hwset4solution

# Assuming that the lter is symmetrical around fif 1

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Unformatted text preview: IF 2 − ∆f2 /2, where ∆f2 is the bandwidth of the last IF ﬁlter. Assuming that the ﬁlter is symmetrical around fIF 1 , the maximum bandwidth is then 4fIF 2 − ∆f2 . The same result is obtained if the secondary image is at fIF 2 − 2fIF 2 . b. From part a.: ∆f1 ≤ 4fIF 2 (ignoring ∆f2 ). The maximum fractional bandwidth of the ﬁrst IF ﬁlter is ∆f1 /fIF 1 = 4fIF 2 /fIF 1 . The requirement that ∆f1 /fIF 1 ≥ 0.02 leads to 4fIF 2 /fIF 1 ≥ 0.02, or fIF 1 ≤ 4(fIF 1 )/0.02 = 4(0.455)/0.02 = 91 MHz. 2-17 Solution |3fLO − 3fi | = fIF 3fLO − 3fi = ±fIF fi = fLO ∓ fIF /3 Use the fact that f...
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## This homework help was uploaded on 03/13/2014 for the course ECE 453 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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