下面函数在区间(0，π）上与函数$y(t)=sin(t)+cos(t)$正交的是
A. $y=cos(2t)$
B. $y=sin(2t)$
C. $y=sin(t)-cos(t)$
D. $y=2sin(t)$
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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} } $
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)] $
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