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NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY
                  Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite




                                   EXPERIMENT # 1


                          Passive Low-Pass and High-Pass Filter




Maala, Michelle Anne C.                                           June 28, 2011
Signal Spectra and Signal Processing/ BSECE 41A1                  Score:




                                  Eng’r. Grace Ramones
                                       Instructor
OBJECTIVES
1.   Plot the gain frequency response of a first-order (one-pole) R-C low-pass filter.
2.   Determine the cutoff frequency and roll-off of an R-C first-order (one-pole) low-
     pass filter.
3.   Plot the phase-frequency of a first-order (one-pole) low-pass filter.
4.   Determine how the value of R and C affects the cutoff frequency of an R-C low-
     pass filter.
5.   Plot the gain-frequency response of a first-order (one-pole) R-C high pass filter.
6.   Determine the cutoff frequency and roll-off of a first-order (one-pole) R-C high
     pass filter.
7.   Plot the phase-frequency response of a first-order (one-pole) high-pass filter.
8.   Determine how the value of R and C affects the cutoff frequency of an R-C high
     pass filter.
SAMPLE COMPUTATION

(Step 4)




(Step 6)
(Question – Step 6)




Question – (Step 7 )
            –


Step 15




(Step 17)




Question (Step 17)




Question (Step 18 )
DATA SHEET

MATERIALS
One function generator
One dual-trace oscilloscope
Capacitors: 0.02 µF, 0.04µF
Resistors: 1 kΩ, 2 kΩ


THEORY
In electronic communication systems, it is often necessary to separate a specific
range of frequencies from the total frequency spectrum. This is normally
accomplished with filters. A filter is a circuit that passes a specific range of
frequencies while rejecting other frequencies. A passive filter consists of passive
circuit elements, such as capacitors, inductors, and resistors. There are four basic
types of filters, low-pass, high-pass, band-pass, and band-stop. A low-pass filter is
designed to pass all frequencies below the cutoff frequency and reject all frequencies
above the cutoff frequency. A high-pass is designed to pass all frequencies above the
cutoff frequency and reject all frequencies below the cutoff frequency. A band-pass
filter passes all frequencies within a band of frequencies and rejects all other
frequencies outside the band. A band-stop filter rejects all frequencies within a band
of frequencies and passes all other frequencies outside the band. A band-stop filter
rejects all frequencies within a band of frequencies and passes all other frequencies
outside the band. A band-stop filter is often is often referred to as a notch filter. In
this experiment, you will study low-pass and high-pass filters.


The most common way to describe the frequency response characteristics of a filter
is to plot the filter voltage gain (Vo/Vi) in dB as a function of frequency (f). The
frequency at which the output power gain drops to 50% of the maximum value is called
the cutoff frequency (fC). When the output power gain drops to 50%, the voltage gain
drops 3 dB (0.707 of the maximum value). When the filter dB voltage gain is plotted
as a function of frequency on a semi log graph using straight lines to approximate the
actual frequency response, it is called a Bode plot. A bode plot is an ideal plot of filter
frequency response because it assumes that the voltage gain remains constant in the
passband until the cutoff frequency is reached, and then drops in a straight line. The
filter network voltage in dB is calculated from the actual voltage gain (A) using the
equation
AdB = 20 log A
where A = Vo/Vi
A low-pass R-C filter is shown in Figure 1-1. At frequencies well below the cut-off
frequency, the capacitive reactance of capacitor C is much higher than the resistance
of resistor R, causing the output voltage to be practically equal to the input voltage
(A=1) and constant with the variations in frequency. At frequencies well above the
cut-off frequency, the capacitive reactance of capacitor C is much lower than the res-
istance of resistor R and decreases with an increase in frequency, causing the output
voltage to decrease 20 dB per decade increase in frequency. At the cutoff frequency,
the capacitive reactance of capacitor C is equal to the resistance of resistor R,
causing the output voltage to be 0.707 times the input voltage (-3dB). The expected
cutoff frequency (fC) of the low-pass filter in Figure 1-1, based on the circuit
component value, can be calculated from
XC = R



