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Abstract—This report explores the frequency, impulse, unit-
ramp, and unit-step responses of a passive and an active RLC
bandpass filter when given a signal. Design rationale for the active
filter is discussed.
I. INTRODUCTION
HIS report is to provide documentation of the steps taken
in the design and creation of both a passive and active
band-pass filter. This filter allows frequencies within a
specific range to pass, but rejects frequencies not within the
range.
Before designing the circuits, the capacitance and
inductance of the components were measured for use in both
the passive and active filter. This is displayed in table 1.
I. Component Rated versus Actual Values
Component Rated Value Actual Values
Capacitor (2) 0.1µF 0.08 µF
Inductor (2) 100 mH 104.7, 105.4 mH
Resistor 220 Ω 215.7 Ω
Resistor 1k Ω 0.993k Ω
Resistor 2.2k Ω 2.18k Ω
By adding a high pass filter with a low pass, a band pass
filter is created, as shown in Fig. 1.
Fig. 1 An explanation of the Band Pass filter, with cutoff
frequency at the -3dB point. [1]
To avoid using integral and differential equations, circuits are
converted from the time domain to the s domain; to do this,
the Laplace transform is used:
As such, when the Laplace transform is used, the circuit
elements also change as described in table 2.
II. Components in Domain Equivalents
Time Domain Phasor S Domain
This document was submitted on June 11, 2015 to Professor Mehmet Vurkac
by the authors; Benedict J. Fawver, Kyla J. Marino and Eric J. Tipler.
Benedict J. Fawver, author, is currently a student and at the Oregon Institute
of Technology, 2301 Campus Dr. Klamath Falls, OR 97601.
R R R
L jωL sL
C
II. OSCILLOSCOPE MEASUREMENTS
Setting up the function generator to simulate useful inputs
like an impulses or ramps is relatively straightforward. For
this project we used a Tektronix AFG 2021 arbitrary function
generator, but other generators in the lab appeared to have
very similar menu formats so the process should be the same.
For an impulse response, the FG should be set to a pulse on
continuous mode. For all options inputs, 1 V amplitude was
used. The frequency was then adjusted until the image was
clear. The square wave function was used to simulate the unit-
step, and the ramp function was used to simulate the ramp
input. In all cases, the inputs only serve as an approximation
of the ideal inputs since signals in wave form repeat. A
psuedo-frequency response was also found by running a sine
wave sweep from 10 Hz to 10k Hz. This gave a wave form
that had the largest amplitude at the center frequency. This
waveform served as a good approximation of the output
voltage vs the input voltage. The graph essentially shows the
magnitude portion of the bode plot.
III. MEASURING R, L, C
The RC circuit in Fig. 2 was modeled to find the calculated
and measured values of two 0.1 µF rated capacitors.
By treating the circuit as a voltage divider, the following
derivation found the frequency cutoff of the circuit:
1𝑉 (
−𝑗 𝜔𝑐 𝐶⁄
𝑅 − (𝑗 𝜔𝑐 𝐶)⁄
) =
1
√2
Kyla J. Marino, author, is currently a student and at the Oregon Institute of
Technology, 2301 Campus Dr. Klamath Falls, OR 97601.
Eric J. Tipler, author, is currently a student and at the Oregon Institute of
Technology, 2301 Campus Dr. Klamath Falls, OR 97601.
Design and analysis of active bandpass filter
Benedict J. Fawver, Kyla J. Marino, Eric J. Tipler
T
Fig. 2. RC Low Pass Filter.
2
‖
−𝑗 𝜔𝑐 𝐶⁄
𝑅 − (𝑗 𝜔𝑐 𝐶)⁄
‖ =
1
√2
1 𝜔𝑐 𝐶⁄
√𝑅2 + (
1
𝜔𝑐 𝐶
)2
=
1
√2
𝑅 =
1
𝜔 𝑐 𝐶
(1)
𝜔𝑐 =
1
𝑅𝐶
= 2𝜋𝑓
𝑓𝑐 =
1
2𝜋𝑅𝐶
(2)
1.602 𝑘𝐻𝑧 =
1
(2𝜋)(993 Ω)(0.1x10−6)
The calculated values were verified by simulating a
transient analysis on the circuit in LTspice. A comparison of
the calculated and measured values of the LTspice simulation
are found in table 1.
