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]
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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.
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