vlsi designcourse lecture notes uyemura textbook

... Response CMOS VLSI Design 4th Ed Pass Transistor Ckts VDD VDD VDD VDD VDD VDD Vs = VDD-Vtn VDD-Vtn VDD-Vtn VDD VDD-Vtn VDD-Vtn Vs = |Vtp| VDD VDD-2Vtn VSS 5: DC and Transient Response CMOS VLSI Design ... Transient Response CMOS VLSI Design 4th Ed Current vs Vout, Vin Vin0 Vin1 Vin4 Vin2 Vin3 Vin3 Vin4 Idsn, |Idsp| Vin5 Vin2 Vin1 Vout 5: DC and Transient Response CMOS VLSI Design 4th Ed VDD 10 ... Response CMOS VLSI Design 4th Ed 26 Inverter Delay Estimate  Estimate the delay of a fanout-of-1 inverter 2C R A Y 2C 2C 2C 2C Y R C R C C C C d = 6RC 5: DC and Transient Response CMOS VLSI Design

Ngày tải lên: 19/03/2014, 10:20

... Lecture 6: Power Outline  Power and Energy  Dynamic Power  Static Power 7: Power CMOS VLSI Design 4th Ed Power and Energy  Power is drawn ... Dissipated as heat in the nMOS transistor 7: Power CMOS VLSI Design 4th Ed Switching Waveforms  Example: VDD = 1.0 V, CL = 150 fF, f = GHz 7: Power CMOS VLSI Design 4th Ed Switching Power T Pswitching ... Gate leakage – Junction leakage – Contention current 7: Power CMOS VLSI Design 4th Ed 10 Switching Probability 7: Power CMOS VLSI Design 4th Ed 15 Example  A 4-input AND is built out of two levels

Ngày tải lên: 19/03/2014, 10:20

... µE lat CMOS VLSI DesignCMOS VLSI Design 4th Ed. 4: Nonideal Transistor Theory 7 Coffee Cart Analogy  Tired student runs from VLSI lab to coffee cart  Freshmen are pouring out of the physics lecture ... region CMOS VLSI DesignCMOS VLSI Design 4th Ed. 4: Nonideal Transistor Theory 8 Mobility Degradation  High E vert effectively reduces mobility – Collisions with oxide interface CMOS VLSI DesignCMOS VLSI ... Lecture 4: Nonideal Transistor Theory CMOS VLSI DesignCMOS VLSI Design 4th Ed. 4: Nonideal Transistor Theory 2 Outline  Nonideal

Ngày tải lên: 19/03/2014, 10:20

... Digital Communications I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 10 Lecture 10 2 Last time, we talked about:  Channel coding  Linear block codes  The error ... word) Lecture 10 9 A Rate ½ Convolutional encoder Encoder )101(=m )1110001011(=U 0 0 1 5 t 1 u 2 u 11 21 uu 0 0 0 6 t 1 u 2 u 00 21 uu Time Output Time Output (Branch word) (Branch word) Lecture ... 0 /10 0/01 S3 = 11 1 /10 ti ti +1 Lecture 10 Time 18 Trellis –cont’d... adder, and “0” otherwise Example: g1 = (111) g 2 = (101 ) u1 m u1 u 2 u2 Lecture 10 11 Encoder representation

Ngày tải lên: 25/10/2013, 06:15

... performances Lecture Today, we are going to talk about: „ Channel coding „ Linear block codes „ „ „ „ The error detection and correction capability Encoding and decoding Hamming codes Cyclic codes Lecture ... Discrete input, continuous output Lecture Linear block codes „ Let us review some basic definitions first which are useful in understanding Linear block codes Lecture 9 Some definitions „ Binary ... Bandpass modulate Channel Digital modulation Digital demodulation Format Source decode Channel decode Lecture Detect Demod Sample What is channel coding? „ Channel coding: Transforming signals to improve

Ngày tải lên: 25/10/2013, 06:15

... LECTURE NOTES ON ELECTRICAL & ELECTRONIC TECHNIQUE file:///D:/ TCBinh_Laptop_Y0809/_COURSEs_Y0809S1/1_Electrica LECTURE NOTES ON ELECTRICAL & ELECTRONIC ... Phong ωt ωt 1999 ĐHBK This lectures notes was composed by Nguyễn Như Phong in 1999 with the of file:///D:/ TCBinh_Laptop_Y0809/_COURSEs_Y0809S1/1_Electrica This lectures notes was composed ... amplifiers Sine-wave Oscillators Pulse Circuits Digital Circuits 10 Power Supply Circuits About the Lecture notes of 9/23/2008 11:53 PM Mạch Điện 1/20 KTĐ & ĐT CHƯƠNG MẠCH ĐIỆN • • • • • Khái niệm