Solving for fC produces the equation



A high-pass R-C filter is shown in figure 1-2. At frequencies well above the cut-off
frequency, the capacitive reactance of capacitor C is much lower than the resistance
of resistor R causing the output voltage to be practically equal to the input voltage
(A=1) and constant with the variations in frequency. At frequencies well below the cut-
off frequency, the capacitive reactance of capacitor C is much higher than the
resistance of resistor R and increases with a decrease in frequency, causing the
output voltage to decrease 20 dB per decade decrease in frequency. At the cutoff
frequency, the capacitive reactance of capacitor C is equal to the resistance of
resistor R, causing the output voltage to be 0.707 times the input voltage (-3dB). The
expected cutoff frequency (fC) of the high-pass filter in Figure 1-2, based on the
circuit component value, can be calculated from
Fig 1-1 Low-Pass R-C Filter




When the frequency at the input of a low-pass filter increases above the cutoff
frequency, the filter output drops at a constant rate. When the frequency at the
input of a high-pass filter decreases below the cutoff frequency, the filter output
voltage also drops at a constant rate. The constant drop in filter output voltage per
decade increase (x10), or decrease ( 10), in frequency is called roll-off. An ideal low-
pass or high-pass filter would have an instantaneous drop at the cut-off frequency
(fC), with full signal level on one side of the cutoff frequency and no signal level on the
other side of the cutoff frequency. Although the ideal is not achievable, actual filters
roll-off at -20dB/decade per pole (R-C circuit). A one-pole filter has one R-C circuit
tuned to the cutoff frequency and rolls off at -20dB/decade. At two-pole filter has
two R-C circuits tuned to the same cutoff frequency and rolls off at -40dB/decade.
Each additional pole (R-C circuit) will cause the filter to roll-off an additional -
20dB/decade. Therefore, an R-C filter with more poles (R-C circuits) more closely
approaches an ideal filter.



In a pole filter, as shown the Figure 1-1 and 1-2 the phase (θ) between the input and
the output will change by 90 degrees and over the frequency range and be 45 degrees
at the cutoff frequency. In a two-pole filter, the phase (θ) will change by 180 degrees
over the frequency range and be 90 degrees at the cutoff frequency.
Fig 1-2 High-Pass R-C Filter




PROCEDURE


Low-Pass Filter


Step 1       Open circuit file FIG 1-1. Make sure that the following Bode plotter
settings are selected: Magnitude, Vertical (Log, F=0 dB, I=–40dB), Horizontal (Log,
F=1 MHz, I=100 Hz)
Step 2       Run the simulation. Notice that the voltage gain in dB has been plotted
between the frequencies 200 Hz and 1 MHz by the Bode plotter. Sketch the curve
plot in the space provided.


           AdB




                                                                     f
Question: Is the frequency response curve that of a low-pass filter? Explain why.

=Yes, Low-pass allow the frequencies below cutoff frequency and blocks the
frequencies above the cutoff frequency.


Step 3        Move the cursor to a flat part of the curve at a frequency of
approximately 100 Hz. Record the voltage gain in dB on the curve plot.
AdB = -0.001 dB


Step 4              Calculate the actual voltage gain (A) from the dB voltage gain (AdB)
A = 0.99988 1


Question: Was the voltage gain on the flat part of the frequency response curve what
you expected for the circuit in Fig 1-1? Explain why.
= I expected it, at below cutoff frequency, the VI is approximately equal to Vo making
the voltage gain approximately equal to 1.



Step 5        Move the cursor as close as possible to a point on the curve that is 3dB
down from the dB at 100 Hz. Record the frequency (cut-off frequency, fC) on the
curve plot.
fC = 7.935 kHz


Step 6        Calculate the expected cutoff frequency (fC) based on the circuit
component values in Figure 1-1.
fC = 7.958 kHz



Question: How did the calculated value for the cutoff frequency compare with the
measured value recorded on the curve plot?
= Almost equal and has 0.29% difference.


