求单边余弦函数的傅里叶变换

求单边余弦函数的傅里叶变换

最近在学习信号与系统课程,输入信号为 x ( t ) = cos ⁡ ( ω 0 t ) u ( t ) x\left( t \right) = \cos \left( {{\omega _0}t} \right)u\left( t \right) x(t)=cos(ω0t)u(t)需要求其傅里叶变换
已知余弦函数的傅里叶变换为 F [ cos ⁡ ω 0 t ] = π [ δ ( ω − ω 0 ) + δ ( ω + ω 0 ) ] F\left[ {\cos {\omega _0}t} \right] = \pi \left[ {\delta \left( {\omega - {\omega _0}} \right) + \delta \left( {\omega + {\omega _0}} \right)} \right] F[cosω0t]=π[δ(ωω0)+δ(ω+ω0)]
单边阶跃函数的傅里叶变换为 F [ u ( t ) ] = π δ ( ω ) − 1 ω j F\left[ {u\left( t \right)} \right] = \pi \delta \left( \omega \right) - {1 \over \omega }j F[u(t)]=πδ(ω)ω1j
根据傅里叶变换的性质 F [ cos ⁡ ( ω 0 t ) u ( t ) ] = 1 2 π [ F [ cos ⁡ ω 0 t ] ∗ F [ u ( t ) ] ] F\left[ {\cos \left( {{\omega _0}t} \right)u\left( t \right)} \right] = {1 \over {2\pi }}\left[ {F\left[ {\cos {\omega _0}t} \right] * F\left[ {u\left( t \right)} \right]} \right] F[cos(ω0t)u(t)]=2π1[F[cosω0t]F[u(t)]]
= π 2 [ δ ( ω − ω 0 ) + δ ( ω + ω 0 ) ] − ( ω 0 j ω 2 − ω 0 2 ) = {\pi \over 2}\left[ {\delta \left( {\omega - {\omega _0}} \right) + \delta \left( {\omega + {\omega _0}} \right)} \right] - \left( {{{{\omega _0}j} \over {{\omega ^2} - {\omega _0}^2}}} \right) =2π[δ(ωω0)+δ(ω+ω0)](ω2ω02ω0j)

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