Ngày tải lên: 25/10/2013, 13:15

... Digital Communications I: Modulation and Coding Course Period - 2007 Catharina Logothetis Lecture 12 Last time, we talked about: How the decoding is performed for Convolutional codes? „ ... decoder? „ What are the soft decisions and hard decisions? „ How does the Viterbi algorithm work? „ Lecture 12 Trellis of an example ½ Conv code Tail bits Input bits 1 0 00 10 11 0/00 0/00 0/00 0/11 ... Output bits 11 10 0/00 0/00 1/11 1/11 1/11 0/10 0/11 1/00 1/01 1/01 0/10 0/10 0/01 t1 t t 1/01 Lecture 12 0/01 t t t Block diagram of the DCS Information source Rate 1/n Conv encoder Modulator

Ngày tải lên: 29/10/2013, 13:15

... Logothetis Lecture 11 Last time, we talked about: Another class of linear codes, known as Convolutional codes We studied the structure of the encoder and different ways for representing it Lecture ... up in a state Lecture 11 State diagram – cont’d 0/00 Input Output (Branch word) Current state 0/11 S0 00 S2 S1 10 01 S1 01 0/01 S2 10 1/11 S0 00 1/00 0/10 1/01 S3 11 S3 11 1/10 Lecture 11 input ... 0/01 t4 Lecture 11 0/10 t5 t6 20 Example of Hard decision Viterbi decoding-cont’d Label al the branches with the branch metric (Hamming distance) Γ(S (ti ), ti ) 2 1 1 0 2 1 t1 t2 t3 t4 Lecture

Ngày tải lên: 29/10/2013, 13:15

... from course homepage: „ „ „ „ „ News Lecture slides (.ppt, pdf) Laboratory syllabus (Lab PM) Set of exercises and formulae Old exams Lecture Schedule „ 13 lectures: „ „ 10 tutorials: „ „ week ... communication system Lecture Scope of the course „ Example of a (digital) communication systems: Cellular wireless communication systems BS Base Station (BS) UE UE UE User Equipment (UE) Lecture Scope ... from week to week Week Final written exam on 12th of March 2007 kl 9.00-14.00 Lecture Staff „ Course responsible and lecturer and giving tutorials: „ „ „ „ Catharina Logothetis Office: Hus (våning

Ngày tải lên: 08/11/2013, 18:15

... signal bandwidth Lecture 2 Today, we are going to talk about: „ The first important step in any DCS: „ Transforming the information source to a form compatible with a digital system Lecture Formatting ... Baseband transmission Lecture Sampling Time domain Frequency domain xs (t ) = xδ (t ) × x(t ) X s ( f ) = Xδ ( f ) ∗ X ( f ) x(t ) | X(f )| | Xδ ( f ) | xδ (t ) xs (t ) | Xs( f )| Lecture Aliasing ... capability Interference and noise immunity Implementation cost and complexity Lecture 19 Spectra of PCM waveforms Lecture 20

Ngày tải lên: 08/11/2013, 18:15

... channel and receiver, symbols are “smeared” Lecture Example: Impact of the channel Lecture Example: Channel impact … hc (t ) = δ (t ) − 0.5δ (t − 0.75T ) Lecture 10 Receiver job Demodulation and ... ‘01’ T T -3B Lecture Example of M-ary PAM … Ts 2Ts 2.2762 V Tb 1.3657 V 2Tb 3Tb 4Tb 5Tb 6Tb 1 Rb=1/Tb=3/Ts R=1/T=1/Tb=3/Ts T 2T 3T 4T 5T 6T Rb=1/Tb=3/Ts R=1/T=1/2Tb=3/2Ts=1.5/Ts T 2T 3T Lecture Today ... , M Lecture 21 Implementation example of matched filter receivers s1 (t ) Bank of matched filters A T T z1 (T ) A T t r (t ) T s2 (t ) −A T T t T ⎡z1 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢z2 ⎥ ⎣ ⎦ =z z z (T ) −A T Lecture

Ngày tải lên: 27/01/2014, 08:20

... Digital Communications I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 4 Lecture 4 2 Last time we talked about:  Receiver structure  Impact of AWGN and ISI on the ... ) − z (t ) = (ai1 − z1 ) 2 + (ai 2 − z 2 ) 2 i = 1,2,3 Lecture 4 12 Orthogonal signal space N-dimensional orthogonal... orthonormal Lecture 4 13 Example of an orthonormal bases Example: ... recovered waveform  Detection:  Estimate the transmitted symbol based on the received sample Lecture 4 4 Receiver structure Frequency down-conversion Receiving filter Equalizing filter Threshold

Ngày tải lên: 27/01/2014, 08:20

... = i j =1 where Ek is the energy of sk (t ) Lecture 10 Maximum likelihood detector block diagram 〈⋅,s1 〉 − E1 z Choose the largest ˆ m 〈⋅, s M 〉 − EM Lecture 11 Schematic example of ML decision ... )dz i =1 Z i z Lecture i 14 Example for binary PAM pz (z | m2 ) pz (z | m1 ) s2 − Eb s1 ψ (t ) Eb ⎛ s1 − s / ⎞ ⎟ Pe (m1 ) = Pe (m2 ) = Q⎜ ⎜ N /2 ⎟ ⎝ ⎠ ⎛ Eb PB = PE (2) = Q⎜ ⎜ N ⎝ Lecture ⎞ ⎟ ⎟ ... N / W N Rb Lecture Rb W : Bit rate : Bandwidth 20 Example of Symbol error prob For PAM signals Binary PAM s2 s1 − Eb s4 −6 Eb ψ (t ) Eb 4-ary PAM s3 s2 E −2 b E b s1 Eb ψ1 (t ) T Lecture T t

