sampling time x(nTs): sampled values xq(nTs): quantized values boundaries Quant. levels 111 3.1867 110 2.2762 101 1.3 657 100 0. 455...
Ngày tải lên: 08/11/2013, 18:15
Tài liệu Modulation and coding course- lecture 3 pptx
... Detect Lecture 3 13 Baseband and bandpass Bandpass model of detection process is equivalent to baseband model because: The received bandpass waveform is first transformed to a baseband waveform. ... with a digital system Sampling Aliasing Quantization Uniform and non-uniform Baseband modulation Binary pulse modulation M-ary pulse modulation M-PAM (M-ay Pul...
Ngày tải lên: 27/01/2014, 08:20
Tài liệu Modulation and coding course- lecture 4 pdf
... A 0 1 s 2 s T A T A− 0 0 Tt )( 1 t ψ T 1 0 0)()()()( )()()(),( /)(/)()( )( 122 0 1212 1111 0 2 2 11 =−−= −=>=< == == ∫ ∫ tAtstd Adtttstts AtsEtst AdttsE T T ψ ψψ ψ 1 2 Digital Communications I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 4 Lecture 4 20 Implementation ... filter to maximize SNR Matched filter receiver and Correlator receiver Le...
Ngày tải lên: 27/01/2014, 08:20