F ( ω ) = F ( f ( t ) ) = ∫ − ∞ + ∞ f ( t ) e − j ω t d t F\left( \omega \right) =\mathscr{F} \left( f\left( t \right) \right) =\int_{-\infty}^{+\infty}{f\left( t \right) \mathrm{e}^{-\mathrm{j}\omega t}\mathrm{d}t} F(ω)=F(f(t))=∫−∞+∞f(t)e−jωtdt
f ( t ) = F − 1 ( F ( ω ) ) = 1 2 π ∫ − ∞ + ∞ F ( ω ) e j ω t d ω f\left( t \right) =\mathscr{F} ^{-1}\left( F\left( \omega \right) \right) =\frac{1}{2\pi}\int_{-\infty}^{+\infty}{F\left( \omega \right) \mathrm{e}^{\mathrm{j}\omega t}\mathrm{d}\omega} f(t)=F−1(F(ω))=2π1∫−∞+∞F(ω)ejωtdω
此部分使用简记记号
f ( t ) ↔ F F ( ω ) , f\left( t \right) \xleftrightarrow{\mathscr{F}}F\left( \omega \right), f(t)F F(ω),其中箭头左侧为时域表达式,右侧为频域表达式.
若 f ( t ) ↔ F F ( ω ) f\left( t \right) \xleftrightarrow{\mathscr{F}}F\left( \omega \right) f(t)F F(ω),则有如下性质.
若
f 1 ( t ) ↔ F F 1 ( ω ) , f 2 ( t ) ↔ F F 2 ( ω ) , f_1\left( t \right) \xleftrightarrow{\mathscr{F}}F_1\left( \omega \right) ,f_2\left( t \right) \xleftrightarrow{\mathscr{F}}F_2\left( \omega \right), f1(t)F F1(ω),f2(t)F F2(ω),则
a f 1 ( t ) + b f 2 ( t ) ↔ F a F 1 ( ω ) + b F 2 ( ω ) . af_1\left( t \right) +bf_2\left( t \right) \xleftrightarrow{\mathscr{F}}aF_1\left( \omega \right) +bF_2\left( \omega \right) . af1(t)+bf2(t)F aF1(ω)+bF2(ω).
f ( t − t 0 ) ↔ F F ( ω ) e − j ω t 0 f\left( t-t_0 \right) \xleftrightarrow{\mathscr{F}}F\left( \omega \right) \mathrm{e}^{-\mathrm{j}\omega t_0} f(t−t0)F F(ω)e−jωt0
f ( t ) e j ω 0 t ↔ F F ( ω − ω 0 ) f\left( t \right) \mathrm{e}^{\mathrm{j}\omega _0t}\xleftrightarrow{\mathscr{F}}F\left( \omega -\omega _0 \right) f(t)ejω0tF F(ω−ω0)
f ′ ( t ) ↔ F j ω F ( ω ) f'\left( t \right) \xleftrightarrow{\mathscr{F}}\mathrm{j}\omega F\left( \omega \right) f′(t)F jωF(ω)
Tips: 可结合电路理论中电感的感抗理解,即
u = L d i d t ⇒ U = j ω L I ⇒ X L = U I = j ω L u=L\frac{\mathrm{d}i}{\mathrm{d}t}\Rightarrow U=\mathrm{j}\omega LI\Rightarrow X_L=\frac{U}{I}=\mathrm{j}\omega L u=Ldtdi⇒U=jωLI⇒XL=IU=jωL.
进一步,有
f ( n ) ( t ) ↔ F ( j ω ) n F ( ω ) f^{\left( n \right)}\left( t \right) \xleftrightarrow{\mathscr{F}}\left( \mathrm{j}\omega \right) ^nF\left( \omega \right) f(n)(t)F (jω)nF(ω)
− j t f ( t ) ↔ F F ′ ( ω ) -\mathrm{j}tf\left( t \right) \xleftrightarrow{\mathscr{F}}F'\left( \omega \right) −jtf(t)F F′(ω)
Tips: 注意左侧有负号!
