Electronic Circuits EE359A
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1 Electronic Circuits EE359A Bruce McNair B Lecture
2 Second order section Ts () = s as + as+ a ω + s+ ω Q 2 2 ω 1 p, p = ± 1 Q 4 Q
3 Second order section Ts () = s as + as+ a ω + s+ ω Q 2 2 ω 1 p, p = ± 1 Q 4 Q Numerator zeroes determine filter response: LP, HP, BP, BS, AP 571
4 Second order sections 572
5 Second order sections 573
6 Second order sections 574
7 Second order LCR resonator Basic resonator 575
8 Second order LCR resonator Transfer function Current drive V o I Voltage drive V o V i 576
9 Second order LCR resonator-poles V o I 1 1 = = Y sc + sl R s = C 2 s 1 s + + CR LC s = C 2 sω 2 s + + ω Q ω 2 ω Q 1 = LC 1 = CR 1 ω = LC Q = ω CR 577
10 Second order LCR resonator-poles 1 ω = LC Q = ω CR ω 1 = 2Q 2CR 578
11 Filter designs approximations to ideal response Butterworth response Tradeoffs of Butterworth + Maximally flat passband + Good phase shift characteristics (more on this later) - Poor attenuation in stopband - Large N (complex filter) needed for reasonable stopband attenuation, rolloff 579
12 Filter designs approximations to ideal response Butterworth vs. Chebychev response Tradeoffs of Butterworth + Maximally flat passband + Good phase shift characteristics (more on this later) - Poor attenuation in stopband - Large N (complex filter) needed for reasonable stopband attenuation, rolloff Tradeoffs of Chebychev - Ripple in passband - Not very good phase shift characteristics (more on this later) + Excellent attenuation in stopband + Smaller N (less complex filter) needed for reasonable stopband attenuation, rolloff 58
13 Filter designs approximations to ideal response Chebychev response Ideal filter Chebychev response 581
14 Filter designs approximations to ideal response Chebychev response 1 T( jω) = for ω ωp ω 1+ ε cos N cos ω p 1 T( jω) = for ω ωp ω 1+ ε cosh N cosh ω p 1 T( jω p ) = 2 1+ ε 582
15 Filter designs approximations to ideal response Chebychev response p k 2k 1π 1 1 N 2 N ε 1 = ωpsin sinh sinh 2k 1π = N 2 N ε 1 jω p cos cosh sinh for k 1,2,..., N Ts () = ε kω N P N 1 2 ( s p1)( s p2)...( s pn ) 583
16 Filter designs approximations to ideal response Chebychev response 1 1 ( ( ) ) 2 log T j ω i ω i 2 Im( p k ).5.5 Pole locations for N=8 ε = 3 db ( ) Re p k 584
17 Choosing filter parameters: Butterworth Example D16.6 Butterworth filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db 585
18 Choosing filter parameters: Butterworth Example D16.6 Butterworth filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db 1 db 3 db
19 Choosing filter parameters: Butterworth Example D16.6 Butterworth filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db A max 2 = 2 log 1+ ε ε = Amax /1 1 1 ε =.59 (16.14) 587
20 Choosing filter parameters: Butterworth Example D16.6 Butterworth filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db A max 2 = 2 log 1+ ε ε = Amax /1 1 1 ε =.59 (16.14) 1 A( ωs ) = 1log 2N 2 ω s 1+ ε ω p (16.15) N = A( ωs ) log 2 ε ω s 2log ω p N = à
21 Choosing filter parameters: Butterworth Example D16.6 Butterworth filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db ε =.59 N = à 11 1 A( ωs ) = 1log 2N 2 ω s 1+ ε ω p A(1.5) = db 589
22 Choosing filter parameters: Butterworth Example D16.6 Butterworth filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db ε =.59 N = à 11 If A min is to be exactly 3 db, what will A max be? 59
23 Choosing filter parameters: Butterworth Example D16.6 Butterworth filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db ε =.59 N = à 11 If A min is to be exactly 3 db, what will A max be? 1 A( ωs ) = 1log 2N 2 ω s 1+ ε ω p A( ωs ) ω 1 s ε = 1 1 ω p 2N ε =. A max =.544 db 591
24 Choosing filter parameters: Butterworth Example D16.6 Butterworth filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db 1 A( ω) ε =.59 N = à 11 If A min is to be exactly 3 db, what will A max be? ε = A max =.544 db ω 2 592
25 Choosing filter parameters: Chebychev Example 16.2 (modified) Chebychev filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db ε = Amax 1 /1 1 (16.21) ε =
