Filter Design

A signal is any useful information and noise is the not so useful information, invariably, a part of the signal. The signals, seemingly arbitrary in nature, may be thought of as sinusoidal functions (the fundamental and the harmonics), as described by Fourier. You can imagine what would happen to signal frequencies, if they were passed through a system of energy storage elements (such as the linear time invariant systems, which have within them an impulse response in the form of sinusoidal functions), obviously, the matching frequencies in the signal and the system would be enhanced, and the unmatched frequencies would subside. The result is a filter, separating the desired frequencies from the undesired. We can tailor the response of such systems to our wishes, by physically choosing the components (in case of an analog system) or the coefficients (in case of a digital system). This is the basics of the filter design: suppressing or enhancing a specific frequency or a set of frequencies by passing a signal source through a system of energy storage elements. Filters modify signals in ways specified by the filter’s frequency response.
Analog filters are a precursor of digital filters and have roots behind the filter design mathematics. The criteria of any filter design are to establish the boundary of the desirable frequencies and then define how much suppression of the undesired frequencies is acceptable. Remember, frequencies can never be eliminated, they can only be suppressed. The first criterion provides the cutoff frequency, and the second one establishes how many poles and zeros are required in the filter design. The design methods are well established in both the analog and the digital domain, and the design procedures are a simple matter of selecting the appropriate components or coefficients. In this section, some commonly used designs of analog filters are discussed that become the prototype for our next section on digital filter design.
We begin our discussion with the explanation of terms used in both the analog and the digital filters.

Filter Terminologies
The following terminologies explain different aspects of filter characteristics. See Figure 1a and 6.1b for a reference.

Pass band: Frequency range preserved in the output
output

Figure 1 a) Different characteristics of a filter response. b) The cutoff frequency and Q of a band-pass filter.

The filter design is essentially implementing solutions of differential equations, either through the Laplace Transform or the convolution process, but the modeling is done through the Transfer Function that formed the input and output relationship.

The higher frequencies are suppressed from the voltage across the capacitor in a network of resistors and capacitors. Similarly, the lower frequencies are suppressed from the voltage across the resistor in the network of a resistors and capacitors, and a band-limited response is achieved with a second order section.

The filter response generally falls into the following four basic types: low-pass, high-pass, band-pass, and band-stop filters. There are variations of filters such as notch filters, narrow band-pass, etc. that are special implementations of the band-pass and band-stop filters.

As the name suggest, the low-pass filters allow low frequency to pass through but stop high frequencies, the high-pass filters allow high frequencies to pass through but stop low frequencies, the band-pass filters allow a specific band of frequencies to pass through but stop all others, and the band-stop filters allow all frequencies to pass through except a certain band.

The fact that the LTI systems only alter the amplitude or phase but not the frequency itself, leaves us with only one choice and that is to suppress the undesired frequencies. Ideally, we would like to have the desired frequencies untouched while the undesired frequencies are eliminated, as shown in Figure 2, but this is not practical, as it requires infinite filter blocks, so we will leave the quest for achieving the ideal response only as a goal.

Figure 2 Ideal filter response: a) low pass, b) high pass, c) band pass, d) band stop.
The filter design procedure has been standardized around a set of low-pass filters having the normalized cutoff frequency . From this prototype low-pass filter all other types may be derived with a simple substitution scheme as shown bellow


Low-pass to low-pass transform ( as the cutoff frequency) $$s_N=s/\omega_c$$
Low-pass to high-pass transform ($$\omega_c$$ as the cutoff frequency) $$s_N=\omega_c/s$$
Low-pass to band-pass (bandwidth)
$$s_N=(s^2+(\omega_2\omega_1))/(s+(\omega_2-\omega_1))$$

Low-pass to band-stop (bandwidth$$\omega_1,\omega_2) $$s_N=(s+((\omega_2-\omega_1) )/(\omega^2+\omega_1\omega_1))$$


For example, a normalized low-pass filter whose transfer function is given as $$H(s)=1/(1+s_N)$$, may be converted into another low-pass filter with the cutoff frequency $$\omega_c=50rad/sec$$ by substituting with . The resulting filter transfer function is $$H(s)=50/(50+s)$$. Similarly, a second order normalized low-pass filter transfer function such as, $$H(s)=1/(s^2+\sqrt{2s}+1)$$ may be converted into a band-stop filter having a bandwidth of 10rad/sec centered around 50rad/sec, using the low-pass to band-stop mapping as
$$H(s)={5760000+4800s^2+s^4)}\over{(5828282+3434s+4829s^2+1.4142136s^2+s^4)}$$



All these calculations could be performed easily with the help of scilab instructions as follows ,

s=poly(0,’s’);
nlp=1/(1+sqrt(2)*s+s^2);
bsmap=(s+(60-40))/(s^2+(60*40));
horner(lp,lpmap)

The Scilab function ‘horner’ performs the required transformation.


