Discrete Time Rect Function(4B) Discrete Time Rect Functions
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Fourier Transform Types Discrete Time Fourier Transform X e j = n = Discrete Fourier Transform x[n] e j n x[n] = 1 +π j 2 π π X (e ω ) e + j ω n X [k] = n = x [n] e j 2 / N k n x [n] = 1 N k = j 2 / N k n X [k ] e 3
DTFT and DTFS L = 2 N +1 1 ( L 1) zero crossings L 2π N +N DTFT (Discrete Time Fourier Transform) X (e j ω ) = sin( ω L/2) sin( ω/2) = L diric( ω, L) = L D L (e j ω ) L = 2 N +1 1 N L/ N (L 1) zero crossings N N +N DTFS (Discrete Time Fourier Series) X [k ] = 1 N sin(π k L/ N ) sin(π k / N ) = L N drcl (k / N, L) 4
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Rect N [n] DTFT Discrete Time Fourier Transform DTFT X e j = n = x[n] e j n x[n] = 1 +π 2 π π X (e j ω ) e + j ω n X (e j ω ) = +N n= N e j ωn x[n] = {e + j ω N + + e j ω N } = e + j ω N {1 + + e j ω 2 N } j ω(2 N+1) = e + j ω N 1 e 1 e j ω j ω(2 N+1)/2 = e + j ω N e e j ω/2 e + j ω( 2N +1)/2 e j ω(2 N+1)/2 e + j ω/2 e j ω/2 = e+ j ω(2 N +1)/2 j ω(2n +1)/2 e = e + j ω/2 e j ω/2 X (e j ω ) = sin( ω L/2) sin( ω/2) = L diric( ω, L) sin( ω(2 N +1)/2) sin( ω/2) = L D L (e j ω ) 1 L = 2 N +1 Dirichlet Function N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 6
Dirichlet Functions D 9 (e j ω ) 2 π D 9 (e j ω ) = sin( ω9/2) 9sin ( ω/2) 8 zero crossings D 11 (e j ω ) = sin ( ω11/2) 11sin( ω/2) 1 zero crossings 8 zero crossings D 13 (e j ω ) = sin ( ω13/2) 13sin( ω/2) 12 zero crossings D 1 (e j ω ) 2 π D 1 (e j ω ) = sin( ω1/2) 1sin ( ω/2) 9 zero crossings 9 zero crossings D 12 (e j ω ) = D 14 (e j ω ) = sin ( ω12/2) 12sin( ω/2) sin( ω14/2) 14 sin ( ω/2) 11 zero crossings 13 zero crossings 7
Magnitude Response.45.4.35.3.25.2.15.1.5-2 -15-1 -5 5 1 15 2 8
Phase Response 3.5 3 2.5 2 1.5 1.5-2 -15-1 -5 5 1 15 2 9
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Rect N [n] * δ N [n] DTFS (1) Discrete Time Fourier Series DTFS X [ k ] = 1 N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = 1 N 1 x[n]e j(2π/ N )k n N n= = 1 +N N n= N x[n]e j(2π/ N ) k n N X [k ] = e + j(2 π N / N ) k + + e j(2 π N / N ) k = e + j (2π/N ) N k 1 e j(2π/ N )( 2N +1) k 1 e j(2π/ N ) k L = 2 N +1 N 1 j(m)(2 N +1) k = e + j (m) N k 1 e m = (2π/ N 1 e j( m)k ) j(m)( 2N +1) k /2 = e + j (m) N k e e+ j(m)(2 N +1) k/2 e j(m)(2 N+1) k /2 e j(m) k /2 e + j(m) k/2 j(m) k/2 e = sin((m)(2 N +1)k /2) sin((m)k /2) X [k ] = 1 N sin((2 π/ N )(2 N +1) k /2) sin((2π/ N ) k /2) Dirichlet Function N +N drcl(t, L) = sin(π Lt ) Lsin(π t) 11
Rect N [n] * δ N [n] DTFS (2) Discrete Time Fourier Series DTFS X [ k ] = 1 N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = 1 N sin((2 π/ N )(2 N +1) k /2) sin((2π/ N ) k /2) drcl (k / N, (2 N +1)) = sin (π k (2 N +1)/ N ) (2 N +1)sin (π k / N ) = 1 N sin (π k (2 N +1)/ N ) sin(π k / N ) X [k ] = (2 N +1) N drcl (k / N, (2 N +1)) X [k ] = 1 N sin(π k L/ N ) sin(π k / N ) X [k ] = L N drcl (k / N, L) L = 2 N +1 1 Dirichlet Function N drcl(t, L) = sin(π Lt ) Lsin(π t) N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 12
