BESSEL 4TH ORDER BAND PASS FILTER

In this post we will use two multiple feedback (MFB) 2nd order band pass filters (see the previous post) to build a 4th order band pass filter. See Fig. 1 for the schemtic and the figure below for the SPICE AC analysis results (click on the figures for a larger view.) This method uses two filters with staggered tunning so that a flat band pass can be obtained. We will use Bessel parameters so that we obtain linear phase shift performance. Refer to Texas Instruments Active Filter Design Techniques Literature Number SLOA088 for a good description of the complete theory for this approach.

Lets design the 4th order band pass Bessel filter with the following specifications:

Center frequency Fm = 1000
Bandwidth B = 100
Q Q = Fm / B = 10
Center gain Km = 1 absolute value

The Bessel coefficients for a Q of 10 are

a1 = 1.3617
b1 = 0.6180

A quantity called alpha is also associated with the Bessel coefficients for a Q of 10. We will call alpha the quantity "a" in the following.

a = 1.0324

The tuning frequencies of the 2nd order sections is

Fm1 = Fm / a = 968.6

Fm2 = Fm * a = 1032

The Q or quality factor for both 2nd order sections is the same:

Q12 = Q * ( 1 + a^2 ) * b1 / ( a * a1) = 9.08

The gain of both sections is the same and the absolute value is

K12 = ( Q12 / Q ) * ( Km / b1 )^0.5 = 1.155

We will define the same capacitor value for all the capacitors in both sections as

C = 0.01 * 10^-6 Farad

So now we can compute the resistor values shown in Fig. 1.

For filter section no. 1 (see Fig. 1)

R2 = Q12 / ( Pi * Fm1 * C ) = 298.4K

R1 = R2 / (2 * K12 ) = 129.2K

R3 = K12 * R1 / ( 2 * Q12^2 - K12 ) = 911 ohms

And for filter section no. 2 we have


R6 = Q12 / ( Pi * Fm2 * C) = 280K

R5 = R6 / ( 2 * K12 ) = 121.2K

R7 = K12 * R12 / ( 2 * Q12^2 -K12 ) = 855 ohms

We have now specified the filter, assigned the value for the capacitors, and computed the resistor values as shown in Fig. 1 (click on the figure for a larger view.) The simulation results are shown below Fig. 1. The simulation results show that the filter does indeed have a center frequency of 1KHZ and a bandwidth of 100HZ at -3db points according to our specifications. Note that a high Q analog filter like the one we designed will not in general be stable for step or pulse transient signals but it will work fine for steady state or slowly varying sine wave input signal.

MFB 2ND ORDER BAND PASS FILTER

Fig. 1 shows a schematic of a multiple feedback type 2nd order band pass filter. Click on the figures for a larger view. For a complete description, transfer function and other information you can refer to Texas Instrument Literature Number SLOA088, "Active Filter Design Techniques". In this post we will show the equations and information that is usually needed for a practical design.

To start with the design engineer should know or specify the center frequency, the bandwidth, the Q of the filter, and the gain at the center frequency. This circuit is very good for this application because of its low parts count and the ability to separately adjust the center frequency, the gain, and the Q of the filter. Note that Q is always the ratio of center frequency to bandwidth for a bandpass filter.

Q = Fm / (F2 - F1)

where Fm is the center frequency, F2 is the -3db high side cut-off and F1 is the -3db cut off on the low frequency side of the response curve. See the simulation response curve above of the filter we will design here.

Although the center frequency is specified, it can be adjusted according to its relation to the component values:


Fm = (1 / (2 * Pi * C)) * ( ( R1 + R3) / (R1 * R2 * R3 ) )^0.5

Note that the actual center frequency is inversely proportional to the value of the capacitors C.

The gain of the filter at the center frequency is

Ko = R2 / (2 * R1)

The Q is related to the center frequency as

Q = Pi * Fm * R2 * C

and the bandwidth is

BW = 1 / ( Pi * R2 * C )

Both capacitors in the circuit are usually set to the same value.

To design the filter we pick both capacitors to be equal to a practical value. Then since we have already picked the center frequency Fm, the Q, and the gain Ko at Fm, we can now calculate the three resistors values R1, R2, and R3 shown in Fig. 1.

R2 = Q / ( Pi * Fm * C )

R1 = R2 / ( 2 * Ko )

R3 = Ko * R1 / ( 2 * Q^2 + Ko )

Note that Ko is the absolute value of the gain even though the op-amp is an inverting configuration.