The circuit was built and the input frequency was adjusted
until the amplitude output was 0.707. This frequency was used
to find the true capacitance of the capacitor using (2).
𝐶 =
1
2𝜋𝑓𝑅
0.08 µ𝐹 =
1
2𝜋(1900)(993)
The same method described for the RC circuit was followed
to find the true values of two 100 mH inductors in LC circuit
in Fig 3. The derivation found the frequency at half power to
be (3).
𝑓𝑐 =
𝑅
2𝜋𝐿
(3)
1.580 𝑘𝐻𝑧 =
993
2𝜋(100𝑥10−3)
IV. RC AND LC VALUES
AC Analysis
Capacitor Inductor
LTspice 1.628 kHz 1.580 kHz
Calculated 1.603 kHz 1.584 kHz
Transient Analysis
Amplitude (Vo) at ½ power
Capacitor Inductor
0.707 0.662
The actual frequencies of the circuit were used to calculate
the true value of the capacitors and inductors using (1) and (2)
respectively. These values are found in table 4.
V. True Values of Capacitor and Inductors
Measured Frequency Capacitor Inductor
1.99 kHz 0.08 µF
1.90 kHz 0.08 µF
1.51 kHz 0.1047 H
1.50 kHz 0.0105 H
VI. RLC-CIRCUIT FREQUENCY RESPONSE
Two bandpass filters were built using the 220-kΩ resistor and
the 2.2-kΩ resistor shown in Fig. 4. By treating the circuit as a
voltage divider and using a theoretical maximum of 1, the
following derivation found the frequency cutoff of the circuit:
𝐻(𝑠) =
𝑉𝑜(𝑠)
𝑉𝑖𝑛(𝑠)
=
𝑅
𝑅 + 𝑠𝐿 +
1
𝑠𝐶
=
𝑅𝑠
𝑅𝑠 + 𝑠2 𝐿 +
1
𝐶
= (
𝑅
𝐿
) (
𝑠
𝑅
𝐿
𝑠 + 𝑠2 +
1
𝐿𝐶
)
𝐻(𝑗𝜔) = (
𝑗𝜔
𝑗𝜔 + (
1
𝐿𝐶
− 𝜔2)
) = 1
1
𝐿𝐶
− 𝜔2
= 0
𝜔 𝑜 =
1
√𝐿𝐶
2𝜋𝑓𝑜 =
1
√𝐿𝐶
𝑓𝑜 =
1
2𝜋√𝐿𝐶
(4)
Fig. 3. RL Low Pass Filter.
Fig. 4. RLC Passive Bandpass Filter Schematics using Measured Values
of 220-kΩ and 2.2-kΩ Rated Resistors
3
The predicted maximum response frequency was calculated
using (4):
𝑓𝑜 =
1
2𝜋√(0.1047)(0.08𝑥10−6)
= 1733.17 𝐻𝑧
The expected frequency responses were sketched then
simulated in SPICE. As seen in (4), the value of the resistor
will have no effect on the maximum frequency. However,
increasing the resistor will cause the magnitude response to
become increasingly round.
Fig.5. Predicted Passive Bandpass Filter Frequency Response Sketch
Fig. 6. SPICE Simulation of RLC Passive Bandpass Filter Frequency
Response using 220-Ω Rated Resistor
Fig. 7. SPICE Simulation of RLC Passive Bandpass Filter Frequency
Response using 2.2-kΩ Rated Resistor
Fig. 8. Vo/Vin Plot in SPICE and Actual Circuit
Fig. 9. Wolfram Alpha Plot of Theoretical Impulse Response for Passive
Bandpass Filter using 216-Ω Resistor
Fig. 10. Wolfram Alpha Plot of Theoretical Impulse Response for Passive
Bandpass Filter using 2180-Ω Resistor
4
Fig. 11. MATlab Plot of Theoretical Unit-Step Response for Passive
Bandpass Filter using 216-Ω Resistor
Fig. 12. MATlab Plot of Theoretical Unit-Ramp Response for Passive
Bandpass Filter using 216-Ω Resistor
Fig. 13. MATlab Plot of Theoretical Unit-Ramp Response for Passive
Bandpass Filter using 2180-Ω Resistor
Fig. 14. SPICE Plot of Frequency Response using FFT Function for Passive
Bandpass Filter using 216-Ω Resistor
Fig. 15. Actual Passive Filter with 2180-Ω Resistor
Fig. 16. Actual Passive Filter with 216-Ω Resistor
Fig. 17. Actual Passive Filter Ramp Response with 216-Ω Resistor
5
VII. DESIGN AND ANALYSIS OF ACTIVE BANDPASS FILTER
The bandpass filter with an amplification of 10 and band of
frequencies between 100 Hz and 10 kHz was created by
cascading a lowpass filter, highpass filter, and inverter. This
allows the upper corner frequency (10 kHz) to be set by the
lowpass filter and the lower corner frequency (100 Hz) to be
set by the highpass filter. The inverter provides the gain of the
filter.
