Chapter 6

# Chapter 6 - Chapter 6 solutions See figures of output...

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Chapter 6 solutions See figures of output signals, next page Æ

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6.3, continued -0.04 -0.02 0 0.02 0.04 -0.5 0 0.5 1 1.5 a t -0.04 -0.02 0 0.02 0.04 -0.5 0 0.5 1 1.5 b t -0.04 -0.02 0 0.02 0.04 -0.5 0 0.5 1 1.5 c t -0.04 -0.02 0 0.02 0.04 -0.5 0 0.5 1 1.5 d t -0.04 -0.02 0 0.02 0.04 -0.5 0 0.5 1 1.5 e t -0.04 -0.02 0 0.02 0.04 -0.5 0 0.5 1 1.5 f t
Transfer Fcn 1 0.0016 s+1 Scope 1 Scope Pulse Generator Part (a)

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Part (b) Part (c)
Part (d) Part (e)

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Part (f)

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6.9 Continued Æ

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6.9(a), continued (b) (c)
6.11 (a) (Note that you don’t need the “Analog Butterworth LP Filter” block; just use a Transfer Function block with the coefficients derived from the ‘ butter(N, Wn, ‘s’) ’ command.) We should select a cutoff frequency for the low-pass filter so that the oscillations in the signal are eliminated as much as possible. This doesn’t specify a precise criterion, however. Here is the signal before and after filtering with a 2 nd order Butterworth low-pass filter with ω c =100 π : The next output plot uses ω c =20 π , giving a smoother result, although it takes longer to get there: (b) Here is the signal after filtering with a 4 th order Butterworth filter with ω c =20 π :

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6.11, (c)
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## This note was uploaded on 06/10/2010 for the course ECES 302 taught by Professor Carr during the Spring '08 term at Drexel.

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Chapter 6 - Chapter 6 solutions See figures of output...

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