12 次元信号とフーリエ変換演習問題解答 - …12 2...

12 2 次元信号とフーリエ変換演習問題解答

1. � 2次元連続空間信号

光学的カメラの写真

� 2次元離散空間信号

ディジタルカメラの写真

2. 図 12.1に示すように，等間隔で並べられたセンサアレイが記録する 1次元の地震波を一つにまとめ

て 2次元信号として表現する場合には，x(t, n) のように表現される信号となる．

図 12.1 アレイ信号

3. (a) 線形性 ∑n1

∑n2

(a1x1(n1, n2) + a2x2(n1, n2)) e−jω1n1e−jω2n2 (12.1)

= a1

∑n1

∑n2

x1(n1, n2)e−jω1n1e−jω2n2 + a2

∑n1

∑n2

x2(n1, n2)e−jω1n1e−jω2n2(12.2)

= a1X1(ejω1 , ejω2) + a2X2(ejω1 , ejω2) (12.3)

(b) 分離性 ∑n1

∑n2

x1(n1)x2(n2)e−jω1n1e−jω2n2 (12.4)

=

(∑n1

x1(n1)e−jω1n1

)·(∑

n2

x2(n2)e−jω2n2

)(12.5)

= X1(ejω1)X2(ejω2) (12.6)

(c) たたみこみ

Y (ejω1 , ejω2) =∑n1

∑n2

y(n1, n2)e−jω1n1e−jω2n2

2 12 2 次元信号とフーリエ変換演習問題解答

=∑n1

∑n2

∑k1

∑k2

h(k1, k2)x(n1 − k1, n2 − k2)e−jω1n1e−jω2n2

=∑k1

∑k2

h(k1, k2)e−jω1k1e−jω2k2

·∑n1

∑n2

x(n1 − k1, n2 − k2)e−jω1(n1−k1)e−jω2(n2−k2) (12.7)

ここで，n′1 = n1 − k1，n′

2 = n2 − k2 とおくと

Y (ejω1 , ejω2) =

(∑k1

∑k2

h(k1, k2)e−jω1k1e−jω2k2

)·∑

n′1

∑n′

2

x(n′1, n

′2)e

−jω1n′1e−jω2n′

2

= H(ejω1 , ejω2)X(ejω1 , ejω2) (12.8)

(d) 周期性

任意の整数 k1 と k2 に対して

X(ejω1 , ejω2) =∑n1

∑n2

x(n1, n2)e−jω1n1e−jω2n2

=∑n1

∑n2

x(n1, n2)e−j(ω1+2k1π)n1e−j(ω2+2k2π)n2

= X(ej(ω1+2πk1), ej(ω2+2πk2)) (12.9)

(e) 対称性 ∑n1

∑n2

x∗(n1, n2)e−jω1n1e−jω2n2

=

(∑n1

∑n2

x(n1, n2)ejω1n1ejω2n2

)∗

=(X(e−jω1 , e−jω2)

)∗= X∗(e−jω1 , e−jω2) (12.10)

したがって，実数の信号 x(n1, n2) = x∗(n1, n2) に対して

X(ejω1 , ejω2) = X∗(e−jω1 , e−jω2) (12.11)

|X(ejω1 , ejω2)| = |X(e−jω1 , e−jω2)| (12.12)

� X(ejω1 , ejω2) = −� X(e−jω1 , e−jω2) (12.13)

4. (a) ∑n1

∑n2

δ(n1, n2)e−jω1n1e−jω2n2 = e−jω1·0e−jω2·0

= 1 (12.14)

図 12.2に振幅スペクトルを示す．

(b) ∑n1

∑n2

δ(n1)e−jω1n1e−jω2n2 =∑n2

e−jω2n2

= 2πδ(ω2) (12.15)

3

-20

2

-20

2

0

0.5

1

ω1 [rad]ω

2 [rad]