The digital and analog filters may be identical in their response up to certain frequencies, but the implementation of each is an entirely different matter. For analog filters, the Transfer Function is implemented as a circuit design with the coefficient values translated to the component values of the resistors, capacitors, and inductors (op-amps may be added in place of inductors and to isolate the blocks of circuits from one another). On the other hand, the digital filters are realized by transforming the Transfer Function into a difference equation that is to be solved with convolution using an iterative algorithm. There is no limit to the range of input frequencies for analog filters, but the digital filters are limited in their frequency response, they can only process frequencies up to the Nyquist limit (which is defined as half the rate of the sampling frequency).

Every filter has a corresponding Transfer Function, but it is easier to implement filters in terms of the building blocks of Transfer Functions. It not only reduces the complexity of the overall system but gives a modular approach to the implementation process.

Each pole contributes to a 20 dB drop in the roll-off rate of the frequency response. The response is improved as we add more poles to the Transfer Function of the system, but then the order of the polynomial grows as we add more poles, making the system more complex to design. One way to simplify the complexity is to cascade blocks of simple Transfer Functions. These blocks are made up of the first and second order systems of single real or single complex poles and zeros. Complex filters are implemented as cascades of the basic building blocks.
The following Transfer Functions provide the four basic types of filters, namely the low-pass, high-pass, band-pass, and band-stop filters. The equivalent circuit matching the Transfer Function is described in the section about analog filters, and the equivalent difference equations for the digital filters are discussed in the next chapter.

First Order Low-Pass Filter Transfer Function
The Transfer Function of Equation 1 is a low-pass single pole filter with the cutoff frequency of p1 and the gain of K at 0 frequency.
$S{V_0\over V_1}\over {1+s/p_1}$$
(1)

The magnitude response is
$$|H(j\omega)|=|{K\over{j(\omega/p_1)}|=K\over{\sqrt{1+(\omega/p_1)}^2}


The phase function from the Equation 1 is
$$\phi=-tan^{-1}{\omega\over{p_1}}$$



The frequency and the phase response of the low-pass transfer function (for ) are shown in the Figure 3.

Figure 3 First order low-pass filter response. a) Magnitude response. b) Phase response .

Second Order Low-Pass Filter Transfer Function
A 40 dB roll-off rate may be achieved with the addition of an extra pole resulting in a second order section as shown in the Transfer Function of Equation 2,
$$\\H(s)={k\omega^2_c\over {s^2+(\omega_c/Q)s+\omega^2_c}\\$$
(2)

When $$\omega_c/Q=\sqrt2$$in Equation 2 becomes a second order Butterworth polynomial, discussed later in the section, the normalized frequency response of the second order section compared with the first order is shown in Figure 4. Notice the improvement in roll-off rate of the second order Transfer Function.

Figure 4 Second order low-pass filter magnitude response compared to the first order filter response.

First Order High-Pass Filter Transfer Function
The Transfer Function in Equation 3 is for a high-pass single-pole filter with the cutoff frequency of p1 and the gain of K at 0 frequency.

$$\\H(s)={V_0}\over{V_1}={k}\over{1+1/sp_1}\\$$
(3)

The magnitude response is

$$\\|H(j\omega)|=|{K}\over{1-j(1/\omegap_1)}| ={K}\over \sqrt{1+j(1/\omegap_1)}\\$$


The phase function from Equation 3 is
$$arg(K)+tan_1{1}\over{\omega p_1}=tan^{-1}1\over{\omega p_1}$$



The magnitude response is shown in Figure 5.

Figure 5 First order high-pass filter magnitude response.

The Second Order High-Pass Filter Transfer Function
The second order gain of a high-pass filter is given as shown in Equation 4.

$$\\H(s)={V_0}\over{V_1}={K_3s^2}\over {s^2+b_1s+b_0}\\$$
(4)

The magnitude response is
$$\\|H(j\omega)|=|{-K_3\omega^2}\over {(b_0-\omega^2)+(j\omega b_1)}|
={-K_3\omega^2}\over {sqrt{(b_0-\omega^2)^2+(\omega b_1)^2}}\\$$



The phase function from Equation 4 is
$$\\tan^{-1}{{b_0\omega}\over{(b_1-\omega^2)}\\$$


The normalized frequency and phase response of the second order high-pass transfer function is shown in Figure 6.

Figure 6 Second order high-pass filter response, the magnitude, and the phase response.

