active filter design 2 order butterworth filter - b0strup filter design 2 order... · klaus...

NAPIER.

University

School of Engineering

Engineering Applications

Module : SE32101

Active Filter Design

2nd order Butterworth response

Sallen and key.

Multi Feedback.

By

Klaus Jørgensen.

Napier No. 04007824

Teacher.

Dr. Mohammad Y Sharif

15V

-15V

R1 7.04k R2 14.09k

R3 10k

R4 10k

C1 4.7n

C2 4.7n

-

+ +OP1 uA741

+

InOut

-15V

15V

R1 41.58k

R2 83.16k

R3 28.87k

C1 390pF

C2 2.34nF

-

+ +

OP1 uA741out+

in

Klaus Jørgensen Active Filter Design

2nd order Butterworth filter

/21

Abstract.:

In this paper there is explained how to calculate a 2nd order low pass Butterworth

response by using a Sallen and key or a MFB (Multi Feedback) filter.

There are calculations on how to design the 2nd order Butterworth filters, there is also

simulations of the calculated filters using the program Tina, and the filters is furthermore

tested in practical, to se how the theoretical and the practical measurement fits together.



/21

Table of contents.:

Introduction.:....................................................................................................................... 4

Specification for the assignment.:....................................................................................... 4

Order of the Butterworth filter. ........................................................................................... 5

Calculation of the 2nd order Sallen and key low-pass filter. ............................................... 6

Simulations of 2nd order Sallen and key filter in Tina. ....................................................... 7

Poles in the Butterworth filter using Sallen and key........................................................... 9

Practical test of 2nd order Sallen and key filter. ................................................................ 10

Calculation of the 2nd order MFB (Multi Feedback) low-pass filter. .............................. 11

Simulations of 2nd order Multi Feedback filter in Tina. .................................................. 12

Poles in the Butterworth filter using MFB (Multi Feedback)........................................... 14

Practical test of 2nd order Multi Feedback filter................................................................ 15

Conclusion. ....................................................................................................................... 16

Reference. ......................................................................................................................... 17

Appendix........................................................................................................................... 18

Appendix 1.

The measurement from the sallen and key filter........................................................... 18

Appendix 2.

Frequency response for 2nd order Sallen and key filter................................................. 19

Appendix 3.

The measurement from the MFB filter. ........................................................................ 20

Appendix 4.

Frequency response for 2nd order MFB (Multi Feedback) filter................................... 21



/21

Introduction.: Active filters are used many places, for example in the telecommunication, in the audio

frequency range (0 kHz to 20 kHz), and for modems to the internet connection, and for

many more things.

There are many types of active filter response, some of them are Butterworth, Bessel, and

Chebyshev, and the filter response can be used in Low-pass, High-pass, Band-pass and

Band-stop filter setups.

Butterworth response is also named the maximally flat filter, because it is maximally flat

in the pass-band response, the Butterworth response is perhaps the mote used type of

filter response. The Chebyshev can has a ripple in the pass-band, this ripple can be from

0,01 dB and up to 3dB ripple, in return for the ripple in the pass-band the Chebyshev

response has very good cutoff at the edge of the pass-band as it is showed in figure 2. [4]

Figure 1 shows the poles in Butterworth, Bessel, and Chebyshev filters, there is no zeroes

in this filter types.

Specification for the assignment.: • Butterworth response.

• -3dB frequency = 3.4KHz.

• Stop-band frequency =6KHz.

• Stop-band attenuation ≥ 10dB.

• Pass-band gain = 2.

• Op-amp 741.

Figure 2 [4]

jω

Butterworth

σ Chebyshev

Bessel

Figure 1



/21

Order of the Butterworth filter. Before the values off the components in the Butterworth filter can be calculated the order

off the filter must be found.

Order off the filter = N [1]

•

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−

≥

C

Log

SLog

N

ωω2

22

*2

11

ώC = 3,4KHz.

ώ2 = 6KHz.

