Lesson+19 - dilbert 1 Lesson 19 Lesson 19 Challenge 18...

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dilbert 1 Lesson 19
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Challenge 18 Filters - Chapter 14:1-3 Challenge 19 Exam #1 average ~ 80 (with curve) Lesson 19 Lesson 19 2
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Challenge 18 The input v i (t) is an impulse. What is v 0 (t)? ) ( 100 ) ( 100 100 / 10 10 / 10 (s) V divider Voltage 100 0 6 4 6 0 t u e t v s s s t v 0 (t) called the circuit’s impulse response, If v i (t) had been a unit step, then v 0 (t)=k(1-e -100t )u(t ) V 0 v(t)= (t) R=10 4 C=10 -6 V(t) R C 3 Lesson 19 v i (t)
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V1(s)=100/s V2(s)=50 R=10 R=15 ZL=3s ZL=2s LI1=30 Impulse Circuit Circuits can spawn impulses. Analyze a circuit in the usual manner and let the math find the impulses. V 0 4 Lesson 19 V1 V2 R=10 R=15 L=3 L=2 S 0 ) 0 ( 10 ) 0 ( 100 ) ( ) ( 50 ) ( 2 1 2 1 i i t V t t V i 1 i 2 V 0 (s) t<0 t> 0 I 2 (s) 1 1 (s) At 0 - only the 100V source is non-zero Section 13.1, IC is given by Li 1 (0 - ) = 3(10) = 30
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V1(s)=100/s V2(s)=50 R=10 R=15 ZL=3s ZL=2s LI1=30 5 Lesson 19 I 1 (s) V 0 (s) ) ( 4 12 ) ( 4 5 12 5 25 30 ) / 100 ( 50 ) ( 5 1 1 t u e t i s s s s s I t No impulse s s s I s s V 60 5 60 32 ) ( ) 2 15 ( ) ( 1 0 Voltage impulse appears ) ( ) 60 60 ( ) ( 32 ) ( 5 0 t u e t t v t I 2 (s)
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V1(s)=100/s V2(s)=50 R=10 R=15 ZL=3s ZL=2s LI1=30 6 Lesson 19 I 1 (s) V 0 (s) ) ( ) ( 4 5 12 5 25 30 ) / 100 ( 50 ) ( 2 1 1 s I s I s s s s s I Final value = 4 Post Mortem i(0 + )=12+4=16A i 1 (0 - )=10A i 1 (0 + )=16A 1 =6A I 2 (s) i 2 (0 - )=0A i 2 (0 + )=16A 2 =16A Remember V=(L)di(t)/dt
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Lesson 19 Filter overview Frequency selective filters provide gain at some frequencies (called passband), attenuation at other frequencies (called stopband). Like steady-state frequency response having a magnitude and phase response. 0 0 0 0 Lowpass Highpass Bandpass Bandstop 7 Lesson 19
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>> n=5; r=80; wp =[8e6 12e6]; (really order 10) >> 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 (fiilter order) -80 dB stopband @ 80 & 120M r/s, How are these filters actually designed? - by machines 8 Lesson 19 From Lesson 18
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From Lesson 18 >> tf(b,a) % transfer function 10 th order transfer function: - (complicated) 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 Machine computes H(s)=|H(s)| (s) at s=j , 0  < . 9 Lesson 19
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Lesson 19 RL Lowpass Filters (Author) V L R What does the circuit look like at j =0? What does the circuit look like at j = ? Where are the filters poles and zero. L R s L R s H / / ) ( The half power point (-3dB) is the frequency that satisfies P(j -3 ) = P max /2 or |V 0 (j -3 )| = 0.707H max ( ) . Z L =sL V 0 -R/L j 10 Lesson 19
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RL Lowpass Filters (Author) V L R ) / ( tan ) ( / / | ) ( | / / ) ( R L j L R L R j H L R j L R j s H ω ω θ ω ω ω ω 1 2 2 10 -2 10 -1 10 0 10 1 -100 -80 -60 -40 -20 0 Frequency (radians) Phase (degrees) 10 -2 10 -1 10 0 10 1 10 -1 10 0 Frequency (radians) Magnitude 11 Lesson 19 V 0 -3B frequency
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In-Class Problem V L R V 0 R=L=1.0 What is the filter’s complex gain at the -3dB frequency if R=L=1?
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