Figure The Second Order Band-Pass Filter Transfer Function
A band pass filter may be obtained from a low pass filter transfer function with a simple substitution of $$s_N$$ with $$(s^2+(\omega_2\omega_1))/(s+(\omega_2-\omega_1))$$ as shown in the Figure 1a. The band-pass filter may also be defined in terms of the center frequency and its two side frequencies where the gain falls off to the –3 dB of the center frequency. The Transfer Function is given as shown in Equation 5.
$$H(s)={V_0}\over {V_1}={K_3s}\over {(s^2+(\omega_2/Q)s\omega^2_0))$$
(5)
$$Q=(\omega_0/\omega_{3d})$$
Where .

The width of the bandwidth is controlled by the quantity Q. If a narrow bandwidth is desirable, set the Q to a high value such as 10 or more.

The magnitude response is








$$\\|H(j\omga)|=|{ {K_3s}\over {(\omega^2_0-\omega^2)+j(\omega_0/Q)}=
{\sqrt{(\omega^2_0-\omega^2)+j(\omega_0/Q)}}\\$$


The phase function from Equation 5 is

$$\\tan^{-1}(j\omega)-tan^{-1}{j(\omga/\Q)}\over{\omega^2-\omega^2)\\$$



The magnitude response is shown in Figure 7.

Figure 7 Second order band-pass filter magnitude response.

The Second Order Band-Stop Filter Transfer Function
The band-stop filter is also known as the notch filter. Similar to the band-pass filter, the Transfer Function is defined in terms of the center frequency that would be removed from the system and the two side frequencies with–3 dB gain. The Transfer Function is given as shown in Equation 6.

$$\\|H(s)=V_0\over V_1={ {K_4(s^2+\omega_c)}\over {s^2+(\omega_0/Q)s+\omega^2}\\$$
(6)

Where
$$\\Q=(\omega_0/\omega_{3db})\\$$
.









The magnitude response is
$$\\|H(j\omega)|=|{K_4\sqrt{(\omega^2+\omega^2)^2}}\over{(\omega^2-\omega^2)+j(\omega_0/Q)}|= {K_4(\omega^2+\omega^2)}\over{\sqrt{(\omega^2+\omega^2)^2}}\over{(\omega^2-\omega^2)+j(\omega_0/Q)}}\\$$


The phase function from the Equation 6 is
$$\\tan^{-1}(j\omega)-tan^{-1}{j(\omga/\Q)}\over{\omega^2-\omega^2)\\$$



The magnitude and phase response is shown in Figure 8a and 6.8b.

Figure 8 Second order band-stop filter magnitude response.

Systems often require more sophisticated filter response than the 3 dB gain and 20 dB per decade drop in the roll-off rate as provided by the first order Transfer Function polynomials. There are different methods that enhance different aspects of the filter characteristics. The Chebyshev--CE]filters improve upon the roll-off rate but add ripples in the pass-band region, whereas the Butterworth filters give flatter response in the pass-band region but are not very efficient in the roll-off rate. We will discuss the Butterworth filters in the next section and the Chebyshev filter in the coming chapter.

The goal in any filter design is to get as close as possible to the ideal filter response. We would like to see a flatter response in the pass-band region and a steeper roll-off in the transition band. The Butterworth polynomials offer such a response. The response function is given as
$$\\H(s)=1\over{1+s^n}\\$$

And the magnitude squared is shown in Equation 7.
$$\\|H(j\omega)|^2=1\{1+(j\omega){^2n}}\\$$

(7)

Notice the coefficient of has been chosen as 1, this is equivalent of normalized cutoff frequency of
$$\omega_0=\omega\over{\omega_{coutoff}}=1$$
. Having the normalized response simplifies the design process by designing the circuit for unit response, and once all the components are identified, in the last step the values are scaled for the desired response. At cutoff frequency when $$\omega=\omega{cutoff}$$, the magnitude response of Equation 7 becomes 0.707 for all values of n. The contribution becomes less significant for $$\omega_0 \lt 1$$, and for $$\omega_0 \gt 1$$, the magnitude approaches 0 faster with increasing n. The result is a flatter response for frequencies less than the cutoff frequency and steeper roll for the frequencies greater than the cutoff frequency with increasing n, as shown in the Figure 9. Through the normalized low-pass filter (as defined in Equation 7) all other types of filters may be derived with the help of the frequency-frequency mapping being mentioned earlier.

Figure 9 The Butterworth response with different values of n.

The Butterworth expansion for N = 1 is given as
$$|H(j\omega)|^2=1\over{1+(\omega/\omega_c)^2}\\$$


Where is the cutoff frequency, as $$\omega \rightarrow \infty$$ the response approaches 0, but when $$\omega$$ gets closer to 0, the response becomes 1. In between, there is a gradual loss of magnitude. Precisely at the cutoff frequency of , the magnitude is –3 dB. This is equivalent to the first order Butterworth response as shown in Figure 4. The expansion of coefficients $$(\omega)^{2N}$$ may be explained as follows.