A design example of a fairly high Q MFB filter is as follows:

Fm = 10KHZ
Ko = 1.0
Q = 10
BW = 1000HZ

C = 0.01ufd ( set for both capacitors as a designer's choice. )

Using the above equations to calculate the resistor values we have

R2 = 31.83K
R1 = 15.92K
R3 = 79.18 ohms

I am using the exact computed resistor and capacitor values here in order to test the accuracy of the design by SPICE simulation. Fig. 1 is the actual simulation schematic.

The simulation results are shown for the AC analysis run in the plot above. The simulation provided the following results:

Upper -3db cut off frequency 10.549KHZ

Lower -3db cut off frequency 9.554KHZ

From above the band width is 995HZ. ( The design band width is 1000HZ.)

The Q is also then found to be 10.05. The design value was 10.0.

So the simulation results are within 1% of the design targets ( as long as the exact component values are used.)

With the high Q of the design, the filter is only conditionally stable in the presence of transient inputs, but for constant sine wave inputs the circuit would function according to its specifications. If a transient step is applied with a 0.1 microsecond rise-time, the circuit will ring about 1000 uS as observed at its output. If transient stability is required, a lower Q would have to be used with of course less frequency selectivity or a wider band width.

4TH ORDER BESSEL HIGH PASS FILTER

A 4th order Bessel high pass filter can be constructed as shown in Fig. 1 (click on the image to see a larger view.) The design example is for a 1KHZ cut off high pass filter. The transient and AC analysis SPICE plots are shown also. Note the smooth variation in phase in the AC analysis and the stability shown by lack of ringing after a pulse edge in the transient analysis.

For convenience we will let the value of the complex operator s be transformed to a substitute complex operator as

u = 1 / s

Then we can write the transfer function for the high pass filter as

A(u) = K / (( 1 + a1 * u + b1 * u^2) * ( 1 + a2 * u + b2 * u^2 )...)

A(u) is valid as long as the filter is a time-invariant linear circuit.

The coefficients for various filters are already known and we will use the Bessel coefficients for a 4th order filter. We choose the Bessel coefficients because it will give us a linear phase response and good pulse response. The Bessel coefficients for 4th order filters that fit the above transfer function are

a1 = 1.3397
b1 = 0.4889
a2 = 0.7743
b2 = 0.3890

To design the filter we will first pick a standard capacitor value that we will use for both 2nd order stages and then calculate the corresponding resistor values using the four Bessel coefficients.

R1 = 1 / ( Pi * Fo * C* ai ) and R2 = ai / (4 * Pi * Fo * C * bi)

where ai and bi are a1, a2 and b1, b2 coefficients depending on which stage the resistors are calculated for, that is a1 and b1 for the first stage and a2 and b2 for the second stage.

Doing the calculations we find

R1 = 2.376K
R2 = 2.181K
R3 = 4.111K
R4 = 1.584K

with C = 0.1UFD and Fo = 1KHZ

Note that R1 and R2 are for the first stage, and R3 and R4 represent the R1 and R2 for the second stage. See Fig. 1.

The Q or quality factor can also be calculated for each stage from


Q1 = ( b1 )^0.5 / a1

Q2 = ( b2 )^0.5 / a2

SALLEN-KEY BESSEL FILTER 4TH ORDER

The transfer function for filters constructed from cascading a 2nd order filter can be calculated with the product of the 2nd oder polynomials from the equation

A(s) = Ao /(( 1 + a1 * s + b1 * s^2)( 1 + a2 * s + b2 * s^2) ... )

We are going to cascade two 2nd order filters that are not identical to obtain a 4th order filter as shown in Fig. 1. The result produces the Bessel filter characteristics overall as long as we use the Bessel coefficients in the calculations of R and C values. The Ac response curve below the schematic shows the results of simulating the circuit on SPICE with the ouput levels and the phase shifts vs. frequency of the first stage and the second stages combined. The circuit is designed to be a Bessel type filter so that the phase shifts are linear and pulse response is good. The circuit values shown produce a 1KHZ low pass filter. We calculate the values below.

First note that since we are not cascading two identical filters, we must use a1, a2, b1, and b2 coefficents. a1 and b1 are used for the first stage, and a2 and b2 are used for the second stage to conform to the theory.

We choose the coefficents in accordance with known coefficients for the major filter types as tabulated here:

4th order Bessel coefficients:

a1 1.3397
b1 0.4889
a2 0.7743
b2 0.3890

Q is calculated from:

Q1 = ( b1 )^0.5 / a1 for the first stage = 0.52

q2 = ( b2 )^0.5 / a2 for the second stage = 0.805

using the Bessel coefficients.

In Fig. 1 we will rename the first resistor on the left as R1 and the next resistor R2. The lower capacitor we will call C1 and the capacitor in the feedback loop as C2 for both stage calulations for reasons of simplicity.

