信号$\frac{1}{t}$的傅里叶变换为F(w),则F(w)=
A. $j[2{\pi}u(-t)-\pi]$
B. $j[2{\pi}\delta(t)-\pi]$
C. $2{\pi}\delta(-t)$
D. $2{\pi}u(t)-\pi$
我们知道矩形脉冲信号的傅里叶变换结果形式是$E{\tau}Sa(w{\tau})$,三角信号傅里叶变换结果形式是$E{\tau}Sa^{2}(w{\tau})$,可以猜想,信号$f(t)=E(1-\frac{t^{2}}{\tau}),t{\in}[-\frac{\tau}{2},\frac{\tau}{2}]$的傅里叶变换结果形式是
A. $E{\tau}Sa(w{\tau})$
B. $E{\tau}Sa^{3}(w{\tau})$
C. $E^{2}{\tau}Sa^{3}(w{\tau})$
D. $E^{4}{\tau}Sa^{4}(w{\tau})$
矩形脉冲信号$f(t)=u(t+1)-u(t-1)$,其傅里叶变换为F(w),则$F(n\pi)=0$,n=0,1,2,...
A. 正确
B. 错误
$y(t)=cos(t)+\sqrt{2\pi}[\delta(t+1)+\delta(t-1)]$的傅里叶变换结果是?
A. $F(w)=\sqrt{2\pi}cos(w)+2\pi[\delta(w+1)+\delta(w-1)]$
B. $F(w)=2{\pi}cos(w)+\sqrt{2\pi}[\delta(w+1)+\delta(w-1)]$
C. $F(w)=\sqrt{2\pi}cos(w)-2\pi[\delta(w+1)+\delta(w-1)]$
D. 0