信号$ e^{-at}cos(w_{0}t)u(t) $的傅里叶变换结果是
A. $ \frac{a + jw}{ (a+jw)^{2} + w_{0}^{2} } $
B. $ \frac{a + jw_{0}}{ (a+jw)^{2} + w_{0}^{2} } $
C. $ \frac{a + jw}{ (a+jw_{0})^{2} + w^{2} } $
D. $ \frac{a + jw_{0}}{ (a+jw_{0})^{2} + w^{2} } $
查看答案
信号$ te^{-|t|} $的傅里叶变换结果为
A. $ -\frac{4jw}{(1+w^{2})^{2}} $
B. $ \frac{2w}{(1+w^{2})^{2}} $
C. $ \frac{2jw}{(1+w^{2})^{2}} $
D. $ -\frac{2jw}{1+w^{2}} $
信号$ f_{1}(t)=cos(4\pi t) $, $ f_{2}(t)= u(t+\tau)-u(t-\tau) $, $ f(t)= f_{1}(t)f_{2}(t) $,则f(t)的傅里叶变换结果为
A. $ Sa(w+4\pi)+Sa(w-4\pi) $
B. $ Sa(w+4\pi) - Sa(w-4\pi) $
C. $ Sa(w+2\pi)+Sa(w-2\pi) $
D. $ Sa(w+2\pi) - Sa(w-2\pi) $
抽样信号$ Sa(2\pi t) $的傅里叶变换结果为
A. $ u(w+2\pi)-u(w-2\pi) $
B. $ u(w+2\pi) + u(w-2\pi) $
C. $ \frac{1}{2}[u(w+2\pi)-u(w-2\pi)] $
D. $ \frac{1}{2}[u(w+2\pi)+u(w-2\pi)] $
信号$f(t)=Asin(\omega_0t)\varepsilon(t)$的傅里叶变换结果是
A. $\frac{jA\pi}{2}[\delta(\omega+\omega_0)-\delta(\omega-\omega_0)]-\frac{\omega_0A}{\omega^2+\omega_0^2}$
B. $\frac{jA\pi}{2}[\delta(\omega+\omega_0)-\delta(\omega-\omega_0)]-\frac{\omega_0A}{\omega^2-\omega_0^2}$
C. $\frac{jA\pi}{2}[\delta(\omega+\omega_0)-\delta(\omega-\omega_0)]+\frac{\omega_0A}{\omega^2+\omega_0^2}$
D. $\frac{A\pi}{2}[\delta(\omega+\omega_0)-\delta(\omega-\omega_0)]-\frac{\omega_0A}{\omega^2-\omega_0^2}$
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