傅里叶变换的性质及证明
1. 线性性质 F[αf1(t)+βf2(t)]=αF1(ω)+βF2(ω)\qquad F[\alpha f_1(t)+\beta f_2(t)]=\alpha F_1(\omega)+\beta F_2(\omega)F[αf1(t)+βf2(t)]=αF1(ω)+βF2(ω)
证明:
F[αf1(t)+βf2(t)]=∫−∞+∞[αf1(t)+βf2(t)]e−jwtdt=α∫−∞+∞f1(t)e−jwtdt+β∫−∞+∞f2(t)e−jwt=αF1(ω)+βF2(ω)(证毕) \begin{aligned} F[\alpha f_1(t)+\beta f_2(t)] &=\int_{-\infty}^{+\infty}[\alpha f_1(t)+\beta f_2(t)]e^{-jwt}dt\\ &=\alpha \int_{-\infty}^{+\infty}f_1(t)e^{-jwt}dt + \beta \int_{-\infty}^{+\infty}f_2(t)e^{-jwt}\\ &=\alpha F_1(\omega)+\beta F_2(\omega)\\ & (证毕) \end{aligned} F[αf1(t)+βf2(t)]=∫−∞+∞[αf1(t)+βf2(t)]e−jwtdt=α∫−∞+∞f1(t)e−jwtdt+β∫−∞+∞f2(t)e−jwt=αF1(ω)+βF2(ω)(证毕)
2. 位移性质 F[f(t±t0)]=e±jwt0F(ω)\qquad F[f(t\pm t_0)]=e^{\pm jwt_0}F(\omega)F[f(t±t0)]=e±jwt0F(ω)
证明:
F[f(t±t0)]=∫−∞+∞f(t±t0)e−jwtdt(令u=t±t0)=∫−∞+∞f(u)e−jw(u∓t0)du=e±jwt0∫−∞+∞f(u)e−jwudu=e±jwt0F(ω)(证毕) \begin{aligned} F[f(t\pm t_0)] &=\int_{-\infty}^{+\infty}f(t\pm t_0)e^{-jwt}dt\qquad(令u=t\pm t_0)\\ &=\int_{-\infty}^{+\infty}f(u)e^{-jw(u\mp t_0)}du\\ &=e^{\pm jwt_0}\int_{-\infty}^{+\infty}f(u)e^{-jwu}du\\ &=e^{\pm jwt_0}F(\omega)\\ & (证毕) \end{aligned} F[f(t±t0)]=∫−∞+∞f(t±t0)e−jwtdt(令u=t±t0)=∫−∞+∞f(u)e−jw(u∓t0)du=e±jwt0∫−∞+∞f(u)e−jwudu=e±jwt0F(ω)(证毕)
3. 对称性质 F[f(−t)]=F(−ω)\qquad F[f(-t)]=F(-\omega)F[f(−t)]=F(−ω)
证明:
F[f(−t)]=∫−∞+∞f(−t)e−jwtdt(令u=−t)=∫+∞−∞f(u)e−jw(−u)d(−u)=∫−∞+∞f(u)e−j(−w)tdu=F(−ω)(证毕) \begin{aligned} F[f(-t)] &=\int_{-\infty}^{+\infty}f(-t)e^{-jwt}dt\qquad(令u=-t)\\ &=\int_{+\infty}^{-\infty}f(u)e^{-jw(-u)}d(-u)\\ &=\int_{-\infty}^{+\infty}f(u)e^{-j(-w)t}du\\ &=F(-\omega)\\ & (证毕) \end{aligned} F[f(−t)]=∫−∞+∞f(−t)e−jwtdt(令u=−t)=∫+∞−∞f(u)e−jw(−u)d(−u)=∫−∞+∞f(u)e−j(−w)tdu=F(−ω)(证毕)
4. 尺度性质 F[f(at)]=1∣a∣F(ωa)(a≠0)\qquad F[f(at)]=\frac{1}{|a|}F(\frac{\omega}{a})\qquad(a\neq0)F[f(at)]=∣a∣1F(aω)(a=0)
证明:
F[f(at)]=∫−∞+∞f(at)e−jwtdt(令u=at)={∫−∞+∞f(u)e−jwuadua,a>0∫+∞−∞f(u)e−jwuadua,a<0(注意上下限)=1∣a∣∫−∞+∞f(u)e−jwauda=1∣a∣F(ωa)(a≠0)(证毕) \begin{aligned} F[f(at)] &= \int_{-\infty}^{+\infty}f(at)e^{-jwt}dt\qquad(令u=at)\\ &= \begin{cases} \int_{-\infty}^{+\infty}f(u)e^{-jw\frac{u}{a}}d\frac{u}{a} & ,a>0\\ \int_{+\infty}^{-\infty}f(u)e^{-jw\frac{u}{a}}d\frac{u}{a} & ,a<0\\ \end{cases}\qquad{(注意上下限)}\\ &=\frac{1}{|a|}\int_{-\infty}^{+\infty}f(u)e^{-j\frac{w}{a}u}da\\ &=\frac{1}{|a|}F(\frac{\omega}{a})\qquad(a\neq0)\\ & (证毕) \end{aligned} F[f(at)]=∫−∞+∞f(at)e−jwtdt(令u=at)={∫−∞+∞f(u)e−jwaudau∫+∞−∞f(u)e−jwaudau,a>0,a<0(注意上下限)=∣a∣1∫−∞+∞f(u)e−jawuda=∣a∣1F(aω)(a=0)(证毕)