The transfer function of the bandpass filter (14) can be
found by multiplying the highpass (10), lowpass (12), and
inverter’s (13) transfer functions together.
Fig. 18. Actual Passive Filter Ramp Response with 2180-Ω Resistor
Fig. 19. Actual Passive Filter Impulse Response with 216-Ω Resistor
Fig. 20. Actual Passive Filter Impulse Response with 2180-Ω Resistor
Fig. 21. Actual Passive Filter Unit-Step Response with 216-Ω Resistor
Fig. 22. Actual Passive Filter Unit-Step Response with 2180-Ω Resistor
Fig. 23. Actual Passive Filter Frequency Response with 216-Ω Resistor
Fig. 24. Actual Passive Filter Frequency Response with 2180-Ω Resistor
Fig. 25. Active Bandpass Filter
6
𝐻(𝑠) = −
1
1+𝑠𝐶1 𝑅
(10)
𝐻(𝑗𝜔) = −
1
1 + 𝑗𝜔𝐶1 𝑅
𝜔1 =
1
𝐶1 𝑅
(11)
𝐻(𝑠) = −
𝑠𝐶2 𝑅
1+𝑠𝐶1 𝑅
(12)
𝐻(𝑠) = −
𝑅 𝑓
𝑅 𝑖
(13)
𝐻(𝑠) = (−
1
1+𝑠𝐶1 𝑅
) (−
𝑠𝐶2 𝑅
1+𝑠𝐶1 𝑅
) (−
𝑅 𝑓
𝑅 𝑖
) (14)
ℒ−1
{
𝐻(𝑠)
𝑠
} = (−
1
𝑠+𝑠2 𝐶1 𝑅
) (−
𝐶2 𝑅
𝑠+𝑠 𝑠 𝐶1 𝑅
) (−
𝑅 𝑓
𝑠𝑅 𝑖
)
ℒ−1
{
𝐻(𝑠)
𝑠2 } = (−
1
𝑠2+𝑠3 𝐶1 𝑅
) (−
𝐶2 𝑅
𝑠2+𝑠3 𝐶1 𝑅
) (−
𝑅 𝑓
𝑠2 𝑅 𝑖
)
The values of the filter resistors were calculated using the
upper and lower corner frequencies as follows:
𝜔1 =
1
𝐶1 𝑅
10000 =
1
(0.08𝑥10−6)𝑅
𝑅 = 198.94 Ω
Simarly, the resistors set by the lower corner frequency was
found to be:
𝑅 = 19894.37 Ω
These values were entered in the SPICE simulation then
manipulated until the AC analysis of the filter showed the
correct frequency response values. MATlab was used to
simulated the transfer function, impulse, step and bode
response. The physical circuit component values were based
on fig. (24)
Fig. 25. SPICE AC Analysis of Active Bandpass Filter Frequency Response
Fig. 26. MATlab Actve Bandpass Filter Impulse Response
Fig. 27. MATlab Actve Bandpass Filter Step Response
Fig. 28. MATlab Actve Bandpass Filter Bode Plot
Fig. 29. MATlab Actve Bandpass Filter Unit-Ramp Response
Fig. 30. Actual Actve Bandpass Filter
7
VIII. CONCLUSION
Active filter are easier to fine tune and produce a better quality
output signal than passive filters. There is more control of the
bandwidth and an ability to adjust the gain. Using higher
resistors produce a more accurate output, overall. Lower
resistors tend to round the outputs. This is obvious in the
MATlab and oscilloscope readings for the various responses.