Step 7        Move the cursor to a point on the curve that is as close as possible to ten
times fC. Record the dB gain and frequency (f2) on the curve plot.
AdB = -20.108 dB
Question: How much did the dB gain decrease for a one decade increase (x10) in
frequency? Was it what you expected for a single-pole (single R-C) low-pass filter?
= The circuit’s roll-off is 17.11 dB decrease per decade increase in frequency.
It is expected because above frequency the output voltage decreases 20dB/decade
increase in frequency; 17.11 dB is almost 20 dB per decade.


Step 8        Click “Phase” on the Bode plotter to plot the phase curve. Make sure that
the vertical axis initial value (1) is -90 and the final value (F) is 0. Run the simulation
again. You are looking at the phase difference (θ) between the filter input and output
as a function of frequency (f). Sketch the curve plot in the space provided.


              θ




                                                                            f




Step 9        Move the cursor to approximately 100 Hz and 1 MHz and record the
phase (θ) in degrees on the curve plot for each frequency (f). Next, move the cursor
as close as possible on the curve to the cutoff frequency (fC) and phase (θ) on the
curve plot.
@100 Hz: θ = –0.72o
@1MHz: θ = –89.544o
@fC: θ = –44.917o
Question: Was the phase at the cutoff frequency what you expected for a singles-
pole (single R-C) low-pass filter?     Did the phase change with frequency? Is this
expected for an R-C low-pass filter?
= The phase at the cut-off frequency is expected. The phase changes between the
input and output. It is expected for R-C low-pass filter because the input and the
output change 88.824 degrees or 90 degrees on the frequency range and 44.917
degrees or 45 degrees.


Step 10        Change the value of resistor R to 2 kΩ in Fig 1-1. Click “Magnitude” on
the Bode plotter. Run the simulation. Measure the cutoff frequency (fC) and record
your answer.
fC = 4.049 kHz



Question: Did the cutoff frequency changes? Did the dB per decade roll-off changes?
Explain.
= The cutoff changes it decreases as the resistance increases. The dB per decade
roll-off did not change. The single pole’s roll-off will always approach 20 dB per
decade in the limit of high frequency even if the resistance changes.


Step 11        Change the value of capacitor C is 0.04 µF in Figure 1-1. Run the
simulation. Measure the new cutoff frequency (fC) and record your answer.
fC = 4.049 kHz


Question: Did the cutoff frequency change? Did the dB per decade roll-off change?
Explain.
= The cutoff changes, it decreases as the capacitance increases The dB per decade
roll-off did not change. The single pole’s roll-off will always approach 20 dB per
decade in the limit of high frequency even if the capacitance changes.
High-Pass Filter



Step 12        Open circuit file FIG 1-2. Make sure that the following Bode plotter
settings are selected: Magnitude, Vertical (Log, F=0 dB, I=–40dB), Horizontal (Log,
F=1 MHz, I=100 Hz)


Step 13        Run the simulation. Notice that the gain in dB has been plotted between
the frequencies of 100Hz and 1 MHz by the Bode plotter. Sketch the curve plot in the
space provided.


             AdB




                                                                          f



Question: Is the frequency response curve that of a high-pass filter? Explain why.
= I expected it. This filter pass the high frequency and blocks the low frequency
depending on the cutoff frequency.