¦X(ejω1,ejω2)¦

図 12.2 振幅スペクトル


-20

2

-20

2

0

2

4

6

ω1 [rad]ω

2 [rad]

¦X(ejω1,ejω2)¦


(c) ∑n1

∑n2

δ(n1 − n2)e−jω1n1e−jω2n2 =∑n2

e−j(ω1+ω2)n2

= 2πδ(ω1 + ω2) (12.16)


-20

2

-20

2

0

2

4

6

ω1 [rad]ω

2 [rad]

¦X(ejω1,ejω2)¦



(d)

∑n1

∑n2

an1bn2e−jω1n1e−jω2n2 =

(∑n1

an1e−jω1n1

)·(∑

n2

bn2e−jω2n2

)

=1

1 − ae−jω1· 11 − be−jω2

(12.17)

図 12.5に a = 0.5, b = 0.75 のときの振幅スペクトルを示す．

-20

2

-20

2

0

2

4

6

ω1 [rad]ω

2 [rad]

¦X(ejω1,ejω2)¦


(e)L1−1∑n1=0

L2−1∑n2=0

e−jω1n1e−jω2n2 =

(L1−1∑n1=0

e−jω1n1

)·(

L2−1∑n2=0

e−jω2n2

)

=1 − e−jω1L1

1 − e−jω1· 1 − e−jω2L2

1 − e−jω2

= e−j(L1−12 ω1+

L2−12 ω2) sin L1

2 ω1

sin ω12

· sin L22 ω2

sin ω22

(12.18)

図 12.6に L1 = 2, L2 = 4 のときの振幅スペクトルを示す．

-20

2

-20

2

0

2

4

6

ω1 [rad]ω

2 [rad]

¦X(ejω1,ejω2)¦


(f) ∑n1

∑n2

cos(αn1 + βn2)e−jω1n1e−jω2n2 (12.19)

5

=∑n1

∑n2

12

(ej(αn1+βn2) + e−j(αn1+βn2)

)e−jω1n1e−jω2n2

=12

(∑n1

∑n2

ej(αn1+βn2)e−j(ω1n1+ω2n2)

)+

12

(∑n1

∑n2

e−j(αn1+βn2)e−j(ω1n1+ω2n2)

)

=12

∑n1

ej(α−ω1)n1 ·∑n2

ej(β−ω2)n2 +12

∑n1

e−j(α+ω1)n1 ·∑n2

e−j(β+ω2)n2

= 4π2δ(ω1 − α) · δ(ω2 − β) + 4π2δ(ω1 + α) · δ(ω2 + β) (12.20)

図 12.7に α = 0.25π, β = 0.5π のときの振幅スペクトルを示す．

-20

2

-20

2

0

10

20

30

ω1 [rad]ω

2 [rad]

¦X(ejω1,ejω2)¦


(g) ∑n1

∑n2

(−1)n1+n2e−jω1n1e−jω2n2 =∑n1

∑n2

ejπ(n1+n2)e−jω1n1e−jω2n2

=∑n1

ej(π−ω1)n1 ·∑n2

ej(π−ω2)n2

= 2πδ(π − ω1) · 2πδ(π − ω2) (12.21)


-20

2

-20

2

0

10

20

30

ω1 [rad]ω

2 [rad]

¦X(ejω1,ejω2)¦


5. (a) X(ejω1 , ejω2) は，ω1 と ω2 のそれぞれに関して 2π で周期的であるから，次の式が成り立つ．


x(n1, n2) =1

4π2

∫ωc<=|ω1|<π

∫ωc<=|ω2|<π

1 · ejω1n1ejω2n2dω1dω2

=1

4π2

∫ 2π−ωc

ωc

∫ 2π−ωc

ωc

1 · ejω1n1ejω2n2dω1dω2 (12.22)

よって，x(n1, n2) は以下のように求められる．

x(n1, n2) =(

12π

∫ 2π−ωc

ωc

ejω1n1dω1d

)·(

12π

∫ 2π−ωc

ωc

ejω2n2dω2

)

