Lesson_33 - Lesson33 Challenge32...

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Lesson 33 Lesson 33 Challenge 32 Lesson 33 – Discrete Fourier Transform Challenge 33
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  Lesson 33 Challenge 32 A 2 nd  order Butterworth filter has been designed and has a transfer function: Determine the filter’s frequency response and determine its value at the  -3dB frequency. ( 29 6 3 2 6 10 10 2 10 + + = s s s H
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  Lesson 33 Challenge 32 The DC gain of the filter is computed to be H(s=j0)=1.0. The -3dB gain point  corresponds to the gain decreasing to a value of H(s=j ϖ c )=0.707.  This can be  achieved by setting the filter’s the denominator to: D(s)| s=j ϖ c  =s 2 +1.414 10 3 s +10 6 | s=j ϖ c  = 1.414 10 6 .   Such that H(s)| s=j ϖ c =10 6 / D(s)| s=j ϖ c  =0.707  For  ϖ c =10 3 , D(j10 3 )=1.414x10 6   and H(j10 3 )=0.707 ( 29 6 3 2 6 10 10 2 10 ) ( ) ( + + = = s s s D s N s H
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  Lesson 33 Challenge 32 Evaluating.   MATLAB study over  ϖ ∈ [0,104] r/s. >> w=1:10^4; >> s=complex(zeros(1,10^4),w); >> d= s.*s + 1414 .*s + 1000000; >> n=1000000; >> h=n./d; >> plot(abs(h)); >> plot(180*angle(h)/pi); ( 29 ° - = = + + - = + + = = = 90 707 . 0 10 2 10 10 10 2 10 10 10 10 2 10 6 6 6 6 6 6 10 6 3 2 6 10 3 3 j j s s s H s s
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  Lesson 33 Challenge 32 0.707 or -3dB -90 ° ϖ c =10 3
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  Lesson 33 Lesson 33 Communications and modulation Information Transmitter Channel Receiver Information Table 1: Commercial Radio Item AM FM Carrier frequency 540-1600 kHz 88.1-107.9 MHz Transmission Bandwidth 10kHz 200kHz Audio bandwidths 3-5kHz 15kHz
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  Lesson 33 Lesson 33 AM-DSB  (DSBAM) Modulation: x(t) cos( ϖ c t) y(t) = x(t) cos( ϖ c t) Y(j ϖ ) = (1/2)(X(j( ϖ - ϖ c ) + X(j( ϖ + ϖ c )) 
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  Lesson 33 Lesson 33
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  Lesson 33 Lesson 33 Modulating
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This note was uploaded on 08/21/2010 for the course EEL 3135 taught by Professor ? during the Spring '08 term at University of Florida.

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Lesson_33 - Lesson33 Challenge32...

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