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Unformatted text preview: (1)Answer: It is easily calculated from the specification of the digital lowpass filter that: 2 / 24 / 12 2 ˆ 3 / 24 / 8 2 ˆ π π π ϖ ϖ π π π ϖ ϖ = = = = = = = = k k T f T k k T f T s s s p p p . And the value of the filter’s frequency response at the above edge frequency can be calculated as follows: 01 . 10 ) ( 40 ) ) ( ( log 20 89126 . 10 ) ( 1 ) ) ( ( log 20 2 10 20 / 1 10 = = Ω ⇒ = Ω = = Ω ⇒ = Ω s s p p j H j H j H j H . Then after prewarped the edge frequencies with normalized sampling period T=1, we can specify the requirement for the Butterworth analog filter as follows. ∞ ≤ Ω ≤ = = Ω ≤ Ω Ω = = ≤ Ω ≤ ≤ Ω ≤ 2 ) /2 2 tan( 2 01 . ) ( 1.1547 /2) 3 2tan( 1 ) ( 89126 . s p π π j H j H Using the two frequency edges, we have 4 2 2 2 2 10 ) 01 . ( ) / 2 ( 1 2589 . 1 ) 89126 . ( ) / 1547 . 1 ( 1 = ≥ Ω + = = Ω + N c N c . By dividing one equation by the other, we obtain the order and cutoff frequency of the filter as 2354 . 1 ) 1 2589 . 1 /( 1547 . 1 10 9.6159 N 1 10 1 2589 . 1 ) 2 / 1547 . 1 ( 20 / 1 c 4 2 = = Ω ≈ ≥ ⇒ ≤ N ....
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This note was uploaded on 01/11/2011 for the course EE 4015 taught by Professor Shuhungleung during the Fall '10 term at City University of Hong Kong.
 Fall '10
 ShuHungLeung
 Digital Signal Processing, Frequency, Signal Processing

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