=12π

[ejω1n1

jn1

]2π−ωc

ωc

· 12π

[ejω2n2

jn2

]2π−ωc

ωc

(12.23)

i. n1, n2 = 0のとき

x(n1, n2) =(

12π

∫ 2π−ωc

ωc

1dω1

)(12π

∫ 2π−ωc

ωc

1dω2

)

=12π

[ω1]2π−ωc

ωc· 12π

[ω2]2π−ωc

ωc(12.24)

=12π

(2π − 2ωc) · 12π

(2π − 2ωc) (12.25)

=(

π − ωc

π

)2

(12.26)

ii. n1 = 0, n2 �= 0 のとき

x(n1, n2) =1

4π2

∫ 2π−ωc

ωc

1dω1

∫ 2π−ωc

ωc

ejω2n2ω2

=(

π − ωc

π

)· 12π

[ejω2n2

jn2

]2π−ωc

ωc

=(

π − ωc

π

)· 12π

ej(2π−ωc)n2 − ejωcn2

jn1

=(

π − ωc

π

)· 12π

e−jωcn2 − ejωcn2

jn2

=(

π − ωc

π

)·(− 1

2π

sin ωcn2

n2

)

= −(

π − ωc

π

)ωc

πsinc

ωcn2

π(12.27)

iii. n1 �= 0, n2 = 0 のとき

前述と同様にして

x(n1, n2) = −(

π − ωc

π

)ωc

πsinc

ωcn1

π(12.28)

iv. n1, n2 �= 0 のとき12π

[ejω1n1

jn1

]2π−ωc

ωc

=12π

ej(2π−ωc)n1 − ejωcn1

jn1

=12π

e−jωcn1 − ejωcn1

jn1

= − 12π

sin ωcn1

n1

7

= −ωc

πsinc

ωcn1

π(12.29)

同様に12π

[ejω2n2

jn2

]2π−ωc

ωc

= −ωc

πsinc

ωcn2

π(12.30)

よって，式 (12.23)から

x(n1, n2) =(−ωc

πsinc

ωcn1

π

)·(−ωc

πsinc

ωcn2

π

)(12.31)

図 12.9に ωc = π/2 のときの信号 x(n1, n2) を示す．

-100

10

-100

10

-0.2

0

0.2

n1

n2

xa(n1,n2)

図 12.9 信号 x(n1, n2)

(b) いま，次のような 2次元離散フーリエ変換 X ′(ejω1 , ejω2) を用いる．

X ′(ejω1 , ejω2) =

1, |ω1| < ωcかつ |ω2| < ωc

0, その他(12.32)

このとき，与えられた X(ejω1 , ejω2) は次のように表される．

X(ejω1 , ejω2) = 1 − X ′(ejω1 , ejω2), ω1, ω2 = −π ∼ π (12.33)

よって，X(ejω1 , ejω2) の逆変換は以下のように求められる．

x(n1, n2) =1

4π2

∫ π

−π

∫ π

−π

X(ejω1 , ejω2)ejω1n1ejω2n2dω1dω2

=1

4π2

∫ π

−π

∫ π

−π

(1 − X ′(ejω1 , ejω2)

)ejω1n1ejω2n2dω1dω2

=1

4π2

∫ π

−π

∫ π

−π

1 · ejω1n1ejω2n2dω1dω2 (12.34)

− 14π2

∫ π

−π

∫ π

−π

X ′(ejω1 , ejω2)ejω1n1ejω2n2dω1dω2

= δ(n1, n2) − 14π2

∫ π

−π

∫ π

−π

X ′(ejω1 , ejω2)ejω1n1ejω2n2dω1dω2

= δ(n1, n2) − ωc

πsinc

(ωcn1

π

)· ωc

πsinc

(ωcn2

π

), n1, n2 = −∞ ∼ ∞ (12.35)