Step 14        Move the cursor to a flat part of the curve at a frequency of
approximately 1 MHz Record the voltage gain in dB on the curve plot.
AdB = 0 dB


Step 15        Calculate the actual voltage gain (A) from the dB voltage gain (AdB).
A=1
Question: Was the voltage gain on the flat part of the frequency response curve what
you expected for the circuit in Figure 1-2? Explain why.
= Yes voltage gain on the flat part of the frequency response curve what I expected
frequencies well above the cut-off frequency VO equal to Vi so the voltage gain A
equals 1


Step 16       Move the cursor as close as possible to the point on the curve that is 3dB
down from the dB gain at 1MHz. Record the frequency (cutoff frequency, fC) on the
curve plot.
fC = 7.935 kHz


Step 17       Calculate the expected cut of frequency (fC) based on the circuit
component value in Figure 1-2
fC = 7.958 kHz


Question: How did the calculated value of the cutoff frequency compare with the
measured value recorded on the curve plot?
= The circuit has 0.29% difference.


Step 18       Move the cursor to a point on the curve that is as close as possible to
one-tenth fC. Record the dB gain and frequency (f2) on the curve plot.
AdB = -20.159 dB


Question: How much did the dB gain decrease for a one-decade decrease ( ) in
frequency? Was it what you expected for a single-pole (single R-C) high-pass filter?
= The dB gain decreases 18.161 dB per decrease in frequency. It is expected for a
single-pole (single R-C) high-pass filter because the frequencies below the cutoff
frequency have output voltage almost decrease 20dB/decade decrease in frequency.


Step 19       Click “Phase” on the Bode plotter to plot the phase curve. Make sure that
the vertical axis initial value (I) is 0o and the final value (f) is 90o. Run the simulation
again. You are looking at the phase difference (θ) between the filter input and output
as a function of frequency (f). Sketch the curve plot in the space provided
θ




                                                                           f


Step 20        Move the cursor to approximately 100 Hz and 1 MHz and record the
phase (θ) in degrees on the curve plot for each frequency (f). Next, move the cursor
as close as possible on the curve to the cutoff frequency (fC). Record the frequency
(fC) and phase (θ).

@ 100 Hz: θ = 89.28
@ 1MHz: θ = 0.456o
@ fC(7.935 kHz): θ = 44.738o


Question: Was the phase at the cutoff frequency (fC) what you expected for a single-
pole (single R-C) high pass filter?
= Yes, the phase at the cutoff frequency (fC) what I expected for a single-pole (single
R-C) high pass filter the input and the output change 89.824 degrees almost 90
degrees on the frequency range and 44.738 degrees almost degrees.

Did the phase change with frequency? Is this expected for an R-C high pass filter?
= Yes the phase between the input and output changes. It is expected in R--C high
pass filter.


Step 21        Change the value of resistor R to 2 kΩ in Figure 1-2. Click “Magnitude”
on the Bode plotter. Run the simulation. Measure the cutoff frequency (fC) and record
your answer.
fC = 4.049 kHz

Question: Did the cutoff frequency change? Did the dB per decade roll-off change?
Explain.
= The cutoff changes it decreases as the resistance increases. The dB per decade
roll-off did not change. The single pole’s roll-off will always approach 20 dB per
decade in the limit of low frequency even if the resistance changes.


Step 22     Change the value of the capacitor C to 0.04µF in Figure 1-2. Run the
simulation/ measure the cutoff frequency (fC) and record you answer.
fC = 4.049 kHz

Question: Did the cutoff frequency change? Did the dB per decade roll-off change?
Explain.
= The cutoff changes it decreases as the resistance increases. The dB per decade
roll-off did not change. The single pole’s roll-off will always approach 20 dB per
decade in the limit of low frequency even if the resistance changes.
CONCLUSION

I conclude that filters are named according to the frequency of signals they allow to
pass through them. There are Low-pass filters that allow only low frequency signals to
pass, High-pass filters that allow only high frequency signals to pass through.
Moreover, in Low-Pass Filter and High-Pass Filter the voltage gain of 1. When it comes
to roll-off, in Low-Pass Filter frequencies at well above cutoff the dB per decade roll-
off decreases by 20 dB per decade increase in frequency but in High-Pass Filter
frequencies below fC the dB per decade roll-off decreases by 20 dB per decade
decrease in frequency. The phase response of low-pass and high-pass between the
input and the output changes 90 degrees and 45 degrees at the cutoff frequency. If
the resistance or capacitance changes, the cutoff frequency also changes. Cutoff is
inversely proportional to the resistance and capacitance. So if the resistance or
capacitance decreases the cutoff increases and when resistance and capacitance
increases the cut-off decreases. Roll-off for Low-Pass and High-Pass Filter is not
affected by the resistance and the capacitance. It remain constant even the
resistance and capacitance changes.

