Use a convolution integral to find f*f, if f(t)=sint.τ=tau -So far, I set it up as h(t)=(f*f)(t)=∫ [from 0 to t of] sin(t-τ) sin(τ) dτ -Integrate by parts u=sinτ , du=cosτ dτ , dv=sin(t-τ) dτ , v=cos(t-τ) h(t)=sinτ cos(t-τ) - ∫ cos(t-τ) cos(τ) dτ -Integrate by parts again u=cosτ , du=-sinτ dτ , dv=cos(t-τ) dτ , v=-sin(t-τ) h(t)=sinτ cos(t-τ) - [sin(t-τ) cosτ - ∫ (-sin(t-τ) ) (-sin(τ) ) dτ] h(t)=∫ sin(t-τ) sin(τ) dτ and so...both sides cancel and you're left with: 0=sinτ cos(t-τ) - sin(t-τ) cosτ ??

Question

Use a convolution integral to find f*f, if f(t)=sint.τ=tau   -So far, I set it up as h(t)=(f*f)(t)=∫ [from 0 to t of] sin(t-τ) sin(τ) dτ   -Integrate by parts u=sinτ , du=cosτ dτ , dv=sin(t-τ) dτ , v=cos(t-τ) h(t)=sinτ cos(t-τ) - ∫ cos(t-τ) cos(τ) dτ   -Integrate by parts again u=cosτ , du=-sinτ dτ , dv=cos(t-τ) dτ , v=-sin(t-τ) h(t)=sinτ cos(t-τ) - [sin(t-τ) cosτ - ∫ (-sin(t-τ) ) (-sin(τ) ) dτ] h(t)=∫ sin(t-τ) sin(τ) dτ   and so...both sides cancel and you're left with: 0=sinτ cos(t-τ) - sin(t-τ) cosτ ??

anonymous · Answer

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