进一步,有
( − j t ) n f ( t ) ↔ F F ( n ) ( ω ) \left( -\mathrm{j}t \right) ^nf\left( t \right) \xleftrightarrow{\mathscr{F}}F^{\left( n \right)}\left( \omega \right) (−jt)nf(t)F F(n)(ω)
∫ − ∞ t f ( τ ) d τ ↔ F 1 j ω F ( ω ) + π δ ( ω ) F ( 0 ) \int_{-\infty}^t{f\left( \tau \right) \mathrm{d}\tau}\xleftrightarrow{\mathscr{F}}\frac{1}{\mathrm{j}\omega}F\left( \omega \right) +\pi \delta \left( \omega \right) F\left( 0 \right) ∫−∞tf(τ)dτF jω1F(ω)+πδ(ω)F(0)
Tips: 可以通过记忆常用变换对
u ( t ) ↔ F 1 j ω + π δ ( ω ) u\left( t \right) \xleftrightarrow{\mathscr{F}}\frac{1}{\mathrm{j}\omega}+\pi \delta \left( \omega \right) u(t)F jω1+πδ(ω) 来辅助记忆该性质.
f ( a t ) ↔ F 1 ∣ a ∣ F ( ω a ) f\left( at \right) \xleftrightarrow{\mathscr{F}}\frac{1}{\left| a \right|}F\left( \frac{\omega}{a } \right) f(at)F ∣a∣1F(aω)
Tips: t t t变为原来的 a a a倍,则频率变为原来的 1 / a 1/a 1/a.
f ( t ) ↔ F F ( ω ) ⇔ F ( t ) ↔ F 2 π f ( − ω ) f\left( t \right) \xleftrightarrow{\mathscr{F}}F\left( \omega \right) \Leftrightarrow F\left( t \right) \xleftrightarrow{\mathscr{F}}2\pi f\left( -\omega \right) f(t)F F(ω)⇔F(t)F 2πf(−ω)
Tips: 注意乘 2 π 2\pi 2π,频率取“负” .
(1)使用冲激函数筛选性质容易看出
δ ( t ) ↔ F 1. \delta \left( t \right) \xleftrightarrow{\mathscr{F}}1. δ(t)F 1.
(2)对(1)利用对偶性质,有
1 ↔ F 2 π δ ( − ω ) = δ ( t ) 为偶函数 2 π δ ( ω ) . 1\xleftrightarrow{\mathscr{F}}2\pi \delta \left( -\omega \right) \xlongequal{\delta \left( t \right) \text{为偶函数}}2\pi \delta \left( \omega \right). 1F 2πδ(−ω)δ(t)为偶函数2πδ(ω).
(3)对(1)利用频域平移性质,有
e j ω 0 t ↔ F 2 π δ ( ω − ω 0 ) . \mathrm{e}^{j\omega _0t}\xleftrightarrow{\mathscr{F}}2\pi \delta \left( \omega -\omega _0 \right). ejω0tF 2πδ(ω−ω0).
(4)对(1)使用时域积分性质,有
u ( t ) ↔ F 1 j ω + π δ ( ω ) . u\left( t \right) \xleftrightarrow{\mathscr{F}}\frac{1}{\mathrm{j}\omega}+\pi \delta \left( \omega \right). u(t)F jω1+πδ(ω).
[注] 我一般直接记住这个变换对,从而记忆时域的积分性质.
(5)根据三角函数的负指数形式,结合(3),可以得到
sin ( ω 0 t ) = e j ω 0 t − e − j ω 0 t 2 j ↔ F 1 2 j ( 2 π δ ( ω − ω 0 ) − 2 π δ ( ω + ω 0 ) ) . \sin \left( \omega _0t \right) =\frac{\mathrm{e}^{\mathrm{j}\omega _0t}-\mathrm{e}^{-j\omega _0t}}{2\mathrm{j}}\xleftrightarrow{\mathscr{F}}\frac{1}{2\mathrm{j}}\left( 2\pi \delta \left( \omega -\omega _0 \right) -2\pi \delta \left( \omega +\omega _0 \right) \right) . sin(ω0t)=2jejω0t−e−jω0tF 2j1(2πδ(ω−ω0)−2πδ(ω+ω0)).
cos ( ω 0 t ) \cos\left( \omega _0t \right) cos(ω0t)也同理,此处省略.