図 12.10に ωc = π/2 のときの信号 x(n1, n2) を示す．

(c) x(n1, n2) =1

4π2

∫∫C

1 · exp(jn1ω1 + jn2ω2)dω1dω2 (12.36)

となる．ここで


-100

10

-100

10

-0.2

0

0.2

0.4

0.6

n1

n2

xb(n1,n2)

図 12.10 信号 x(n1, n2)

C ={

(ω1, ω2)|√

ω21 + ω2

2<= ωc

}(12.37)

式 (12.36)の右辺を求めるために，ω1 と ω2 を次のように極座標の形式で表す．

ω1 = ω cos φ, ω2 = ω sin φ (12.38)

このとき，ヤコビ行列式は ω となるので

x(n1, n2) =1

4π2

∫ ωc

0

∫ π

−π

exp[jω(n1 cos φ + n2 sin φ)]ωdφdω

=1

4π2

∫ ωc

0

∫ π

−π

ω exp[jω√

n21 + n2

2 cos(θ − φ)]

dφdω (12.39)

となる．ただし

n2/n1 = tan θ (12.40)

とおいた．したがって

x(n1, n2) =12π

∫ ωc

0

ωJ0

(ω√

n21 + n2

2

)dω

=ω2

c

2π·J1

(ωc

√n2

1 + n22

)ωc

√n2

1 + n22

, −∞ <= n1, n2 <= ∞ (12.41)

ここで，J0(x) と J1(x) は，それぞれ 0 次と 1 次の第 1 種ベッセル関数である．図 12.11に

ωc = π/2 のときの信号 x(n1, n2) を示す．

-100

10

-100

10

-0.2

0

0.2

n1

n2

xc(n1,n2)

図 12.11 信号 x(n1, n2)

9

6. (a) 線形性

N1−1∑n1=0

N2−1∑n2=0

(a1x(n1, n2) + a2x2(n1, n2)) W k1n1N1

W k2n2N2

= a1

N1−1∑n1=0

N2−1∑n2=0

x(n1, n2)W k1n1N1

W k2n2N2

+ a2

N1−1∑n1=0

N2−1∑n2=0

x2(n1, n2)W k1n1N1

W k2n2N2

= a1X1(k1, k2) + a2X2(k1, k2) (12.42)

(b) 分離性N1−1∑n1=0

N2−1∑n2=0

x1(n1)x2(n2)W k1n1N1

W k2n2N2

=

(N1−1∑n1=0

x1(n1)W k1n1N1

)·(

N2−1∑n2=0

x2(n2)W k2n2N2

)

= X1(k1)X2(k2) (12.43)

(c) 循環たたみこみ

Y (k1, k2) =N1−1∑n1=0

N2−1∑n2=0

y(n1, n2)W k1n1N1

W k2n2N2

=N1−1∑n1=0

N2−1∑n2=0

(N1−1∑p1=0

N2−1∑p2=0

h(p1, p2)x((n1 − p1)N1 , (n2 − p2)N2)

)W k1n1

N1W k2n2

N2

=N1−1∑p1=0

N2−1∑p2=0

h(p1, p2)Wk1p1N1

W k2p2N2

·N1−1∑n1=0

N2−1∑n2=0

x((n1 − p1)N1 , (n2 − p2)N2)

·W k1(n1−p1)N1N1

Wk2(n2−k2)N2N2

= H(k1, k2)X(k1, k2) (12.44)

(d) 周期性

任意の整数 p1 と p2 に対して

X(k1 + p1N1, k2 + p2N2) =N1−1∑n1=0

N2−1∑n2=0

x(n1, n2)W(k1+p1N1)n1N1

W(k2+p2N2)n2N2

=N1−1∑n1=0

N2−1∑n2=0

x(n1, n2)W k1n1N1

W k2n2N2

W p1N1n1N1

W p2N2n2N2

=N1−1∑n1=0

N2−1∑n2=0

x(n1, n2)W k1n1N1

W k2n2N2

= X(k1, k2) (12.45)