Next we need to assign a value for C1 so that we can calculate a value for C2 and the resistor values R1, R2 The value of C2 must be set so that the R values are positive real numbers. First,

C1 = 0.1ufd (defined as a starting point). Then

C2 > = 4* C1 * b1 / a1^2 = 0.1333ufd

We will set C2 to 0.15ufd for the first stage and we will need a hihger value of 0.33ufd for C2 in the second stage as an available standard value. Now we can calculate the resistor values from the solutions to the quadratic equation for the first stage

R1, R2 =( a1 * C2 -/+ (a1^2 * C2^2 - 4 * b1 * C1 * C2)^0.5) / (4 * Pi * Fo * C1 *C2)

The solutions are

R1, R2 = 508, 1.624K ohms (find the closest standard values.)

In the theory it is correct to use R1 as the value resulting from the calculation with the negative of the square root of the above equation, and R2 as the result of the calculations from the positive square root. It is a good idea to always verify your design with a simulation and testing a prototype circuit.

Similar calculations are done for the second stage to get resistor values of R1 = 331 ohms and 901 ohms.

The 2nd order stages have now been completely designed and in Fig. 1 we show two stages cascaded to achieve a 4th order filter.

The AC response at the output of each stage is shown if the figure below the schematic above. The advantage of the linear phase shift is that the group delay is constant and the pulse response is excellent for the Bessel filter. Simulation with a pulse input shows that the pulse response at the first stage output is faster than that of the 2nd output stage. So the trade off of the faster cut-off of the 4th order vs. the 2nd order filter is a slower pulse response of the higher order filter.

Note that you cannot apply the same identical stage for both stages as the results will not be accurate and it does not conform to the theory of the correct transfer function.

60HZ TWIN-T NOTCH FILTER

A Twin-T RC filter can be buffered with one or two op-amps to provided a usable notch filter for audio circuits or other applications.

Fig. 1 shows the AC response of the circuit of Fig. 2, a Twin-T filter with LTC6244 rail-to-rail 5MHZ op-amps. We used the following equations to calculate the values in the circuit elements R1 - R3 and C1 - C3. R4 and R5 could be set as shown or a 50K potentiometer could be used to adjust the Q of the circuit. We set the values shown to give us a high Q notch with approximately 60db of attenuation at 60HZ.

The following calculations were used:

Fo = 1 / ( 2 * Pi * R1 * C1 )

For this design we will set Fo to 60HZ and calculate the basic RC time constants of the filter using the normal value relationships for this type of filter

R1 = R2 = 2 * R3

C1 = C2 = C3 / 2

R1*C1 = 1 / ( 2 * Pi * Fo ) = 0.0026526

Let C1 = 0.1 * 10^-6

Then,

R1 = 0.0026526 / ( 0.1 * 10^-6 ) = 26526

R2 = 26526 C2 = 0.1 * 10^-6

C3 = 0.2 * 10^-6 R3 = 13263

In regard to setting values for R4 and R5, we found that better Q and response was obtained by setting R4 less than R5. Fig. 1 shows the results of simulation of the circuit of Fig. 2.

Note that the source has to be able to drive the input impedance of the filter which is approximately 160 ohms at 20KHZ. If a higher input impedance is needed you must reduce the value of C1 and C2 which mainly control the input impedance of the filter. Then you will have to calculate new values for R1 - R3.

2ND ORDER NARROW BAND FILTER WITH GAIN

A high Q narrow band filter with gain is shown in Fig. 1. This filter can be used to filter steady state signals but it is not suitable for audio use or with signals that have frequent transient signals. Its high Q and gain will cause it to ring or be unstable with such transients. The high Q, high gain circuit is with R5 not populated. The circuit can be used with audio signals if Q and gain is reduced with the " Q spoiler" resistor R5 set to approximately 12.5K maximum. With R5 in place the Q is approximately 1.3 and the gain is approximately 22.5db at the pass frequency, in the case shown, of 2KHZ. Without R5 populated, the Q is set to 10 and the gain is approximately 40db or 100 at the pass frequency. A 0.05 volt peak 2KHZ input signal will produce a 5 volt peak output signal.

The design shown has been simulated for AC response and transient response using the LTC6244 rail-to-rail 5MHZ op-amp. Fig.1 is the actual simulation schematic.

To design the circuit, we must first decide on the pass band center frequency Fo.
Next set the Q value. Then set a practical value for C1 and C2, e.g., 0.01UFD. Next we will calculate a value called Beta which is the ratio of the center frequency in radians to the Q value. Set Fo to 2KHZ and set Q to 10.

