Lesson+18 - 1 Lesson 18 Lesson 18 Lesson 18 Challenge 17...

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Lesson 18 1 Lesson 18
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Challenge 17 Continuation of Lesson 17 and impulsive circuits - Chapter 13.8 Challenge18 Lesson 18 2 Lesson 18
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Challenge 17 Given: What is the steady-state response of y(t) = h(t)*x(t)? You need to compute the system’s magnitude and phase responses at = 2 (i.e., at the input frequency). Therefore ) ( ) ( ) ( ) ( ) ( ) ( t u t t x and s s s H 60 2 cos 2 3 2 3 . 56 45 . 0 45 2 8 7 . 33 13 ) 2 2 ( ) 3 2 ( ) 2 ( ; 2 @ 2 2 j j j H   ) 2 cos( 5 . 0 ~ ) 4 2 ( cos 45 . 0 ) 3 . 56 60 2 ( cos 1 45 . 0 ) ( t t t t y 3 Lesson 18
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Review: Steady-State Analysis >> n=5; r=80; wp=[8e6 12e6]; >> ftype='bandpass'; >> [b,a]=cheby2(n,r,wp,ftype,'s'); % filter >> [H,w]=freqs(b,a); % frequency response >> plot(w,20*log10(abs(H))); N=10 (filter order), not 5! -80 dB stopband How is a system’s steady-state frequency response actually synthesized and/or analyzed? - by machines % b=num; a=den 4 Lesson 18
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>> tf(b,a) % transfer function H(s)=b(s)/a(s) Transfer function: (10 th order) 2000 s^9 - 2.939e-005 s^8 + 8.96e017 s^7 - 9.43e009 s^6 + 1.368e032 s^5 - 7.097e023 s^4 + 8.258e045 s^3 - 2.466e036 s^2 + 1.699e059 s + 7.654e050 --------------------------------------------------------------------------- s^10 + 3.546e006 s^9 + 4.863e014 s^8 + 1.369e021 s^7 + 9.398e028 s^6 + 1.974e035 s^5 + 9.022e042 s^4 + 1.261e049 s^3 + 4.302e056 s^2 + 3.012e062 s + 8.154e069 Machines can compute H(s)=|H(s)| (s) at s=j , 0  < . The problem is not designing the filter but implementing the filter in hardware or software. 5 Lesson 18
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On the practical side SAW RF analog filter Notch RF filter 6 Lesson 18
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General use (analog filters) Supply external capacitors. 7 Lesson 18
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TI – Internet Radio ($4.00) Analog components 8 Lesson 18
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To those few students here who have actually operated a vacuum cleaner and couldn’t resist the temptation to “tug the power plug” to disconnect the machine while it still running, you inadvertently invented a ozone generator. What is the physical explanation on why the machine drew that long blue arc? 9 Lesson 18
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Vi(t) R L R C v 0 (t) 1 2 ½ F 1 H v 1 (t) Some system’s are naturally interpreted in the context of an impulse. Consider the proper system studied in Challenge 16. 4 4 3 ) ( 2 2 s s s s H Zeros = 0, 0. Poles = -0.667 j0.943 Heaviside: H(s) = K 1 +K 2 /(s+0.667+j0.943)+ K 3 +K2/(s+0.667+j0.943) K 1 0 because the system is proper.
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