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# HW1-ans - 2.2.1 Given Sa(2)=sinc(f and = 2f Sat=sintt at t0...

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2.2.1) Given ( )= ( ) Sa ωϵ2 sinc fϵ and ω = 2πf = Sat sintt at t 0 = = 1 at t 0 Therefore Saωϵ2 ⟹Sa2πfϵ2⟹Saπfϵ as ω = 2πf But Saπfϵ⟹sinπfϵπfϵ ---(1) since = Sat sintt = sinct sinπtπtat t 0 = = 1 at t 0 Therefore = sincfϵ sinπfϵπfϵ -----(2) Since (1) = (2) it implies that ( )= ( ) Sa ωϵ2 sinc fϵ 2.2.3) Code: >> t=-2:0.1:20; >> x=sin(t); >> x(t<=0)=0; >> plot(t,x) 2.2.5)

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(ii) -∞∞ = -∞∞ 4cosπλδλdλ 4cosπλ δλdλ But the impulse function i.e δλ = 0 -∞< < at λ 0 = = 1 at λ 0 = < <∞ 0 at 0 λ -∞∞ = × = -∞∞ = ⟹4cosπλ δλdλ 4cosπ0 1 as at λ 0
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HW1-ans - 2.2.1 Given Sa(2)=sinc(f and = 2f Sat=sintt at t0...

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