An AC tachometer produces a sinusoidal signal with a frequency and amplitude both proportional to rotational rate.  While rotational rate information is encoded both into frequency and amplitude, it is often necessary to have a signal that is a DC signal voltage proportional to speed.  The diode rectifier circuit - shown below - is one way to convert an AC input signal (from the tachometer) to a DC voltage.  You need to investigate using this circuit.

• Before you connect the circuit, calibrate the tachometer and observe the tachometer waveform.
• Connect the tachometer voltage to an oscilloscope and adjust the voltage sensitivity and the time scale.
• Connect a power supply to the motor and get the motor spinning slowly.  After getting a feel for things, adjust the motor speed as low as you can get it reliably.
• Observe the waveform on the scope.
• Is the waveform sinusoidal?
• Calibrate the tqchometer using a wristwatch or a stopwatch.  Count the number of revolutions in a predetermined time period (30 seconds?) to determine the rotational rate.
• You should get four (4) cycles per revolution of the tachometer.  It should be an even integer number of cycles per revolution.
• Note the range of signal frequency at the voltage limits.  Determine the lowest and highest frequencies for the allowed voltage.
• Calibrate the tachometer, giving an expression for both the frequency and the output voltage (AC peak value) as a function or ratational rate.  Run the motor+tachometer at several different voltages to check this.
• Connect the circuit above.
• Use either a 0.1mf or a 1.0mf capacitor.
• You need to choose the resistor (R).
• The input voltage to the motor should not exceed 10 volts.
• Observe the waveform on an oscilloscope when the tachometer generates the input voltage.
• Compare the waveforms you observe with the waveforms in the rectifier simulator.
• You may want to note that the ratio of time constant to period of the sinusoidal input really determines the shape of the response.  In other words, as long as that ratio is constant (or equivalently, the product of time constant and frequency is constant) the graph of the output voltage (across the capacitor) will look the same (after adjusting the time scale).

• The filter should filter out the AC component.
• You should expect the filtering problem to be worst for low frequencies.  Since the lowest frequency is 0 Hz, you have an impossible problem.  Instead, assume that the rotational rate is such that you will have at least 20Hz at the output of the tachometer.
• Choose the time constant of the filter based on 20 Hz.  Aim for noise fluctuationas that are less than 2% of the DC voltage measured. (Or 2% of the peak voltage of the filtered signal as seen on the scope)  At 20 Hz, what should the time constant be?  What value of resistance do you need if you use a 1.0mf capacitor?
• For whatever time constant you use, record the AC voltage signal in, and the filtered output.  That will mean that you need to have two traces on the oscilloscope.
• Do one measurement at 20 Hz.
• Do the other measurement at whatever speed and signal frequency you get when you apply 10 volts to the motor.
• Is the measurement at the higher speed what it should be for the time constant you chose?  You need to justify that with a calculation.
• Finally, you want to get two transient measurements.
• Jump the input from zero to 10 v and observe the filtered output on the scope.  You might want to try to get the starry night along with it.  If that doesn't work out, get the output of the filter and record it.  Do you get a good measurement of the transient?  Explain.
• Do the same going from 10 v to zero v.  (i.e. turn off the power supply to the motor). Do you get a good measurement of the transient?  Explain.

• Simulators (no threshold, with threshold) in a note/interactive problem
• Simulators for up-ramps and down-ramp AC signals in a note