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Exp passive filter (4)

  • 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT # 1 Passive Low-Pass and High-Pass Filter Maala, Michelle Anne C. June 28, 2011 Signal Spectra and Signal Processing/ BSECE 41A1 Score: Eng’r. Grace Ramones Instructor
  • 2. OBJECTIVES 1. Plot the gain frequency response of a first-order (one-pole) R-C low-pass filter. 2. Determine the cutoff frequency and roll-off of an R-C first-order (one-pole) low- pass filter. 3. Plot the phase-frequency of a first-order (one-pole) low-pass filter. 4. Determine how the value of R and C affects the cutoff frequency of an R-C low- pass filter. 5. Plot the gain-frequency response of a first-order (one-pole) R-C high pass filter. 6. Determine the cutoff frequency and roll-off of a first-order (one-pole) R-C high pass filter. 7. Plot the phase-frequency response of a first-order (one-pole) high-pass filter. 8. Determine how the value of R and C affects the cutoff frequency of an R-C high pass filter.
  • 4. (Question – Step 6) Question – (Step 7 ) – Step 15 (Step 17) Question (Step 17) Question (Step 18 )
  • 5. DATA SHEET MATERIALS One function generator One dual-trace oscilloscope Capacitors: 0.02 µF, 0.04µF Resistors: 1 kΩ, 2 kΩ THEORY In electronic communication systems, it is often necessary to separate a specific range of frequencies from the total frequency spectrum. This is normally accomplished with filters. A filter is a circuit that passes a specific range of frequencies while rejecting other frequencies. A passive filter consists of passive circuit elements, such as capacitors, inductors, and resistors. There are four basic types of filters, low-pass, high-pass, band-pass, and band-stop. A low-pass filter is designed to pass all frequencies below the cutoff frequency and reject all frequencies above the cutoff frequency. A high-pass is designed to pass all frequencies above the cutoff frequency and reject all frequencies below the cutoff frequency. A band-pass filter passes all frequencies within a band of frequencies and rejects all other frequencies outside the band. A band-stop filter rejects all frequencies within a band of frequencies and passes all other frequencies outside the band. A band-stop filter rejects all frequencies within a band of frequencies and passes all other frequencies outside the band. A band-stop filter is often is often referred to as a notch filter. In this experiment, you will study low-pass and high-pass filters. The most common way to describe the frequency response characteristics of a filter is to plot the filter voltage gain (Vo/Vi) in dB as a function of frequency (f). The frequency at which the output power gain drops to 50% of the maximum value is called the cutoff frequency (fC). When the output power gain drops to 50%, the voltage gain drops 3 dB (0.707 of the maximum value). When the filter dB voltage gain is plotted as a function of frequency on a semi log graph using straight lines to approximate the actual frequency response, it is called a Bode plot. A bode plot is an ideal plot of filter frequency response because it assumes that the voltage gain remains constant in the passband until the cutoff frequency is reached, and then drops in a straight line. The
  • 6. filter network voltage in dB is calculated from the actual voltage gain (A) using the equation AdB = 20 log A where A = Vo/Vi A low-pass R-C filter is shown in Figure 1-1. At frequencies well below the cut-off frequency, the capacitive reactance of capacitor C is much higher than the resistance of resistor R, causing the output voltage to be practically equal to the input voltage (A=1) and constant with the variations in frequency. At frequencies well above the cut-off frequency, the capacitive reactance of capacitor C is much lower than the res- istance of resistor R and decreases with an increase in frequency, causing the output voltage to decrease 20 dB per decade increase in frequency. At the cutoff frequency, the capacitive reactance of capacitor C is equal to the resistance of resistor R, causing the output voltage to be 0.707 times