# hw2_p4 - (a ∞ g(n)e−iωn G(ω = n=−∞ ∞ = ∞ 1 1...

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Unformatted text preview: (a) +∞ g (n)e−iωn G(ω ) = n=−∞ +∞ = +∞ 1 1 δ (n)e−iωn − δ (n − 1)e−iωn 2 2 n=−∞ n=−∞ 1 1 −iω −e 22 1 1 − e−iω = 2 1 = e−iω/2 eiω/2 − e−iω/2 2 1 −iω/2 ω =e 2j · sin 2 2 ω −iω/2 −iπ =e e sin 2 ω = e−i(ω/2+π) sin 2 [using sifting property] = This ﬁlter is a high-pass ﬁlter, as is evident from the magnitude plot. See Figure 2 for the magnitude plot and Figure 3 for the phase plot. (b) (i) See Figure + x[n] y [n] − D D 1 2 Figure 1: Block Diagram for Problem 4.b.i (ii) The impulse response h(n) is the output of the ﬁlter when the input is δ (n), that is, h(n) = y (n) if x(n) = δ (n). Therefore, h(n) = 1 [δ (n) − δ (n − 2)] 2 (iii) +∞ h(n)e−iωn H (ω ) = n=−∞ +∞ = +∞ 1 1 δ (n)e−iωn − δ (n − 2)e−iωn 2 2 n=−∞ n=−∞ 1 1 −i2ω −e 22 1 1 − e−i2ω = 2 1 = e−iω eiω − e−iω 2 1 = e−iω 2j · sin (ω ) 2 = e−iω e−iπ sin (ω ) = [using sifting property] = e−i(ω+π) sin (ω ) This ﬁlter is a band-pass ﬁlter, as is evident from the magnitude plot. See Figure 2 for the magnitude plot and Figure 3 for the phase plot. (c) ¢ ¢ Magnitude Responses for Problem 4 1.0 G( ) H( ) 0.8 0.6 0.4 0.2 ¡ 0.0 4 3 2 1 0 1 Figure 2: Magnitude Plots for Problem 4 2 3 4 ¥ ¥ Phase Responses for Problem 4 2.0 G( ) H( ) 1.5 1.0 0.5 ¤££ £££ £ £ £ 0.0 0.5 1.0 1.5 2.0 4 3 2 1 0 1 Figure 3: Magnitude Plots for Problem 4 2 3 4 ...
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## This note was uploaded on 12/24/2011 for the course ELECTRICAL 20 taught by Professor Babak during the Spring '11 term at Berkeley.

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hw2_p4 - (a ∞ g(n)e−iωn G(ω = n=−∞ ∞ = ∞ 1 1...

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