Ngày tải lên: 27/01/2014, 08:20

... Lecture Notes in Control and Information Sciences Editors: M Thoma · M Morari 316 R.V Patel ž F ... 32, 96 tool orientation control 113 tracking 205 orientation 54, 69, 70 position 53, 69, 70 twist Lecture Notes in Control and Information Sciences Edited by M Thoma and M Morari Further volumes

Ngày tải lên: 05/03/2014, 15:20

... Lecture Notes in Computer Science Edited by G Goos, J Hartmanis and J van Leeuwen Advisory Board: ... o n ; M i l a n ; Paris ; S a n t a C l a r a ; S i n g a p o r e ; T o k y o : Springer, 1995 (Lecture notes in computer science ; 1016) ISBN 3-540-60642-4 NE: GT CR Subject Classification (1991): ... available to a moving observer from spccularitics Image and Vision Computing, 7(1):3842, 1989 Lecture Notes in Computer Science For information about Vols 1-949 please contact your bookseller

Ngày tải lên: 05/03/2014, 15:20

... Biochemical Pharmacology Lecture Notes Michael Palmer, Department of Chemistry, University of Waterloo, Canada Third edition, ... colorful lab coats of all times (he also had one of the most paper-jammed offices ever). His Nobel lecture (available on the web) is an inter- esting read – a mix of brilliant and utterly ‘naive’ideasthat makes

Ngày tải lên: 15/03/2014, 14:20

... signal bandwidth Lecture 2 Today, we are going to talk about:  The first important step in any DCS:  Transforming the information source to a form compatible with a digital system Lecture Formatting ... Baseband transmission Lecture Sampling Time domain xs (t ) = xδ (t ) × x(t ) x(t ) Frequency domain X s ( f ) = Xδ ( f ) ∗ X ( f ) | X( f )| | Xδ ( f ) | xδ (t ) xs (t ) | Xs( f ) | Lecture Aliasing ... capability Interference and noise immunity Implementation cost and complexity Lecture 19 Spectra of PCM waveforms Lecture 20 M-ary pulse modulation  M-ary pulse modulations category:    

Ngày tải lên: 23/03/2014, 10:21

... = i j =1 where Ek is the energy of sk (t ) Lecture 10 Maximum likelihood detector block diagram 〈⋅,s1 〉 − E1 z Choose the largest ˆ m 〈⋅, s M 〉 − EM Lecture 11 Schematic example of the ML decision ... =1 Z i z i Lecture 14 Example for binary PAM pz (z | m2 ) pz (z | m1 ) s2 − Eb s1 ψ (t ) Eb  s1 − s /   Pe (m1 ) = Pe (m2 ) = Q  N /2     Eb PB = PE (2) = Q  N      Lecture 15 ... i≠k Lecture 18 Example of upper bound on av Symbol error prob based on union bound ψ (t ) s i = Ei = Es , i = 1, ,4 d i ,k = Es i≠k Es d = Es s3 − Es s2 d1, d 2,3 s1 d 3, d1, Es ψ (t ) s4 − Es Lecture

Ngày tải lên: 23/03/2014, 10:21

... Digital Communications I: Modulation and Coding Course Period 3 - 2007 Catharina Logothetis Lecture 10 Lecture 10 2 Last time, we talked about:  Channel coding  Linear block codes  The error ... word) Lecture 10 9 A Rate ½ Convolutional encoder Encoder)101(=m )1110001011(=U 0 0 1 5 t 1 u 2 u 11 21 uu 0 0 0 6 t 1 u 2 u 00 21 uu Time Output Time Output (Branch word) (Branch word) Lecture ... 00 1/11 S 2 = 10 0/11 S1 = 01 1/01 S3 = 11 1/00 0 /10 0/01 1 /10 ti ti +1 Lecture 10 Time 18 Trellis... −1 Lecture 10 15 State diagram – cont’d  A state diagram is a way to represent

Ngày tải lên: 30/03/2014, 10:20

... XOR-based FA • HA-based FA • Other FA Circuits – a few others options are covered in the textbook ECE 410, Prof A Mason Lecture Notes 12.4 Full Adder Circuits • AOI Structure FA – implements following ... pi 1 0 1 gi 0 0 – carry-kill is not needed ECE 410, Prof A Mason Lecture Notes 12.19 Static CMOS Manchester Implementation • Textbook Circuit Implementation – ci+1 = if gi=0 – ci+1 = if gi=1 ... an n-sized adder ECE 410, Prof A Mason Lecture Notes 12.23 16b Adder Using 4b CLA Blocks • Create SUMs from outputs of this circuit ECE 410, Prof A Mason Lecture Notes 12.24 Other Adder Implementations...

Ngày tải lên: 28/04/2014, 11:04