(e) 対称性N1−1∑n1=0

N2−1∑n2=0

x∗(n1, n2)W k1n1N1

W k2n2N2

=

(N1−1∑n1=0

N2−1∑n2=0

x(n1, n2)(W k1n1

N1W k2n2

N2

)∗)∗


(12.46)

ここで

(W knN )∗ =

(exp

(−j

2π

Nkn

))∗

= exp(

j2π

Nkn

)

= exp(−j

2π

N(N − k)n

)

= W(N−k)nN (12.47)

よって (N1−1∑n1=0

N2−1∑n2=0

x(n1, n2)(W k1n1

N1W k2n2

N2

)∗)∗

=

(N1−1∑n1=0

N2−1∑n2=0

x(n1, n2)(W

(N1−k1)n1N1

W(N2−k2)n2N2

))∗

= X∗(N1 − k1, N2 − k2) (12.48)

したがって，実数の信号 x(n1, n2) = x∗(n1, n2) に対して

X(k1, k2) = X∗(N1 − k1, N2 − k2) (12.49)

|X(k1, k2)| = |X(N1 − k1, N2 − k2)| (12.50)

� X(k1, k2) = −� X(N1 − k1, N2 − k2) (12.51)

7. (a)N1−1∑n1=0

N2−1∑n2=0

δ(n1, n2)W k1n1N1

W k2n2N2

= W k1·0N1

W k2·0N2

= 1 (12.52)

図 12.12に N1 = N2 = 16 のときの振幅スペクトルを示す．

05

1015

05

10150

1

2

k1

k2

¦X(k1,k2)¦


(b)N1−1∑n1=0

N2−1∑n2=0

δ(n1)W k1n1N1

W k2n2N2

= W k1·0N1

N2−1∑n2=0

W k2n2N2

11

=N2−1∑n2=0

W k2n2N2

(12.53)

k2 = 0 のときN2−1∑n2=0

W k2n2N2

= N2 (12.54)

k2 �= 0 のときN2−1∑n2=0

W k2n2N2

=1 − W k2N2

N2

1 − W k2N2

= 0 (12.55)


05

1015

05

10150

5

10

15

k1

k2

¦X(k1,k2)¦


(c)N1−1∑n1=0

N2−1∑n2=0

δ(n1 − n2)W k1n1N1

W k2n2N2

=N−1∑n=0

W k1nN1

W k2nN2

=N−1∑n=0

W(k1+k2)nN1

(12.56)

i. k1 + k2 = 0 すなわち k1 = k2 = 0 のときN−1∑n=0

W(k1+k2)nN1

=N−1∑n=0

1

= N (12.57)

ii. k1 + k2 = N のときN−1∑n=0

W(k1+k2)nN1

=N−1∑n=0

WNnN1

=N−1∑n=0

1

= N (12.58)

iii. k1 + k2 がその他の値のときN−1∑n=0

W(k1+k2)nN1

=1 − W

(k1+k2)NN

1 − W k1+k2N


=1 − 1

1 − W k1+k2N

= 0 (12.59)


05

1015

05

10150

5

10

15

k1

k2

¦X(k1,k2)¦


(d)

N1−1∑n1=0

N2−1∑n2=0

1W k1n1N1

W k2n2N2

=

(N1−1∑n1=0

W k1n2N1

)·(

N2−1∑n2=0

W k2n2N2

)(12.60)

i. k1 = k2 = 0 のとき(N1−1∑n1=0

W k1n1N1

)·(

N2−1∑n2=0

W k2n2N2

)=

N1−1∑n1=0

1 ·N2−1∑n2=0

1

= N1N2 (12.61)

ii. k1 = 0, k2 �= 0 のときN2−1∑n2=0

W k2n2N2

=1 − W k2N2

N2

1 − W k2N2

=1 − 1

1 − W k2N2

= 0 (12.62)

