ECE4330 Lecture 19 The Fourier Transform (cont.) Prof. Mohamad Hassoun

This lecture covers the following topics:  The significance of phase information: Signal transmission example  Applications of the Fourier transform o Fourier transform in a basic communication system o Low-pass filters: ideal vs practical (Butterworth filters) o Filter Design: low-pass, high-pass, band-pass, band-reject, notch, and all-pass  Analog filter design as a numerical optimization problem  Appendix: Experimental demonstration of amplitude modulation.

The Significance of Phase Information: Signal Transmission Example As we saw in the previous lecture, an ideal transmission system has uniform attenuation 푘 for all frequencies. Also, the transmitted signal is received delayed by 푡푑, but with no distortion; i.e., 푦(푡) = 푘푓(푡 − 푡푑).

The above system has the transfer function 퐻(휔) = 푘푒−푗푡푑휔 from which the magnitude and angle are obtained as: |퐻(휔)| = 푘, ∠퐻(휔) = −푡푑휔. Example. Determine the steady-state output for the above ideal transmission system to the input, 푓(푡) = cos(푡) + cos(2푡). Here, the signal has two component frequencies (휔1 = 1, 휔2 = 2). 푦푠푠(푡) = |퐻(휔1)| cos(푡 + ∠퐻(휔1)) + |퐻(휔2)| cos(2푡 + ∠퐻(휔2))

= 푘cos(푡 − 푡푑) + 푘cos(2푡 − 2푡푑)

= 푘cos(푡 − 푡푑) + 푘cos(2(푡 − 푡푑))

= 푘푓(푡 − 푡푑) ←No distortion

Let us repeat this problem assuming a constant angle, ∠퐻(휔) = −푡푑.

푦푠푠(푡) = 푘cos(푡 − 푡푑) + 푘cos(2푡 − 푡푑) 푡 = 푘cos(푡 − 푡 ) + 푘cos (2 (푡 − 푑)) ≠ 푘푓(푡 − 푡 ) 푑 2 푑

Notice that the second component is delayed by half the delay of the first component. So, different parts of the signal arrive at different times! This leads to a distorted signal.

The following Mathcad simulations illustrate this phase-based distortion phenomenon.

Your turn: Consider the following passive linear circuit with 퐿 = 1 and 퐶 = 1.

 Determine (by inspection) the steady-state output for the following inputs: o 푣푖푛(푡) = 1 o 푣푖푛(푡) = cos(휔푡), as 휔 → ∞  Determine the transfer function 퐻(휔).  Does the circuit distort the amplitude of its input?  Does the circuit distort the phase of its input? 3  Let 푣푖푛(푡) = sin (푡). Determine the steady-state output, 푣푠푠(푡).  Repeat for 푣푖푛(푡) = sin(10휋푡).

Ans. 1 − 푠2 1 + 휔2 퐻(푠) = → 퐻(휔) = 1 + 푠2 1 − 휔2

푣푠푠(푡) = ∞ + 0.244 sin(3푡) = ∞

Note: The circuit resonates at 휔 = 1. This circuit is supposed to operate at 휔 ≫ 1. For those high frequencies, the frequency response of the filter is given by, 퐻(휔) = −1 = 1푒푗휋

Therefore, for 휔 ≫ 1 the filter is an all-pass filter with constant phase shift of 휋. Refer to the section entitled “All-Pass Filter” later in this lecture for additional analysis and application.

Fourier Transform in a Basic Communication System Consider the following communication system.

There are two main factors affecting the design of a communication system. First, the transmission antenna must be of practical size. The size of the transmission antenna (퐿) is proportional to the wave-length, 푐 2휋푐 m 퐿 ∝ 휆 = = , 푐 ≅ 3 x 108 푓 휔 s For a (talk radio channel) speech signal, 푚(푡), the frequency components are in the range 100Hz < 푓 < 20KHz. And the effective signal bandwidth for intelligible speech is about 5KHz. So, if we are to transmit speech signals directly then the length of required transmission antenna would be of the order of several kilometers: (3)108 퐿 ∝ 휆 > ≅ 15Km (20)103 Therefore, we need to shift the signal frequency to a higher range; i.e., we need to modulate 푚(푡) so that it rides on a high-frequency signal 푦(푡) in order to make the size of the antenna practical. Other requirements include the ability of the receiver to select one specific transmission channel from the superposition of several concurrent transmissions (by different stations), as illustrated in the following figure.

The Fourier transform is a convenient tool for designing communication systems. The frequency spectrum representation of a signal renders it easier to understand the processing and design aspects of the system as compared to the time-domain picture. We start by Fourier transforming the signals, 푚1(푡) and 푚2(푡).

Next, we need to shift the spectra of 푀1(휔) and 푀2(휔) to high frequencies, but with minimal spectral overlap. This is achieved by applying frequency division multiplexing (FDM) as illustrated in the following figure.

Note: For simplicity, we assumed that 푚1(푡) and 푚2(푡) are even and real test signals, which leads to even and real Fourier transforms 푀1(휔) and 푀2(휔), respectively. The carrier frequencies for AM radio range from 540KHz to 1700KHz and are spaced at 10KHz intervals. Recall that the radian frequency 휔 is 2휋 times the Hertz frequency, 푓. From the above figure, we notice that in order to minimize spectral overlap, we should require that the sum of the signals’ (rad/sec) bandwidth to be less than the carrier frequency separation:

휔퐵2 + 휔퐵1 < 휔푐2 − 휔푐1 or 퐵2 + 퐵1 < 푓푐2 − 푓푐1 For AM radio, the intelligible signal bandwidth is about 5KHz, which would require neighboring carrier frequencies to be separated by 10KHz. The above requirement is important so that the receiving station can employ a proper narrow band-pass filter to select the spectra of the signal of interest. Modulation at the transmitter stage: Consider the Fourier transform pair 푚(푡) ↔ 푀(휔). The frequency shift of the signal can be obtained by multiplying the time-domain signal by 푒푗휔푐푡 (recall the frequency-shift property of the Fourier transform) as shown below.

푗휔푐푡 Since 푒 = cos(휔푐푡) + 푗sin(휔푐푡) is complex (non-physical), let us see what happens if the modulator used cos(휔푐푡) as the modulating signal.

So, the transmitted signal has a duplicated spectrum centered at ±휔푐 1 1 푦(푡) ↔ 푀(휔 − 휔 ) + 푀(휔 + 휔 ) 2 푐 2 푐 with uniform attenuation of 50% for each component (refer to the spectral plot shown below).