If the power vectors of the pole frequency $$(\omega_c)^{2N}$$ were to be drawn on a circle (with the radius $$(\omega_c)$$()) that can be normalized and set equal to 1, all vectors would appear as equally spaced on a semicircle. Since multiplying a unit vector to itself simply shifts the vector to an angle, the roots of Equation 7 are all shifted by an angle $$\pi/N$$with respect to the real axis. This could be evident from the following analogy.

The poles are actually N roots of the denominator polynomial of the transfer function, as shown in Equation 9.

$$\\H(s)=1\over{1+s^N}\\$$
$$\\1+s^N=0\\$$
$$\\s^N=-1,s=\sqrt^N{e^j\pi}=\sqrt^N{-1}\\$$
$$s=cos(\pi/N)+jsin(\pi/N)\\$$
$$H(j\omega)H(-j\omega)=|H(j\omega)|^2=1+(-1)^ns^{2N}=0\\$$




(9)

Where s is a complex value $$s=e^{j\pi}$$evaluated on a unit circle.

Figure 10 indicates pole locations for various values of N. Only the poles on the left side of the circle are taken as the roots of the transfer function polynomial. Except for the first pole, all others are complex conjugates. If N is even, all poles are complex conjugates and when N is odd, there is only one real and the rest are complex conjugate poles. An nth order Butterworth Filter is defined as
$$\\|H(s)|^2=1\over{\prod^N_{i=1}(s-s_i)}=1\over{(s-s_1) (s-s_2)\hdots(s-s_N)}\\$$


Where$$\\s_i=e{j\pi[2i+n-1/2N]}=
Cos(\pi{2i+n-1}\over 2n)}+jsin({\pi{2i+n-1}\over{2n})\\$$


The following calculations indicate the pole vectors for different values of N:

$$\\N=1 s_1=cos(\pi)+jsin(\pi)=1\\$$

$$\\H(s)=1\over{(s+1)}\\SS

$$\\N=2 s_1=cos(\pi3/4)+jsin(\pi3/4)=-.707+j.707\\$$


$$\\s_2=cos(\pi5/4)+jsin(\pi5/4)=-.707+j.707\\$$

$$\H(s)=1\over {(s^2_2+\sqrt2s_2+1)}\\$$

$$\\N=3 s_1=cos(\pi2/3)+jsin(\pi2/3)=-.5+j.866\\$$


$$\\s_2=cos(\pi)+jsin(\pi)=1\\$$


$$\\s_3=cos(\pi2/3)+jsin(\pi2/3)=-.5+j.866\\$$

$$\\H(s)=1\over{s^3_3+2s^2_3+s_3+1)}= 1\over{(s_3+1)(2s^2_3+s_3+1)}\\$$


$$\\N=4 s_1=cos(\pi5/8)+jsin(\pi5/8)=-.382+j.924\\$$


$$\\s_2=cos(\pi7/8)+jsin(\pi7/8)=-.924+j.383\\$$


$$\\s_3=cos(\pi9/8)+jsin(\pi9/8)=-.924+j.383\\$$


$$\\s_4=cos(\pi 11/8)+jsin(\pi 11/8)=-.382+j.924\\$$


$$\\H(S)=1\over{s^4_4+2.61s^3_4+2.61s^2_4+s_4+1)}=1\over
{(s^2_4+0.765s_4+1)(s^2+0.765s_4+1)}\\$$


The design parameter N or the filter order is calculated by the specifications of the magnitude at the cutoff frequency and the rejection frequency. For example, if the magnitude at the 20Hz is specified as –3d and the attenuation at 30Hz is –40 db then the filter order N is,



$$\\-3db\rightarrow -3\over20 \rightarrow |10^{-3\over20}|^2=1\over{1+\epsilon^2}=-.5,\epsilon=0.707\\$$

$$\\-40db\rightarrow -40\over20 \rightarrow |10^{-40\over20}|^2=1 \over{1+A^2}=-.0001,A=0.01\\$$

$$\\N \ge{log(A/\epsilon)}\over {log(\omega_r/\omega_c)}\ge10.5,N=11\\$$

Figure 10 The pole locations of the Butterworth expansion.

Next, we discuss the implementation details of the Transfer Functions in terms of the electronic circuit designs for the analog filters (the digital filters are discussed in the next chapter).

Analog Filters
In an analog domain, a filter is a circuit that produces the expected response specified by the corresponding filter Transfer Function. We have seen earlier how a first and second order differential equation’s response acts as a frequency selector. The higher frequencies are suppressed from the voltage across the capacitor in a network of resistors and capacitors. Similarly, the lower frequencies are suppressed from the voltage across the resistor in the network of a resistor and capacitor.