the input voltage (-3dB). The expected cutoff frequency (fC) of the low-pass filter in Figure 1-1, based on the circuit component value, can be calculated from XC = R Solving for fC produces the equation A high-pass R-C filter is shown in figure 1-2. At frequencies well above the cut-off frequency, the capacitive reactance of capacitor C is much lower than the resistance of resistor R causing the output voltage to be practically equal to the input voltage (A=1) and constant with the variations in frequency. At frequencies well below the cut- off frequency, the capacitive reactance of capacitor C is much higher than the resistance of resistor R and increases with a decrease in frequency, causing the output voltage to decrease 20 dB per decade decrease in frequency. At the cutoff frequency, the capacitive reactance of capacitor C is equal to the resistance of resistor R, causing the output voltage to be 0.707 times the input voltage (-3dB). The expected cutoff frequency (fC) of the high-pass filter in Figure 1-2, based on the circuit component value, can be calculated from
  • 7. Fig 1-1 Low-Pass R-C Filter When the frequency at the input of a low-pass filter increases above the cutoff frequency, the filter output drops at a constant rate. When the frequency at the input of a high-pass filter decreases below the cutoff frequency, the filter output voltage also drops at a constant rate. The constant drop in filter output voltage per decade increase (x10), or decrease ( 10), in frequency is called roll-off. An ideal low- pass or high-pass filter would have an instantaneous drop at the cut-off frequency (fC), with full signal level on one side of the cutoff frequency and no signal level on the other side of the cutoff frequency. Although the ideal is not achievable, actual filters roll-off at -20dB/decade per pole (R-C circuit). A one-pole filter has one R-C circuit tuned to the cutoff frequency and rolls off at -20dB/decade. At two-pole filter has two R-C circuits tuned to the same cutoff frequency and rolls off at -40dB/decade. Each additional pole (R-C circuit) will cause the filter to roll-off an additional - 20dB/decade. Therefore, an R-C filter with more poles (R-C circuits) more closely approaches an ideal filter. In a pole filter, as shown the Figure 1-1 and 1-2 the phase (θ) between the input and the output will change by 90 degrees and over the frequency range and be 45 degrees at the cutoff frequency. In a two-pole filter, the phase (θ) will change by 180 degrees over the frequency range and be 90 degrees at the cutoff frequency.
  • 8. Fig 1-2 High-Pass R-C Filter PROCEDURE Low-Pass Filter Step 1 Open circuit file FIG 1-1. Make sure that the following Bode plotter settings are selected: Magnitude, Vertical (Log, F=0 dB, I=–40dB), Horizontal (Log, F=1 MHz, I=100 Hz) Step 2 Run the simulation. Notice that the voltage gain in dB has been plotted between the frequencies 200 Hz and 1 MHz by the Bode plotter. Sketch the curve plot in the space provided. AdB f
  • 9. Question: Is the frequency response curve that of a low-pass filter? Explain why. =Yes, Low-pass allow the frequencies below cutoff frequency and blocks the frequencies above the cutoff frequency. Step 3 Move the cursor to a flat part of the curve at a frequency of approximately 100 Hz. Record the voltage gain in dB on the curve plot. AdB = -0.001 dB Step 4 Calculate the actual voltage gain (A) from the dB voltage gain (AdB) A = 0.99988 1 Question: Was the voltage gain on the flat part of the frequency response curve what you expected for the circuit in Fig 1-1? Explain why. = I expected it, at below cutoff frequency, the VI is approximately equal to Vo making the voltage gain approximately equal to 1. Step 5 Move the cursor as close as possible to a point on the curve that is 3dB down from the dB at 100 Hz. Record the frequency (cut-off frequency, fC) on the curve plot. fC = 7.935 kHz Step 6 Calculate the expected cutoff frequency (fC) based on the circuit component values in Figure 1-1. fC = 7.958 kHz Question: How did the calculated value for the cutoff frequency compare with the measured value recorded on the curve plot? = Almost equal and has 0.29% difference. Step 7 Move the cursor to a point on the curve that is as close as possible to ten times fC. Record the dB gain and frequency (f2) on the curve plot. AdB = -20.108 dB
