Show that the limit of 1/2T * integral of m²(t)cos(2wct)dt from -T/2 to T/2 as T approaches infinity is 0Power efficiency in amplitude modulation Consider the amplitude-modulated signal ϕₐₘ (t)=A cosω t+m(t)cosω t. The carrier is A cosω t, while m(t)cosω t represents the sidebands. The power in the carrier is P =A²/2Let Pₛ be the power in the sidebands and Pₘ the power in the message signal. Then Pₛ= lim T→[infinity]1/T∫ₜ/₂ᵀ/² m² (t)cos² (ωt)dt=1/2 lim T→[infinity] 1/T ∫ ʳ²₋ₜ/₂ m²(t)[1+cos(2ω t)]dt But ∫ᵗ²/²ₜ/₂ m²(t)cos(2ωt)dt=0 Therefore, Pₛ=1/2 lim T→[infinity]1/T∫ₜ/₂ᵀ/²m² (t)dt The right-hand side of the last equation is half the power Pₘ in the message signal. Thus, Pₛ = 1/2Pₘ
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