The filter circuitry in our designs will have basically two types of components, the passive components (the resistors and capacitors) and the active components (the op-amps), hence the name active RC filters. The op-amps play dual roles in the analog filters. They not only simulate inductors (if needed as a second order section) but also act as non-interacting blocks to prevent loading effects on the passive components. Since op-amps play an important role in setting up the network equations, a brief overview is presented next as a refresher.

Op-amp as Differential Amplifiers
A differential amplifier’s function is to amplify the difference between two signals. The basic schematic of the op-amp is represented in Figure 11a, and the corresponding circuit model is shown in Figure 11b. The Ri is the differential input impedance of an extremely high value with practically no current flowing through it. The Rs is the output impedance, a negligible quantity as if it does not exist in the circuit. The amplifier gain of Figure 11a is expressed as

$$\\v_0=a_0(v_1-v_2)\\$$


Where a0 is the open loop gain of the op-amp when there is no feedback and is usually very large in the range of 105 to 106, considered infinite for an ideal op-amp. We will use the equivalent of voltage controlled voltage source, as shown in Figure 11b, in the loop analysis of the network when we use op-amps in our filter design.

Two things need to be remembered when dealing with op-amps:

Infinite input resistance means the current into the inverting input is zero:
$$\\i=0\\$$



Infinite gain means the difference between v+ and v– is zero:

$$\\v_+ -v_-=0\\$$

Figure 11 a) Op-amp as a differential amplifier. b) Equivalent circuit model.
c) Input and output voltage across the op-amp.

The two most common configurations are provided next to help identify the loop equation created by a filter circuit.

Figure Inverting Amplifier
Figure 12a shows an inverting amplifier configuration and Figure 12b shows the summation of current at the inverting input.

The current passing through the two resistors is equal as no current should be going through the op-amp,

$$\\i_1 -i_2=i\\$$


Since all the current is passing through the source resistor R1 and the feedback resistor R2, we have the following voltage drop
$$\\v_i=R_1i and v_0=-R_2i\\$$


and


Resulting in the amplifier voltage gain of
$$\\v_\over v_i=R_2\overR_1\\$$



The op-amp will provide whatever output voltage is necessary so that both the input voltages are equal.

Figure 12 a) The inverting configuration of an op-amp. b) Summation of the current at the node.

Non-Inverting Amplifier
Figure 13 is a non-inverting amplifier configuration showing the summation of the current at the inverted input.

Since no current is passing through any of the op-amp input, the two input voltages are equal:
$$\\v_+=v_-\\$$


The voltage across $$\\v_=R_1\over{R_1+R_2} and v_+=v_-=v_i\\$$
and .
The voltage gain is
$$\\v_0\over v_i={R_1+R_2}\over{R_1}=1+R_2\over R_1\\$$



Resulting in the voltage gain of
$$\\v_0={R_1+R_2}\over{R_1}v_i\\$$

Figure 13 The non-inverter configuration of an op-amp.

With the basic configurations of op-amps in place, we are ready to design some simple building block analog filters using resistors, capacitors, and op-amps.

Active RC Filters The design of the analog filters requires establishing the loop equations of the four basic types of the filter Transfer Functions, namely, the low-pass, high-pass, band-pass, and band-stop filters. The requirements are specified in terms of the cutoff frequencies, establishing the poles and zeros of the filter’s Transfer Function and improvement is achieved by cascading several blocks in a series. The network loop equation parameters directly correspond to the component values, and the design requirement is simply a matter of choice between the different component compositions. In 1955, R. P. Sallen and E. L. Key described these filter circuits that have become the de facto standard for the filter designs.

The circuit shown in Figure 14 is a generalized form of the Sallen-Key circuit, where generalized impedance terms, Z, are used for the passive filter components, and R3 and R4 are the non-inverting op-amp gain.

**** Insert Figure 14 here *****
Figure 14 a) The generalized Sallen-Key circuit. b) Loops for node current analysis.


KCL at Vf

$$\\v_f({1\over {Z1}}+{1\over {Z2}}+{1\over {Z4}}=V_i(1\over{Z_1})+ V_p(1\over{Z_2})+ V_0(1\over{Z_4})\\$$




KCL at Vp
$$\\v_p({1\over {Z2}}+{1\over {Z3}}=V_f(1\over{Z_2})\\$$
$$\\v_f=V_p({i+Z_2\over {Z3})\\$$




