3) Go back to the circuit in Fig. 2, but replace the square wave input with a 3 volt p-p sinewave input. Looking at both the input and output, with the two scope probes, slowly increase theinput frequency from 50 Hz to 500 Hz. This circuit is called a "low pass filter". This means thatas you move to higher frequencies the output voltage decreases. Higher frequenciescorrespond to shorter times.

At the higher frequencies the impedance of the capacitor decreases, so the outputbecomes a smaller fraction of the input. We say the "attenuation" of the circuit in Fig. 2 getsgreater at higher frequencies. In addition, a phase shift develops between the output and the inputwaveforms. Measure the output of the circuit for a 1 volt p-p input for about 10 frequenciesbetween 50 Hz and 500 Hz. At what frequency is the signal reduced to 1/√2, or about 0.7, of itsoriginal amplitude? This frequency is called the "break point" frequency, and is related to thetime constant, RC, by the formula

f= 1/ 2πRC

What is the phase shift at the break point frequency? What is it at much higherfrequencies?

4) Repeat the frequency measurements with the circuit in Fig. 3. This is a "high passfilter," so that the response is lower at lower frequencies. Lower frequencies correspond tolonger times. At the lower frequencies the impedance of the capacitor increases, so theattenuation of the circuit in Fig. 3 gets greater. In addition, a phase shift develops between theoutput and the input waveforms. As in Part 3, measure the output of the circuit for a 1 volt p-pinput for about 10 frequencies between 50 Hz and 500 Hz. At what frequency is the signalreduced to 0.7 of its original amplitude? What is the phase shift at this frequency? What is it atmuch lower frequencies?

ASSIGNMENT

Use your measurements in Parts 3 and 4 to plot the attenuation, or ratio of the outputvoltage to the input voltage (Y axis), as a function of the frequency (X axis) from 50 Hz to 500Hz for the low pass and high pass filters.