The following figure depicts the modulation operation in the time domain:

Now, having received the modulated transmitted signal 푦(푡) at the receiver, how can one extract 푚(푡)?

Let us assume that the receiver can generate a cos(휔푐푡) that is in phase with the transmitter’s modulating sinusoid. Multiplying 푦(푡) by this sinusoid generates:

1 Using the trigonometric identity: cos2(푥) = (1 + cos 2푥), we may write 2 the demodulated signal 푒(푡) as 1 1 푒(푡) = 푚(푡) + 푚(푡) cos(2휔 푡) 2 2 푐 and write its corresponding Fourier transform as 1 1 퐸(휔) = 푀(휔) + [푀(휔 + 2휔 ) + 푀(휔 − 2휔 )] 2 4 푐 푐 Here is the associated spectrum,

The spectrum suggests that 푀(휔) can be extracted using a low-pass filter whose cutoff frequency is equal to the bandwidth of 푚(푡).

The above analysis suggests the following receiver circuit [휔푐 represents the carrier frequency in rad/sec],

Your turn: Assume a signal 푚(푡) is modulated by cos (휔푐푡) and then demodulated by cos (휔푐푡 + 휃). Determine the expression for the signal 푒(푡). Discuss the effect of 휃. The following is a Mathcad validation of the above communication system for the even signal 푚(푡) = 푒−2|푡|. It assumes that the modulation and demodulation signal is cos (휔푐푡), where 휔푐 = 40 rad/sec.

A low-pass filter cut-off frequency is the frequency 휔0 at which the filter 2 attenuation is |퐻(휔 )| = √ |퐻(0)|. The cut-off frequency leads to 50% 0 2 2 2 1 power attenuation [(√ ) = ], with a dB (decibels) value of 2 2

2 2 1 10log ((√ ) ) = 10log ( ) ≅ −3dB 10 2 10 2

The negative sign means attenuation. When a non-ideal low-pass filter is used, the output signal will be distorted as can be seen from the above simulation with the first-order low-pass filter.

Simulations Repeated with a 2nd-order LP filter

Note that the shift 푡푑 is due to the delay introduced by the filter. This delay is about 0.17 sec and can be verified by measuring the slope of the 2nd- order phase response in the vicinity of the origin. Note how the 2nd order filter leads to less distortion but slightly more delay. We may repeat the above simulation of the communication system using Matlab’s Simulink (employing a 2nd-order filter with cut-off frequency

휔0 = 10 and dc gain of 2):

Refer to the appendix for an experimental demonstration of amplitude modulation Low-Pass Filters: Ideal vs Practical (Butterworth Filters) Ideal “Brick-Wall” Low-Pass Filter (no signal distortion)

The transfer function for this filter can be easily obtained from the magnitude 휔 response, |퐻(휔)| = rect ( ), and phase response, angle(퐻(휔)) = 푒−푗휔푡푑, as 2휔0 휔 퐻(휔) = rect ( ) 푒−푗휔푡푑 2휔표 The unit-impulse response ℎ(푡) for this filter can be obtained using the inverse Fourier transform of 퐻(휔) (employ Fourier Pair #18 and the time-shift property)

휔 휔표 −1 −푗휔푡푑 ℎ(푡) = 퐹 {rect ( ) 푒 } = sinc[휔표(푡 − 푡푑)] 2휔표 휋 So, from the following plot, we observe that ℎ(푡) is non-causal; the response of the filter for an impulse 훿(푡) applied at 푡 = 0 exists for 푡 < 0. The above ideal low- pass filter characteristics cannot be realize physically because ℎ(푡) ≠ 0 for 푡 < 0.

In the frequency domain, the Paley-Wiener Criterion can be used to test to see if a mathematical transfer function 퐻(휔) can be realized physically. The criterion is necessary and sufficient and is mathematically stated in terms of the amplitude response as ∞ |ln|퐻(휔)|| ∫ 2 푑휔 < ∞ −∞ 1 + 휔 If |퐻(휔)| does not satisfy the above criterion then it is physically unrealizable. Note that if |퐻(휔)| is zero over any finite band then |ln|퐻(휔)|| = ∞ over that finite band and the integral is infinite; meaning that the filter is unrealizable. However, it is possible for the above integral to be finite if |퐻(휔)| is zero at a single frequency (or over a set of discrete frequencies).

2 Your turn: Show that the transfer function 퐻(휔) = 푒−푎휔 is physically unrealizable (assume 푎 > 0). Employ two methods: (1) By showing that ℎ(푡) is non-causal, and (2) by showing that |퐻(휔)| violates the Paley-Wiener Criterion. For Part (1) you may use Mathcad to solve the integral. The following low-pass filter characteristics are typical of realizable filters. Note the distortion introduced by the amplitude and phase responses.

Higher-Order Realizable (Practical) Low-Pass Filters We desire a low-pass filter that is maximally-flat |퐻(휔)| over

[−휔0, +휔0]. (Refer to the textbook, Section 7.5.) A class of maximally flat filters can be obtained by utilizing the Butterworth function where 푛 is the filter order, 1 퐴(휔, 푛) = |퐻(휔, 푛)| = 휔 2푛 √1 + ( ) 휔0 The following is a plot showing the magnitude response of the

Butterworth filter for a normalized cut-off frequency, 휔표 = 1.

The transfer function for a LP Butterworth filter (has no zeros) is given by 1 퐻(푠) = (푠 − 푝1)(푠 − 푝2) … (푠 − 푝푛)

푗휋(2푘+푛−1) 푗휃푘 with poles 푝푘 = 푒 = 푒 2푛 , 푘 = 1, 2 , … 푛, where all poles 푝푘 belong to the LHP and are symmetrically located on the unit circle. For example, for 푛 = 1, the pole is at −1. For 푛 = 2, the poles are complex 3휋 5휋 푗 푗 conjugate and are located at 푝1 = 푒 4 and 푝2 = 푒 4 .