1. ( ) mLogSdBSLog 23,316201010*20 1

22 =⎟⎠⎞

⎜⎝⎛ −=⇒−= −

2. nd

C

mm

KKLog

mLog

Log

SLog

N 293,135,49324,954

4,36*2

123,3161

*2

112

2

22 ≈=⇒

⎟⎠

⎞⎜⎝

⎛

⎟⎠

⎞⎜⎝

⎛−

⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−

≥

ωω

The filter is calculated to be a 2nd order filter, from the calculation above.

There will be tow poles in this filter.

ώC ώ2

S2

ώ

|H(dώ)|

Figure 3 [2]

Passband Transition-band

Stopband



/21

Calculation of the 2nd order Sallen and key low-pass filter. 2nd order Sallen and key filter calculations. [1]

• C1 = C2 and is chosen to 4,7nF

• C1n = C2n and is chosen to 1F

• C

n

MCCω*

1 1= and C

n

MCCω*

2 2=

• KnKCfcC

MC

96,97,4*4,3*2

11**2

1*11

=⇒⇒=ππω

• 128,6043,796,9*2

1*2

11 EKKKMR Ω≈Ω=⇒=

• 1215086,1496,9*2*22 EKKKMR Ω≈Ω=⇒=

• R3 = R4 and is chosen to be 10KΩ

• ( ) ( ) dBLoggainLogAu 62*20*20 =⇒=

The diagram there is shown in figure 4 was simulated in the program Tina, with the

calculated value, and the outcome is shown in figure 5, the -3dB frequency 3,4KHz and

-10,52dB at 6KHz.

15V

-15V

R1 7.04k R2 14.09k

R3 10k

R4 10k

C1 4.7n

C2 4.7n

-

+ +OP1 uA741

+

InOut

Figure 4



/21

Simulations of 2nd order Sallen and key filter in Tina. Figure 5 shows the gain in dB to the frequency, on the output of the filter (figure 4)

Figure 6 shows the gain between the signal In and the signal Out, and the gain is 2.

Input signal : 1KHz / 2Vpeek-peek sine wave.

224=⇒=

−

−

peekpeek

peekpeek

UinUout

Au

T

6dB / gain 2

3.4KHz / -3dB

6KHz / -10.52dB

Frequency (Hz)10.00 100.00 1.00k 10.00k 100.00k 1.00M

Gai

n (d

B)

-60.00

-50.00

-40.00

-30.00

-20.00

-10.00

0.00

10.00

6KHz / -10.52dB

3.4KHz / -3dB

6dB / gain 2

a b

Figure 5

Figure 6

T

Signal In Signal Out

Time (s)0.00 2.00m 4.00m 6.00m 8.00m 10.00m

Vol

tage

(V)

-2.00

-1.00

0.00

1.00

2.00

Signal Out Signal In



/21

Signal In : 300Hz / 2Vpeek-peek Square wave.

Signal In : 3KHz / 2Vpeek-peek Square wave.

In figure 7 the output signal is similar to the input signal on 300Hz because the frequency

is much lower then the -3dB point at 3,4KHz and the harmonic frequency is getting

through the filter.

In figure 8 the input signal is a square wave on 3KHz and the harmonic frequency to

3KHz square wave signal is not getting through the filter, and therefore is the signal out

of the filter is a sine wave near by 3KHz.

T


Time (s)0.00 2.00m 4.00m 6.00m 8.00m 10.00m 12.00m 14.00m 16.00m 18.00m 20.00m

Vol

tage

(V)

-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00


Figure 7

T


Time (s)0.00 1.00m 2.00m 3.00m 4.00m 5.00m

Vol

tage

(V)

-3.00

-2.00

-1.00

0.00

1.00

2.00

3.00


Figure 8



/21

Poles in the Butterworth filter using Sallen and key. In a 2nd order Butterworth filter is there no zeros only poles and there is as many poles as

the filter order.

