EE 230 second-order filters – �1

It is reasonable to attempt to build second-order circuits using two first-order circuits.

low-Q second-order circuits

Tlp = Go ⋅ωc

s + ωcThp = Go ⋅

ss + ωc

Tlp = [Go1 ⋅ωc1

s + ωc1 ] ⋅ [Go2 ⋅ωc2

s + ωc2 ]

GoT = Go1Go2 ω2o = ωc1ωc2 QP =

ωc1ωc2

ωc1 + ωc2

= (Go1Go2)ωc1ωc2

s2 + (ωc1 + ωc2) s + ωc1ωc2

= (GoT) ω2o

s2 + ( ωo

QP ) s + ω2o

EE 230 second-order filters – �2

Thp = [Go1 ⋅s

s + ωc1 ] ⋅ [Go2 ⋅s

s + ωc2 ]= (Go1Go2) s2

s2 + s (ωc1 + ωc2) + ωc1ωc2

= (GoT) s2

s2 + ( ωo

QP ) s + ω2o

GoT = Go1Go2

ω2o = ωc1ωc2

QP =ωc1ωc2

ωc1 + ωc2

EE 230 second-order filters – �3

Tbp = [Go1 ⋅ωc1

s + ωc1 ] ⋅ [Go2 ⋅s

s + ωc2 ]

= (Go1Go2)ωc1s

s2 + s (ωc1 + ωc2) + ωc1ωc2

= GoT ⋅( ωo

QP ) s

s2 + ( ωo

QP ) s + ω2o

GoT = Go1Go2ωc1

ωc1 + ωc2

ω2o = ωc1ωc2

QP =ωc1ωc2

ωc1 + ωc2

EE 230 second-order filters – �4

R

C

–

+

Vo

–

+

Vi L

R

–

+

Vo

–

+

Vi

R

–

+

Vo

–

+

Vi

C

R

–

+

Vo

–

+

Vi

L

–+

R1

R2

Vi

C2

Vo

–+

R1

Rw

Vo

C1

EE 230 second-order filters – �5

As noted earlier, in order to achieve filters with higher QP (> 0.5), some sort of resonance effect is required.

The simplest approach with passive circuits is to use capacitors and inductors together to create the energy-trading, back-and-forth tug-of-war that occurs with inductors and capacitors together. At the resonance frequency, the AC impedances of the two components effect cancel, leaving behind only the resistive part.

As we saw in EE 201, the fact the impedances cancel does not mean that they are gone. In fact the sinusoidal voltages across the reactive components can have a very big magnitude — even bigger than the source. This is not some violation of basic physics. The extra voltage (and hence extra energy) were built up during the transient time that the sinusoids were evolving towards their steady-state values. Since AC analysis intentionally ignores the initial transient, the large resonance voltages (or currents) seem to have come out of nowhere. If we try to use the “extra” energy, we would see a transient flow of energy out of the reactive components, matching the transient that built up the energy in the first place.

EE 230 second-order filters – �6

CLRZP

We recognize this as the impedance of a capacitor multiplied by a high-pass function.

1ZP

=1R

+1sL

+ sC

ZP =s2 ( 1

sC )s2 + s ( 1

RC ) + 1LC

=ZC ⋅ s2

s2 + s ( 1RC ) + 1

LC

To synthesize passive versions the different second-order filter functions consider the simple RLC parallel combination shown. We know that this circuit will exhibit resonance.

The impedance of the parallel combination is:

After a bit of algebraic finagling:

ZP = ZC ⋅s2

s2 + s ( ωo

QP ) + ω2o

ω2o =

1LC

QP = ωoRC

Resonance frequency of the circuit.

EE 230 second-order filters – �7

Low-passGiven the characteristics of the parallel combination, it seems likely that second-order transfer functions could be realized using it as a starting point. Consider the circuit at right. It can be viewed as a simple voltage divider with ZL and ZP = ZR||ZC.

C

L

R+–

vi–

+vo

ZP = R ( 1sC ) =

R1 + sRC

Vo (s) =ZP

ZP + ZLVi (s)

T (s) =Vo (s)Vi (s)

=ZP

ZP + ZL=

R1 + sRCR

1 + sRC + sL

After some algebraic jujitsu:

T (s) =1

LC

s2 + s ( 1RC ) + 1

LC

=ω2

o

s2 + s ( ωo

QP ) + ω2o

ω2o =

1LC

QP = ωoRC

Go = 1

EE 230 second-order filters – �8

High-passIt is probably not surprising that swapping the capacitance and inductor in the low-pass circuit changes the transfer function to a high-pass. Again, the circuit is a simple voltage divider with ZC and ZP = ZR||ZL.

