Electronic Circuits EE359A

Transcription

1 Electronic Circuits EE359A Bruce McNair B Lecture

2 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 579

3 Second order LCR resonator-poles 1 ω = LC Q = ω CR ω 1 = 2Q 2CR 58

4 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 581

5 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 582

6 Filter designs approximations to ideal response Chebychev response Ideal filter Chebychev response 583

7 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 584

8 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 ) 585

9 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 586

10 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 587

11 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

12 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) 589

13 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 11 59

14 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 591

15 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? 592

16 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 593

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 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 2 594

18 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 =

19 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 596

20 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 597

21 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 598

22 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 599

23 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 6

24 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 61

25 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 62

26 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 63

27 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 64

28 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 65

29 Second order LCR resonator Basic resonator 66

30 Second order LCR resonator Transfer function Current drive V o I Voltage drive V o V i 67

31 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 68

32 Second order LCR resonator-poles 1 ω = LC Q = ω CR ω 1 = 2Q 2CR 69

33 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

34 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 611

35 Second order LCR lowpass V o Zero locations: sl 1 sc + 1 R 612

36 Second order LCR lowpass V o Zero locations: sl 1 sc + 1 R 613

37 Second order LCR highpass V o Zero locations: sc sl 614

38 Second order LCR highpass V o Zero locations: sc sl 615

39 Practical limitations of LCR resonators V o 1 ω = LC Q = ω CR 616

40 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 617

41 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 618

42 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 619

43 Alternatives to inductors V o 62

44 Alternatives to inductors V o 621

45 Alternatives to inductors V o Impedance of inductor is sl Z in looks like inductor with L=C 4 R 1 R 3 R 5 /R 2 622

46 Alternatives to inductors 623

47 Alternatives to inductors V o LPF sl 624

48 Alternatives to inductors +K V o LPF 3 op-amps 1 capacitor 4-6 resistors versus 1 inductor 625

49 Two integrator filter (high-pass example) V V hp i = s Ks ω + s+ ω Q Biquardratic circuit (ratio of two quadratic polynomials) 626

50 Two integrator filter (high-pass example) V V hp i = s Ks ω + s+ ω Q 1 ω ω V V V KV Q s s 2 hp + hp + 2 hp = i 2 1 ω ω Vhp = KVi Vhp V 2 hp Q s s 627

51 Two integrator filter (high-pass example) V V hp i = s Ks ω + s+ ω Q 1 ω ω V V V KV Q s s 2 hp + hp + 2 hp = i 2 1 ω ω Vhp = KVi Vhp V 2 hp Q s s Integrators 628

52 Two integrator filter (high-pass example) 2 1 ω ω Vhp = KVi Vhp V 2 hp Q s s 629

53 Two integrator filter (high-pass example) 2 1 ω ω Vhp = KVi Vhp V 2 hp Q s s 63

54 Two integrator filter (simultaneous LP, BP, HP) T T hp bp ( s) ( s) = = s s 2 Ks ω + s + ω Q 2 2 Kωs ω + s + ω Q 2 2 ( ) ( ( ) ) ( ( ) ) 2 log T lp ( j ω) 2 log T bp j ω 2 log T hp j ω 1 T lp ( s) = s 2 Kω ω + s + ω Q ω 631

55 Alternate filter elements Y 1 Y 2 B 6 B < 6 2 in DC block DC block out 6 db B 6 6 db B 6 w Y 2 Y 1 Y 1 Y 2 Piezoelectric crystal filters 632

56 Alternate filter elements Output transducer Input transducer Mechanical filters 633

57 Alternate filter elements Equivalent LC filter R S L 1 L 3 C 2 R L Transmission line filters l/8 l/8 l/8 l/8 l/8 Microstrip 634

58 Alternate filter elements Input Samples x(kt) x n-1 x n-2 x n x 2 x 1 x n 1 i = cx i i Output Samples y(kt) c n-1 c n-2 c n c 2 c 1 c (,,,,1,,,, ) an impulse (,,,,c n-1,c n-2,c n-3,,c 2,c 1,c,,,, ) impulse response of filter Digital filters (FIR) 635

59 Alternate filter elements Input Samples x(kt) + Output Samples y(kt) x n-1 x n-2 x n x 2 x 1 x n 1 i = cx i i c n-1 c n-2 c n c 2 c 1 c Digital filters (IIR) 636

60 Alternate filter elements Input Samples x(kt) + Output Samples y(kt).99 1 x cx c y i i 999 Digital filters (IIR) 637