よって，(N1−1∑n1=0

W k1n1N1

)·(

N2−1∑n2=0

W k2n2N2

)=

(N1−1∑n1=0

W k1n1N1

)· 0

= 0 (12.63)

iii. k1 �= 0, k2 = 0 のとき

上と同様にして(N1−1∑n1=0

W k1n1N1

)·(

N2−1∑n2=0

W k2n2N2

)= 0 ·

(N2−1∑n2=0

W k2n2N2

)

= 0 (12.64)

iv. k1, k2 �= 0 のとき

13


W k1n1N1

)·(

N2−1∑n2=0

W k2n2N2

)= 0 · 0

= 0 (12.65)


05

1015

05

10150

100

200

k1

k2

¦X(k1,k2)¦


(e)

x(n1, n2) =

1 0 <= n1 <= L1 − 1, 0 <= n2 <= L2 − 1

0 その他(12.66)

X(k1, k2) =L1−1∑n1=0

L2−1∑n2=0

W k1n1N1

W k2n2N2

=L1−1∑n1=0

W k1n1N1

·L2−1∑n2=0

W k2n2N2

(12.67)

k1 = 0, k2 �= 0 のとき

X(k1, k2) = L1 ·1 − W k2L2

N2

1 − W k2N2

= L1 · W k2(L2−1)/2N2

· sin(πk2L2/N2)sin(πk2/N2)

(12.68)

k1 �= 0, k2 = 0 のとき

X(k1, k2) = L2 · W k1(L1−1)/2N1


(12.69)

上記以外のとき

X(k1, k2) =1 − W k1L1

N1

1 − W k1N1

· 1 − W k2L2N2

1 − W k2N2

= Wk1(L1−1)/2N1


·W k2(L2−1)/2N2


(12.70)

図 12.16に N1 = N2 = 16, L1 = 2, L2 = 4 のときの振幅スペクトルを示す．


05

1015

05

10150

5

10

k1

k2

¦X(k1,k2)¦


(f)

X(k1, k2) =N1−1∑n1=0

N2−1∑n2=0

cos(αn1 + βn2)W k1n1N1

W k2n2N2

=N1−1∑n1=0

N2−1∑n2=0

12(ej(αn1+βn2) + e−j(αn1+βn2))W k1n1

N1W k2n2

N2

=12

N1−1∑n1=0

N2−1∑n2=0

ej(αn1+βn2)W k1n1N1

W k2n2N2

+12

N1−1∑n1=0

N2−1∑n2=0

e−j(αn1+βn2)W k1n1N1

W k2n2N2

=12

N1−1∑n1=0

ejαn1W k1n1N1

N2−1∑n2=0

ejβn2W k2n2N2

+12

N1−1∑n1=0

e−jαn1W k1n1N1

N2−1∑n2=0

e−jβn2W k2n2N2

(12.71)

ここで，α = 2πp/N1, β = 2πq/N2（p, q は整数）とすると，

X(k1, k2) =12

N1−1∑n1=0

ej 2πpN1

n1W k1n1N1

N2−1∑n2=0

ej 2πqN2

n2W k2n2N2

+12

N1−1∑n1=0

e−j 2πpN1

n1W k1n1N1

N2−1∑n2=0

e−j 2πqN2

n2W k2n2N2

=12

N1−1∑n1=0

W−n1pN1

W k1n1N1

N2−1∑n2=0

W−n2qN2

W k2n2N2

+12

N1−1∑n1=0

Wn1(p−N1)N1

W k1n1N1

N2−1∑n2=0

Wn2(q−N2)N2

W k2n2N2

=12

N1−1∑n1=0

W−n1(k1−p)N1

N2−1∑n2=0

Wn2(k2−q)N2

+12

N1−1∑n1=0

Wn1(k1+p−N1)N1

N2−1∑n2=0

Wn2(k2+q−N2)N2

=N1N2

2δ(k1 − p)δ(k2 − q) +

N1N2

2δ(k1 + p − N1)δ(k2 + q − N2)