The transfer function can then be written as (your turn: show it) 1 1 퐻(푠) ≅ = 푗3휋 푗5휋 2 (푠 − 푒 4 ) (푠 − 푒 4 ) 푠 + √2푠 + 1

For 푛 = 4, the poles are at

leading to the transfer function (your turn: show it) 1 퐻(푠) = 푠4 + 2.61푠3 + 3.41푠2 + 2.61푠 + 1

The following are the magnitude and angle spectra associated with the 2nd and 4th order Butterworth filters, respectively, with cutoff frequency of 푗휔 휔 = 10 [obtained by setting 푠 = in the above normalized 퐻(푠)], 0 10

Matlab’s Computation of Butterworth Filter Poles

Your turn: Repeat (employ Mathcad and then repeat employing Simulink) the earlier analyzed communication system simulation rd employing a 3 order Butterworth LP filter with 휔0 = 10. Repeat with 휔0 = 15. The answer is shown below. Ans. [Employing Mathcad. Left (right) plot has 휔0 = 10 (휔0 = 15)]

The Significance of Locating Filter Poles on the Unit-Circle Locating the filter poles on the unit-circle in the LHP of the complex frequency plane leads to a flat frequency response in the region

[−휔0 + 휔0]. The following example illustrates the effect of moving the poles off the unit circle. Let us consider a second-order low-pass filter (훼 < 0 for stability) 1 1 퐻(푠) = ∗ = (푠 − 푝)(푠 − 푝 ) (푠 − (훼 + 푗휔1))(푠 − (훼 − 푗휔1)) The magnitude response for this filter is 1 1 |퐻(휔)| = = |푗휔 − (훼 + 푗휔1)||푗휔 − (훼 − 푗휔1)| 퐿1퐿2

2 2 Let the poles of the system for the first filter be 푝, 푝∗ = − √ ± 푗 √ and for 2 2 2 2 the second filter be 푝, 푝∗ = −0.6 √ ± 푗 √ . 2 2

The (normalized) magnitude response for both filters is compared in the following Mathcad generated plot. Note: 퐻2(0) ≅ 1.47 and its magnitude response is normalized by this same constant in the plot for comparison purposes.

By inspecting the above plots, we observe that the Butterworth filter (labelled H1) with the poles on the unit-circle has a relatively flat, non- increasing magnitude response in the passband region. On the other hand, the filter (labelled H2) with poles inside the unit-circle exhibits a peak inside the passband region.

Your turn: Show that the extreme points of |퐻(휔)| occur at 휔 = 0 and 휔 = 2 2 ±√휔1 − 훼 . Also, show that when the poles are on the unit-circle (i.e., when 2 2 √휔1 + 훼 = 1) the only extreme point occurs at 휔 = 0 and has amplitude 1. Finally, show that when 훼 ≪ 휔1the peaks are at ±휔1with value |퐻(±휔1)| ≅ 1 1 ≅ . 2 2 2훼휔1 훼√훼 +4휔1 Your turn: The following are the transfer functions of the normalized 4th- order low-pass filters that are normally discussed and compared in textbooks on analog filter design. Evaluate the filters based on their magnitude plots.

1 Butterworth: 퐻1(푠) = (푠2+0.76537푠+1)(푠2+1.84776푠+1)

105 Bessel: 퐻2(푠) = (푠2+5.79242푠+9.14013)(푠2+4.20758푠+11.4878)

0.35785 Chebyshev: 퐻3(푠) = (푠2+0.35071푠+1.06352)(푠2+0.84668푠+0.356412)

0.015397(푠2+2.53555)(푠2+12.09931) Elliptic: 퐻4(푠) = (푠2+0.25496푠+1.06044)(푠2+0.92001푠+0.47183)

[Reference: Principles of Active Network Synthesis and Design, Gobind Daryanani (Wiley, 1976)]

Your turn: Employ Mathcad to compute and compare (by plotting) the phase response (0 < 휔 < .9) and the step response (0 < 푡 < 15) of each of the above four filters. Your answers should look like the ones in the figures below. Evaluate the filters based on the shape of their corresponding responses.

Filter Design Example. A direct method for designing, say a second- order, low-pass filter would be to start with the simple generic form of a 2nd-order low-pass filter, written in terms of its coefficients, namely 1 퐻(푠) = 2 푠 + 푎1푠 + 푎0 and then solve for the coefficient such that the filter meets the required specifications. The frequency response for such filter is given by 1 퐻(휔) = 2 −휔 + 푗푎1휔 + 푎0 Assume that the filter specifications are as follows:  Unit gain at dc 2  Gain of √ at 휔 = 1 (cutoff frequency) 2 The first specifications is met by setting 푎0 = 1. For the second specification we need to solve for 푎1 according to the equation

√2 1 1 1 |퐻(1)| = = 2 | = = 2 |(1−휔 )+푗푎1휔| 휔=1 |푗푎1| 푎1

Solving the algebraic equation we obtain 푎1 = √2, leading to the second- order Butterworth filter 1 퐻(푠) = 푠2 + √2푠 + 1 Let us repeat the design by requiring that the gain is 0.9 for 휔 = 1. This leads to 1 1 1 |퐻(1)| = 0.9 = 2 | = = |(1 − 휔 ) + 푗푎1휔| 휔=1 |푗푎1| 푎1 or, 푎1 = 1.11. The following plot compares the magnitude response for both filters. Notice that the second design is not maximally flat in the passband region.

Your turn: Design a 3rd-order low-pass filter with the 퐻(푠) shown below. Plot the magnitude response and compare it to that of a 3rd-order low-pass Butterworth filter. Also, plot the poles of 퐻(푠) in the complex frequency plain and compare to those of the 3rd-order Butterworth filter. 1 퐻(푠) = 3 2 푎3푠 + 푎2푠 + 푎1푠 + 푎0 with the following specifications:

Note: In the above figure, H2 is the 3rd-order Butterworth filter, 1 퐻(푠) = 푠3 + 2푠2 + 2푠 + 1

Example. Design an analog low-pass Butterworth filter to meet the specifications below. |퐻(휔)| = 0.9 for 휔 = 1000 and |퐻(휔)| = 0.1 for 휔 = 3000. Plot the spectra of the filter’s frequency response,

퐻(ω). You need to find the filter’s order 푛 and cutoff frequency 휔표. Solution: Employing the magnitude response for a low-pass Butterworth filter and applying the two requirements we obtain a set of two equations 1 |퐻(1000)| = = 0.9 1000 2푛 √1 + ( ) 휔0 1 |퐻(3000)| = = 0.1 3000 2푛 √1 + ( ) 휔0 Which can be simplified as

1000 2푛 1 ( ) = ( 2) − 1 = 0.235 휔0 0.9 3000 2푛 1 ( ) = ( 2) − 1 = 99 휔0 0.1

Taking the natural logarithm of both sides leads to

2푛[6.908 − ln(휔0)] = −1.448

2푛[8.006 − ln(휔0)] = 4.495

We can now solve the first equation for 2푛 and substitute it in the second equation to get 1.448 [− ] [8.006 − ln(휔0)] = 4.495 6.908 − ln(휔0) or, after some manipulations we get

ln(휔0) = 7.17 or 휔0 ≅ 1,300 Substituting in the first equation and solving gives, 2푛[6.908 − 7.17] = −1.448 or 푛 ≅ 2.7. So, we round to 푛 = 3.