26 Choosing filter parameters: Chebychev Example 16.2 (modified) Chebychev filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db Amax ε = 1 /1 1 (16.21) A ε =.59 ω S = 1 log 1+ cosh cosh ω (16.22) p A(ω s ) 1 cosh ε 2 N = N = 5.22 ω cosh 1 s ( ω ) ε N S ω p 594
27 Choosing filter parameters: Chebychev Example 16.2 (modified) Chebychev filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db Amax ε = 1 /1 1 (16.21) ε = ω S A( ωs ) = 1 log 1+ ε cosh N cosh ω (16.22) p Or, use a tool like Mathcad to calculate A(ω s ) for various N N A(ω s ) db db db 595
28 Choosing filter parameters: Chebychev Example 16.2 (modified) Chebychev filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db Amax ε = 1 /1 1 (16.21) A ε =.59 ω S = 1 log 1+ cosh cosh ω p ( ω ) ε N S (16.22) N A(ω s ) db db db An 11 th order Butterworth would have been necessary 596
29 Choosing filter parameters: Chebychev Example 16.2 (modified) Chebychev filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db th order Butterworth A B ( ω) 1 A C ( ω) th order Chebychev ω 597
30 Comparing filter performance Butterworth vs. Chebychev filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db Phase response th order Butterworth ϕ.b ( ω) ϕ.c ( ω) 1 5 th order Chebychev ω 2 598
31 Comparing filter performance Butterworth vs. Chebychev filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db Group delay response ( ) = φ( ω) τ ω d dω th order Butterworth τ B ( ω) τ C ( ω) th order Chebychev ω 599
32 Comparing filter performance Butterworth vs. Chebychev filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db Group delay response ( ) = φ( ω) τ ω d dω th order Butterworth τ B ( ω) τ C ( ω) 6 4 Peak-peak group delay variation ~ ω 6
33 Comparing filter performance Butterworth vs. Chebychev filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db Group delay response ( ) = φ( ω) τ ω d dω τ B ( ω) τ C ( ω) Peak-peak group delay variation ~55 5 th order Chebychev ω 61
34 Comparing filter performance Butterworth vs. Chebychev filter with A max = 1 db, ω s /ω p = 1.5 and A min = 3 db Group delay response ( ) = φ( ω) τ ω d dω th order Butterworth τ B ( ω) τ C ( ω) th order Chebychev ω 62
35 Comparing filter performance A B ( ω) 1 A C ( ω) 2 1 Amplitude variation ω Phase variation 8 τ B ( ω) τ C ( ω) Group delay variation ω 63
36 Second order LCR resonator Basic resonator 64
37 Second order LCR resonator Transfer function Current drive V o I Voltage drive V o V i 65
38 Second order LCR resonator-poles V o I 1 1 = = Y sc + sl R s = C 2 s 1 s + + CR LC s = C 2 sω 2 s + + ω Q ω 2 ω Q 1 = LC 1 = CR 1 ω = LC Q = ω CR 66
39 Second order LCR resonator-poles 1 ω = LC Q = ω CR ω 1 = 2Q 2CR 67
40 Second order LCR resonatorzeroes Break any ground connection to inject V i ( ) T s Generic structure ( ) ( ) o = = i 2 ( ) ( ) + ( ) V s Z s V s Z s Z s
41 Second order LCR resonatorzeroes Break any ground connection to inject V i ( ) T s Generic structure ( ) ( ) o = = i 2 ( ) ( ) + ( ) V s Z s V s Z s Z s 1 2 Zeroes occur if Z 2 (s)=, as long as Z 1 (s) is not also zero 69
42 Second order LCR lowpass V o Zero locations: sl 1 sc + 1 R 61
43 Second order LCR lowpass V o Zero locations: sl 1 sc + 1 R 611
44 Second order LCR highpass V o Zero locations: sc sl 612
45 Second order LCR highpass V o Zero locations: sc sl 613
46 Practical limitations of LCR resonators V o 1 ω = LC Q = ω CR 614
47 Practical limitations of LCR resonators V o Consider a maximally flat LPF with f = 4 khz R = 5 Ω 1 ω = LC Q = ω CR 615
48 Practical limitations of LCR resonators V o Consider a maximally flat LPF with f = 4 khz R = 5 Ω Q C 1 = 2 Q = ω R C = 5627 pf 1 ω = LC Q = ω CR 616
49 Practical limitations of LCR resonators V o Consider a maximally flat LPF with f = 4 khz R = 5 Ω Q = 1 2 C = Q ω R 1 ω = LC Q = ω CR C = 5627 pf L = 1 ω C 2 L =.281 H 617
Electronic Circuits EE359A
Electronic Circuits EE359A Bruce McNair B26 bmcnair@stevens.edu 21-216-5549 Lecture 22 578 Second order LCR resonator-poles V o I 1 1 = = Y 1 1 + sc + sl R s = C 2 s 1 s + + CR LC s = C 2 sω 2 s + + ω
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