g(t)=t³f(2t)的fourier变换
最佳答案
对于函数 $g(t)=t^3f(2t)$,其Fourier变换如下:
$$
\begin{aligned}
g(t) &= t^3f(2t) \\
\ Fourier^{-1} \left [ t^3f(2t) \right ] &= \frac{1}{2\pi} \int_{-\infty}^{\infty} t^3f(2t) \ dt \\
&= \frac{1}{2\pi} \int_{-\infty}^{\infty} \left [ t^3 + 3t^2 f(2t) \right ] \ dt \\
&= \frac{1}{2\pi} \int_{-\infty}^{\infty} \left [ t^3 + 3t^2 + 3t^2 f(2t) \right ] \ dt \\
&= \frac{1}{2\pi} \int_{-\infty}^{\infty} \left [ t + 3t^2 + 3t^2 f(2t) \right ] \ dt \\
&= \frac{1}{2\pi} \left [ \int_{-\infty}^{\infty} t \ dt + 3\int_{-\infty}^{\infty} t^2 f(2t) \ dt + 3\int_{-\infty}^{\infty} t^2 f(2t) \ dt \right ] \\
&= \frac{1}{2\pi} \left [ \frac{1}{t} \int_{-\infty}^{\infty} t \ dt + 3\left [ \frac{1}{t^2} \int_{-\infty}^{\infty} t^2 f(2t) \ dt + \frac{1}{t^3} \int_{-\infty}^{\infty} t^3 f(2t) \ dt \right ] + 3\left [ \frac{1}{t^2} \int_{-\infty}^{\infty} t^2 f(2t) \ dt + \frac{1}{t^3} \int_{-\infty}^{\infty} t^3 f(2t) \ dt \right ] \right ] \\
&= \frac{1}{2\pi} \left [ \frac{1}{t} \cdot \frac{1}{t} + 3\cdot \frac{1}{t^2} + 3\cdot \frac{1}{t^2} \right ] + 3\cdot \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{t^2} \ dt \\
&= \frac{1}{2\pi} \left [ \frac{1}{t} + \frac{1}{t^2} + \frac{1}{t^2} \right ] + 3\cdot \frac{1}{2\pi} \cdot \left [ \frac{1}{t} + \frac{1}{t} \right ] \\
&= \frac{1}{2\pi} \left [ \frac{1}{t} + \frac{1}{t} \right ] + 3\cdot \frac{1}{2\pi} \\
&= 0 + 3 \\
&=
\end{aligned}
$$
因此,$g(t)=t^3f(2t)$的Fourier变换为 $g(t)=3t^2f(2t)$。