Mathcad verification (numerical solution):

The filter’s transfer function is then (the coefficients can be obtained from the formula for the poles or from the Butterworth Table of filter coefficients presented later in this lecture), 1 퐻(푠) = 푠 3 푠 2 푠 ( ) + 2 ( ) + 2 ( ) + 1 1300 1300 1300 The frequency response spectra is plotted below.

Your turn: Design an analog low-pass Butterworth filter to meet the specifications below. Plot the spectra of the filter’s frequency response, 퐻(ω). |퐻(휔)| = 0.9 for 휔 = 0.2휋 and |퐻(휔)| = 0.2 for 휔 = 0.3휋

Simulink Simulation of an 8th Order Butterworth Low-Pass Filter

The filter’s input is 푓(푡) = sin(1100푡) + sin (2000푡).

Filter passes the lower frequency only; cut-off at 휔0 = 1,300 rad/sec:

(The filter input is blue, sin(1100푡) is yellow and the filter output is red)

Filter passes both frequencies; cut-off at 휔0 = 2,200 rad/sec:

(Filter input: blue; filter output: red. Note the distortion. Try larger 휔0)

Filter Design: Low-Pass, High-Pass, Band-Pass, Band-Reject, All-Pass In this section we consider the design of (analog) electric filters based on the Butterworth characteristics. We can summarize the earlier findings of the 푡ℎ normalized (cutoff frequency 휔0 = 1) 푛 -order low-pass Butterworth filter transfer function as, 1 1 퐻(푠) = = 푛 푛−1 (푠 − 푝1)(푠 − 푝2) … (푠 − 푝푛) 푎푛푠 + 푎푛−1푠 + ⋯ + 푎1푠 + 푎0

푗휋(2푘+푛−1) 푗휃푘 with poles 푝푘 = 푒 = 푒 2푛 , 푘 = 1, 2 , … 푛. For convenience, we can tabulate the values of the poles and denominator coefficients of 퐻(푠) as shown in the following table (note that 푎0 and 푎푛 are always 1).

휃푘 n 푎0 푎1 푎2 푎3 푎4 푎5 푎6 푎7 푎8 휋 1 1 1 3휋 5휋 , 2 1 √2 1 4 4 2휋 4휋 , 휋, 3 1 2 2 1 3 3 5휋 7휋 9휋 11휋 , , , 4 1 2.61 3.41 2.61 1 8 8 8 8 3휋 4휋 6휋 7휋 , , 휋, , 5 1 3.24 5.24 5.24 3.24 1 5 5 5 5 7휋 9휋 11휋 13휋 15휋 17휋 , , , , , 6 1 3.87 7.46 9.14 7.46 3.87 1 12 12 12 12 12 12 8 1 5.13 13.14 21.85 25.69 21.85 13.14 5.13 1

The denormalized transfer function for a low-pass Butterworth filter with order 푛, dc gain 퐾 and cutoff-frequency 휔표 can be obtained according to the formula 퐾 퐻퐿푃(푠) = 푠 푛 푠 푛−1 푠 푎푛 ( ) + 푎푛−1 ( ) + ⋯ + 푎1 ( ) + 1 휔표 휔표 휔표 Example. Obtain a second-order Butterworth low-pass filter with cutoff frequency 1000 rad/sec and dc gain of 10.

From the above table we have the denominator coefficients √2 and 1. Hence, the normalized transfer function is 1 퐻(푠) = 푠2 + √2푠 + 1 The denormalized filter transfer function (with 퐾 = 10) is then

10 107 퐻퐿푃(푠) = = 푠 2 푠 2 3 6 1 ( ) + √2 ( ) + 1 푠 + 10 √2푠 + 10 103 103 Your turn: Obtain the transfer function for a third-order Butterworth low-pass filter with cutoff frequency of 1000 rad/sec and dc gain of 10. Plot the magnitude and angle of 퐻(휔).

1010 Ans. 퐻 (푠) = 퐿푃 푠3+(2)103푠2+(2)106푠+109

Active Circuit Realization of Second-Order Low-Pass Filters The following second-order op-amp circuit (known as the Sallen-Key circuit)

has the transfer function (your turn: derive it)

1 ( ) 푅1푅2퐶1퐶2 퐻 푠 = 1 1 1 푠2 + ( + ) 푠 + 푅1퐶1 푅2퐶1 푅1푅2퐶1퐶2 Example. Design the above active circuit such that it realizes the low-pass Butterworth filter,

107 퐻퐿푃(푠) = 푠2 + 103√2푠 + 106 We may express the above transfer function as

106 퐻퐿푃(푠) = (10) 푠2 + 103√2푠 + 106 which suggests that a Sallen-Key circuit can be used followed by an op-amp amplifier stage of gain 10. So, by matching coefficients in the expression

1 6 10 푅1푅2퐶1퐶2 = 1 1 1 푠2 + 103√2푠 + 106 푠2 + ( + ) 푠 + 푅1퐶1 푅2퐶1 푅1푅2퐶1퐶2 1 1 1 we arrive at the requirements = 106 and + = 103√2. We have 푅1푅2퐶1퐶2 푅1퐶1 푅2퐶1 ample flexibility to set the four element values since we have only two constraints

(equations) to satisfy. So, we choose 푅1 = 푅2 = 10KThis choice leads to 퐶1 ≈ 0.14F and 퐶2 = 70.7nF. In practice, the capacitors are picked from what is available in production and would be chosen as close as possible to the theoretical values obtained. Alternatively, we can choose practical capacitor values first and then solve for the resistor values. Potentiometers can then be used to (precisely) set the resistor values in the circuit. Your turn: Set 퐶1 = 0.1F and 퐶2 = 50nF and solve for 푅1 and 푅2.

Higher order filters (with even order) can be implemented by cascading two or more Sallen-Key stages, with each stage designed to match one conjugate pairs of poles. In other words if a 4th-order low-pass filter with poles 푒±푗휃1 and 푒±푗휃2 is to be realized, then the first Sallen-Key circuit stage is designed to match poles 푒±푗휃1 and the second stage matches 푒±푗휃2 . A final stage consisting of an op-amp voltage amplifier circuit is used to set the dc gain, 퐾. The structure is shown in the figure below.