Solving for Vp






$$\\V_p=V_i{{Z2Z3Z4}\over{Z2Z3Z4+Z1Z2Z4+Z1Z2Z3+Z2Z2Z4+Z2Z2Z1}}+
V_0{{Z1Z1Z3}\over{Z2Z3Z4+Z1Z2Z4+Z1Z2Z3+Z2Z2Z4+Z2Z2Z1}}\\$$


The op-amp gain K


$$\\V_n=V_0({R3}\over{R3+R4})\\$$


Since op-amp amplifies the difference between the two input terminals,
$$\\V_0=K(V_0-V_n)\\$$



We can obtain the following generalized transfer function,
$$V_0\overV_i={(K\{{Z1Z2\overZ3Z4}+{Z1\overZ3}+{Z2\overZ3}+{Z1(1-K)\overZ4}+1)}


Please see the following link for design guidelines: http://focus.ti.com/lit/an/sloa024b/sloa024b.pdf.

Figure Single Pole Low-Pass Filter
The circuit of Figure 14 is a single pole low-pass filter, corresponding to the Transfer Function of Equation 10. The output Vi measured on capacitor C is given as

$$\\H(s)=V_0\over\V_1=Z_c{Z_r+Z_c}\\$$


Substituting for the values of $$Z_c=1/{j\omega C}$$ and
$$\\Z_R=R\\$$, we get Equation 10.
$$H(j\omega)=V_0\V_1=1\{1+j\omegaRC}\\$$
(10)

Figure 15: First order low-pass active RC filter.

Figure Design Considerations
From Equation 10, we can draw the following filter characteristics:

The cutoff frequency is shown in Equation 11.
$$\\\omega_c=1/{RC}\\$$
(11)

The frequency response magnitude is shown in Equation 12.

$$\\|H(j\omega)|=1\over{RC}\times 1\over \sqrt{(1/RC)^2+\omega^2\\$$
(12)

The gain in decibels is shown in Equation 13.
$$\\20log|H(j\omega)|=20log|1\over {RC}|-20log|\sqrt {(1/RC)^2+\omega^2}\\$$
(13)

The phase change:
$$\\\Theta = -arctan(\omega RC)\\$$


A plot of the frequency response magnitude and phase change is presented in Figure 6.

Example 1
Design a low-pass filter with the cutoff frequency of .

From Equation 10, we have
$$\\1/RC = 2\pi1000rad/s^{-1}\\$$



Since we have two element values to choose from, by selecting R=10 K, we get
$$\\C =1/ 2\pi1000*10000 = 1.57\times 10^{-7}\mue F\\$$



Equation Example 2
Define the cutoff frequency for R=10 K and C=0.001 in Figure 14.

From Equation 11,
$$\\\omega_c=1/(10\times10^3\times0.001\time 10^{-6})=100 krad/sec\\$$



Equation Example 3
Define the magnitude in dB at frequency 10 KHz, for R=1K and $$C=0.01\mueF$$ in Figure 1.

$$\\\omega=2\pif=2\pi\times10^4 rad/s\\$$


From Equation 11,
$$\\1/RC=1/10^{-8}\times10^3=10^5 rad/sec\\$$


From Equation 12,
$$\\|H(j\omega)|=10^5\over {\sqrt (2\pi \times 10^4)^2+(10^5)^2}=0.97\\$$



From Equation 13,

Magnitude in dB

Figure Second Order Low-Pass Filter
The roll-off rate could be improved to 40 dB with the implementation of a second order circuit, as shown in Figure 16. The cutoff frequency is normalized to 1 so the design could be standardized, and only in the final analysis we scale up the component values to represent the true cutoff point. The generalized Transfer Function is given as shown in Equation 14.

$$\\H(s)={k\omega^2_c}\over {s^2 + (\omega_c/Q)s+\omega^2_c\\$$

(14)

For a second order Butterworth response
$$\\\omega_c}/Q=\sqrt{2}\\$$


$$N=2 s_1=cos(\pi3/4)+jsin(\pi3/4)=-.707-j.707\\$$

$$\\ s_2=cos(\pi5/4)+jsin(\pi5/4)=-.707-j.707\\$$

$$\\H(s)=1\over{s^2_2+\sqrt2s_2+1}\\$$


The loop equation is given as shown in Equation 15.
$$\\H(s)=V_0\over V_1={K/R_1C1R_2C_2}\over{s^2+s(1/R_1C_1+1/R_2C_1+1/R_2C_2-K/R_2C_2)+1}\\$$
(15)

Figure 16 Second order low-pass active RC filter.

Equation Example 4
Design a Butterworth filter with a –3 dB gain at $$2\pi10000rad/sec$$. The circuit of Figure 15 has input resistor R1=1K.

