Exp passive filter (7)
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This document describes an experiment to analyze the frequency response of passive low-pass and high-pass filters. The objectives are to plot the gain and phase responses of first-order RC filters, determine cutoff frequencies, and observe how component values affect cutoff frequency. Simulation results show that low-pass filters pass low frequencies and attenuate high frequencies above the cutoff frequency. High-pass filters do the opposite, passing high frequencies and attenuating low frequencies below the cutoff. For both filters, the cutoff frequency is determined by the RC time constant, and increasing or decreasing resistance and capacitance values lowers the cutoff frequency as expected. The phase response also shifts as expected, with about a 45 degree phase shift at the cutoff frequency for both single
Exp passive filter (7)
- 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT # 1 Passive Low-Pass and High-Pass Filter Bani, Arviclyn 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. COMPUTATION Step 4 Step 6 Question – Step 6 Question – Step 7 – Step 15
- 3. 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
- 4. 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 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
- 5. 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.
- 6. 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
- 7. Question: Is the frequency response curve that of a low-pass filter? Explain why. =I expected it. This filter pass the low frequency and blocks the high frequency depending on 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 the same and only has 0.29% difference. 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? = The circuits roll-off is at rate of 17.11 dB decrease per decade increase in frequency. I expected it because, above frequency the output voltage decreases 20dB/decade increase in frequency; 17.11 dB is approximately equal to 20 dB per decade.
- 8. 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? = I expected it. The phase changes between the input and output. I expected it 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
- 9. Question: Did the cutoff frequency changes? Did the dB per decade roll-off changes? Explain. = The cutoff changes, as a matter of fact 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. = The cutoff changes, as a matter of fact 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. 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.
- 10. 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 I expected it, 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. I expected it, 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
- 11. 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 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
- 12. Question: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain. = The cutoff changes, as a matter of fact 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 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, as a matter of fact 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 low frequency even if the capacitance changes.
- 13. CONCLUSION The following conclusions are made after experimenting. This will compare the low-pass filter and the high-pass filter: The frequencies that allowed by the filter: (Low-Pass Filter) Allows the frequencies below the cut-off frequency and blocks the frequencies above the cut-off frequency. (High-Pass Filter) Allows the frequencies above the cut-off frequency and blocks the frequencies below the cut-off frequency. Voltage Gain: Both Low Pass and High Pass has a voltage gain of 1. VO = VI (Low-Pass Filter) The voltage gain at well below the cutoff frequency is almost equal to 1; (High-Pass Filter) The voltage gain becomes 1 if it is well above the fC Roll-off: (Low-Pass Filter) decreases by 20 dB per decade increase in frequency. (High-Pass Filter) decreases by 20 dB per decade decrease in frequency. Phase (Low-Pass and High Pass Frequency) The phase of low-pass and high-pass between the input and the output changes 90 degrees over the frequency range and 45 degrees at the cutoff frequency. When Resistance and Capacitance Changes: (Effect on Cutoff for Low-Pass and High-Pass Filter) If the resistance or capacitance changes, the cutoff frequency also changes. Cutoff is inversely proportional to the resistance and capacitance. (Effect on Roll-off for Low-Pass and High-Pass Filter) Roll-off is not affected by the resistance and the capacitance.