2nd order filter = 2 poles

4th order filter = 4 poles

The position of the poles can be calculated by using a program like MathCAD.

1. The value of the components in the filter has to be defined in MathCAD.

R1 7.4 103⋅:= R2 14.09 103

⋅:=

R3 10 103⋅:= R4 10 103

⋅:=

C1 4.7 10 9−⋅( ):= C2 4.7 10 9−

⋅( ):= 2. Then the transfer function has to be found, and it can be found by using Tina.

W s( )R3− R4−

R4− C1 R3⋅ R1⋅ C2 R4⋅ R1⋅− C2 R4⋅ R2⋅−( ) s⋅+ C2 C1⋅ R4⋅ R2⋅ R1⋅ s2⋅−

:=

3. IT is only the part under the division linie of transfer function there has to be used

because there are no zeros in a Butterworth filter. The equation from under the

linie has to be set equal to 0.

R4− C1 R3⋅ R1⋅ C2 R4⋅ R1⋅− C2 R4⋅ R2⋅−( ) s⋅+ C2 C1⋅ R4⋅ R2⋅ R1⋅ s2⋅−⎡⎣ ⎤⎦ 0

4. And from that MathCAD can calculate the expression for s.

s

12 C2 C1 R4 R2 R1⋅⋅⋅⋅⋅

C1 R3 R1⋅⋅ C2 R4 R1⋅⋅− C2 R4 R2⋅⋅− C12 R32 R12⋅⋅ 2 C1 R3 R12 C2 R4⋅⋅⋅⋅⋅− 2 C1 R3 R1 C2 R4 R2⋅⋅⋅⋅⋅⋅− C22 R42 R12

⋅⋅+ 2 C22 R42 R1 R2⋅⋅⋅⋅+ C22 R42 R22⋅⋅ 4 C2 C1 R42 R2 R1⋅⋅⋅⋅⋅−+( )

1

2+

⎡⎢⎢⎣

⎤⎥⎥⎦⋅

12 C2 C1 R4 R2 R1⋅⋅⋅⋅⋅

C1 R3 R1⋅⋅ C2 R4 R1⋅⋅− C2 R4 R2⋅⋅− C12 R32 R12⋅⋅ 2 C1 R3 R12 C2 R4⋅⋅⋅⋅⋅− 2 C1 R3 R1 C2 R4 R2⋅⋅⋅⋅⋅⋅− C22 R42 R12

⋅⋅+ 2 C22 R42 R1 R2⋅⋅⋅⋅+ C22 R42 R22⋅⋅ 4 C2 C1 R42 R2 R1⋅⋅⋅⋅⋅−+( )

1

2−

⎡⎢⎢⎣

⎤⎥⎥⎦⋅

⎡⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎦

:=

5. And from that the imaginary and the real part can be calculated.

In Tina a visual plot of the zeros can be simulated by using an ideal op-amp

shown in figure 9.

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅⋅

⋅+⋅=

44

44

101.508i -101.438-101.508i101.438-

s

KKKK

08.1538,14Im08,1538,14Re

−−=+−=

T

Real part-20.00k -15.00k -10.00k -5.00k 0.00 5.00k 10.00k

Imag

inar

y pa

rt

-20.00k

-15.00k

-10.00k

-5.00k

0.00

5.00k

10.00k

15.00k

20.00k

Figure 9



/21

Practical test of 2nd order Sallen and key filter. The circuit on page 6 figure 4 was build on a breadboard, to see how it would work in

practices, and the result is shown in figure 10.

The practical result is far from the result as the simulation in Tina gave.

At about -15dB the curve should have been about 10KHz and NOT about 50KHz as it is

in figure 10.

The reason for the bump on figure 10 can be several things first of all the resistance there

is used in the practical test doesn’t have the same, value as the resistance there is used in

the simulation in Tina, in the practical test there is used the E12 value of the resistance.

Another reason is the capacitors there is used to test with has an accuracy of ±20%.

