1. COMPUTATION
(Step 4) (Step 15)
(Step 6)
(Step 17)
(Question – Step 6)
Question (Step 17)
Question – (Step 7 )
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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 (f C). 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

2. 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 resistance 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

3. 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.
Ad
B
f
Question: Is the frequency response curve that of a low-pass filter? Explain why.
= Yes it is. This filter allows the low frequency to pass and blocks the high frequency based on it 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 d
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.
= Yes, because VI 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

4. Question: How did the calculated value for the cutoff frequency compare with the measured value recorded on
the curve plot?
= The difference is exactly 0.29%.
Step 7 Move the cursor to a point on the curve that is as close as possible to ten times f C. 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?
= It is what I expected. dB gain decreases 17.11 dB per decade increase in frequency. Above frequency the
output voltage decreases 20dB/decade increase in frequency; 17.11 dB is approximately equal to 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 (f C)
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?
= It is what I expected. The phase between the input and output changes. 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.
= Yes,because it decreases. 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.
= Yes because it decreases. 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.

5. 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.
= It is what I expected, it passes all the frequencies above the cutoff frequency and rejects all the frequencies
below 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.
= At frequencies well above the cut-off frequency VO = Vi therefore 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?
= There is a difference of 0.29%.
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 what I expected, 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 0 and the final value (f) is 90 o. Run the simulation again. You are looking at the phase difference (θ)
o
between the filter input and output as a function of frequency (f). Sketch the curve plot in the space provided

6. 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).
at 100 Hz: θ = 89.28
at 1MHz: θ = 0.456o
at 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?
It is what I expected, 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.
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.
= Yes ,it decreases. The roll-off did not change. Roll-off is still the same even if 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.
= Yes it decreases. The roll-off did not change. Roll-off is still the same even if capacitance changes.
CONCLUSION
After this experiment, I could say that the cut-off fC is the basis of a filter.It filters the frequency if
it will ve allowed or rejected. In low-pass filter it only allows the frequencies below the cutoff frequency. On the
other hand, the high-pass filter only allows the frequencies above the cutoff frequency.
The voltage gain of low-pass filter below the cutoff frequency is almost equal to 1 because V o = Vi. The voltage
gain in high-pass filter becomes 1 if it is well above the fC because Vo = Vi.
Frequencies above cutoff (for the low-pass filter) the dB per decade roll-off decreases by 20 dB per
decade increase in frequency. Frequencies below f C (for high-pass filter) ) the dB per decade roll-off decreases
by 20 dB per decade decrease in frequency. The Phase response for a first-order low-pass filter and high-pass
filter, vOUT always lags vIN by some phase angle betweeen 0 and 90°.
If the resistance or capacitance changes, the cutoff frequency also changes. Cutoff is inversely
proportional to the resistance and capacitance. However, the roll-off is not affected by the resistance and the
capacitance.