Assignment 2. Tyler Shendruk February 19, 2010

Transcription

1 Assgnment yler Shendruk February 9, 00 Kadar Ch. Problem 8 We have an N N symmetrc matrx, M. he symmetry means M M and we ll say the elements of the matrx are m j. he elements are pulled from a probablty densty { /a for a < m j < a pm j ) ) 0 otherwse.. Problem 8 a) he characterstc functon s just the fourer transform of the probablty densty so for each element pk) pm j )e kmj dm j a a 0 + a a dm a a e kmj j a a k e kmj a [ e ka e ka] ak pm j )e kmj dm j + a 0 pk) snka) ka ). Problem 8 b) o fnd the probablty of gettng a specfc trace, we could use the defnton p ) d N {m })p{m }) N dm pm )

2 then take p exp km ) but that s a lot of work. Instead we should recognze p ) N p k) N snka) ka Kadar Ch. Problem 0 hs problem s a practce n changng varables. he current s a functon of voltage by [ ] N snka) p ) 3) ka IV ) [exp ev/k B ) ] 4) he nstantaneous voltage s drawn from a gaussan dstrbuton of zero mean and varance σ. We call ths probablty densty p V.. Problem 0 a) What s the probablty densty of current, p I? It may not be obvous how the probablty denstes are related but t may be clearer to you how the probabltes are. If there s some probablty P of gettng some current V what s the probablty of gettng the correspondng current from Eq. 4. It must be the same, of course. herefore, usng the defnton of probablty from probablty densty, we see P p V dv p I di p I p V dv di. 5) We want to gve p I as a functon of current not voltage so for future use we rearrange Eq. 4 to be V I) k [ ] B I + I0 ln. 6) e

3 We are now ready to fnd the probablty densty for the current: dv p I p V di [ ]) d kb I + I0 p V ln di e k B p V V ) e I + k B e I + I exp V )) 0 πσ σ k B e I + I exp kb 0 πσ σ e [ ]) )) I + I0 ln. Problem 0 b) What s the mean value for the current? he average current s the current that corresponds to the average voltage. We don t want to deal wth p I snce t s so ugly. < I > Ip I di ) dv I p V di di IV )p V dv [exp ev/k B ) ] p V dv exp ev/k B )p V dv p V dv }{{} πσ πσ exp ev/k B ) exp V πσ σ I 0 π πσ β exp I 0 πσ exp πσ e exp k B V ) σ V dv exp αv βv ) dv ) α 4β σ ) ) e 4 k B )) dv ) ] eσ < I > [exp kb 7) 3

4 .3 Most Probable Current he most probable current s the current at the maxmum of the probablty densty so dp I di 0 d di k B e k B e πσ I + [ d di exp πσ σ I + exp σ [ ]) ))) kb I + I0 ln e [ ]) )] kb I + I0 ln e Droppng the constants and settng A kb ) σ e gves [ 0 d [ ]) )] I + I0 exp A ln di I + [ ) ] [ ]) ) I + I0 I + I0 I + ) A ln exp A ln Drop the exponent and the nconsequental fracton to see ) I + I0 0 A ln ) I + A ln I0 whch after all that work s I + exp I [exp I [exp A ) A 3 Sethna Ch. 5 Problem { ) ] ) } ] eσ. 8) k B We have a chan made up of n + steps to the rght and n steps to the left. hat means we have a total of N n + + n steps and f each lnk n the chan has length d then the chan s length s L n + n ) d. 3. Problem a) We can rearrange the number of steps and the length to gve n + and n n terms of the other varables n + N + L 9) n N L. 0) 4

5 he total number of confguratons s Ω N! n +!n! N! N + ) L! N ) L. )! he entropy s just the logarthm of the number of confguratons S band k B ln Ω ) N! k B ln N + ) L! N ) L! N S band k B [ln N! ln + L ) N! ln L ) ]!. ) Of course, for ths to be useful we wll probably have to use Strlng s formula but ths s the exact answer. 3. Problem b) We know F de bath 3) dl and we also know S bath E. 4) Puttng the two together results n F S bath. 5) he above only dealt wth the entropy of the bath. But what s the change n length dong? It s maxmzng the total/unversal entropy. So f S S bath + S band then 0 S S bath + S band F + S band S band F 6) 5

