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Ask seic se MATLAB

                                           HU-370

                                   26 OktwbrÐou 2007


1.     a. Estw p ena dianusma st lh, diastasewn 1 × N . Dhmiourgeiste mia sunatrhsh
          pou ja pairnei san ìrisma to p kai ja epistrefei enan N ×N pÐnaka me stoiqeia
                                          (A)ij = p(|i − j| + 1)
          To +1 qreiˆsthke gia na eÐmaste sumbatoi me thn arijmhsh tou MATLAB.
       b. O pinakac me thn parapˆnw idiothta legetai pÐnakac Toeplitz. Yaqte sto
          documentation tou MATLAB an uparqei mia tètoia sunarthsh.
2. Dhmiourgeiste mia sunarthsh pou ja deqetai san ìrisma tic rÐzec r enìc poluwnÔmou
   kai ja epistrefei touc suntelestèc p tou poluwnÔmou. An 1 × N einai oi diˆstaseic
   tou r, poiec eÐnai oi diastˆseic tou p?
     Parˆdeigma: an r = [1, −1] tìte p = [1, 0, −1].
     Bo jeia: MporeÐte na qrhsimopoieisete thn sunarthsh xcorr. “doc xcorr” gia
     leptomereiec.
3. Dhmiourgeiste mia sunˆrthsh pou ja pairnei san orismata dÔo dianusmata x kai h
   kai ja epistrefei ena dianusma y . Ta x kai y eqoun diastash 1 × M enw to h eqei
   diastash 1 × N . Ta stoiqeÐa tou y dÐnontai apì ton tÔpo
                                  N −1
                        y(n) =           h(k)x(n + k)     n = 1, 2, ..., M
                                  k=0

     Prosoq : Problhma dhmiourgeite ìtan to n parei timec konta sto M , opote to
     k + n > M . Se mia tetoia periptwsh to MATLAB ja petaxei lˆjoc. Lush se auto
     to prìblhma eÐnai na makrunoume katˆllhla to x
       a. eÐte me mhdenika
       b. eÐte me anˆklash tou x
       g. eÐte me epanalhyh tou x.
4. DhmioureÐste mia elikoeid c kampÔlh kai zwgrafÐste thn me th sunˆrthsh “plot3”.
   Orismoc elikoeid c kampÔlhc
                             C : (Rcos(t), Rsin(t), ct)    t ∈ [0, 10π]

                                               1
5. JewrÐa Arijmwn (Anoiqta Probl mata)

    a. DhmiourgeÐste mia sunarthsh pou ja deqetai san ìrisma ènan artio arijmì N
       kai ja epistrefei ta zeugaria pr¸twn (prime) arijmwn pou to ajroismˆ touc
       dÐnei N . P.q. N = 16 = 3 + 13 = 5 + 11.
    b. Estw N enac jetikoc akeraioc arijmoc kai sunarthsh

                                           3N + 1   N odd
                            f (N ) =
                                           N/2      N even

       BreÐte pìsec forec prepei na efarmostei h f (N ) anadromika etsi ¸ste na
       d¸sei thn tim  1 pr¸th fora, dhladh, f (f (...f (N )...) = 1.




                                       2

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  • 1. Ask seic se MATLAB HU-370 26 OktwbrÐou 2007 1. a. Estw p ena dianusma st lh, diastasewn 1 × N . Dhmiourgeiste mia sunatrhsh pou ja pairnei san ìrisma to p kai ja epistrefei enan N ×N pÐnaka me stoiqeia (A)ij = p(|i − j| + 1) To +1 qreiˆsthke gia na eÐmaste sumbatoi me thn arijmhsh tou MATLAB. b. O pinakac me thn parapˆnw idiothta legetai pÐnakac Toeplitz. Yaqte sto documentation tou MATLAB an uparqei mia tètoia sunarthsh. 2. Dhmiourgeiste mia sunarthsh pou ja deqetai san ìrisma tic rÐzec r enìc poluwnÔmou kai ja epistrefei touc suntelestèc p tou poluwnÔmou. An 1 × N einai oi diˆstaseic tou r, poiec eÐnai oi diastˆseic tou p? Parˆdeigma: an r = [1, −1] tìte p = [1, 0, −1]. Bo jeia: MporeÐte na qrhsimopoieisete thn sunarthsh xcorr. “doc xcorr” gia leptomereiec. 3. Dhmiourgeiste mia sunˆrthsh pou ja pairnei san orismata dÔo dianusmata x kai h kai ja epistrefei ena dianusma y . Ta x kai y eqoun diastash 1 × M enw to h eqei diastash 1 × N . Ta stoiqeÐa tou y dÐnontai apì ton tÔpo N −1 y(n) = h(k)x(n + k) n = 1, 2, ..., M k=0 Prosoq : Problhma dhmiourgeite ìtan to n parei timec konta sto M , opote to k + n > M . Se mia tetoia periptwsh to MATLAB ja petaxei lˆjoc. Lush se auto to prìblhma eÐnai na makrunoume katˆllhla to x a. eÐte me mhdenika b. eÐte me anˆklash tou x g. eÐte me epanalhyh tou x. 4. DhmioureÐste mia elikoeid c kampÔlh kai zwgrafÐste thn me th sunˆrthsh “plot3”. Orismoc elikoeid c kampÔlhc C : (Rcos(t), Rsin(t), ct) t ∈ [0, 10π] 1
  • 2. 5. JewrÐa Arijmwn (Anoiqta Probl mata) a. DhmiourgeÐste mia sunarthsh pou ja deqetai san ìrisma ènan artio arijmì N kai ja epistrefei ta zeugaria pr¸twn (prime) arijmwn pou to ajroismˆ touc dÐnei N . P.q. N = 16 = 3 + 13 = 5 + 11. b. Estw N enac jetikoc akeraioc arijmoc kai sunarthsh 3N + 1 N odd f (N ) = N/2 N even BreÐte pìsec forec prepei na efarmostei h f (N ) anadromika etsi ¸ste na d¸sei thn tim  1 pr¸th fora, dhladh, f (f (...f (N )...) = 1. 2