Equating the Transfer Function of Equation 14 and the loop Equation 15, we get three equations (Equations 6.16 through 6.18) and five unknowns R1, R2, C1, C2 and K,

$$\\H(0)=10=K/{R_1C_1R_2C_2}\\$$
(16)
$$\\\sqrt2\omega_c=(1/R_1C_1+1/R_2C_1+1/R_2C_2-K/R_2C_2)\\$$

(17)

$$\\\omega^2=(1/R_1C_1R_2C_2)\\$$
(18)

Let’s use normalized values of R1 = 1, C1 = 1,and solve for the other unknowns using Equations 6.16, 6.17, and 6.18.
$$\\H(s)=1\over{s^2+(1/Q)s+1}=1\over{(s^2+(2/C_1)s+1)C_1C_2}\\$$


$$\\C1=2Q\\$$
$$\\C2=1\over2Q\\$$


To realize a second order, we equate the coefficients,
$$\\s^2+(11/Q)s+1=s^2+\sqrt2s+1\\$$
$$\\Q=1.414\\$$


Using the resistor value and the frequency scaling ,

Multiply each capacitor by
$$\\1\over1000\times 1{\2\pi10000}\\$$
.
$$\\C1=2\times1.414\over{1000\times 2\pi10000}=.045\mueF\\$$


$$\\C2=1\over{2\times 1.414\times 1000\2\pi10000}=.056\mueF\\$$



$$\\R1=1KI\\$$
$$R_2=1K\\$$


Figure The Band-Pass Filter
All resonant circuits act as band-pass filters. The circuit of Figure 16 corresponds to the Transfer Function of Equation 19, which allows a certain band of frequencies to pass through while suppressing the rest.

$$\\H(s)={H_2s}\over{s^2+(\omega_0/Q)\omega_0+\omega_0^2}\\$$
(19)

There is one frequency (the resonant frequency, also called the center frequency) that has the maximum output voltage Vmax. The frequency to the left of the center frequency with the magnitude 0.707 Vmax is the low cutoff frequency, and the frequency to the right with 0.707Vmax is the high cutoff frequency. The bandwidth is between the low and high cutoff points.

The loop equation of the circuit of Figure 16 is given as shown in Equation 20.
$$\\H(s)={V_0\over V_I}={-(s/C_1R_1)}\over{s^2+s(C_1+C_2)/(C_1C_2R_2+1/C_1R_1C_2R_2}\\$$

(20)

Equating the parameters of Equations 6.19 and 6.20 and simplifying Equation 20 by letting C1 = 1, C2 = 1, and R2 = 1, we can reduce the choice to just one variable R1 to be specified,

$$\\H(s)={V_0\over V_I}={-(s/R_1)}\over{s^2+2s+1/R_1}\\$$


The cutoff frequency is shown in Equation 21.
$$\\\omega_0^2=1/R_1\\$$

(21)

The bandwidth quality factor is shown in Equation 22.
$$\\(\omega_0/Q)=2\\$$

$$\\Q=1/\sqrt{R_1}\\$$
(22)

Figure 17 Second order band-pass active RC filter.

Equation Example 5
Design a band-pass filter with Q = 2 and $$\omega=2\pi10000rad/s^{-1} with R1=10K$$.

The choice could be narrowed down by making C1 = C2 = C and R1 = R2 = R and substituting the values into Equations 6.21 and 6.22, resulting in Equations 6.23 and 6.24.

$$\\\omega^2_0=2/(CR)^2=[2\pi1000]^2\\$$
(23)
$$\\\omega_0/Q=1/(CR)=2\pi1000\\$$

(24)

Let R = 10K and from Equations 6.23 and 6.24 we get

$$\\C1=C2=1.6\times10{-6}\mueF R_1=R_2=10K\\$$


Figure Single Pole High-Pass Filter
Implementing the Transfer Function of Equation 3 gives a single pole high-pass filter. Notice that swapping the resistor and capacitor of the low-pass filter of Figure 14 converts it into a high-pass filter, as shown in Figure 17.

The loop equation is described as shown in Equation 25.

$$\\H(j\omega)={V_0\overV_1}=1\over{1-j(1/\omegaRC)}\\$$
(25)
Figure Design Considerations
From Equation 25, we can draw the following filter characteristics,

The cutoff frequency is shown in Equation 26.
$$\\\omega_c=1/RC
(26)

The frequency response magnitude is shown in Equation 27.
$$\\|H(j\omega)|=1\over{RC}\times1\over{\sqrt{(1/RC)^2+\omega^2}}\\$$
(27)

The gain in decibels is shown in Equation 28.
$$\\20log|H(j\omega)|=20log|1\over{RC}-20log|\sqrt{(1/RC)^2+\omega^2}\\$$

(28)

The phase change is shown in Equation 29.
$$\\\Theta=-arctan(1/\omegaRC)\\$$

(29)

A plot of the frequency response is presented in Figure 18.