The breadboard there is used in the test is not that good.

The test setup is showed in figure 11.

The measurement for figure 10 can be found in appendix 1.

Figure10

-20

-15

-10

-5

0

5

10

100 1000 10000 100000

Frequency.

dB.

Figure 11

Frequency Generator

Oscilloscope

Test board Filter.

Power-supply ±15V

Signal In

Signal Out

+15V -15V



/21

Calculation of the 2nd order MFB (Multi Feedback) low-pass filter.

The Multi Feedback filter in figurer 12 has the same specifications as the Sallen and key

filter.

2nd order Multi Feedback filter calculations [4].

C1 is chosen to be ≤ 0.5nF chosen to 390pF

11414,11

==

ba

( ) ( )( )622 2,234.2

414,121*1*4*390

11*1*4*12 EnFnFp

aAubCC ≈=

−−⇒

−≥

( )( ) ( )[ ]

( )( )( ) ( )[ ]12

22

22

8216,8334,2*390*4,3**4

34,2*414,121*34,2*390*1*434.2*414,1

2*1***42*11*2*1*1*42*12

Ekknpk

nnpn

CCfcCaAuCCbCaR

Ω≈Ω=−−−−

⇒−−−

=

π

π

123948,41216.8321 Ekkk

AuRR Ω≈Ω=⇒=

122222 2717,2816,83*34,2*390*4,3**4

12*2*1***4

13 EkkknpkRCCfc

bR Ω≈Ω=⇒=ππ

-15V

15V

R1 41.58k

R2 83.16k

R3 28.87k

C1 390pF

C2 2.34nF

-

+ +

OP1 uA741out+

in

Figure 12



/21

Simulations of 2nd order Multi Feedback filter in Tina.

Figure 13 shows the gain in dB to the frequency, on the output of the filter.

Input signal : 1KHz / 2Vpeek-peek sine wave.

Figure 14 shows the gain between the signal In and the signal Out, and the gain is 2.

• 224=⇒=

−

−

peekpeek

peekpeek

UinUout

Au

T

6dB / gain 2

3,4KHz / -3dB

6KHz / -10.38dB

Frequency (Hz)1 10 100 1k 10k 100k 1M

Gai

n (d

B)

-90.00

-77.50

-65.00

-52.50

-40.00

-27.50

-15.00

-2.50

10.00

6KHz / -10.38dB

3,4KHz / -3dB

6dB / gain 2

a b

Figure 13

T

Signal In Sinal Out

Time (s)0.00 2.50m 5.00m 7.50m 10.00m

Vol

tage

(V)

-2.00

-1.00

0.00

1.00

2.00

Sinal OutSignal In

Figure 14



/21

Signal In : 300Hz / 2Vpeek-peek Square wave.

Signal In : 3KHz / 2Vpeek-peek Square wave.

In figure 15 the output signal is similar to the input signal on 300Hz because the

frequency is much lower then the -3dB point at 3,4KHz and the harmonic frequency is

getting through the filter.

In figure 16 the input signal is a square wave on 3KHz and the harmonic frequency to

3KHz square wave signal is not getting through the filter, and therefore is the signal out

of the filter is a sine wave on 3KHz.

T


Time (s)0.00 5.00m 10.00m 15.00m 20.00m 25.00m 30.00m

Vol

tage

(V)

-3.00

-1.50

0.00

1.50

3.00

Signal OutSignal In

Figure 15

T


Time (s)0.00 500.00u 1.00m 1.50m 2.00m 2.50m 3.00m 3.50m 4.00m 4.50m 5.00m

Vol

tage

(V)

-3.00

-2.25

-1.50

-750.00m

0.00

750.00m

1.50

2.25

3.00

Signal OutSignal In

Figure 16



/21

Poles in the Butterworth filter using MFB (Multi Feedback). The poles is calculated the same way as with the sallen and key filter on page 9