High-Pass Butterworth Filter Design

A Butterworth high-pass filter with cutoff frequency 휔표 and pass-band frequency gain 퐾 can be obtained from a Butterworth low-pass filter with the same cutoff frequency by performing a three-step process: Step 1: For a specified value of 푛, obtain the low-pass filter normalized transfer function 퐻(푠), 1 퐻(푠) = 푛 푛−1 푎푛푠 + 푎푛−1푠 + ⋯ + 푎1푠 + 1 1 Step 2: Generate a high-pass filter normalized transfer function by replacing 푠 by 푠 to obtain 1 퐻′(푠) = 1 푛 1 푛−1 1 푎 ( ) + 푎 ( ) + ⋯ + 푎 ( ) + 1 푛 푠 푛−1 푠 1 푠 푠푛 = 푛−1 푛 푎푛 + 푎푛−1푠 + ⋯ + 푎1푠 + 푠 Step 3. Obtain the denormalized high-pass filter from the normalized transfer 푠 function (in Step 2) by replacing 푠 by and introducing a gain factor 퐾, if needed. 휔표

푠 푛 퐾 ( ) 휔표 퐻퐻푃(푠) = 푠 푠 푛−1 푠 푛 푎푛 + 푎푛−1 ( ) + ⋯ + 푎1 ( ) + ( ) 휔표 휔표 휔표

Example. Obtain the transfer function of a 3rd-order high-pass Butterworth filter with cutoff frequency 103 rad/sec and a high-frequency gain of 10. From the table of denominator coefficients, the 3rd-order normalized low-pass filter transfer function is 1 퐻(푠) = 푠3 + 2푠2 + 2푠 + 1 Transforming from low-pass to high-pass by replacing 푠 by 1/푠 we obtain,

1 푠3 퐻′(푠) = = 1 3 1 2 1 1 + 2푠 + 2푠2 + 푠3 ( ) + 2 ( ) + 2 ( ) + 1 푠 푠 푠 푠 Now, we denormalize by replacing 푠 by to obtain, 휔표

푠 3 퐾 ( ) 103 퐻퐻푃(푠) = 푠 푠 2 푠 3 1 + 2 ( ) + 2 ( ) + ( ) 103 103 103 which can also be expressed as

퐾푠3 퐻 (푠) = 퐻푃 푠3 + (2)103푠2 + (2)106푠 + 109 Finally, since the high-frequency gain is specified to be 10, we may obtain the value of 퐾 as the solution to

lim 퐻퐻푃(푠) = 10 푠→∞ to obtain 퐾 = 10.

Your turn: Plot the magnitude and angle response for the above high-pass filter.

Your turn: Show that the following op-amp circuit implements a second-order high-pass filter. Determine the filter’s cutoff frequency 휔표.

Matlab-Based Analog Filter Design Toolbox: afd The Analog Filter Design (AFD) Toolbox bundles a suite of tools for analog filter design and analysis. Given a filter type (high-pass or low-pass) it can calculate pole and zero placements, display time and frequency-domain system responses, calculate needed resistor and capacitor values for various active-circuit implementations and complete several other useful functions. This toolbox is not part of Matlab and need to be installed as an Add-On. A compressed folder containing the software is available from the course web page. Download by clicking the “Analog Filter Design for Matlab” link. Then extract the files from the downloaded zipped folder. Double-click the file to install. This will create the Add-On in Matlab, as shown in the following figure.

The executable script afd.m need to be dragged from the unzipped folder and pasted into the metadata directory as you can see in the above figure. The toolbox is executed by typing afd at the Matlab prompt. A tutorial is available on the course webpage. The following example illustrates the design and active-circuit generation of a (two-stage) fourth-order, low-pass Butterworth filter with 10KHz cutoff frequency and dc gain of 1. First, enter the values as shown below.

From the “Analyze” menu we choose “List Transfer Function” and then “Plot Poles and Zeros” to obtain the following:

The frequency response is obtained by choosing “Plot Frequency Response” from the “Analyze” menu to obtain,

Next, we may design an active Sallen-Key circuit that implements the filter by selecting “Build Circuit” from the “Build” menu. The two required stages are shown below along with component values. Here, we choose 5% components.

The design is verified using Multisim as shown in the figure below.

Note to instructor. Demonstrate in class.

Input frequency = 1KHz

Input frequency = 15KHz

Band-Pass and Band-Reject Filter Design

A band-pass filter 퐻퐵푃(푠) can be achieved by cascading a low-pass filter and a high-pass filter, 퐻퐵푃(푠) = 퐻퐿푃(푠)퐻퐻푃(푠), as illustrated in the figure below. It is important to note that the band-pass lower cutoff frequency is the high-pass filter’s cutoff frequency 휔퐻푃. Also, the band-pass upper cutoff frequency is that of the low-pass filter’s cutoff frequency 휔퐿푃 (with 휔퐿푃 > 휔퐻푃). Similarly, a band-reject (band-stop) filter can be implemented by adding the outputs of a low-pass and a high-pass filters (here, 퐻퐵푆(푠) = 퐻퐿푃(푠) + 퐻퐻푃(푠) with 휔퐿푃 < 휔퐻푃) as depicted in the following figure (on the right hand side).

Your turn: Employ Simulink to generate a signal consisting of the sum of three sinusoids oscillating at frequencies 800 rad/sec, 1000 rad/sec and 1200 rad/sec, respectively. Design a bandpass analog filter that only passes the 1000 rad/sec sinusoid. Employ the built-in analog filter block (search the Simulink Library Browser for “analog filter”) to customize the filter. Solve the problem in two different ways: (1) Employing a bandpass filter, and (2) Employing a cascade of a low-pass and a high-pass filters. Which implementation is more efficient in terms of filter order?

Note to instructor: Demonstrate Simulink solution in class.

Sallen-Key Band-Pass and Band-Reject 2nd-Order Active Filter Circuits (Note: The two resistor arrangement connecting the output to ground is used to set the passband gain)

Your turn. Consider the circuit in the following figure. Assume 푅 = 2Ω, 퐿 = 10mH and 퐶 = 1휇F.

