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 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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