【信号与系统 - 9】傅里叶变换的性质习题

1 习题

已知 $f(t)$ 的傅里叶变换为 $F(jw)$ ，求如下信号的傅里叶变换

• （1）$t\cdot f(3t)$
  解：
  $f(3t)\leftrightarrow \frac{1}{3}F(j\frac{w}{3})$
  $t\cdot f(3t)\leftrightarrow j\frac{1}{3}\cdot \frac{d}{dw}[F(j\frac{w}{3})]$
  其中：$d(\frac{w}{3})=\frac{1}{3}dw$，则 $dw=3d(\frac{w}{3})$，所以 $\frac{d}{dw}[F(j\frac{w}{3})]=\frac{1}{3}\frac{d}{d(\frac{w}{3})}[F(j\frac{w}{3})]$，则 $t\cdot f(3t)\leftrightarrow j\frac{1}{9}\cdot F^{\prime }(j\frac{w}{3})$

• （2）$(t-1)\frac{d[f(t)]}{dt}$
  解：

$$\frac{d[f(t)]}{dt}\leftrightarrow jwF(jw)$$
$$t\cdot \frac{d[f(t)]}{dt}\leftrightarrow j\frac{d}{dw}[jwF(jw)]=-\frac{d}{dw}[wF(jw)]=-[F(jw)+w{F}^{\prime }(jw)]$$
$$(t-1)\cdot \frac{d[f(t)]}{dt}\leftrightarrow j\frac{d}{dw}[jwF(jw)]=-\frac{d}{dw}[wF(jw)]=-[F(jw)+w{F}^{\prime }(jw)]-j{F}^{\prime }(jw)=-[F(jw)+(w+1)F^{\prime }(jw)]$$

• （3）$(2-t)f(2-t)$
  解：

$$f(2-t)=f[-(t-2)]\leftrightarrow F(-jw)e^{-j2w}$$
$$t\cdot f(2-t)\leftrightarrow j\frac{d}{dw}[F(-jw)e^{-j2w}=-j\frac{d}{d(-w)}[F(-jw)e^{-j2w}]=-j[F^{\prime }(-jw)e^{-j2w}-j2F(-jw)e^{-j2w}]$$
$$(2-t)\cdot f(2-t)\leftrightarrow 2F(-jw)e^{-j2w}+[j{F}^{\prime }(-jw)e^{-j2w}-2F(-jw)e^{-j2w}]=j{F}^{\prime }(-jw)e^{-j2w}$$

2 补充：二倍角以及积化和差公式

$$\{\begin{array}{l}cos\alpha \cdot cos\beta =\frac{1}{2}[cos(\alpha +\beta )+cos(\alpha -\beta )]【调制】\\ sin\alpha \cdot sin\beta =\frac{1}{2}[cos(\alpha +\beta )-cos(\alpha -\beta )]\\ cos\alpha \cdot sin\beta =\frac{1}{2}[sin(\alpha +\beta )-sin(\alpha -\beta )]\\ sin\alpha \cdot cos\beta =\frac{1}{2}[sin(\alpha +\beta )+sin(\alpha -\beta )]\end{array}$$

$$sin(2\alpha )=2sin\alpha \cdot cos\alpha$$
$$\{\begin{array}{l}cos(2\alpha )=2co{s}^2\alpha -1=1-2si{n}^2\alpha =co{s}^2\alpha -si{n}^2\alpha \\ co{s}^2\alpha =\frac{cos(2\alpha )+1}{2}\\ si{n}^2\alpha =\frac{1-cos(2\alpha )}{2}\end{array}$$

$$tan(2\alpha )=\frac{2tan\alpha }{1-ta{n}^2\alpha }$$