1. The value of the components in the filter has to be defined in MathCAD.

R1 41.58 103⋅:= R2 83.16 103

⋅:=

R3 28.87 103⋅:=

C1 390 10 12−⋅:= C2 2.34 10 9−

⋅:= 2. Then the transfer function has to be found, and it can be found by using Tina.

w s( )R2

R1− R2− R1⋅ R3 R1⋅− R3 R2⋅−( ) C1⋅ s⋅+ C2 C1⋅ R3⋅ R2⋅ R1⋅ s 2⋅−

:=

3. IT is only the part under the division linie of transfer function there has to be used

because there are no zeros in a Butterworth filter. The equation from under the

linie has to be set equal to 0.

R1− R2− R1⋅ R3 R1⋅− R3 R2⋅−( ) C1⋅ s⋅+ C2 C1⋅ R3⋅ R2⋅ R1⋅ s2⋅− 0

4. And from that MathCAD can calculate the expression for s.

s

12 C2 C1 R3 R2 R1⋅⋅⋅⋅⋅

C1− R2 R1⋅⋅ C1 R3 R1⋅⋅− C1 R3 R2⋅⋅− C12 R22 R12⋅⋅ 2 C12 R2 R12 R3⋅⋅⋅⋅+ 2 C12 R22 R1 R3⋅⋅⋅⋅+ C12 R32 R12

⋅⋅+ 2 C12 R32 R1 R2⋅⋅⋅⋅+ C12 R32 R22⋅⋅ 4 C2 C1 R3 R2 R12

⋅⋅⋅⋅⋅−+( )1

2+

⎡⎢⎢⎣

⎤⎥⎥⎦⋅

12 C2 C1 R3 R2 R1⋅⋅⋅⋅⋅

C1− R2 R1⋅⋅ C1 R3 R1⋅⋅− C1 R3 R2⋅⋅− C12 R22 R12⋅⋅ 2 C12 R2 R12 R3⋅⋅⋅⋅+ 2 C12 R22 R1 R3⋅⋅⋅⋅+ C12 R32 R12

⋅⋅+ 2 C12 R32 R1 R2⋅⋅⋅⋅+ C12 R32 R22⋅⋅ 4 C2 C1 R3 R2 R12

⋅⋅⋅⋅⋅−+( )1

2−

⎡⎢⎢⎣

⎤⎥⎥⎦⋅

⎡⎢⎢⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎦

:=

5. And from that the imaginary and the real part can be calculated.

In Tina a visual plot of the zeros can be simulated by using an ideal op-amp.

shown in figure 17.

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅⋅

⋅+⋅=

44

44

101.51i -101.511-101.51i101.511-

s

KKKK

1,1511,15Im1,1511,15Re

−−=+−=

T

Real part-20.00k -15.00k -10.00k -5.00k 0.00 5.00k 10.00k

Imag

inar

y pa

rt

-20.00k

-15.00k

-10.00k

-5.00k

0.00

5.00k

10.00k

15.00k

20.00k

Figure 17



/21

Practical test of 2nd order Multi Feedback filter. The circuit shown in figure 12 on pages 11 was build on a breadboard with E12 values.

And the result is shown in figure 18 and the result of the practical measurement is close

to the Tina simulation in figure 13 on page 12.

The test setup is showed in figure 19

The measurement for figure 18 can be found in appendix 3

Figure 18

-35

-30

-25

-20

-15

-10

-5

0

5

10

100 1000 10000 100000

Frequency.

dB.

Figure 19

Frequency Generator

Oscilloscope

Test board Filter.

Power-supply ±15V

Signal In

Signal Out

+15V -15V



/21

Conclusion. In the test of the sallen and key filter there is showed on page 10 figure 10 is the cut-off

frequency on 3,4KHz not at -3dB and on of the reasons can also be that the capacitor is

chosen to be 4,7nF, and the value of the capacitor maybe to large, in setoff 4,7nF the

value shout perhaps been chosen to a value on 1nF, that maybe had given another output

of the sallen and key circuit.