  • 10. Question: How much did the dB gain decrease for a one decade increase (x10) in frequency? Was it what you expected for a single-pole (single R-C) low-pass filter? = The circuit’s roll-off is 17.11 dB decrease per decade increase in frequency. It is expected because above frequency the output voltage decreases 20dB/decade increase in frequency; 17.11 dB is almost 20 dB per decade. Step 8 Click “Phase” on the Bode plotter to plot the phase curve. Make sure that the vertical axis initial value (1) is -90 and the final value (F) is 0. Run the simulation again. You are looking at the phase difference (θ) between the filter input and output as a function of frequency (f). Sketch the curve plot in the space provided. θ f Step 9 Move the cursor to approximately 100 Hz and 1 MHz and record the phase (θ) in degrees on the curve plot for each frequency (f). Next, move the cursor as close as possible on the curve to the cutoff frequency (fC) and phase (θ) on the curve plot. @100 Hz: θ = –0.72o @1MHz: θ = –89.544o @fC: θ = –44.917o
  • 11. Question: Was the phase at the cutoff frequency what you expected for a singles- pole (single R-C) low-pass filter? Did the phase change with frequency? Is this expected for an R-C low-pass filter? = The phase at the cut-off frequency is expected. The phase changes between the input and output. It is expected for R-C low-pass filter because the input and the output change 88.824 degrees or 90 degrees on the frequency range and 44.917 degrees or 45 degrees. Step 10 Change the value of resistor R to 2 kΩ in Fig 1-1. Click “Magnitude” on the Bode plotter. Run the simulation. Measure the cutoff frequency (fC) and record your answer. fC = 4.049 kHz Question: Did the cutoff frequency changes? Did the dB per decade roll-off changes? Explain. = The cutoff changes it decreases as the resistance increases. The dB per decade roll-off did not change. The single pole’s roll-off will always approach 20 dB per decade in the limit of high frequency even if the resistance changes. Step 11 Change the value of capacitor C is 0.04 µF in Figure 1-1. Run the simulation. Measure the new cutoff frequency (fC) and record your answer. fC = 4.049 kHz Question: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain. = The cutoff changes, it decreases as the capacitance increases The dB per decade roll-off did not change. The single pole’s roll-off will always approach 20 dB per decade in the limit of high frequency even if the capacitance changes.
  • 12. High-Pass Filter Step 12 Open circuit file FIG 1-2. Make sure that the following Bode plotter settings are selected: Magnitude, Vertical (Log, F=0 dB, I=–40dB), Horizontal (Log, F=1 MHz, I=100 Hz) Step 13 Run the simulation. Notice that the gain in dB has been plotted between the frequencies of 100Hz and 1 MHz by the Bode plotter. Sketch the curve plot in the space provided. AdB f Question: Is the frequency response curve that of a high-pass filter? Explain why. = I expected it. This filter pass the high frequency and blocks the low frequency depending on the cutoff frequency. Step 14 Move the cursor to a flat part of the curve at a frequency of approximately 1 MHz Record the voltage gain in dB on the curve plot. AdB = 0 dB Step 15 Calculate the actual voltage gain (A) from the dB voltage gain (AdB). A=1
