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案例分析 Inverse Transform of Vector Padded Inverse Transform of Matrix Conjugate Symmetric Vector + e' E" P# w% r% N9 n j* R- m5 M) ?/ D
" _+ Z0 j, b$ T- [* ?2 w* l案例分析
6 ?6 s" c1 j) c8 ^0 w9 [Inverse Transform of Vector5 u* |0 x5 a( y/ O7 b, {
# Y% R9 p7 M7 }) z4 U- % The Fourier transform and its inverse convert between data sampled in time and space and data sampled in frequency.
- %
- % Create a vector and compute its Fourier transform.
- X = [1 2 3 4 5];
- Y = fft(X)
- % Y = 1×5 complex
- %
- % 15.0000 + 0.0000i -2.5000 + 3.4410i -2.5000 + 0.8123i -2.5000 - 0.8123i -2.5000 - 3.4410i ⋯
- %
- % Compute the inverse transform of Y, which is the same as the original vector X.
- ifft(Y)
- % ans = 1×5
- %
- % 1 2 3 4 51 v, k0 f1 p/ y
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) p) H. |5 i' g4 R& |. h$ S结果如下：% H* i. m, t! K" m; U9 i! G
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ifft_vector
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Y =6 h2 W2 J) t' z* Q
/ {$ ~2 y" r3 X/ O
1 至 4 列. V9 [1 i' @& I
7 }0 v7 O/ Q, Z7 A& i/ O0 R& l# k
15.0000 + 0.0000i -2.5000 + 3.4410i -2.5000 + 0.8123i -2.5000 - 0.8123i. b( u( d& r9 z4 O% W' ]# C Q6 {/ E
, y. u1 @$ N0 p5 S* x% `5 f 5 列
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4 J5 P. @1 _& @: c6 L. Y" Y -2.5000 - 3.4410i
$ A/ ]2 s {5 s( ~7 X; A8 \6 O. {% |2 |1 A" h0 d/ p- M z5 J, c
ans =
( S. G2 X: q7 U
8 J8 G7 K1 t! D$ W- t0 y 1 2 3 4 5
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% K9 h7 _% y' w* Y" ?9 p+ A
Padded Inverse Transform of Matrix. m- r- H7 O& p
- clc
- clear
- close all
- % The ifft function allows you to control the size of the transform.
- %
- % Create a random 3-by-5 matrix and compute the 8-point inverse Fourier transform of each row.
- % Each row of the result has length 8.
- Y = rand(3,5)
- n = 8;
- X = ifft(Y,n,2)
- size(X)
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1 s9 J: a9 S2 d/ H8 O: ^% c
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& U1 K# p1 F! i5 y2 o3 u结果如下：
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Y =
: k( H S. F; `2 D, r' @/ ~- g# l2 J2 H3 z$ E& w
0.8147 0.9134 0.2785 0.9649 0.9572
. _- \5 z3 m, {6 O: u' O2 X S 0.9058 0.6324 0.5469 0.1576 0.4854
# f) o$ P# j+ Y- q! {) Z 0.1270 0.0975 0.9575 0.9706 0.8003
4 L" Q7 o& C+ `3 l$ R+ s5 I9 b
' k' V/ O& [ y. k6 uX =+ T! S5 h" x- o0 E) G( ~6 Y
8 x F) q$ l) I# f
1 至 4 列' @; F7 Q( P6 G6 O
7 Y' P* r! h+ K& T* g, j. i) [ 0.4911 + 0.0000i -0.0224 + 0.2008i 0.1867 - 0.0064i -0.0133 + 0.1312i( G7 p- p- K9 k( Y0 a/ m
0.3410 + 0.0000i 0.0945 + 0.1382i 0.1055 + 0.0593i 0.0106 + 0.0015i
# E& \( _) K/ O* r3 l 0.3691 + 0.0000i -0.1613 + 0.2141i -0.0038 - 0.1091i -0.0070 - 0.0253i. L+ G, T7 G, ?+ D* l' G4 P7 i- m
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5 至 8 列
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0.0215 + 0.0000i -0.0133 - 0.1312i 0.1867 + 0.0064i -0.0224 - 0.2008i; @/ B' s5 J8 X5 y' W
0.1435 + 0.0000i 0.0106 - 0.0015i 0.1055 - 0.0593i 0.0945 - 0.1382i3 k' y' n: Z4 v8 E5 {2 I: R
0.1021 + 0.0000i -0.0070 + 0.0253i -0.0038 + 0.1091i -0.1613 - 0.2141i1 ~ U5 {* u2 Z4 U \/ h" P3 J
# ^1 [2 N8 m2 ]
ans =
. K" {7 t Z1 y! N6 q
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& s. ?: }) M/ M" t- I. j上面的程序是计算矩阵每一行的8点ifft，故结果是每一行的ifft有8个元素，而计算矩阵 Y 每一行的 ifft，关键语句为：8 F6 S2 G2 R0 s w6 g! R
& ?0 N- h3 n' G% {1 }X = ifft(Y,n,2)，里面的2，如果去掉2，则是对矩阵Y的每一列计算ifft，测试如下：4 G/ }9 ^% I( f' ^6 B/ J$ Y, @5 w
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- clc
- clear
- close all
- % The ifft function allows you to control the size of the transform.