푉 a. Determine 퐻(휔) = 표. 푉푠 b. Plot |퐻(휔)| versus 휔 for 휔 ∈ [9,800 10,200]. Identify the filter type.

c. Solve for 휔0 such that |퐻(휔0)| is maximized. d. Solve for the cutoff frequencies 휔푐1 and 휔푐2. 휔 e. Determine the quality factor = 0 . 휔푐2−휔푐1 (You may employ Mathcad to solve for Parts b, c and d.)

Ans. (Part e) 푄 = 135 Notch Filter A notch filter is a band-reject filter with a very narrow (selective) band stop. Applications of notch filters include filtering out the 60Hz interference picked up by electronic circuits from power lines inside buildings. Here, in addition to exhibiting nulls at 휔푛 = (2휋)60 ≅ 377 rad/sec, we would like the magnitude of 퐻(휔) to be flat at all other frequencies.

One way to design such a filter is to introduce a zero 푠푧1 at 푗휔푛 (and at −푗휔푛 , because of the complex conjugate roots required by physical systems). But in order to cancel the nulling effects of the zeros away from the null frequency we must introduce poles 푠푝1, 푠푝2 = 훼 ± 푗휔푛 (in the LHP; i.e., 훼 < 0) that are located very close to the zeros of 퐻(휔). Here, 훼 should be set to a small fraction of 휔푛 (say, 훼 = −0.1휔푛). Therefore, a notch filter transfer function takes the form, (푠 − 푗휔 )(푠 + 푗휔 ) 퐻(푠) = 푛 푛 [푠 − (훼 + 푗휔푛)][푠 − (훼 − 푗휔푛)]

The following plot depicts the magnitude spectrum of a (60 Hz) notch filter with 휔푛 = 120휋 and 훼 = −0.1휔푛,

Your turn: 1. Show that the transfer function for the above notch filter is given by,

2훼푠 − 훼2 ( ) 퐻 푠 = 1 + 2 2 (푠 − 훼) + 휔푛 2. Show that the impulse response of this filter is given by,

훼푡 ℎ(푡) = 훿(푡) − 퐴(훼, 휔푛)푒 cos(휔푛푡 + 휃(휔푛)) 푢(푡)

4 2 훼 −1 −훼 where, 퐴(훼, 휔푛) = √4훼 + 2 and 휃(휔푛) = tan ( ). 휔푛 2휔푛

3. Plot ℎ(푡) vs 푡 (excluding the impulse) and |퐻(휔)| vs 휔 for 휔푛 = 120휋 and 훼 = −30. Repeat for 훼 = −10. What are your conclusions about the effect of 훼 on the time it takes for the response of the filter to reach its steady-state? What about the effects of 훼 on the selectivity of the filter?

Your turn: Consider the following notch filter (휔푛 = 5), with 훼 = −1.2 or −0.2. 2훼푠 − 훼2 퐻(푠) = 1 + (푠 − 훼)2 + 25 a. Compare the plots of the magnitude and angle responses of these two filters. b. Employ Mathcad to solve for and plot the zero-state responses of the filter to the input 푓(푡) = cos(8푡) + cos(5푡) for 훼 = −1.2 and 훼 = −0.2. Your plots should look like the following plots,

Your turn: Consider the following RLC circuit. Assume that 퐶 = 100 휇F, 퐿 = 70.3 mH and 푅 = 10 

a. Find the steady-state response, 푣표푢푡(푡) for the input 푣푖푛(푡) = cos(120휋푡) + 0.2cos(2000휋푡). b. Plot 푣푖푛(푡) and 푣표푢푡(푡) on the same set of axis (use 0 < 푡 < 40 ms and −1.5 < 푣 < 1.5). c. Plot the magnitude spectrum of 퐻(휔), for 0 < 휔 < 2000. What type of filter does the circuit implement? Determine the frequency that is completely blocked by the circuit (you may use the trace function of Mathcad)? Simulink Example of a Notch Filter http://www.mathworks.com/help/dsp/examples/removing-an-interfering-tone-from-a-streaming-audio-signal.html

A notch filter is used to eliminate a specific frequency from a given signal. In their most common form, the filter design parameters for notch filters are center frequency for the notch and the 3 dB bandwidth. The center frequency is the frequency point at which the filter has a gain of zero. The 3 dB bandwidth measures the frequency width of the notch of the filter computed at the half-power or -3 dB attenuation point. In this example, you tune a notch filter in order to eliminate a 250 Hz sinusoidal tone corrupting an audio signal. You can control both the center frequency and the bandwidth of the notch filter and listen to the filtered audio signal as you tune the design parameters. The command audioToneRemovalExampleApp launches a user interface designed to interact with the simulation. It also launches a spectrum analyzer to view the spectrum of the audio with and without filtering along with the magnitude response of the notch filter.

All-Pass Filters Circuits with transfer functions of the form, (푠 − 푧 )(푠 − 푧 ) … (푠 − 푧 ) 퐻(푠) = 1 2 푛 (푠 − 푝1)(푠 − 푝2) … (푠 − 푝푛) that have the same number of poles as zeros, with each pole (located in the LHP for stability) having a corresponding zero that is symmetrically (with respect to the 푗휔 axis) located in the RHP are referred to as all-pass filters. Here, 푝푖 = −푎푖 + 푗푏푖 (푎푖 > 0) and 푧푖 = 푎푖 + 푗푏푖. An all-pass filter has a constant attenuation at all frequencies but the relative phase between input and output varies with frequency.

An example of a first-order all-pass filter is, 푗휔 − 푎 퐻(휔) = 푗휔 + 푎 It can be shown that

2푎휔 휔2−푎2 |퐻(휔)| = 1 and ∠퐻(휔) = atan2 ( , ) 휔2+푎2 휔2+푎2

Your turn: Plot the phase response for the above first-order, all-pass filter (assume 푎 = 1). The transfer functions for a first-order and a second-order all-pass filters take the following general forms, respectively,

푠−푎 푠2−푎푠+푏 퐻(푠) = and 퐻(푠) = (푎, 푏 > 0) 푠+푎 푠2+푎푠+푏 The following circuit (presented in a “your turn” problem earlier in this lecture) is an example of a second-order passive all-pass filter (also known as a lattice phase equalizer)

The lattice filter has an important application for stereo audio feeds. Phase distortion on a monophonic audio line does not have a serious effect on the quality of the sound unless it is very large. The same is true of the absolute phase distortion on each leg (left and right channels) of a stereo pair of lines. However, the differential phase between legs has a very dramatic effect on the stereo perception. This is because the formation of the stereo image in the brain relies on the phase difference information from the two ears. A phase difference translates to a delay, which in turn can be interpreted as a direction the sound came from. Consequently, landlines used by broadcasters for stereo transmissions are equalized to very tight differential phase specifications. In other words, an all-pass filter 퐻1(푠) (with proper placement of its poles) can be used in cascade with the main filter 퐻(푠) so as to affect (shape) the overall phase response.

