Therefore, Z((w) is as follows (j) 190-170 100-80 10 80100 1701900 (b)y(t)is the output from the band-pass filter, HGw), with input z(t)as derived in part (a). We know, Gw)= hgu)zg Let us consider Yw)l and zY(w) separately. Y(a)I is the band-pass filtered version of ZG)l with frequency components between 90 to 180 and-180 to-90 rad/sec YGu) 170 100 100 170 ∠Y(j ∠H(1u)+∠Z(u aU 200 200 Combining the magnitude and angle, Y(w)=Y(jw)lejzrgu Consider Y(w)as the Fourier transform of the sum of two sinusoidal signals; one with Wo= 100 and another with Wo= 170. Using the time-shifting property of Fourier transform,a(t-to)+e-3uto X(u) cos(100(t 200+=cos(170(t 45 17 cos(100t 2)+zc(17ot (c)Now the sampling function s(t)is changed with T= 90
Therefore, Z(jw) is as follows: Z(jw) (45) ··· ··· −190−170 −100 −80 −10 10 80 100 170 190 w (b) y(t) is the output from the band-pass filter, H(jw), with input z(t) as derived in part (a). We know, Y (jw) = H(jw)Z(jw) Let us consider |Y (jw)| and ∠Y (jw) separately. |Y (jw)| is the band-pass filtered version of |Z(jw)| with frequency components between 90 to 180 and -180 to -90 rad/sec. |Y (jw)| (45) −170 −100 0 100 170 w ∠Y (jw) = ∠H(jw) + ∠Z(jw) πw πw = − + 0 = − 200 200 Combining the magnitude and angle, Y (jw) = |Y (jw)|ej∠Y (jw) . Consider Y (jw) as the Fourier transform of the sum of two sinusoidal signals; one with wo = 100 and another with wo = 170. Using the time-shifting property of Fourier ←→ e transform, x(t − to) −jwtoX(jw), FT 45 π 45 π y(t) = cos(100(t − )) + cos(170(t − )) π 200 π 200 45 π 45 17π = cos(100t − ) + cos(170t − ) π 2 π 20 (c) Now the sampling function s(t) is changed with T = 2π 90 , 6
s(t) 6(t-kn)-∑6( kT Taking the Fourier transform 2 6( k=-0 k=-0 2 TT k=-0 k=-0 2丌 6( k=-0 2丌 ∑( k=-0 Seperating the odd and even terms of k T 6(-k -even k=even +>(0-k)+>60-62F ∑6(m r(t)=cos(10t)as before. To find Z(w), we need to convolve X(o) with the impulse train in S(ju)and scale the result by 2T SGa) is as sketched below
� � � � s(t) 0 (1) (−1) −T 2 T 2 −T T −3T 2 3T 2 −2T 2T ··· ··· t � � T s(t) = δ(t − kT) − δ(t − kT − ) ∞ ∞ 2 k=−∞ k=−∞ Taking the Fourier transform, ∞ ∞ 2π 2π 2π 2π e−jw T S(jw) = δ(w − k ) − 2 T T δ(w − k ) T T k=−∞ k=−∞ 2π �∞ 2π 2π ∞ T 2π 2 δ(w − k π e− T jk 2 T δ(w − k ) − T T = ) T k=−∞ k=−∞ � 2π 2π ∞ = 2π ∞ δ(w − k ) − (e−jπ) kδ(w − k 2π) T T T T k=−∞ k=−∞ � 2π 2π ∞ 2π ∞ � 2π = δ(w − k ) − (−1)kδ(w − k ) T T T T k=−∞ k=−∞ Seperating the odd and even terms of k, 2π � 2π 2π � 2π S(jw) = δ(w − k ) − δ(w − k ) T T T T k=even k=even 2π � 2π 2π � 2π + δ(w − k ) + δ(w − k ) T T T T k=odd k=odd 4π � 2π = δ(w − k ) T T k=odd x(t) = cos( 10t) as before. To find Z(jw), we need to convolve X(jw) with the impulse train in S(jw) and scale the result by 2 1 π . S(jw) is as sketched below, 7