Electronic Circuits EE359A Bruce McNair B26 bmcnair@stevens.edu 21-216-5549 Lecture 22 569
Second order section Ts () = s as + as+ a 2 2 1 ω + s+ ω Q 2 2 ω 1 p, p = ± 1 Q 4 Q 1 2 2 57
Second order section Ts () = s as + as+ a 2 2 1 ω + s+ ω Q 2 2 ω 1 p, p = ± 1 Q 4 Q 1 2 2 Numerator zeroes determine filter response: LP, HP, BP, BS, AP 571
Second order sections 572
Second order sections 573
Second order sections 574
Second order LCR resonator Basic resonator 575
Second order LCR resonator Transfer function Current drive V o I Voltage drive V o V i 576
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 Q ω 2 ω Q 1 = LC 1 = CR 1 ω = LC Q = ω CR 577
Second order LCR resonator-poles 1 ω = LC Q = ω CR ω 1 = 2Q 2CR 578
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
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
Filter designs approximations to ideal response Chebychev response Ideal filter Chebychev response 581
Filter designs approximations to ideal response Chebychev response 1 T( jω) = for ω ωp 2 2 1 ω 1+ ε cos N cos ω p 1 T( jω) = for ω ωp 2 2 1 ω 1+ ε cosh N cosh ω p 1 T( jω p ) = 2 1+ ε 582
Filter designs approximations to ideal response Chebychev response p k 2k 1π 1 1 N 2 N ε 1 = ωpsin sinh sinh 2k 1π 1 1 + = 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
Filter designs approximations to ideal response Chebychev response 1 1 ( ( ) ) 2 log T j ω i 6 2 4 6.5 1 1.5 2 2 1 3 ω i 2 Im( p k ).5.5 Pole locations for N=8 ε = 3 db 1 1.8.6.4.2 ( ) Re p k 584
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
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 - 1 1.5 586
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
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 ) 1 1 1 log 2 ε ω s 2log ω p N = 1.183 à 11 588
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 = 1.183 à 11 1 A( ωs ) = 1log 2N 2 ω s 1+ ε ω p A(1.5) = 32.874 db 589
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 = 1.183 à 11 If A min is to be exactly 3 db, what will A max be? 59
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 = 1.183 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
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) 1 1 2 1 1.5.5 ε =.59 N = 1.183 à 11 If A min is to be exactly 3 db, what will A max be? ε =.365 4 3 3 4.5 1 1.5 2 A max =.544 db ω 2 592
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 =.59 593
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 1 1 1 ε 2 N = N = 5.22 ω cosh 1 s 2 2 1 ( ω ) ε N S ω p 594
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 =.59 2 2 1 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 ) 4 21.583 db 5 29.914 db 6 38.269 db 595
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 2 2 1 ( ω ) ε N S (16.22) N A(ω s ) 4 21.583 db 5 29.914 db 6 38.269 db An 11 th order Butterworth would have been necessary 596
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 1 1 1.5 1 11 th order Butterworth A B ( ω) 1 A C ( ω) 2 3 3 5 th order Chebychev 4.5 1 1.5 2 ω 597
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 18 1 1 1.5 11 th order Butterworth ϕ.b ( ω) ϕ.c ( ω) 1 5 th order Chebychev 18.5 1 1.5 2 ω 2 598
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 1 1 3 8 ( ) = φ( ω) τ ω d dω 1 1.5 11 th order Butterworth τ B ( ω) τ C ( ω) 6 4 2 5 th order Chebychev.5 1 1.5 2 ω 599
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 1 1 3 8 ( ) = φ( ω) τ ω d dω 1 1.5 11 th order Butterworth τ B ( ω) τ C ( ω) 6 4 Peak-peak group delay variation ~2 2.5 1 1.5 2 ω 6
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 1 1 3 8 ( ) = φ( ω) τ ω d dω 1 1.5 τ B ( ω) τ C ( ω) 6 4 2 Peak-peak group delay variation ~55 5 th order Chebychev.5 1 1.5 2 ω 61
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 1 1 3 8 ( ) = φ( ω) τ ω d dω 1 1.5 11 th order Butterworth τ B ( ω) τ C ( ω) 6 4 2 5 th order Chebychev.5 1 1.5 2 ω 62
Comparing filter performance 1 1 1.5 A B ( ω) 1 A C ( ω) 2 1 Amplitude variation 3 3 4.5 1 1.5 2 1 1 3 ω 1 1.5 Phase variation 8 τ B ( ω) τ C ( ω) 6 4 2 Group delay variation.5 1 1.5 2 ω 63
Second order LCR resonator Basic resonator 64
Second order LCR resonator Transfer function Current drive V o I Voltage drive V o V i 65
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 Q ω 2 ω Q 1 = LC 1 = CR 1 ω = LC Q = ω CR 66
Second order LCR resonator-poles 1 ω = LC Q = ω CR ω 1 = 2Q 2CR 67
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 68
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
Second order LCR lowpass V o Zero locations: sl 1 sc + 1 R 61
Second order LCR lowpass V o Zero locations: sl 1 sc + 1 R 611
Second order LCR highpass V o Zero locations: sc sl 612
Second order LCR highpass V o Zero locations: sc sl 613
Practical limitations of LCR resonators V o 1 ω = LC Q = ω CR 614
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
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
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