- %
- % Create a random 3-by-5 matrix and compute the 8-point inverse Fourier transform of each row.
- % Each row of the result has length 8.
- Y = rand(3,5)
- n = 8;
- X = ifft(Y,n)
- size(X)9 F% e* j4 w- ^# I" \) e
; D1 ?$ j, Z0 Z: q0 Y
$ T; J# I) I( N. g: Q4 X3 w- }% }9 h5 F. \Y =$ j2 u) r- S: B& a" U/ J% ?8 {
+ x* z% } K2 m 0.1419 0.7922 0.0357 0.6787 0.3922- ?2 m+ z7 a! A1 F
0.4218 0.9595 0.8491 0.7577 0.6555
2 ^( g; O& V3 m" ~: B( I 0.9157 0.6557 0.9340 0.7431 0.1712) a5 Z, o; \, U; ]' [) R3 K; B
3 D+ z. T6 X3 S- C& G3 o
X =2 y0 i* |3 [7 M
0 S( H2 P9 A; j! }' F' R 1 至 4 列% r0 A% Q8 J: @( Q, P
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0.1849 + 0.0000i 0.3009 + 0.0000i 0.2274 + 0.0000i 0.2725 + 0.0000i; [) h4 C- b% j* h
0.0550 + 0.1517i 0.1838 + 0.1668i 0.0795 + 0.1918i 0.1518 + 0.1599i& _ N9 J4 p6 K( f, {- L1 B
-0.0967 + 0.0527i 0.0171 + 0.1199i -0.1123 + 0.1061i -0.0080 + 0.0947i
5 r3 `5 H, \; P0 Q -0.0195 - 0.0772i 0.0142 + 0.0028i -0.0706 - 0.0417i 0.0179 - 0.0259i& Q" y' Q0 f% I, |6 `! m" H
0.0795 + 0.0000i 0.0611 + 0.0000i 0.0151 + 0.0000i 0.0830 + 0.0000i1 V, A5 k8 N% B4 D3 d
-0.0195 + 0.0772i 0.0142 - 0.0028i -0.0706 + 0.0417i 0.0179 + 0.0259i
. A/ l C* B# P' }4 w0 O0 Y -0.0967 - 0.0527i 0.0171 - 0.1199i -0.1123 - 0.1061i -0.0080 - 0.0947i
/ v% \9 p# d" g- d) G; o- w$ D3 t 0.0550 - 0.1517i 0.1838 - 0.1668i 0.0795 - 0.1918i 0.1518 - 0.1599i0 r7 v3 n! v6 j& E5 V$ {2 j
8 Q2 y g! V# D# x" G 5 列
5 U6 {; C$ P1 Z* a$ d& M, a, S. J1 D$ {2 g- B2 O
0.1524 + 0.0000i
& x5 u5 t1 k& i6 V 0.1070 + 0.0793i
j8 O/ M( @3 J/ x$ D/ z 0.0276 + 0.0819i
7 Q# J h+ O5 ]6 ^ -0.0089 + 0.0365i
+ o- F) k6 ]6 ~9 ~& r, q -0.0115 + 0.0000i
% W. e2 @1 K, C e G/ B- @& m -0.0089 - 0.0365i9 V5 ?3 O3 k, \" ?
0.0276 - 0.0819i1 `) E6 ~" W+ ?# ?
