hw3_sol

# At frequency omega 3pi4 homega has amplitude 14 so

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Unformatted text preview: s 3 fs fs (18) Figure 12: W (ω ) = Zd (ω )H (ω ) (19) At frequency omega = pi/4, H(omega) has amplitude 3/4; so 2x3/4 gives us 3/2. At frequency omega = 3pi/4, H(omega) has amplitude 1/4; so -2/3x1/4 gives us -1/6. Figure 13: Y (f ) = W (ω )|ω= 2πf and ﬁltered with LPF of cutoﬀ frequency 4kHz fs 3 1 1 δ (f ) + (δ (f − 1000) + δ (f + 1000)) − (δ (f − 3000) + δ (f + 3000)) 2 4π 12π 1 3 1 y (t) = + cos(2π (1000t)) − cos(2π (3000t)) 2 2π 6π Y (f ) = 7 Figure 14: b) P = 0.5ms, 1 P = 2kHz, fc = 8kHz. Z (f ) = HLP (f )X (f ) Figure 15: n ) fs ω Zd (ω ) = fs repfs [Z ( fs )] 2π z [n] = z (nTs ) = z ( Since fs &lt; 2fc , there will be aliasing in Zd (ω ). 8 Figure 16: W (ω ) = Zd (ω )H (ω ) Figure 17: Y (f ) = W (ω )|ω= 2πf and ﬁltered with LPF of cutoﬀ frequency 4kHz fs Figure 18: 1 1 δ (f ) + (δ (f − 2000) + δ (f + 2000)) 2 3π 1 2 y (t) = + cos(2π (2000t)) 2 3π Y (f ) = 9 Here is an alternative derivation that leads to a simpler expression for H(omega) and allows one to very quickly evaluate H(omega) for omega = pi/2 and pi. Problem 4 a) Let x[n] = δ [n],then h[n] = (δ [n] + δ [n − 1] + δ [n − 2])/3 H (ω ) = (1 + e−jω + e...
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