REFERENCES
[1] Active Band Pass Filter. Available: http://www.electronics-
tutorials.ws/filter/filter_7.html
Fig. 31. Actual Actve Bandpass Filter Impulse Response
Fig. 32. Actual Actve Bandpass Filter Ramp Response
Fig. 33. Actual Actve Bandpass Filter Unit-Step Response
Fig. 34. Actual Actve Bandpass Filter Unit-Impulse Response

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Design and Analysis of Active Bandpass Filter

  • 1. 1  Abstract—This report explores the frequency, impulse, unit- ramp, and unit-step responses of a passive and an active RLC bandpass filter when given a signal. Design rationale for the active filter is discussed. I. INTRODUCTION HIS report is to provide documentation of the steps taken in the design and creation of both a passive and active band-pass filter. This filter allows frequencies within a specific range to pass, but rejects frequencies not within the range. Before designing the circuits, the capacitance and inductance of the components were measured for use in both the passive and active filter. This is displayed in table 1. I. Component Rated versus Actual Values Component Rated Value Actual Values Capacitor (2) 0.1µF 0.08 µF Inductor (2) 100 mH 104.7, 105.4 mH Resistor 220 Ω 215.7 Ω Resistor 1k Ω 0.993k Ω Resistor 2.2k Ω 2.18k Ω By adding a high pass filter with a low pass, a band pass filter is created, as shown in Fig. 1. Fig. 1 An explanation of the Band Pass filter, with cutoff frequency at the -3dB point. [1] To avoid using integral and differential equations, circuits are converted from the time domain to the s domain; to do this, the Laplace transform is used: As such, when the Laplace transform is used, the circuit elements also change as described in table 2. II. Components in Domain Equivalents Time Domain Phasor S Domain This document was submitted on June 11, 2015 to Professor Mehmet Vurkac by the authors; Benedict J. Fawver, Kyla J. Marino and Eric J. Tipler. Benedict J. Fawver, author, is currently a student and at the Oregon Institute of Technology, 2301 Campus Dr. Klamath Falls, OR 97601. R R R L jωL sL C II. OSCILLOSCOPE MEASUREMENTS Setting up the function generator to simulate useful inputs like an impulses or ramps is relatively straightforward. For this project we used a Tektronix AFG 2021 arbitrary function generator, but other generators in the lab appeared to have very similar menu formats so the process should be the same. For an impulse response, the FG should be set to a pulse on continuous mode. For all options inputs, 1 V amplitude was used. The frequency was then adjusted until the image was clear. The square wave function was used to simulate the unit- step, and the ramp function was used to simulate the ramp input. In all cases, the inputs only serve as an approximation of the ideal inputs since signals in wave form repeat. A psuedo-frequency response was also found by running a sine wave sweep from 10 Hz to 10k Hz. This gave a wave form that had the largest amplitude at the center frequency. This waveform served as a good approximation of the output voltage vs the input voltage. The graph essentially shows the magnitude portion of the bode plot. III. MEASURING R, L, C The RC circuit in Fig. 2 was modeled to find the calculated and measured values of two 0.1 µF rated capacitors. By treating the circuit as a voltage divider, the following derivation found the frequency cutoff of the circuit: 1𝑉 ( −𝑗 𝜔𝑐 𝐶⁄ 𝑅 − (𝑗 𝜔𝑐 𝐶)⁄ ) = 1 √2 Kyla J. Marino, author, is currently a student and at the Oregon Institute of Technology, 2301 Campus Dr. Klamath Falls, OR 97601. Eric J. Tipler, author, is currently a student and at the Oregon Institute of Technology, 2301 Campus Dr. Klamath Falls, OR 97601. Design and analysis of active bandpass filter Benedict J. Fawver, Kyla J. Marino, Eric J. Tipler T Fig. 2. RC Low Pass Filter.