根据Fourier正变换定义,有
F ( ω ) = ∫ − ∞ + ∞ e − a t u ( t ) e − j ω t d t = ∫ 0 + ∞ e − ( a + j ω ) t d t = 1 a + j ω . F\left( \omega \right) =\int_{-\infty}^{+\infty}{\mathrm{e}^{-at}u\left( t \right) \mathrm{e}^{-\mathrm{j}\omega t}\mathrm{d}t}=\int_0^{+\infty}{\mathrm{e}^{-\left( a+\mathrm{j}\omega \right) t}\mathrm{d}t}=\frac{1}{a+\mathrm{j}\omega}. F(ω)=∫−∞+∞e−atu(t)e−jωtdt=∫0+∞e−(a+jω)tdt=a+jω1.
同样,根据Fourier正变换定义,有
F ( ω ) = ∫ − ∞ 0 e ( a − j ω ) t d t + ∫ 0 + ∞ e − ( a + j ω ) t d t = 1 a − j ω + 1 a + j ω = 2 a a 2 + ω 2 . F\left( \omega \right) =\int_{-\infty}^0{\mathrm{e}^{\left( a-\mathrm{j}\omega \right) t}\mathrm{d}t}+\int_0^{+\infty}{\mathrm{e}^{-\left( a+\mathrm{j}\omega \right) t}\mathrm{d}t}=\frac{1}{a-\mathrm{j}\omega}+\frac{1}{a+\mathrm{j}\omega}=\frac{2a}{a^2+\omega ^2}. F(ω)=∫−∞0e(a−jω)tdt+∫0+∞e−(a+jω)tdt=a−jω1+a+jω1=a2+ω22a.
F ( ω ) = − ∫ − ∞ 0 e ( a − j ω ) t d t + ∫ 0 + ∞ e − ( a + j ω ) t d t = − 1 a − j ω + 1 a + j ω = − 2 j ω a 2 + ω 2 . F\left( \omega \right) =-\int_{-\infty}^0{\mathrm{e}^{\left( a-\mathrm{j}\omega \right) t}\mathrm{d}t}+\int_0^{+\infty}{\mathrm{e}^{-\left( a+\mathrm{j}\omega \right) t}\mathrm{d}t}=-\frac{1}{a-\mathrm{j}\omega}+\frac{1}{a+\mathrm{j}\omega}=\frac{-2\mathrm{j}\omega}{a^2+\omega ^2}. F(ω)=−∫−∞0e(a−jω)tdt+∫0+∞e−(a+jω)tdt=−a−jω1+a+jω1=a2+ω2−2jω.
注意到符号函数 s g n ( t ) \mathrm{sgn} \left( t \right) sgn(t)可以看作(3)中的双边指数衰减(奇函数)当 a a a趋于0的情形,因此符号函数的Fourier变换
F ( ω ) = lim a → 0 − 2 j ω a 2 + ω 2 = − 2 j ω = 2 j ω . F\left( \omega \right) =\underset{a\rightarrow 0}{\lim}\frac{-2\mathrm{j}\omega}{a^2+\omega ^2}=\frac{-2\mathrm{j}}{\omega}=\frac{2}{\mathrm{j}\omega}. F(ω)=a→0lima2+ω2−2jω=ω−2j=jω2.
单位阶跃函数可以用符号函数表示:
u ( t ) = 1 2 ( s g n ( t ) + 1 ) , u\left( t \right) =\frac{1}{2}\left(\mathrm{sgn}\left( t \right) +1 \right), u(t)=21(sgn(t)+1), 故根据Fourier变换的线性性质和函数 f ( t ) = 1 f\left( t \right) =1 f(t)=1的Fourier变换,可以得到
F ( ω ) = 1 2 ( 2 j ω + 2 π δ ( ω ) ) = 1 j ω + π δ ( ω ) . F\left( \omega \right) =\frac{1}{2}\left( \frac{2}{\mathrm{j}\omega}+2\pi \delta \left( \omega \right) \right) =\frac{1}{\mathrm{j}\omega}+\pi \delta \left( \omega \right) . F(ω)=21(jω2+2πδ(ω))=jω1+πδ(ω).