nd-order filters The general form for the transform function of a second order filter is that of a biquadratic (or biquad to the cool kids). T (s) A p s a s a 0 s b s b 0 As before, the poles of the transfer function (zeroes of the denominator) determine the general characteristics and the location of the zeroes of the function (zeroes of the numerator) determine the type of filter. We will write the denominator in a form that includes parameters that describe the general behavior. D (s) s ω 0 Q s ω 0 ω 0 is the characteristic frequency (determines the corner of LP/HP or center frequency of bandpass). Q P is the pole quality factor. Q P determines the sharpness of the frequency response curve. Q P has no dimensions. EE 30 second-order filters
Q D ( s) s s ω0 Use the quadratic formula to determine the poles. P, ± Q Q ω0 The key is the square-root. If the argument under the square-root is positive, there will be two real roots. If the argument is negative, the roots will be a complex conjugate pair. The dividing line is QP 0.5. jω if QP < 0.5 two real, negative roots. P, Q ± 4Q X P X P σ D (s) (s 800 rad/s) (s 00 rad/s) s (000 rad/s) s 3.6x0 rad /s EE 30 5 600 rad/s Q 0.3 second-order filters
QP 0.5 double real and negative roots. P, Q D (s) (s 000 rad/s) 6 jω σ X X P P s (000 rad/s) s 0 rad /s 000 rad/s Q 0.5 jω QP > 0.5 complex conjugate roots P, ±j Q Q P P 4Q P X σ P X D (s) [s (500 rad/s j000 rad/s)] [s (500 rad/s j000 rad/s)] 8 rad/s 6 s (000 rad/s) s.5 0 rad /s Q.8 EE 30 second-order filters 3
low-pass T (s) N (s) s ω 0 Q s ω 0 To achieve a low-pass response, the zeroes must be out at infinity, meaning that N(s) a 0. (a 0) To make the magnitude of the transfer function go to as s goes to zero, choose a 0 ω 0. T lp (s) ω 0 s ω 0 Q s ω 0 T lp (jω) ω 0 ω 0 ω j ω 0 Q ω EE 30 second-order filters 4
high-pass T (s) N (s) s ω 0 Q s ω 0 Now we want the zeroes to be at s 0 in order to get the high-pass response. In that case choose N(s) s so that T(s 0) 0 and T(s ), as expected. T hp (s) s s ω 0 Q s ω 0 T hp (jω) ω ω 0 ω j ω 0 Q ω EE 30 second-order filters 5
band-pass T (s) N (s) s ω 0 Q s ω 0 Finally, we want T to go to zero at s 0 and at s. To achieve this, we choose N(s) a s. The peak of the function will be when s jω 0. (Check it.) In order to make T at the peak, choose a ω 0 /Q. T bp (s) ω 0 Q s s ω 0 Q s ω 0 T bp (jω) j ω 0 Q ω ω 0 ω j ω 0 Q ω EE 30 second-order filters 6
To synthesize passive versions the different second-order filter functions consider the simple LC parallel combination shown. ZP L C sc Zp Zp s s s s s s sc C sc Q LC LC Q C ω0 ω0 LC The denominator is in exactly the form for second-order filters. If we make voltage-divider circuits using the above combination, we will get transfer functions with that denominator. EE 30 second-order filters 7
low-pass L vi vo C Z ZC sc sc ( ) sc Z ZC V o (s ) V i (s ) Z ZC ZL T (s ) sc sc s LC EE 30 s s LC C LC ω0 s s Q ω0 second-order filters 8
high-pass C vi vo L Z sl ZL Z Z L V o (s ) V i (s ) Z ZL Z C T (s ) sl sl sc s LC s LC EE 30 s s s C LC Q ω0 s s s second-order filters 9
band-pass vi L vo C ZL Z C sc sc () s LC ZL ZC V o (s ) V i (s ) Z L ZC T ( s) s LC s LC s LC EE 30 s s s C C s Q s s Q LC ω0 second-order filters 0
Example This circuit is a simple C ladder circuit. It consists of two passive C combinations strung together. Vi(s) Vx(s) k! C 0. µf k! C 0. µf (s) Using the usual node-voltage approach. Vi Vx Vx sc Vx Vi EE 30 Vx sc s C Vi s C Vx Vx ( s C ) ( s C ) second-order filters
Multiplying out and collecting everything together: Vi s C s C s C C s C T (s) Vi s ( C C ) s ( C C C ) Dividing through by the coefficient on s to get into a standard form: T (s) C C s s C C C C C This is a second-order low-pass: T (s) ωo ωo s s ωo QP ωo C C ωo QP C C C Using the values for the components: ωo 5000 rad/s. EE 30 ωo / QP 7,500 rad/s. QP 0.86. (two real roots.) second-order filters
Example L Another ladder network, but this one has an inductor. Vx(s) 0.05 mh Vi(s) k! (s) k! C 0. µf Same basic approach: Vi Vx Vx EE 30 Vx Vx sc Vi Vx Vx ( s C) Vi Vi LC s C s s LC ( s C) second-order filters 3
T (s) V o V i s LC s L C After some re-arrangement to get to a standard form: T (s) s s LC L( ) C( ) LC Again,this is a second-order low-pass with: ω o LC ω o Q P L ( ) C ( ) Plugging in the numbers ω o 0,000 rad/s. ω o / Q P 5,000 rad/s. Q P 0.333. (two real roots.) EE 30 second-order filters 4
Example 3 C V i (s) C 3 4 3 V o (s) We recognize this as two cascaded op-amp circuits one is low pass and the other high pass. T (s) T lp (s) T hp (s) s C 4 3 s 3 C 3 G o C s C G oω s ω G os s 3 C 3 G o s s ω 3 ω C ω 3 3 C 3 EE 30 second-order filters 5
T (s) [G o G o ] ω s s s (ω ω 3 )ω ω 3 So this is a band-pass, with: ω o ω ω 3 ω o Q P ω ω 3 EE 30 second-order filters 6
Inductor simulation circuit 3 C 4 Z in jωl 5 L C 4 3 5 EE 30 second-order filters 7
Vin I5 Z5 Vy Vin Z4 I4 Z4 Z5 Vin Vy Z3 Vin Iin Z Z Z3 I Vx I I3 Vin Vin Zin Iin Z4 Vin Z3 Z5 Iin I I I3 Vx Vin Z I Vin EE 30 Z Z4 Z3 Z5 Vin Vy Z4 Vin Z5 I4 I5 I3 Vin Vin I4 I5 Z5 Vx Z Z Z4 Iin Vin Z Z3 Z5 Vin Z Z3 Z5 3 5 C4 Zin s Iin Z Z4 second-order filters 8