  • 2. 2 ‖ −𝑗 𝜔𝑐 𝐶⁄ 𝑅 − (𝑗 𝜔𝑐 𝐶)⁄ ‖ = 1 √2 1 𝜔𝑐 𝐶⁄ √𝑅2 + ( 1 𝜔𝑐 𝐶 )2 = 1 √2 𝑅 = 1 𝜔 𝑐 𝐶 (1) 𝜔𝑐 = 1 𝑅𝐶 = 2𝜋𝑓 𝑓𝑐 = 1 2𝜋𝑅𝐶 (2) 1.602 𝑘𝐻𝑧 = 1 (2𝜋)(993 Ω)(0.1x10−6) The calculated values were verified by simulating a transient analysis on the circuit in LTspice. A comparison of the calculated and measured values of the LTspice simulation are found in table 1. The circuit was built and the input frequency was adjusted until the amplitude output was 0.707. This frequency was used to find the true capacitance of the capacitor using (2). 𝐶 = 1 2𝜋𝑓𝑅 0.08 µ𝐹 = 1 2𝜋(1900)(993) The same method described for the RC circuit was followed to find the true values of two 100 mH inductors in LC circuit in Fig 3. The derivation found the frequency at half power to be (3). 𝑓𝑐 = 𝑅 2𝜋𝐿 (3) 1.580 𝑘𝐻𝑧 = 993 2𝜋(100𝑥10−3) IV. RC AND LC VALUES AC Analysis Capacitor Inductor LTspice 1.628 kHz 1.580 kHz Calculated 1.603 kHz 1.584 kHz Transient Analysis Amplitude (Vo) at ½ power Capacitor Inductor 0.707 0.662 The actual frequencies of the circuit were used to calculate the true value of the capacitors and inductors using (1) and (2) respectively. These values are found in table 4. V. True Values of Capacitor and Inductors Measured Frequency Capacitor Inductor 1.99 kHz 0.08 µF 1.90 kHz 0.08 µF 1.51 kHz 0.1047 H 1.50 kHz 0.0105 H VI. RLC-CIRCUIT FREQUENCY RESPONSE Two bandpass filters were built using the 220-kΩ resistor and the 2.2-kΩ resistor shown in Fig. 4. By treating the circuit as a voltage divider and using a theoretical maximum of 1, the following derivation found the frequency cutoff of the circuit: 𝐻(𝑠) = 𝑉𝑜(𝑠) 𝑉𝑖𝑛(𝑠) = 𝑅 𝑅 + 𝑠𝐿 + 1 𝑠𝐶 = 𝑅𝑠 𝑅𝑠 + 𝑠2 𝐿 + 1 𝐶 = ( 𝑅 𝐿 ) ( 𝑠 𝑅 𝐿 𝑠 + 𝑠2 + 1 𝐿𝐶 ) 𝐻(𝑗𝜔) = ( 𝑗𝜔 𝑗𝜔 + ( 1 𝐿𝐶 − 𝜔2) ) = 1 1 𝐿𝐶 − 𝜔2 = 0 𝜔 𝑜 = 1 √𝐿𝐶 2𝜋𝑓𝑜 = 1 √𝐿𝐶 𝑓𝑜 = 1 2𝜋√𝐿𝐶 (4) Fig. 3. RL Low Pass Filter. Fig. 4. RLC Passive Bandpass Filter Schematics using Measured Values of 220-kΩ and 2.2-kΩ Rated Resistors
  • 3. 3 The predicted maximum response frequency was calculated using (4): 𝑓𝑜 = 1 2𝜋√(0.1047)(0.08𝑥10−6) = 1733.17 𝐻𝑧 The expected frequency responses were sketched then simulated in SPICE. As seen in (4), the value of the resistor will have no effect on the maximum frequency. However, increasing the resistor will cause the magnitude response to become increasingly round. Fig.5. Predicted Passive Bandpass Filter Frequency Response Sketch Fig. 6. SPICE Simulation of RLC Passive Bandpass Filter Frequency Response using 220-Ω Rated Resistor Fig. 7. SPICE Simulation of RLC Passive Bandpass Filter Frequency Response using 2.2-kΩ Rated Resistor Fig. 8. Vo/Vin Plot in SPICE and Actual Circuit Fig. 9. Wolfram Alpha Plot of Theoretical Impulse Response for Passive Bandpass Filter using 216-Ω Resistor Fig. 10. Wolfram Alpha Plot of Theoretical Impulse Response for Passive Bandpass Filter using 2180-Ω Resistor
  • 4. 4 Fig. 11. MATlab Plot of Theoretical Unit-Step Response for Passive Bandpass Filter using 216-Ω Resistor Fig. 12. MATlab Plot of Theoretical Unit-Ramp Response for Passive Bandpass Filter using 216-Ω Resistor Fig. 13. MATlab Plot of Theoretical Unit-Ramp Response for Passive Bandpass Filter using 2180-Ω Resistor Fig. 14. SPICE Plot of Frequency Response using FFT Function for Passive Bandpass Filter using 216-Ω Resistor Fig. 15. Actual Passive Filter with 2180-Ω Resistor Fig. 16. Actual Passive Filter with 216-Ω Resistor Fig. 17. Actual Passive Filter Ramp Response with 216-Ω Resistor