5. 微分性质 F[f′(t)]=jwF(ω)\qquad F[f'(t)]=jwF(\omega)F[f′(t)]=jwF(ω)
证明:
前提:f(x)在(−∞,+∞)(-\infty,+\infty)(−∞,+∞)上连续或只有有限个可去间断点,且limx→∞f(x)=0\lim\limits_{x\rightarrow\infty} f(x)=0x→∞limf(x)=0
F[f′(t)]=∫−∞+∞f′(t)e−jwtdt=∫−∞+∞e−jwtdf(t)=0−(−jw)∫−∞+∞e−jwtdt=jwF(ω)(证毕) \begin{aligned} F[f'(t)] &=\int_{-\infty}^{+\infty}f'(t)e^{-jwt}dt\\ &=\int_{-\infty}^{+\infty}e^{-jwt}df(t)\\ &=0-(-jw)\int_{-\infty}^{+\infty}e^{-jwt}dt\\ &=jwF(\omega)\\ & (证毕) \end{aligned} F[f′(t)]=∫−∞+∞f′(t)e−jwtdt=∫−∞+∞e−jwtdf(t)=0−(−jw)∫−∞+∞e−jwtdt=jwF(ω)(证毕)
用数学归纳法易证:当limx→∞f(k)(x)=0 (k=0,1,…,n−1)时,\lim\limits_{x\rightarrow\infty}f^{(k)}(x)=0\ (k=0,1,\dots,n-1)时,x→∞limf(k)(x)=0 (k=0,1,…,n−1)时,
F[f(n)(t)]=(jw)nF(ω)F[f^{(n)}(t)]=(jw)^nF(\omega)F[f(n)(t)]=(jw)nF(ω)
6. 积分性质 F[∫−∞tf(t)dt]=1jwF(ω)\qquad F[\int_{-\infty}^{t}f(t)dt]=\frac{1}{jw}F(\omega)F[∫−∞tf(t)dt]=jw1F(ω)
证明:
前提:limt→+∞∫−∞tf(t)dt=0\lim\limits_{t\rightarrow+\infty}\int_{-\infty}^{t}f(t)dt=0t→+∞lim∫−∞tf(t)dt=0
令g(t)=∫−∞tf(t)dt,有limt→∞g(t)=0\qquad令g(t)=\int_{-\infty}^{t}f(t)dt,\qquad 有\lim\limits_{t\rightarrow\infty}g(t)=0令g(t)=∫−∞tf(t)dt,有t→∞limg(t)=0
∴F[g′(t)]=jwF[g(t)]∵g′(t)=f(t)∴F[f(t)]=1jwF[g(t)](证毕) \begin{aligned} & \therefore F[g'(t)]=jwF[g(t)]\\ & \because g'(t)=f(t)\\ & \therefore F[f(t)]=\frac{1}{jw}F[g(t)]\\ & (证毕) \end{aligned} ∴F[g′(t)]=jwF[g(t)]∵g′(t)=f(t)∴F[f(t)]=jw1F[g(t)](证毕)
7. 卷积性质 F[f∗g]=F(f)⋅F(g)\qquad F[f*g]=F(f)\cdot F(g)F[f∗g]=F(f)⋅F(g)
证明:
F[f∗g]=∫−∞+∞∫−∞+∞f(t)g(τ−t)dt e−jwτdτ=∫−∞+∞∫−∞+∞f(t)g(τ−t)e−jw(τ−t)dτ e−jwtdt(令u=τ−t)=∫−∞+∞∫−∞+∞f(t)g(u)e−jwudu e−jwtdt=∫−∞+∞f(t)e−jwtdt∫−∞+∞g(u)e−jwudu=F(f)⋅F(g)(证毕) \begin{aligned} F[f*g] &=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(t)g(\tau-t)dt\ e^{-jw\tau}d\tau \\ &=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(t)g(\tau-t)e^{-jw(\tau-t)}d\tau\ e^{-jwt}dt\qquad(令u=\tau-t)\\ &=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(t)g(u)e^{-jwu}du\ e^{-jwt}dt\\ &=\int_{-\infty}^{+\infty}f(t)e^{-jwt}dt\int_{-\infty}^{+\infty}g(u)e^{-jwu}du\\ &=F(f)\cdot F(g)\\ & (证毕) \end{aligned} F[f∗g]=∫−∞+∞∫−∞+∞f(t)g(τ−t)dt e−jwτdτ=∫−∞+∞∫−∞+∞f(t)g(τ−t)e−jw(τ−t)dτ e−jwtdt(令u=τ−t)=∫−∞+∞∫−∞+∞f(t)g(u)e−jwudu e−jwtdt=∫−∞+∞f(t)e−jwtdt∫−∞+∞g(u)e−jwudu=F(f)⋅F(g)(证毕)