Lab 04: Function Generator and Oscilloscope

Objective

Reading Materials

Equipment

  • Function Generator
  • Digital Multimeter
  • Resistors: 1M, 1K,

Background

  • Peak Voltage (Vp): The maximum instantaneous value of a function as measured from the zero-volt level.
  • Peak-to-Peak Voltage (Vpp): The full voltage between positive and negative peaks of the waveform; that is, the sum of the magnitude of the positive and negative peaks.
  • Root Mean Square Voltage (Vrms): The root-mean-square or effective value of a waveform. The effective voltage would produce the same power in a resistive load as a DC voltage.
  • Average Voltage (Vavg): The waveform level is defined by the condition that the area enclosed by the curve above this level is exactly equal to the area enclosed by the curve below this level.

Sine Wave

Sine Wave Voltages — Vp, Vpp, Vavg, Vrms

Square Wave Rms Vpp

The formula for Sine Wave:

The instantaneous voltage of a sine wave can be expressed as:

  • \(v(\theta ) = {V_P}\;\sin (\theta )\)
  • \(v(t) = {V_P}{\mkern 1mu} \sin (\omega t) = {V_P}{\mkern 1mu} \sin (2\pi ft)\)

WaveformRoot-Mean-Square Voltage (𝑉rms)Average Voltage (𝑉avg)
Sine \({v_{rms}} = {{{V_P}} \over {\sqrt 2 }} \approx 0.707\;{V_P}\)

\({v_{rms}} = {{{V_{PP}}} \over {2\sqrt 2 }} \approx 0.354\;{V_{PP}}\)
\({v_{avg}} = {{2{V_P}} \over \pi } \approx 0.637\;{V_P}\)

\({v_{avg}} = {{{V_{PP}}} \over \pi } \approx 0.318\;{V_{PP}}\)
Square \({v_{rms}} = {V_{P}}\)

\({v_{rms}} = {{{V_{PP}}} \over 2}\)
\({v_{avg}} = {V_{P}}\)

\({v_{avg}} = {{{V_{PP}}} \over 2} = 0.5\;{V_{PP}}\)
Triangle \({v_{rms}} = {{{V_P}} \over {\sqrt 3 }} \approx 0.577\;{V_P}\)

\({v_{rms}} = {{{V_{PP}}} \over {2\sqrt 3 }} \approx 0.289\;{V_{PP}}\)
\({v_{avg}} = {{{V_P}} \over 2} = 0.5\;{V_P}\)

\({v_{avg}} = {{{V_{PP}}} \over 4} = 0.25\;{V_{PP}}\)

\[{V_{PP}} = 2 \times {V_P}\]

Procedure

Questions

  1. What value of output resistance would the "ideal" Function Generator have? Explain.
  2. A Function Generator is adjusted to give a no-load (open-circuit) voltage of 1.0 V. When a 1 KΩ load is placed across the output terminals of the Function Generator, the output voltage drops to 0.8 V. Find the output resistance of the Function Generator.
  3. Voltage Ratio in decibels \((dB) = 20{\log _{10}}\left( {\frac{{{V_1}}}{{{V_2}}}} \right)\)
    1. Express the following ratios in decibels (dB):
      1. V1 / V2 = 100
      2. V1 / V2 = 0.1
    2. Change these dB values to voltage ratios:
      1. -10 dB
      2. 3 dB
  4. A Function Generator has a specification that the variation in the output voltage with frequency will remain flat to within ±1 dB over the frequency range of 10 Hz to 100 kHz. Find the corresponding maximum percentage variation in the output voltage over this frequency range.

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