C

LR+–

vi–

+vo

ZP = R∥sL =sLR

R + sL

T (s) =ZP

ZP + ZC=

sLRR + sL

sLRR + sL + 1

sC

Jumping directly to the transfer function:

After some algebraic gymnastics:

T (s) =s2

s2 + s ( 1RC ) + 1

LC

=s2

s2 + s ( ωo

QP ) + ω2o

ω2o =

1LC

QP = ωoRC

Go = 1

EE 230 second-order filters – �9

Band-passOne more variation of the basic circuit — moving the resistor so that it forms a voltage divider with the inductor/capacitor pair — gives a band-pass response. The divider uses ZR and ZP = ZL||ZC.

CL

R

+–

vi–

+vo

ZP = sL ( 1sC ) =

sL1 + s2LC

T (s) =ZP

ZP + ZR=

sL1 + s2LCsL

1 + s2LC+ R

The transfer function is:

After some algebraic sleight-of-hand:

T (s) =s ( 1

RC )s2 + s ( 1

RC ) + 1LC

=s ( ωo

QP )s2 + s ( ωo

QP ) + ω2o

ω2o =

1LC

QP = ωoRC

Go = 1

EE 230 second-order filters – �10

Example

C

L

R+–

vi–

+vo

1 mH

15 nF1 k!

What type of filter is this? What are ωo, QP, and Go for this filter? What value of resistance would be needed to make QP = 1?

From the configuration, we see that this is a low-pass filter.

ωo =1�LC

=1�

(1 mH) (15 nF)= 25, 800 rad/s

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QP = ωoRC = (25, 800 rad/s) (1 kΩ) (15 nF) = 0.4275<latexit sha1_base64="az15w4bRi6Ed0zt/fbDQJs0TzHo=">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</latexit><latexit sha1_base64="az15w4bRi6Ed0zt/fbDQJs0TzHo=">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</latexit><latexit sha1_base64="az15w4bRi6Ed0zt/fbDQJs0TzHo=">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</latexit>

Go = 1.

To make QP = 1, R would have to be increased to 2.34 k!.

EE 230 second-order filters – �11

Example

Design the band-pass circuit at right to have a center frequency at 3 kHz and a bandwidth of 300 Hz.

CL

R

+–

vi–

+vo

ωo = 2πfo = 2π (3000Hz) = 18.85 krad/s<latexit sha1_base64="78DWkRm30h9Rxm0RMDm6LuWUUpo=">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</latexit><latexit sha1_base64="78DWkRm30h9Rxm0RMDm6LuWUUpo=">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</latexit><latexit sha1_base64="78DWkRm30h9Rxm0RMDm6LuWUUpo=">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</latexit>

Choose C = 220 nF (somewhat arbitrarily). Then

L =1

ω2oC

=1

(18.85 krad/s)2 (220 nF)= 12.8 mH

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(Might need to fiddle around with values a bit to find the right components.)

BW =ωoQP

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QP =ωoBW = 10

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ωoQP

=1RC

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R =QPωoC

=10

(18.85 krad/s) (220 nF)= 2.41 kΩ

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EE 230 second-order filters – �12

Example

Design a high-pass filter, as shown at right, to be “maximally” flat with corner frequency at 500 Hz.

C

LR+–

vi–

+vo

Maximally flat means that QP =1�2

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With this value of QP, ωc = ωo = 2π(500 Hz) = 3.14 krad/s.

Choose C = 1 µF (again, somewhat arbitrarily). Then

L =1

ω2oC

=1

(3.14 krad/s)2 (1 μF)= 101 mH

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ωoQP

=1RC

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R =QPωoC

=0.707

(3.14 krad/s) (1 μF)= 225 Ω

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EE 230 second-order filters – �13

Inductor simulation circuit

L = C4R1R3R5R2

–+– +

R1 R2 R3

R5

C4

Zin = jωL

EE 230 second-order filters – �14

–+

– +

Z1 Z2 Z3

Z5

Z4

+–

Vin

Iin

Vin

I1 I2 I3 I4

I5

Vx

VyVin

Zin =VinIin

I5 =VinZ5

I4 = I5 =VinZ5

= Vin

�1+

Z4Z5

�Vy = Vin + Z4I4

I3 =Vin � Vy

Z3

= VinZ4Z3Z5

I2 = I3

Vx = Vin + Z2I2

= Vin

�1+

Z2Z4Z3Z5

�

Iin = I1 =Vin � Vx

Z1

Iin = VinZ2Z4

Z1Z3Z5

Zin =VinIin

=Z1Z3Z5Z2Z4

= sR1R3R5C4R2= sL