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 Q ω 2 ω Q 1 = LC 1 = CR 1 ω = LC Q = ω CR 579
Second order LCR resonator-poles 1 ω = LC Q = ω CR ω 1 = 2Q 2CR 58
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
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
Filter designs approximations to ideal response Chebychev response Ideal filter Chebychev response 583
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 584
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 ) 585
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 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 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 1 db 3 db - 1 1.5 588
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 A max 2 = 2 log 1+ ε ε = Amax /1 1 1 ε =.59 (16.14) 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 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 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 1 A( ωs ) = 1log 2N 2 ω s 1+ ε ω p A(1.5) = 32.874 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 ε =.59 N = 1.183 à 11 If A min is to be exactly 3 db, what will A max be? 592
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 593
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 2 594
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 ε = Amax 1 /1 1 (16.21) ε =.59 595
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 (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 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 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 597
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 598
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 599
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 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 11 th order Butterworth τ B ( ω) τ C ( ω) 6 4 2 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 Peak-peak group delay variation ~2 2.5 1 1.5 2 ω 62
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 63
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 64
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 65
Second order LCR resonator Basic resonator 66
Second order LCR resonator Transfer function Current drive V o I Voltage drive V o V i 67
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 68
Second order LCR resonator-poles 1 ω = LC Q = ω CR ω 1 = 2Q 2CR 69
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 61
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
Second order LCR lowpass V o Zero locations: sl 1 sc + 1 R 612
Second order LCR lowpass V o Zero locations: sl 1 sc + 1 R 613
Second order LCR highpass V o Zero locations: sc sl 614
Second order LCR highpass V o Zero locations: sc sl 615
Practical limitations of LCR resonators V o 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 1 ω = LC Q = ω CR 617
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
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
Alternatives to inductors V o 62
Alternatives to inductors V o 621
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
Alternatives to inductors 623
Alternatives to inductors V o LPF sl 624
Alternatives to inductors +K V o LPF 3 op-amps 1 capacitor 4-6 resistors versus 1 inductor 625
Two integrator filter (high-pass example) V V hp i = s Ks 2 2 2 ω + s+ ω Q Biquardratic circuit (ratio of two quadratic polynomials) 626
Two integrator filter (high-pass example) V V hp i = s Ks 2 2 2 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
Two integrator filter (high-pass example) V V hp i = s Ks 2 2 2 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
Two integrator filter (high-pass example) 2 1 ω ω Vhp = KVi Vhp V 2 hp Q s s 629
Two integrator filter (high-pass example) 2 1 ω ω Vhp = KVi Vhp V 2 hp Q s s 63
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 2 2 2.1 1 1 ω 631
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
Alternate filter elements Output transducer Input transducer Mechanical filters 633
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
Alternate filter elements Input Samples x(kt) x n-1 x n-2 x n-3...... x 2 x 1 x n 1 i = cx i i Output Samples y(kt) c n-1 c n-2 c n-3...... 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
Alternate filter elements Input Samples x(kt) + Output Samples y(kt) x n-1 x n-2 x n-3...... x 2 x 1 x n 1 i = cx i i c n-1 c n-2 c n-3...... c 2 c 1 c Digital filters (IIR) 636
Alternate filter elements Input Samples x(kt) + Output Samples y(kt).99 1 x cx c y i 4.361 1 5.5 2 4 6 8 1 1 i 999 Digital filters (IIR) 637