Therefore, p()=(∑ Taking Fourier transform ∑ Following is the plot of P(ja). Sm(PGo)=0 PGo) Note that the impulse at the origin disappeared in the above graph because of the subtraction with 2(w) Now, using the multiplication property again, Y(u)=2(Z(u)* P(u) The real and imaginary part of Y(a) is shown below Ryujo) 2=m-2a2-2a2+om 20 20+00 Myo)/ 20c=� � Therefore, � � � +∞ 2π p(t) = q(t) ∗ δ(t − n wc ) − 1 n=−∞ Taking Fourier transform, ∞ 4 sin( πω ) � P(jω) = 2ωc ωc δ(ω − kωc) − 2πδ(ω) ω k=−∞ Following is the plot of P(jw). m{P(jw)} = 0. P(jω) −2ω c −ωc −4ω c −3ωc −6ω c −5ωc ω c 2ω c 3ω c 4ω c 5ω c 6ω c ω 0 4 4 Note that the impulse at the origin disappeared in the above graph because of the subtraction with 2πδ(ω). Now, using the multiplication property again, Y (jω) = 2 1 π (Z(jω) ∗ P(jw)). The real and imaginary part of Y (jw) is shown below: Re{ Y(jω)} π −2ω 2ω 2 ω −2ω −ωm −2ωc+ωm −ωm ωm 2ωc−ωm 2ωc c c c +ωm 2 π Im{ } Y(j ) ω −2ω c−ωm −2ω c+ωm −ω m ωm 2ω c−ωm 2ω c+ωm ω 2