Figure 18 The circuit for a first order high-pass active RC filter.

Equation Example 6
Design a high-pass filter with cutoff frequency of and the transition band of 20 dB per decade.

The single pole solution of Equation 25 satisfies 20 dB requirements.

Substituting the cutoff frequency$$\omega=2\pi1000rad/s^{-1}$$ into Equation 26, we get

$$\\(1/RC)=2\pi1000\\$$


If we pick R = 1K, the capacitor n$$C=1.57x10{-6} \mueF$$

Equation Example 7
Define the magnitude and phase of the filter output of Example 6 at 100 Hz.

The magnitude from Equation 27:
$$\\|H(j\omega)|=2\pi1000\times1\over{\sqrt{(2\pi1000)^2+100^2}=0.999\\$$



The log magnitude from Equation 28:
$$\\20log|0.999|==-0.01\\$$



The phase from Equation 28:
$$\\Theta=-arctan(2\pi100\times2\pi1000)=0.001 deg\\$$


Figure Second Order High-Pass Filter
Implementing the Transfer Function of Equation 30 provides a 40 dB roll-off rate.

$$\\H(s)={H_2\omega}\over{\omega^2+b_1\omega+b_0}\\$$
(30)

The circuit of Figure 18 has the loop equation matching the Transfer Function

$$\\H(s)={V_0\over V_1}={Ks^2}\over{s^2+s(1/R_1C_1-K/R_2C_1+1/R_2C_2+K/R_2C2)+1/R_1C_1R_2C_2}\\$$
(31)

There are five unknowns—R1, R2, C1, C2, and the gain K—with three equations,

The gain
$$\\H=K\\$$


The cutoff frequency

$$\\\omega^2_c=b_0=1\{R_1R_2C_1_C_2}\\$$


For a second order Butterworth response
$$\\b_1=\sqrt2 \omega_c = {1-K}\over{R_1C_1}+{1}\over{R_2C_2}+{1}\over{R_2C_1}\\$$

Figure 19 The circuit for a second order high-pass active RC filter.

Equation Example 8
Design a second order high-pass Butterworth filter with a –3 dB gain at $$\\2\pi10000rad/sec\\$$ and a gain of 10 at
$$\\2\pi10000rad/sec\\$$ dB. The circuit of Figure 18 has the resistor value R1 = 1K.

Equating the Transfer Function of Equation 1 and the loop of Equation 1, we get three equations and five unknowns: R1, R2, C1, C2, and K. We could normalize the values of R1 = 1,C1 = 1, $$\omega_0=1\\$$ and obtain the three unknown values in normalized form:

$$\\K=10,C_2=9.414 R_2=1/9.414 R_1=1 C_1=1\\$$
,

The actual values are $$\\R_1=1K R_2=1K\times0.11=111\Omega C_1=0.016\mue F C_2=0.15\mue F\\$$



Figure Band-Stop Filter
A band stop response is achieved by canceling the band-pass response with a pure resonant response from the numerator polynomial.

The circuit shown in Figure 19 has a gain, as shown in Equation 32.


$$\\H(s)={H_4(s^2+\omega^2)}\over{s^2+(\omega_p/Q)\omega_0+\omega_p^2}\\$$
(32)

Figure 20 Circuit for a second order band-stop active RC filter.

The loop equations are given as
$$\\H(s)={V_0\overV_1}={K(C_1C_2s^2+1/R_1R_2)}\over
{C_1C_2s^2+s[(C_1+C_2)/R_2+C_2(1/R_1+1/R_2)(1-K)]+1/R_1R_2}\\$$



There are five design parameters, R1, R2, C1, C2, and the gain K. and three equations, making C1 = C2 = C and R1 = R2 = R, we can reduce the number of variables to three and solve for them using,
$$\\H(s)={V_0\overV_1}={K(C^2s^2+1/R_1^2)}\over
{C^2s^2+2Cs/R+1/R^2}\\$$



Equation Summary
In this chapter, we studied the filter design methods of the four basic types of analog filters: the low-pass, high-pass, band-pass, and band-stop filters. The filters were designed as solutions of differential equations that governed the input and output relationship forming the filter transfer function and implemented as electrical circuits of resistors, capacitors, and op-amps whose component values matched the coefficient values of the transfer function. The poles and zeros of the transfer functions describe the characteristics of the filters such as the cutoff frequency and the roll-off rate in the transition band, and the filter response was measured by how well the filter suppresses frequencies beyond the cutoff region. It is possible to improve upon the filter response by adding extra poles at the cutoff point that produce steeper roll-off in the filter transition band.