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# Exp passive filter (9)

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### Exp passive filter (9)

1. 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite EXPERIMENT # 1 Passive Low-Pass and High-Pass Filter Balane, Maycen June 28, 2011 Signal Spectra and Signal Processing/ BSECE 41A1 Score: Eng’r. Grace Ramones Instructor
2. 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.
3. 3. COMPUTATION (Step 4) −0.01 dB = 20log A −0.01 dB 20 = 20log A 20 −0.05 × 10−3 = logA 10−0.05×10−3 = 10log A 10−0.05×10−3 = A A = 0.99988 ≅ 1 (Step 6) 𝑓𝐶 = 1 2𝜋𝑅𝐶 = 1 2𝜋(0.02µ𝐹)(1𝑘Ω) = 7.957747𝑘𝐻𝑧 ≅ 7.958𝑘𝐻𝑧 (Question – Step 6) %𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = | (7.958𝑘𝐻𝑧 − 7.935𝐾𝐻𝑧 7.958𝑘𝐻𝑧 | × 100% = 0.29% Question – (Step 7 ) (−2.998dB) –(20.108 dB) = 17.11dB (Step 15) 0 dB = 20log A 0 dB 20 = 20log A 20 0 = logA 100 = 10log A 100 = A A = 1 (Step 17) 𝑓𝐶 = 1 2𝜋𝑅𝐶 = 1 2𝜋(0.02µ𝐹)(1𝑘Ω) = 7.957747𝑘𝐻𝑧 ≅ 7.958𝑘𝐻𝑧 Question (Step 17) %𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = | (7.958𝑘𝐻𝑧 − 7.935𝐾𝐻𝑧 7.958𝑘𝐻𝑧 | × 100% = 0.29% Question (Step 18 ) 20.159 dB− (−2.998dB) = 18.161dB
4. 4. 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
5. 5. 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 1 2𝜋𝑓𝐶 𝐶 = 𝑅 Solving for fC produces the equation 1 2𝜋𝑅𝐶 = 𝑓𝐶 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 1 2𝜋𝑅𝐶 = 𝑓𝐶
6. 6. 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.
7. 7. 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. f AdB
8. 8. Question: Is the frequency response curve that of a low-pass filter? Explain why. =Yes it is the frequency response curve that of a low-pass filter, Low-pass blocks the frequencies above the cutoff frequency and allows the frequencies below 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. = Yes, the voltage gain on the flat part of the frequency response curve what you expected for the circuit in Fig 1-1 , because 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? = the calculated value for the cutoff frequency compare with the measured value recorded on the curve plot is almost equal and has 0.29% difference.
9. 9. 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 dB gain decrease for a one decade increase (x10) in frequency is 17.11 dB decrease per decade increase in frequency. It is expected for a single-pole (single R-C) low-pass filter 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. 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 f θ
10. 10. 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 cutoff frequency is expected for a singles-pole (single R-C) low- pass filter. 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
11. 11. Question: Did the cutoff frequency change? Did the dB per decade roll-off change? Explain. = The cutoff changes, it decreases as the capacitance increases. On the other hand 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. Question: Is the frequency response curve that of a high-pass filter? Explain why. = It is the frequency response curve that of a high-pass filter. This filter passes the high frequency and blocks the low frequency depending on the cutoff frequency. f AdB
12. 12. 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, the voltage gain on the flat part of the frequency response curve is expected for the circuit in Figure 1-2 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 calculated value of the cutoff frequency compare with the measured value recorded on the curve plot is almost equal with 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?
13. 13. = 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 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. f θ
14. 14. 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 although 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 although if the resistance changes.
15. 15. CONCLUSION I therefore conclude that: a simple passive Low Pass Filter and High Pass Filter, can be easily made by connecting together in series a single resistor with a single capacitor. For Low-Pass Filter:  Low pass filter pass the signals with low frequency and attenuates the signals with high frequency.  The voltage gain at well below the cutoff frequency is approximately equal to 1; Vo = Vi.  Frequencies at well above cutoff the dB per decade roll-off decreases by 20 dB per decade increase in frequency  The phase of low-pass between the input and the output changes 90 degrees over the frequency range 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.  Roll-off not remain constant even the resistance and capacitance changes. For High-Pass Filter  A high pass filter pass high frequency signals and attenuates low frequency signals..  Voltage gain at well above the cutoff frequency is 1  High-Pass Filter frequencies below fC the dB per decade roll-off decreases by 20 dB per decade decrease in frequency.  In addition, the phase of high-pass between the input and the output changes 90 degrees over the frequency range 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.  Roll-off remain constant even the resistance and capacitance changes.