Your turn: Derive the transfer functions for the following all-pass filters.

Your turn: A three-amplifier biquad filter is shown in the circuit below. (a) 푉1(푠) 푉2(푠) 푉3(푠) Determine the transfer functions 퐻1(푠) = , 퐻2(푠) = and 퐻3(푠) = 푉푖푛(푠) 푉푖푛(푠) 푉푖푛(푠) in terms of the symbolic 푅’s and 퐶’s. (b) Plot (use a log scale for 0.1 < 휔 < 10) the magnitude response for the three transfer functions on the same graph (for the plot, assume that all resistor values are 1Ω and all capacitor values are 1F). Identify the filters types. (c) At what frequency does |퐻2(휔)| attain its maximum value?

Answer for 퐻2(푠): −퐶 푅 푅 (푅 + 푅 )푠 ( ) 2 5 6 2 3 퐻2 푠 = 푅 푅 +푅 푅 푠2 + 1 ( 2 3) 푠 + 3 푅2푅4퐶1 푅1+푅6 푅2푅4푅5퐶1퐶2 Your turn: Synthesize the following transfer function using the circuit shown below, and plot its magnitude and angle responses. 푉 (푠) 푠 + 4 퐻(푠) = 표 = − 푉푖푛(푠) 푠 + 6

Your turn: Show that the following circuit has the transfer function shown. Plot the magnitude and angle responses.

푉표(푠) −2푠 퐻(푠) = = 2 푉푖푛(푠) 푠 + 2푠 + 2

Your turn: The following circuit is a capacitance multiplier. Show that the input 퐼 푅 admittance is given by, 1 = (1 + 2) 퐶. 푉1 푅1

푉 (푠) Your turn: Determine the transfer function 퐻(푠) = 표 for the following circuit, 푉푖푛(푠) as a function of 퐿1, 퐿2, 퐿3, 퐶1, 퐶2 and 퐶3. Use nodal analysis (two equations) and then employ Mathcad to solve the equations symbolically. Then, determine integer 푠2(푠2+1) element values such that 퐻(푠) = . Plot the filter’s magnitude and angle 2푠4+5푠2+2 response. Discuss the frequency response.

푉 (푠) Your turn: Determine the transfer function 퐻(푠) = 표 for the following filter. 퐼푖푛(푠) Assume all element values are equal to 1. You may use Mathcad to solve the three nodal equations. Plot the filter’s magnitude and angle response. Discuss the frequency response behavior of the filter.

Your turn: Show that the following filter is a fifth-order Butterworth low-pass filter with dc gain of 0.5 and a -3dB cutoff frequency of about 62.8 MHz. Compare the magnitude spectrum of this filter to the one using 퐶1 = 10nF, 퐶3 = 32nF, 퐶5 = 10nF, 퐿2 = 26nF, 퐿4 = 26nF. Discuss the results.

Analog Filter Design as a Numerical Optimization Problem

Consider an analog filter with the transfer function,

푚 푚−1 푎푚푠 + 푎푚−1푠 + ⋯ + 푎1푠 + 푎0 퐻(s) = 푛 푛−1 푠 + 푏푛−1푠 + ⋯ + 푏1푠 + 푏0

Where 푚 ≤ 푛. We are interested in solving for the 푛 + 푚 + 2 parameters:

푎0, 푎1, 푎2, … , 푎푚, 푏0, 푏1, … 푏푛−1, 푐

Such that a set of specifications on the magnitude and phase response of the 퐻(휔) are met. Let us assume that the following set of 푁 specifications are given on the magnitude response of the filter at frequencies ω푖, 푖 = 1,2 , … , 푁:

|퐻(ω1)| = 퐴1, |퐻(ω2)| = 퐴2, … , |퐻(ω푖)| = 퐴푖, … |퐻(ω푁)| = 퐴푁

Also, based on earlier discussion about minimizing phase distortion, we would like to force the phase response of the filter to be linear over the pass-band region, i.e., ∠퐻(휔) = −푐휔. This adds the following 푀 constraints (where, the 휔푗’s are user specified and are supposed to sample the frequencies of the pass band),

arg(퐻(휔1)) = −푐휔1, arg(퐻(휔2)) = −푐휔2, … , arg (퐻(휔푗))

= −푐휔푗, … , arg(퐻(휔푀)) = −푐휔푀

Typically, the number of specifications (constraints) is greater than the number of parameters, 푁 + 푀 > 푛 + 푚 + 2 (over-determined system of equations) and a perfect solution does not exist. Note that we may approximate the filter parameters by minimizing the sum-of-squared error (SSE) criterion (recall LSE regression problems from your numerical methods course): 퐼(푎0, 푎1, 푎2, … , 푎푚, 푏0, 푏1, … 푏푛−1, 푐) 푁 푀 2 2 = ∑[|퐻(ω푖)| − 퐴푖] + 휆 ∑[arg(퐻(휔푖)) + 푐휔푖] 푖=1 푖=1

This is a nonlinear optimization problem that can have multiple solutions (local minima) with varying degrees of solution accuracy. The 휆 is a positive weighting factor that we can increase to give more emphasis on ∗ ∗ ∗ meeting the phase linearity constraint. A solution (푎0, 푎1, … , 푎푚, ∗ ∗ ∗ ∗ 푏0, 푏1 … 푏푛−1, 푐 ) is referred to as the least-squares-error (LSE solution). The LSE solution is a sort of a “compromise” solution that attempts to satisfy all constraints, as close as possible.

The SSE criterion 퐼 is a nonlinear function of its (regression) parameters. Therefore, special numerical optimization methods must be employed (such as Matlab’s fminsearch and Mathcad’s Minimize optimization functions). The problem can have multiple solutions (local minima). Care must be taken in choosing initial search parameters so that the underlying numerical algorithms converge to a proper design for the digital filter. By “proper design” we mean a design that locates the poles to the left of the 푗휔 axis (ensuring stability). The following example illustrates the LSE- based filter design.

