Lecture #4: Several notes 1. Recommend this book, see Chap and 3 for the basics about Matlab. [1] S. C. Chapra, Applied Numerical Methods with MATLAB

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2 Lecture #4: Several notes 1. Recommend this book, see Chap and 3 for the basics about Matlab. [1] S. C. Chapra, Applied Numerical Methods with MATLAB for Engineers and Scientists. New York: McGraw-Hill, F (ω) = 1 for a > 0. a+jω we can obtain the polar form as: F (ω) = 1 a +ω e j tan 1 ( ω a ) Be familiar with this usage. -> A/D -> -> D/A -> 1) ) ˆx(t) = x(t)δ T (t) = n x(nt )δ(t nt ) Because the impulse train δ T (t) is a periodic signal, we can have the trigonometric Fourier series as δ T (t) = 1 T [1 + cos(ω st) + cos(ω s t) + cos(3ω s t) +...], 1

3 where Ω s = π/t. therefore, ˆx(t) = x(t)δ T (t) = 1 T [x(t)+x(t) cos(ω st)+x(t) cos(ω s t)+x(t) cos(3ω s t)+...], To find ˆX(Ω), we take the Fourier transform of the fight-hand side of the above equation. Note F[x(t)e jω 0t ] = Therefore we have ˆX(Ω) = 1 T n= x(t)e jω 0t e jωt dt = X(Ω nω s ) x(t)e j(ω Ω 0)t dt = X(Ω Ω 0 ) x(t) cos(nω s t) X(Ω nω 0 ) + X(Ω + nω 0 ) -3

4 .1. X(Ω) Ω s Ω c > Ω c, Ω s > Ω c ˆX(Ω) X(Ω) ˆX(Ω). Ω s Ω c Ω c, Ω s Ω c ˆX(Ω) X(Ω) ˆX(Ω) 1. Ω s, 1/T s. 3. ˆX(Ω) 3

5 .1.3 x a (t), f c, f s > f c x a (t) f s = f c Example: x(t) = cos(πf 0 t + π/), f c = f 0, f s = f Ω l Ω h f l f h f s > f h? -6 f l + (n 1)f s < f l f h + nf s > f h i.e.. f h n < f s < f 1 n 1 f h n < f l n 1 (.1) n < f h = f h f h f l B B = f h f l, n m = arg max n ( n < f h B int() f s ) = int ( ) fh. B f s > f h m 1. f h f s > B.. m = 1, x a (t) > > > ˆx a (t). x a (t) > > > > > ˆx a (t). 4

6 1. As far as the spectrum at the right side is consider, we count the pollution from the spectrum from left side. Do we need to count the pollution from its own?. In (.1), do we need to consider upper-limiter of f s? 5

7 Lecture#5:,.1.5. X(Ω) = { X(Ω) Ω < Ω s 0 otherwise ˆX(Ω) = 1 T s X(Ω) Ω < Ω s H(Ω) = { Ts Ω < Ωs 0 otherwise ˆX(Ω)H(Ω) = X(Ω) y(t) = x(t). h(t) = 1 H(Ω)e jωt dω = 1 Ωs/ T s e jωt dω = sinc π π Ω s/ 6 ( ) πt T s

8 y(t) = ˆx(t) h(t) = x(t), y(t) = = = = = ˆx(τ)h(t τ)dτ [ n= n= n= x(τ)δ(τ nt s ) n= ] h(t τ)dτ x(τ)h(t τ)δ(τ nt s )dτ x(nt s )h(t nt s ) x(nt s )sinc(π(t nt s )/T s ) 7

9 . b b M = b M b..1 ˆx(t) > Q(.) > y n Q = {(y i, R i ), i = 1,,..., M} 1. R i. y i R i = (m i 1, m i ]. x min = m 0 < y 1 m 1 < y m <... < y M m M = x max 8

10 Lecture#6: project Q(x) x x Q(x) x Q(x) x Q(x) m X, p(x), D = E[(X Q(X)) ] = M R i (x y i ) p(x)dx E EX: Y = 1,, 3, 4, 5, 6, P (Y = i) = 1, i = 1,, 3, 4, 5, 6, 6 EX: Y = 1,, 3, 4, 5, 6, P (Y = 6) = P (Y = 5) = 1 ; P (Y = 4) = P (Y = 3) = P (Y = ) = P (Y = 1) = 0, EX: E{Y }... x Q(x) /x SNR = 10 log E[(X E(X)) ] D E[(X E(X)) ] D EX: 9

11 x ( v, v), 1.5, x ( v, 1v] 0.5, x ( 1v, 0v] y = 0.5, x (0v, 1v] 1.5, x (1v, v) P (x) = { , x ( v, 0], x (0v, v) E{X} = xp (x)dx = = E[(X E(X)) ] = (x 1 ) P (x)dx = 13 1 D = E[(X Q(X)) ] = = 1 4 R i (x y i ) P (x)dx (x + 1.5) dx + (x + 0.5) (x 0.5) 3 8 dx + (x 1.5) 3 8 dx 1 SNR = 10 log E[(X E(X)) ] D = D y i = 0 we have, D = E[(X Q(X)) ] = m i 1 y i p(x)dx = y i = M m i 1 (x y i ) p(x)dx m i 1 xp(x)dx m i 1 xp(x)dx m i 1 p(x)dx y i x y i = m i+m i 1. [a,b] B = b a, p(x) = 1/B, D 1 = b a (x Q(x))p(x)dx = M M (x Q(x))p(x)dx = E(x i y i ) = E(x i ) E(y i ) = 0. m i 1 10

12 here x i R i Let B M M D = (x y i ) 1 m i 1 B dx = 1 M [ ] 1 B 3 (m3 i m 3 i 1) y i (m i m i 1) + yi (m i m i 1 ) = M [ 1 B 3 (m i + m i m i 1 + m i 1) 1 ] 4 (m i + m i 1 ) = 1 M 3 B 1 = 1 b M, D = = B b Ex: -v,v) b b 5mv. B = 4v. σ = D = / = 4 1 b b = 7.85 b 8 11

13 project 1.. Matlab simulink ; Matlab : simulink Matlab 8. 1

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幻灯片 1 Digital Signal Processing mailfzh@nwpu.edu.cn /gary/ 1. FT FT. 3. 4. DFT 5. 6. DFT 7. 1. FT FT (FS) (FT) ( ) xt () Dirichlet (, ), 1 T () = ( Ω), ( Ω ) = () T T jkωt jkωt xt X k e X k xte dt e jkω t k