  • 13. Question: Was the voltage gain on the flat part of the frequency response curve what you expected for the circuit in Figure 1-2? Explain why. = Yes voltage gain on the flat part of the frequency response curve what I expected frequencies well above the cut-off frequency VO equal to Vi so the voltage gain A equals 1 Step 16 Move the cursor as close as possible to the point on the curve that is 3dB down from the dB gain at 1MHz. Record the frequency (cutoff frequency, fC) on the curve plot. fC = 7.935 kHz Step 17 Calculate the expected cut of frequency (fC) based on the circuit component value in Figure 1-2 fC = 7.958 kHz Question: How did the calculated value of the cutoff frequency compare with the measured value recorded on the curve plot? = The circuit has 0.29% difference. Step 18 Move the cursor to a point on the curve that is as close as possible to one-tenth fC. Record the dB gain and frequency (f2) on the curve plot. AdB = -20.159 dB Question: How much did the dB gain decrease for a one-decade decrease ( ) in frequency? Was it what you expected for a single-pole (single R-C) high-pass filter? = The dB gain decreases 18.161 dB per decrease in frequency. It is expected for a single-pole (single R-C) high-pass filter because the frequencies below the cutoff frequency have output voltage almost decrease 20dB/decade decrease in frequency. Step 19 Click “Phase” on the Bode plotter to plot the phase curve. Make sure that the vertical axis initial value (I) is 0o and the final value (f) is 90o. Run the simulation again. You are looking at the phase difference (θ) between the filter input and output as a function of frequency (f). Sketch the curve plot in the space provided
  • 14. θ f Step 20 Move the cursor to approximately 100 Hz and 1 MHz and record the phase (θ) in degrees on the curve plot for each frequency (f). Next, move the cursor as close as possible on the curve to the cutoff frequency (fC). Record the frequency (fC) and phase (θ). @ 100 Hz: θ = 89.28 @ 1MHz: θ = 0.456o @ fC(7.935 kHz): θ = 44.738o Question: Was the phase at the cutoff frequency (fC) what you expected for a single- pole (single R-C) high pass filter? = Yes, the phase at the cutoff frequency (fC) what I expected for a single-pole (single R-C) high pass filter the input and the output change 89.824 degrees almost 90 degrees on the frequency range and 44.738 degrees almost degrees. Did the phase change with frequency? Is this expected for an R-C high pass filter? = Yes the phase between the input and output changes. It is expected in R--C high pass filter. Step 21 Change the value of resistor R to 2 kΩ in Figure 1-2. Click “Magnitude” on the Bode plotter. Run the simulation. Measure the cutoff frequency (fC) and record your answer.
  • 15. fC = 4.049 kHz Question: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain. = The cutoff changes it decreases as the resistance increases. The dB per decade roll-off did not change. The single pole’s roll-off will always approach 20 dB per decade in the limit of low frequency even if the resistance changes. Step 22 Change the value of the capacitor C to 0.04µF in Figure 1-2. Run the simulation/ measure the cutoff frequency (fC) and record you answer. fC = 4.049 kHz Question: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain. = The cutoff changes it decreases as the resistance increases. The dB per decade roll-off did not change. The single pole’s roll-off will always approach 20 dB per decade in the limit of low frequency even if the resistance changes.
  • 16. CONCLUSION I conclude that filters are named according to the frequency of signals they allow to pass through them. There are Low-pass filters that allow only low frequency signals to pass, High-pass filters that allow only high frequency signals to pass through. Moreover, in Low-Pass Filter and High-Pass Filter the voltage gain of 1. When it comes to roll-off, in Low-Pass Filter frequencies at well above cutoff the dB per decade roll- off decreases by 20 dB per decade increase in frequency but in High-Pass Filter frequencies below fC the dB per decade roll-off decreases by 20 dB per decade decrease in frequency. The phase response of low-pass and high-pass between the input and the output changes 90 degrees and 45 degrees at the cutoff frequency. If the resistance or capacitance changes, the cutoff frequency also changes. Cutoff is inversely proportional to the resistance and capacitance. So if the resistance or capacitance decreases the cutoff increases and when resistance and capacitance increases the cut-off decreases. Roll-off for Low-Pass and High-Pass Filter is not affected by the resistance and the capacitance. It remain constant even the resistance and capacitance changes.