Example. Design the third-order low-pass analog filter

2 푎2푠 + 푎1푠 + 푎0 퐻(s) = 3 2 푠 + 푏2푠 + 푏1푠 + 푏0 according to the following specifications: 1. Its magnitude is 1 at ω = 0, 1 at ω = 0.8, and 1 at ω = 0.9 2 2. Its magnitude at ω = 1 is √ (cut-off frequency) 2 3. Its magnitude is 0.3 at ω = 1.2 and is 0 at ω = 1.8 4. Its phase is 0 at ω = 0, −0.25푐 at ω = 0.25, −0.5푐 at ω = 0.5 and −0.9푐 at ω = 0.9, for some positive 푐. So, we have seven parameters to solve for and ten constraints. The SSE criterion is given by the function (with 휆 chosen as 0.2)

퐼(푎2, 푎1, 푎0, 푏2, 푏1, 푏0, 푐) = [(|퐻(0)| − 1)2 + (|퐻(0.8)| − 1)2 + (|퐻(0.9)| − 1)2 2 + (|퐻(1)| − √2/2) + (|퐻(1.2)| − 0.3)2+(|퐻(1.8)| − 0)2] 2 2 + (휆) [(arg(퐻(0)) + 0) + (arg(퐻(0.25)) + 0.25푐) 2 2 + (arg(퐻(0.5)) + 0.5푐) + (arg(퐻(0.9)) + 0.9푐) ]

Note that you have to experiment numerically with several initial search points and choose the physically feasible solution (e.g., poles in the left half plane) that gives the smallest value for 퐼. The following is the Mathcad worksheet that generates the solution (for the indicated initial search point),

Therefore, the LSE solution to the optimization problem is,

∗ ∗ ∗ ∗ ∗ ∗ (푎2, 푎1, 푎0, 푏2, 푏1, 푏0) = (0.261, 0.061, 0.7, 1.194, 1.303, 0.69) which achieves a value of 퐼 = 0.018 . This solution leads to the transfer function

0.261푠2 + 0.061푠 + 0.7 퐻(푠) = 푠3 + 1.194푠2 + 1.303푠 + 0.69

The poles are (located poles in the LHP) −0.7179, −0.2381 ± 푗0.9511 and the zeros are at −0.1169 ± 푗1.6335. The following plot depicts the location of the poles (blue) and zeros (red).

The following (red trace) is Mathcad’s solution for the magnitude response. The red circles represent the design magnitude data.

The blue response in the above plot is that for the 3rd-order Butterworth low-pass filter

1 퐻(푠) = 푠3 + 2푠2 + 2푠 + 1

The following plot compares the phase response for the two filters over the pass band. The red circles are the angle data.

The following is the LSE solution for the above filter design problem utilizing Matlab’s fminsearch optimization function. Case 1: Magnitude and angle specifications are imposed ( 휆 = 1)

Case 2: Only Magnitude specifications are imposed (휆 = 0)

Your turn: Determine the LSE solution for the third-order analog filter with the following specifications. Also, generate plots similar to the three plots shown above.  Its magnitude is 1 at ω = 0, 1 at ω = 0.9, and 0.85 at ω = 1  Its magnitude is 0.3 at ω = 1.5 and is 0.2 at ω = 2  Its magnitude is 0.1 at ω = 3 and is 0 at ω = 5  Its phase is 0 at ω = 0, −0.25푐 at ω = 0.25, −0.5푐 at ω = 0.5 and −0.9푐 at ω = 0.9. Set 휆 = 0.3 and experiment with different sets of initial conditions. Appendix Experimental Demonstration of Amplitude Modulation

In the following experiment a 1 푉푝푝 positive 1 kHz triangular wave is modulated by a 20 kHz sinusoid of 1 푉푝푝. The modulated signal is then multiplied by the same sinusoidal signal and the resulting signal passes th through a 4 -order low-pass Butterworth filter with cutoff frequency 푓표.

The following picture shows the experimental setup used. It shows the low- pass filter sitting on top of two analog multipliers. One multiplier is used to implement modulation. The other multiplier is used for demodulation.

The figure below depicts the spectrum of the demodulated signal.

The following oscilloscope capture depicts the demodulated signal (blue trace) after passing through the low-pass filter with 푓표 = 2 kHz. Here, the filter attenuates the higher harmonics of the triangular wave causing the output to look sinusoidal.

The following oscilloscope screen capture depicts the demodulated signal after passing through the low-pass filter, with 푓표 = 10 kHz. Here, the filter captures all of the significant harmonics of the triangular wave leading to proper reconstruction. The output signal in this case is 푦(푡) ≅ 0.5푓(푡).

Increasing the filter cutoff frequency to 20 kHz leads to distortion (aliasing) because the 4th-order filter passes an attenuated portion of the image spectrum, centered at 40 kHz. This is depicted in the following oscilloscope image.

Mini Project Design and build a 6th order Butterworth low-pass analog filter with cutoff frequency of 10 KHz. You need to perform the following:

 Build your filter using available components. Make sure you use low tolerance capacitors that come as close as possible to the design values. You might want to use small pots for the resistors.  Measure the actual values of all resistors and capacitors used. Determine the actual transfer function, 퐻(푠). Find its coefficients based on the actual component values used in your circuit.  Use Mathcad/Matlab to determine the frequency response [plot |퐻(휔)| 푣푠 휔] of your filter.  Determine and plot the poles of your filter. Compare them to the design poles.  Use Multisim to simulate your filter (using measured component values) and generate a frequency response plot.  Test your filter experimentally (using a sine wave generator and an oscilloscope) and determine its frequency response (generate plot from data). What is the gain at 1 KHz, 10 KHz, 11 KHz, and 15 KHz? Submit a complete electronic report (with introduction, discussion, simulations, oscilloscope outputs and conclusions) to document your work. Be prepared to demonstrate the hardware setup to your instructor. (Note: This filter can be used as part of a mini project presented at the end of Lecture 23). Hints: A cascade of three stages of Sallen-Key circuits can be used to implement your filter. You may use the AFD Toolbox to design the filter and to determine the circuit component values. Additional op-amp circuits may be used as buffers and amplifiers if needed. Make sure you use op-amps with bipolar supply voltages of ±12 Volt or higher. Make sure the signal processed by the filter does not get clipped (as a result of op-amp saturation).

Sample Student Work (by Alex Pluff, Winter 16)

Comparing the poles for the ideal and the actual (experimental) filters

Actual filter frequency response

In the following oscilloscope pictures, the blue trace represents the input signal and the yellow trace represent the output signal (1KHz, 10KHz, 11KHz and 15KHz).