Lab 05: Oscilloscope Operation


  • To become familiar with using an oscilloscope and function generator.
  • To measure phase angle using Lissajous polar.




The oscilloscope is a device that displays a graph of voltage vs. time (voltage on the vertical axis, time on the horizontal axis). If the voltage is DC, that is, constant in time, then the oscilloscope display is a horizontal line, whose vertical position indicates the voltage.

The oscilloscope screen has 1 cm divisions on both axes. There is a volts per division (volts/div) knob, which sets the vertical (volts) scale and a seconds per division (sec/div) knob which sets the horizontal (time) scale. There are knobs for setting the vertical and horizontal position of the display.

Under the volts/div knob is a 3-position switch that reads (AC - ground - DC).

  • In the ground position, the input to the oscilloscope is grounded (set to 0 volts), and the display becomes a horizontal line whose position (which can be adjusted with the vertical position knob) is the zero volts position. For instance, one could set the middle line of the screen to be 0 volts. Then positions above the middle would be positive voltages, and positions below the middle would be negative voltages.
  • When the switch is in the DC position, the signal is input to the oscilloscope unaltered.
  • When the switch is in the AC position, an internal capacitor is in series with the input to the oscilloscope, and the DC component of the signal is removed.


Lissajous Figures

Lissajous (pronounced "LEE-suh-zhoo") figure is a parametric plot of the harmonic system. Lissajous figures displayed on an oscilloscope can be used to give a quick estimate of the relative phase and the frequency of two signals. 

X-Y Display Mode

Lissajous figures can be obtained with an oscilloscope when it is operated in X-Y display mode. The X-Y display mode converts the oscilloscope from a volts-versus-time display to a volts-versus-volts display using two input channels. Channel 1 is the Y-axis input, and channel 2 is X-axis input.




For most electronic systems the relationship between the 3 dB bandwidth (BW) and the rise time (10% to 90%) is given approximately by RISE_TIME x BW =0.35.

Note: If a signal passes through several systems in series, (i.e. cascaded) the rise time at the output is given approximately as the square root of the sum of the squares of the rise times of the individual systems.

  1. An oscilloscope has a 35 MHz BW. Find the apparent rise time of the oscilloscope display if the signal input to the oscilloscope is an ideal pulse waveform.
  2. An oscilloscope has a 35 MHz BW. The oscilloscope is connected directly to a Pulse Generator and the oscilloscope display shows a rise time of 15 ns. Find the rise time of the Pulse Generator output.
  3. A Pulse Generator produces a pulse waveform output with a 10 ns rise time. After the pulse signal has passed through some electronic system it is displayed on the screen of a 35 MHz oscilloscope. The oscilloscope display shows a rise time of 25 ns. Find the rise time and the 3 dB BW of the electronic system.
  4. The input voltage of a system is displayed on CH 1 of an oscilloscope and the output voltage is displayed on CH 2. Both waveforms are sinusoidal with a period of 1 ms. The CH 2 waveform lags behind the CH 1 waveform by 0.1 ms. Find the phase shift produced by the system under study. Express the answer in degrees and in radians.

Why Rise Time and Bandwidth's Product is 0.35?

This is how it works: A simple R-C circuit can derive the bandwidth and rise time relationship:

  1. The time constant will be \(\tau = RC\). The bandwidth is \(BW = \frac{1}{{2\pi \tau }} = \frac{1}{{2\pi RC}}\)
    Rise time is the time measured from Vout,90% - Vout,10%. The response for an RC network is Vout = Vin (1 – e-t/τ)
  2. For Vout (t) = 0.1 Vin, we can get
    \(0.1{V_{in}} = {V_{in}}(1 - {e^{\frac{{ - {t_{10\% }}}}{\tau }}})\)
    where t10% is the time when the voltage reaches 10% of the final value.
  3. After simplification:
    \(0.9 = {e^{\frac{{ - {t_{10\% }}}}{\tau }}}\)
    Applying the natural log to the above equation, we can get:
    \(\ln (0.9) = \frac{{ - {t_{10\% }}}}{\tau } \Rightarrow {t_{10\% }} = - \tau \times \ln (0.9)\)
    Similarly, for  Vout(t) = 0.9 Vin, we can have 
    \({t_{90\% }} = - \tau \times \ln (0.1)\)
  4. The rise time is then:
    \({t_{RiseTime}} = {t_{90\% }} - {t_{10\% }} = \tau \times (\ln 0.9 - \ln 0.1) \simeq 2.2\tau \Rightarrow \tau = \frac{{{t_{RiseTime}}}}{{2.2}}\)
  5. Substitute the above value of τ into the expression for bandwidth:
    \(BW = \frac{1}{{2\pi \tau }} = \frac{{2.2}}{{2\pi \times {t_{RiseTime}}}} = \frac{{0.35}}{{{t_{RiseTime}}}} \Rightarrow {t_{RiseTime}} \times BW = 0.35\)

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