Rect N [n] * δ N [n] DTFS (3) Discrete Time Fourier Series DTFS X [ k ] = 1 N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = 1 N sin(π k L/ N ) sin(π k / N ) Period : N (odd L), 2N (even L) X [k ] = L N drcl (k / N, L) L = 2 N +1 1 (L-1) zero crossings Dirichlet Function N drcl(t, L) = sin(π Lt ) Lsin(π t) N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 13
Rect N [n] * δ N [n] DTFS (4) t = 2 t = 1 t = t =+1 t =+2 t = 2 t = 1 t = t =+1 t =+2 odd L=9 even L=1 9 zero crossings 8 zero crossings k= 32 k= 16 k= k=+16 k =+32 k= 32 k= 16 k= k=+16 k=+32 (L-1) zero crossings (L-1) zero crossings Dirichlet Function drcl (t, L) = sin(π L t) L sin(π t) X [k ] = 9 drcl (k /16, 9) 16 3, 2, 1,, +1, +2, +3, L 14
Rect N [n] * δ N [n] DTFS (5) Period : N (odd L), 2N (even L) k= 32 k= 16 k= k=+16 k=+32 (L-1) zero crossings X [k ] = 9 16 15 drcl (k /16, 9)
Rect 2 [n] * δ 8 [n] DTFS Example Discrete Time Fourier Series DTFS X [ k ] = 1 N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = 1 N sin(π k (2 N +1)/ N ) sin(π k / N ) X [k ] = 1 8 sin(π k 5/8) sin(π k /8) X [k ] = L N drcl (k / N, L) X [k ] = 5 drcl (k /8, 5) 8 N =8 L = 5 (N = 2) Period : N = 8 (odd L = 5) (L 1) = 4 zero crossings Dirichlet Function L = 2 N +1 1 N =8 drcl(t, L) = sin(π Lt ) Lsin(π t) N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 16
Magnitude Response.45.4.35.3 7 16 drcl ( k 16, 7 ) = 1 sin(π k 7/16) 16 7sin(π k /16).25.2.15.1.5-2 -15-1 -5 5 1 15 2 17
Phase Response 3.5 3 2.5 7 16 drcl ( k 16, 7 ) = 1 sin(π k 7/16) 16 7sin(π k /16) 2 1.5 1.5-2 -15-1 -5 5 1 15 2 18
Rect N [n] * δ N [n] DFT Discrete Fourier Transform X [k] = n = x [n] e j 2 / N k n x [n] = 1 N k = j 2 / N k n X [k ] e X [k ] = sin((2π/ N )(2 N +1)k /2) sin ((2π/ N )k /2) = sin(π k / N (2 N +1)) sin(π k / N ) = sin(π k / N L) sin(π k / N ) drcl(k / N, (2 N +1)) = sin (π k / N (2 N +1)) (2 N +1)sin (π k / N ) X [k ] = (2 N +1) drcl (k / N, (2 N +1)) = L drcl(k / N, L) Dirichlet Function L = 2 N +1 1 N drcl(t, L) = sin(π Lt ) Lsin(π t) N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 19
Rect N [n] * δ N [n] DTFS & DFT Discrete Time Fourier Series DTFS X [ k ] = 1 N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = 1 N sin(π k L/ N ) sin(π k / N ) X [k ] = L N drcl (k / N, L) Discrete Fourier Transform X [k] = n = x [n] e j 2 / N k n x [n] = 1 N k = j 2 / N k n X [k ] e X [k ] = sin(π k / N L) sin(π k / N ) X [k ] = L drcl (k / N, L) 2
References [1] http://en.wikipedia.org/ [2] J.H. McClellan, et al., Signal Processing First, Pearson Prentice Hall, 23 [3] G. Beale, http://teal.gmu.edu/~gbeale/ece_22/fourier_series_2.html [4] C. Langton, http://www.complextoreal.com/chapters/fft1.pdf