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# Exp passive filter (1)

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### Exp passive filter (1)

1. 1. COMPUTATION (Step 4) (Step 15) (Step 6) (Step 17) (Question – Step 6) Question (Step 17) Question – (Step 7 ) – Question (Step 18 )DATA SHEETMATERIALSOne function generatorOne dual-trace oscilloscopeCapacitors: 0.02 µF, 0.04µFResistors: 1 kΩ, 2 kΩTHEORYIn electronic communication systems, it is often necessary to separate a specific range of frequencies from thetotal frequency spectrum. This is normally accomplished with filters. A filter is a circuit that passes a specificrange 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 allfrequencies above the cutoff frequency. A high-pass is designed to pass all frequencies above the cutofffrequency and reject all frequencies below the cutoff frequency. A band-pass filter passes all frequencies withina band of frequencies and rejects all other frequencies outside the band. A band-stop filter rejects allfrequencies within a band of frequencies and passes all other frequencies outside the band. A band-stop filterrejects 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 voltagegain (Vo/Vi) in dB as a function of frequency (f). The frequency at which the output power gain drops to 50% ofthe maximum value is called the cutoff frequency (f C). When the output power gain drops to 50%, the voltagegain drops 3 dB (0.707 of the maximum value). When the filter dB voltage gain is plotted as a function offrequency on a semi log graph using straight lines to approximate the actual frequency response, it is called a
2. 2. Bode plot. A bode plot is an ideal plot of filter frequency response because it assumes that the voltage gainremains constant in the passband until the cutoff frequency is reached, and then drops in a straight line. Thefilter network voltage in dB is calculated from the actual voltage gain (A) using the equationAdB = 20 log Awhere A = Vo/ViA low-pass R-C filter is shown in Figure 1-1. At frequencies well below the cut-off frequency, the capacitivereactance of capacitor C is much higher than the resistance of resistor R, causing the output voltage to bepractically equal to the input voltage (A=1) and constant with the variations in frequency. At frequencies wellabove the cut-off frequency, the capacitive reactance of capacitor C is much lower than the resistance ofresistor R and decreases with an increase in frequency, causing the output voltage to decrease 20 dB per decadeincrease in frequency. At the cutoff frequency, the capacitive reactance of capacitor C is equal to the resistanceof resistor R, causing the output voltage to be 0.707 times the input voltage (-3dB). The expected cutofffrequency (fC) of the low-pass filter in Figure 1-1, based on the circuit component value, can be calculated fromXC = RSolving for fC produces the equationA high-pass R-C filter is shown in figure 1-2. At frequencies well above the cut-off frequency, the capacitivereactance of capacitor C is much lower than the resistance of resistor R causing the output voltage to bepractically equal to the input voltage (A=1) and constant with the variations in frequency. At frequencies wellbelow the cut-off frequency, the capacitive reactance of capacitor C is much higher than the resistance ofresistor R and increases with a decrease in frequency, causing the output voltage to decrease 20 dB per decadedecrease in frequency. At the cutoff frequency, the capacitive reactance of capacitor C is equal to theresistance of resistor R, causing the output voltage to be 0.707 times the input voltage (-3dB). The expectedcutoff frequency (fC) of the high-pass filter in Figure 1-2, based on the circuit component value, can becalculated fromFig 1-1 Low-Pass R-C FilterWhen the frequency at the input of a low-pass filter increases above the cutoff frequency, the filter outputdrops at a constant rate. When the frequency at the input of a high-pass filter decreases below the cutofffrequency, the filter output voltage also drops at a constant rate. The constant drop in filter output voltage perdecade increase (x10), or decrease ( 10), in frequency is called roll-off. An ideal low-pass or high-pass filterwould have an instantaneous drop at the cut-off frequency (fC), with full signal level on one side of the cutofffrequency 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 thecutoff frequency and rolls off at -20dB/decade. At two-pole filter has two R-C circuits tuned to the same cutofffrequency and rolls off at -40dB/decade. Each additional pole (R-C circuit) will cause the filter to roll-off anadditional -20dB/decade. Therefore, an R-C filter with more poles (R-C circuits) more closely approaches an ideal
3. 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 changeby 90 degrees and over the frequency range and be 45 degrees at the cutoff frequency. In a two-pole filter, thephase (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 FilterPROCEDURELow-Pass FilterStep 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 Hzand 1 MHz by the Bode plotter. Sketch the curve plot in the space provided. Ad B fQuestion: 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 cutofffrequency.Step 3 Move the cursor to a flat part of the curve at a frequency of approximately 100 Hz. Record the voltagegain in dB on the curve plot.AdB = -0.001 dStep 4 Calculate the actual voltage gain (A) from the dB voltage gain (AdB)A = 0.99988 1Question: Was the voltage gain on the flat part of the frequency response curve what you expected for thecircuit 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 kHzStep 6 Calculate the expected cutoff frequency (fC) based on the circuit component values in Figure 1-1.fC = 7.958 kHz