Beta = 2 * Pi * Fo / Q

Beta = 1256.6 with an Fo of 2KHZ and a Q of 10.

In the case of the circuit shown we can now calculate the value of R4 as

R4 = 2 / ( Beta * C ) = 159.16K with C=C1=C2= 0.01UFD

The closest standard value would then be selected from the standard resistance decade table.

Also we can compute the bandwidth of the filter measured at the -3db points as

BW = Beta / ( 2 * Pi ) = Fo / Q = 200. The bandwidth for this circuit is 10% of the center frequency with Q set at 10.

It is also known for the circuit of Fig. 1 that the circuit component values are related to the center frequency as

Fo = 1 / ( 2 * Pi * ( R2||R3 * R4 )^0.5 * C )

where R2||R3 is the parallel value of R2 and R3. We can re-write the equation as


R2||R3 = 1 / ( Wo^2 * C^2 * R4 ) = 397.9

where

Wo = 2 * Pi * Fo

Assuming that R2 = R3, we can then say that R2 and R3 are 2 * (R2||R3) = 795.8 each as shown in Fig. 1. The closest standard value would be selected from the standard resistance decade table.

We have now completely designed the circuit of Fig. 1.

The performance of the circuit was verified using a SPICE circuit simulator. The center frequency was at 2KHZ. Bandwidth was 200 HZ. Gain was 40db or 100. With a 50 millivolt peak 2KHZ input signal, the output signal was 5 volts peak. AC analysis showed that symmetry was excellent.

2ND ORDER BAND PASS FILTER

Fig. 1 shows a second order band pass filter that is designed to have a center frequncy of 2KHZ and is stable with transient signals or noise inputs. The schematic is the actual simulation schemtaice used including the biasing circuit as the basic circuit. Note that a practical value of 0.01UFD is set for the capacitors to start with but the R values are calculated as shown below and the calculated values are shown to 5 or 6 digits for use in the simulation but standard resistor decade values that are close will provide acceptable results.

We want this filter to have a maximally flat response in the pass band so the damping factor d is set to the square root of 2, as

d = ( 2 )^2 = 1.1412...

If we wanted a maximally flat time delay instead, such as might be required for an audio cross-over network, we would set d to the square root of 3 instead.

Once we have decided on d we can then calculate a gain that will be stable for the circuit. For an ordinary wide band pass filter, like the circuit in Fig. 1, we don't need a lot of gain and we need to be careful not to use too much gain as the filter may become unstable and ring on any kind of input transient, making it useless. We will use a standard formula to calculate the gain K that will work. We will treat the gain K as a positive value although the real gain is actually negative due to the inverting amplifier circuit we are using.

K = 3 - d = 1.5857

We can round off the gain to 1.59.

Now there is a criteria for setting the minimum value of the quality factor of the filter Q. Again, we don't want too high a Q for this simple wide band filter as the circuit may not then be time stable. The criteria is:

Q > ( K / 2 )^0.5 = 0.89

So what ever Q we choose it needs to be greater than 0.89. So lets try a Q = 1.

Also we set the center of the pass band to 2KHZ for demonstration.

Fo = 2000

Now we can calculate the resistor values for the filter:

R3 = Q / ( 2 * Pi *K * C * Fo ) = 1 / ( 2 * 3.14159 * 1.5857 * 0.01*10^-6 * Fo )

R3 = 5004.9

R4 = Q / (( 2 * Q^2 - K ) * 2 * Pi * C * Fo ) = 19409

From the equation for R4 we can see that 2Q^2 must be greater than the absolute value of K, and 2Q^2 cannot equal K, or R4 must be removed from the filter and then we have a different circuit and performance. I have found that including R4 makes a better filter.

R2 = 2 * Q / ( 2 * Pi * C * Fo ) = 15915.5

R1 is simply a dummy load for the op-amp and has no theoretical effect on calculations.

Simulating the circuit as shown in Fig. 1 with the Linear Technology LTC6244 rail-to-rail 5MHZ op-amp we have the following results:

Quite symmetrical band pass curve on a log of frequency vs voltage db scale with

Fo = 2KHZ

F low at -3db = 1.24KHZ approx.

F high at -3db = 3.3KHZ approx.

-20db at 20KHZ approx.
-20db at 200HZ approx.

Phase shift at Fo is -90 degrees.

With an input signal of 0.62893, the output signal is 1 volt as predicted from the gain K calculation.

Note that

Fo = ( F low * F high )^0.5 = ( 1240 * 3300)^0.5 = 2023HZ , very close to Fo = 2KHZ.

You will note also that the band pass shape for the rates of attenuation is proportional to a percentage of the frequency, not an absolute value of frequency delta from Fo.