  • 5. 5 VII. DESIGN AND ANALYSIS OF ACTIVE BANDPASS FILTER The bandpass filter with an amplification of 10 and band of frequencies between 100 Hz and 10 kHz was created by cascading a lowpass filter, highpass filter, and inverter. This allows the upper corner frequency (10 kHz) to be set by the lowpass filter and the lower corner frequency (100 Hz) to be set by the highpass filter. The inverter provides the gain of the filter. The transfer function of the bandpass filter (14) can be found by multiplying the highpass (10), lowpass (12), and inverter’s (13) transfer functions together. Fig. 18. Actual Passive Filter Ramp Response with 2180-Ω Resistor Fig. 19. Actual Passive Filter Impulse Response with 216-Ω Resistor Fig. 20. Actual Passive Filter Impulse Response with 2180-Ω Resistor Fig. 21. Actual Passive Filter Unit-Step Response with 216-Ω Resistor Fig. 22. Actual Passive Filter Unit-Step Response with 2180-Ω Resistor Fig. 23. Actual Passive Filter Frequency Response with 216-Ω Resistor Fig. 24. Actual Passive Filter Frequency Response with 2180-Ω Resistor Fig. 25. Active Bandpass Filter
  • 6. 6 𝐻(𝑠) = − 1 1+𝑠𝐶1 𝑅 (10) 𝐻(𝑗𝜔) = − 1 1 + 𝑗𝜔𝐶1 𝑅 𝜔1 = 1 𝐶1 𝑅 (11) 𝐻(𝑠) = − 𝑠𝐶2 𝑅 1+𝑠𝐶1 𝑅 (12) 𝐻(𝑠) = − 𝑅 𝑓 𝑅 𝑖 (13) 𝐻(𝑠) = (− 1 1+𝑠𝐶1 𝑅 ) (− 𝑠𝐶2 𝑅 1+𝑠𝐶1 𝑅 ) (− 𝑅 𝑓 𝑅 𝑖 ) (14) ℒ−1 { 𝐻(𝑠) 𝑠 } = (− 1 𝑠+𝑠2 𝐶1 𝑅 ) (− 𝐶2 𝑅 𝑠+𝑠 𝑠 𝐶1 𝑅 ) (− 𝑅 𝑓 𝑠𝑅 𝑖 ) ℒ−1 { 𝐻(𝑠) 𝑠2 } = (− 1 𝑠2+𝑠3 𝐶1 𝑅 ) (− 𝐶2 𝑅 𝑠2+𝑠3 𝐶1 𝑅 ) (− 𝑅 𝑓 𝑠2 𝑅 𝑖 ) The values of the filter resistors were calculated using the upper and lower corner frequencies as follows: 𝜔1 = 1 𝐶1 𝑅 10000 = 1 (0.08𝑥10−6)𝑅 𝑅 = 198.94 Ω Simarly, the resistors set by the lower corner frequency was found to be: 𝑅 = 19894.37 Ω These values were entered in the SPICE simulation then manipulated until the AC analysis of the filter showed the correct frequency response values. MATlab was used to simulated the transfer function, impulse, step and bode response. The physical circuit component values were based on fig. (24) Fig. 25. SPICE AC Analysis of Active Bandpass Filter Frequency Response Fig. 26. MATlab Actve Bandpass Filter Impulse Response Fig. 27. MATlab Actve Bandpass Filter Step Response Fig. 28. MATlab Actve Bandpass Filter Bode Plot Fig. 29. MATlab Actve Bandpass Filter Unit-Ramp Response Fig. 30. Actual Actve Bandpass Filter
  • 7. 7 VIII. CONCLUSION Active filter are easier to fine tune and produce a better quality output signal than passive filters. There is more control of the bandwidth and an ability to adjust the gain. Using higher resistors produce a more accurate output, overall. Lower resistors tend to round the outputs. This is obvious in the MATlab and oscilloscope readings for the various responses. REFERENCES [1] Active Band Pass Filter. Available: http://www.electronics- tutorials.ws/filter/filter_7.html Fig. 31. Actual Actve Bandpass Filter Impulse Response Fig. 32. Actual Actve Bandpass Filter Ramp Response Fig. 33. Actual Actve Bandpass Filter Unit-Step Response Fig. 34. Actual Actve Bandpass Filter Unit-Impulse Response