6 3.3 Problem c) o fnd the force n terms of the number of monomers, we just combne the results of the last two secton and use Strlng s formula: F S band F S band [ N k B ln N! ln + L ) N! ln L ) ]! k B [ N + L ) N ln + L ) N + + L ) [ k B N ln + L ) + + N ln L [ k B N ln + L ) + N ln L )] [ k N B ln + )] L N L F k ) B Nd + L ln Nd L N ) L ) + ] N ln L ) N + L )] 7) o see the sprng constant, we rewrte the term nsde the logarthm and then expand t assume L Nd.e. the nstantaneous length L s far less than t s contour/full length N d). F k B ln + L ) Nd k B LNd ) +... k B Nd L. So we dentfy the sprng constant to be 3.4 Problem d) K k B Nd. 8) What happens f we heat the elastc band whle t s under tenson? he change n length wth temperature s F F F. 9) L he term F L s a Maxwell equaton: F S L F 6

7 whch we found earler. And the other term s ether by defnton f you remember these knds of thngs) or from the last part related to the sprng constant F K Now Eq. 9 becomes F F F K K L. Snce both length and temperature must be postve, the rubber band under tenson contracts when heated. KL 4 Sethna Ch. 5 Problem 5 F L he A bç! language s made up of three letters/sounds hey have probable Sounds hoot A slap B clck C occurances of p A p B /4 and p C /. 4. Problem 5 a) What s the Shannon entropy per letter transmtted or Shannon entropy rate)? he entropy per sound s S k s p) ln p) A bç! [ ) k s 4 ln + ) 4 4 ln + )] 4 ln [ k s ln ln ] ln k s 3 ln 4. Problem 5 b) Show that a communcaton transmttng bts can transmt no more than one unt of Shannon entropy per bt. 7

8 We are lookng for the maxmum Shannon entropy per bt: p j S SH 0 he Shannon entropy wll depend on the probablty dstrbuton whch as always) wll be subject to the constrant p. Maxmze S SH k s p ln p p log p for unts of bts, as n the other queston. o mplement the Lagrange multpler method we defne [ ] S S SH + λ p [ ] p log p + λ p And we say that we are gong to have a message that s N letters long and snce the message s beng transmtted n bts we ll say N n where n s the number of bts used to transmt the message of length N. Anyway, maxmze S for all varables: S [p log p j p p ] λp p j j ln p ln ln + λ 0 ln p ln ln + λ 0 ln p + + λ ln 0 ln p λ ln p exp λ ln ) But exp λ ln ) s a constant. For now let s say exp λ ln ) p. We ddn t maxmze by λ yet: λ S p 0 Whch s just the orgnal condton. But now we know that p p so we can 8

9 say p p p pn p N hs was very expected. Usng ths maxmzng probablty dstrbuton we fnd the Shannon entropy of on transmsson: S k s p ln p ln ) N ln N N ln N ln N N ln N ln N ln N ln ln n ln n ln ln n So the maxmum Shannon entropy of a bnary message of length N n where n the number of bts used, s equal to n. hat s to say that the maxmum s one unt of Shannon entropy per bt. 4.3 Problem 5 d) Fnd the compresson scheme that saturates to S n. 3 From Part a) S k s ln and snce we wll be compressng the language nto bts we set k s / ln whch leads to S 3 hs says that the average number of bts for each letter s 3/ or.5. 9

10 C occurs wth the greatest probablty so let s nsst that t s a sngle null bt: C 0 Now the trck to ths s that me must be able to dstngush any letter from any combnaton of letters. For example, say C 0, B and A 0. hs s not and acceptble compresson snce there s no dstngushng BC from A. We could expect ths snce the number of bts per letter would have been An acceptble compresson would be Notce A could not just be. C 0 B A 0 Now the average number of bts s , as desred. 0