(12.72)


15

05

1015

05

10150

50

100

k1

k2

¦X(k1,k2)¦


(g) (−1)n1+n2 = exp(jπn1) exp(jπn2) であるからN1−1∑n1=0

N2−1∑n2=0

(−1)n1+n2W k1n1N1

W k2n2N2

=N1−1∑n1=0

N2−1∑n2=0

exp(jπn1) exp(jπn2)W k1n1N1

W k2n2N2

=N1−1∑n1=0

N2−1∑n2=0

exp(jπn1) exp(jπn2)W k1n1N1

W k2n2N2

=

(N1−1∑n1=0

exp(jπn1)W k1n1N1

)(N2−1∑n2=0

exp(jπn2)W k2n2N2

)

=

(N1−1∑n1=0

exp(

j2π

N1· N1

2n1

)W k1n1

N1

)·(

N2−1∑n2=0

exp(

j2π

N2· N2

2n2

)W k2n2

N2

)

=

(N1−1∑n1=0

W(k1−N1/2)n1N1

)·(

N2−1∑n2=0

W(k2−N2/2)n2N2

)(12.73)

i. k1 = N1/2, k2 = N2/2 のとき(N1−1∑n1=0

W(k1−N1/2)n1N1

)·(

N2−1∑n2=0

W(k2−N2/2)n2N2

)=

(N1−1∑n1=0

1

)·(

N2−1∑n2=0

1

)

= N1N2 (12.74)

ii. k1 = N1/2, k2 �= N2/2 のとき(N1−1∑n1=0

W(k1−N1/2)n1N1

)·(

N2−1∑n2=0

W(k2−N2/2)n2N2

)= N1 ·

1 − W(k2−N2/2)n2N2N2

1 − W(k2−N2/2)n2N2

= N1 · 1 − 1

1 − W(k2−N2/2)n2N2

= 0 (12.75)

iii. k1 �= N1/2, k2 = N2/2 のとき


W(k1−N1/2)n1N1

)·(

N2−1∑n2=0

W(k2−N2/2)n2N2

)= 0 · N2

= 0 (12.76)


iv. k1 �= N1/2, k2 �= N2/2 のとき


W(k1−N1/2)n1N1

)·(

N2−1∑n2=0

W(k2−N2/2)n2N2

)= 0 · 0

= 0 (12.77)


05

1015

05

10150

100

200

k1

k2

¦X(k1,k2)¦


8. 図 12.19に 2次元 DFTの複素乗算回数と複素加算回数を示す．

� �【MATLAB12.1】%2次元離散フーリエ変換の計算量p = 0:9; %べきの範囲N = 2.^p; %信号のサイズN1 = N; N2=N;

mult_direct = N1.^2 .* N2.^2; %直接計算の乗算回数mult_rcd = 0.5*N1.* N2.*log(N1.*N2); %行列分解の乗算回数

add_direct = N1.*N2.*(N1.*N2-1); %直接計算の加算回数add_rcd = N1.*N2.*log(N1.*N2); %行列分解の加算回数

subplot(2,2,1);loglog(N,mult_direct,’*’, N,mult_rcd,’o’); grid;xlabel(’Signal length N’); ylabel(’Num. of Multiplication’);legend(’Direct’,’R-C decomp’,2)

subplot(2,2,2);loglog(N,add_direct,’*’, N,add_rcd,’o’); grid;xlabel(’Signal length N’); ylabel(’Num. of Addition’);legend(’Direct’,’R-C decomp’,2)� �

17

100

101

102

103

100

105

1010

1015

Signal length N

Num. of Multiplication

DirectR-C decomp

100

101

102

103

100

105

1010

1015

Signal length N

Num. of Addition

DirectR-C decomp

(a) 複素乗算回数 (b) 複素加算回数

図 12.19 2 次元 DFT の複素乗算回数と複素加算回数

12 次元信号とフーリエ変換演習問題解答 - …12 2...

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