0.1070 - 0.0793i
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% y1 }+ R4 u! ?, L2 z( F7 Y6 N4 jans =
1 w; E% m% i% {* i: U* W
( z' S! Y. X& ~8 I3 H, J! Z# O0 _ 8 5
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8 D2 J1 C6 V0 b% k/ \
Conjugate Symmetric Vector
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2 q6 L) c9 ?& ^5 |9 PFor nearly conjugate symmetric vectors, you can compute the inverse Fourier transform faster by specifying the 'symmetric' option,
8 ]: M/ l8 \; ~0 Q; Uwhich also ensures that the output is real. Nearly conjugate symmetric data can arise when computations introduce round-off error.
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Create a vector Y that is nearly conjugate symmetric and compute its inverse Fourier transform. # u% l* \. |: u
Then, compute the inverse transform specifying the 'symmetric' option, which eliminates the nearly 0 imaginary parts.
% X" x, F: T: k6 U, z3 g
3 Z! V% J- j% r$ b' k8 ]" Z对于近似共轭对称矢量，您可以通过指定“symmetric”选项来更快地计算逆傅里叶变换，这也确保了输出是真实的。 当计算引入舍入误差时，可能出现几乎共轭的对称数据。
: L# _9 _ h" y
" M U! Y4 i4 P& Q0 F4 o( K创建几乎共轭对称的向量Y并计算其逆傅里叶变换。然后，计算指定'对称'选项的逆变换，它消除了近0虚部。 |5 r! v% j) x! I- v
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- clc
- clear
- close all
- % For nearly conjugate symmetric vectors, you can compute the inverse Fourier transform faster by specifying the 'symmetric' option,
- % which also ensures that the output is real. Nearly conjugate symmetric data can arise when computations introduce round-off error.
- %
- % Create a vector Y that is nearly conjugate symmetric and compute its inverse Fourier transform.
- % Then, compute the inverse transform specifying the 'symmetric' option, which eliminates the nearly 0 imaginary parts.
- Y = [1 2:4+eps(4) 4:-1:2]
- % Y = 1×7
- %
- % 1.0000 2.0000 3.0000 4.0000 4.0000 3.0000 2.0000 ⋯
- X = ifft(Y)
- % X = 1×7 complex
- %
- % 2.7143 + 0.0000i -0.7213 + 0.0000i -0.0440 - 0.0000i -0.0919 + 0.0000i -0.0919 - 0.0000i -0.0440 + 0.0000i -0.7213 - 0.0000i ⋯
- Xsym = ifft(Y,'symmetric')
- % Xsym = 1×7
- %
- % 2.7143 -0.7213 -0.0440 -0.0919 -0.0919 -0.0440 -0.7213 ⋯
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& S4 X( Z4 \* M; v" P( b0 v' H& S# K# y. }& ^! w# b/ \
8 R1 C, ? ~7 y" F8 P( D! [结果如下：
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Y =' h6 [" H% Q( v, } T! V
# o) E) }8 L% j) |$ B& z 1.0000 2.0000 3.0000 4.0000 4.0000 3.0000 2.0000
7 j7 ]2 o# J) @/ v1 @
; j( A* E$ f2 {8 _X =
9 ~" H8 u2 z/ m* H$ {& C w) Z1 \6 f" \2 |9 m& H& ^
1 至 4 列0 J" C5 ]$ @% ?
% I1 W3 \, }; i+ r: E% w+ b& g* q
2.7143 + 0.0000i -0.7213 + 0.0000i -0.0440 - 0.0000i -0.0919 + 0.0000i/ W! U1 g& Z% n( d
# Z" ]9 Y2 y! T1 V 5 至 7 列2 U, ]0 D" r' B. s, H0 j' X& y* K& _- G
, [2 L4 J1 p+ @: L# p% s' }5 A
-0.0919 - 0.0000i -0.0440 + 0.0000i -0.7213 - 0.0000i5 D P8 r2 K/ X9 ~* g, Z; G* v" ]
, _$ L) e9 K0 \( [/ d- J( e
Xsym =
" ^+ k2 _& \+ H. i) ?! O4 ]3 X6 A0 a- \1 g0 u
2.7143 -0.7213 -0.0440 -0.0919 -0.0919 -0.0440 -0.7213
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发表于 2019-12-11 09:00
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