The output of the MFB (Multi Feedback) filter on page 15 figure 18 looks a lot like the

simulation from Tina on page 12 figure 13 but it is not quite as good as the simulation but

it is acceptable, on of the reasons for this may be that the capacitor was chosen to a value

of 1nF, and not 4,7nF as in the sallen and key circuit.

If the capacitor in the MFB filter was chosen to a value of 4,7nF in setoff 1nF the test

may had looked different.

___________________________________

Klaus Jørgensen 19 November 2004



/21

Reference. 1. Hand out by Mohammad Y Sharif

2. Passive and Active filters, theory and implementations

by Wai-kai Chen, University of Illinois at Chicago.

ISBN: 0471843180

3. Active Filter Design Handbook

by G. S. Moschytz and Petr Horn

ISBN: 0471278505

4. Chapter 16 Active Filter Design Techniques.

PDF document from Texas Instruments.

http://focus.ti.com/general/docs/lit/getliterature.tsp?literatureNumber=sloa088&fil

eType=pdf

5. The Electrical Engineering Handbook.

Chapter 29 Active Filters

http://www.kemt.fei.tuke.sk/predmety/KEMT350_EP/_materialy/ch029.pdf



/21

Appendix.

Appendix 1.

The measurement from the sallen and key filter.

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

pekPek

pekpek

UinUout

LogdB *20

Hz dB Uin pek-pek Uout pek-pek 120 5.93 2 3.92 300 5.93 2 3.92 500 5.93 2 3.92 800 6.15 2 4.12

1000 6.31 2 4.28 1200 6.43 2 4.4 1400 6.67 2 4.64 1600 6.85 2 4.84 1800 7.13 2 5.16 2000 7.45 2 5.56 2200 7.69 2 5.88 2400 7.98 2 6.28 2600 8.12 2 6.48 2800 8.2 2 6.6 2900 8.25 2 6.68 3000 8.25 2 6.68 3100 8.22 2 6.64 3200 8.2 2 6.6 3300 8.12 2 6.48 3400 8.06 2 6.4 3500 7.95 2 6.24 3600 7.78 2 6 3700 7.48 2 5.6 3800 7.19 2 5.24 3900 6.95 2 4.96 4000 6.8 2 4.8 4100 6.35 2 4.32 4200 6.02 2 4 4300 5.75 2 3.76 4400 5.47 2 3.52 4500 5.26 2 3.36 5000 3.94 2 2.48 5500 2.92 2 1.96 6000 1.93 2 1.56 10000 -2.52 2 0.56 20000 -8.42 2 0.144 30000 -10.56 2 0.088 40000 -13.57 2 0.044 50000 -15.23 2 0.03



/21

Appendix 2.

Frequency response for 2nd order Sallen and key filter.



/21

Appendix 3.

The measurement from the MFB filter.

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

pekPek

pekpek

UinUout

LogdB *20

Hz dB Uin pek-pek Uout pek-pek 100 6.24 2 4.1 200 6.24 2 4.1 800 6.24 2 4.1

1000 6.24 2 4.1 1500 6.24 2 4.1 2000 5.85 2 3.92 2500 5.2 2 3.64 2800 4.71 2 3.44 3000 4.4 2 3.32 3200 3.97 2 3.16 3300 3.75 2 3.08 3400 3.64 2 3.04 3500 3.41 2 2.96 3600 3.05 2 2.84 3700 2.92 2 2.8 3800 2.54 2 2.68 4000 2.01 2 2.52 4300 1.29 2 2.32 4600 0.34 2 2.08 5000 -0.54 2 1.88 7000 -5.35 2 1.08 10000 -10.75 2 0.58 20000 -17.72 2 0.26 40000 -30.46 2 0.06



/21

Appendix 4.

Frequency response for 2nd order MFB (Multi Feedback) filter.

active filter design 2 order butterworth filter - b0strup filter design 2 order... · klaus...

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