Op-Amp Output Units With Oscillatory Input Voltage

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SUMMARY

The discussion centers on the confusion regarding the output units of an integrating operational amplifier (op-amp) circuit with an oscillatory input voltage of 5V[sin(100t)]. The key issue identified is the misunderstanding of how the integration of an oscillatory function results in output voltage units of volts rather than volts per second. The correct approach involves expressing the input voltage in terms of angular frequency, leading to the integration yielding a result with appropriate voltage units. The participant emphasizes the importance of using symbolic representation rather than numerical values to avoid confusion.

PREREQUISITES
  • Understanding of operational amplifier circuits
  • Familiarity with integration in the context of signal processing
  • Knowledge of angular frequency and its relationship to frequency
  • Basic proficiency in using MathCAD for circuit analysis
NEXT STEPS
  • Study the principles of integrating op-amps in signal processing
  • Learn about angular frequency and its conversion to frequency (f = ω/2π)
  • Practice symbolic manipulation in circuit analysis to avoid unit confusion
  • Explore the use of MathCAD for solving electrical engineering problems
USEFUL FOR

Electrical engineering students, particularly those studying operational amplifiers and signal processing, as well as professionals seeking to clarify unit conversions in circuit analysis.

cmmcnamara
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Hi all,

ME major here that's very confused about the integrating op-amp. It's pretty simple what I am confused about, just unit problems.

Suppose you have the following integrating op-amp circuit:

Feedback capacitor: 10μF
Input Resistance: 10kΩ
Input Voltage: 5V[sin(100t)]

My problem arises with the units here. For an integrating op-amp the output voltage can be shown to be vo=-τ∫vi*dt . The time constant obviously has units of Hz or s^-1. However for any oscillatory input, the integration results in another oscillatory output which carries the time units inside the trig function which doesn't allow those units to cancel those of s^-1, leading to output "voltage" having units of V/s. What am I missing here? How am I wrong? I appreciate any help in advance!
 
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The 100 in sin(100t) has units of radians per second. When you multiply 100\ \text{rad/s} with t seconds, you get an angle in radians. Then you can take the sine of that. The sine of something that has units of seconds is meaningless.

This is the reason why you should NEVER EVER EVER solve problems with actual numbers. Use variables and plug in the values for those variables at the end. I bet you wouldn't have had this confusion if you or the person who wrote this problem had written the oscillatory input as V_{\text{in}} = V \sin ( \omega t ) with V = 5 \ \text{volts} and \omega = 100 \ \text{rad/s}. Radians are not actual units, so it is probably easier to remember the relationship between angular frequency and plain old frequency, which is \omega = 2 \pi f. In other words, V_{\text{in}} = V \sin ( 2 \pi f t ), where f = \frac{100}{2 \pi}\ \text{Hz}.

Integrate V \sin ( 2 \pi f t ) to get - \frac{V}{2 \pi f} \cos ( \omega t ), which has units of \frac{\text{volts}}{\text{Hz}} = \text{volts} \times \text{seconds}. Now, when you multiply that by a constant that has units of \text{Hz} or 1/\text{seconds}, the result has units of \text{volts}, which is what you want.
 
Last edited:
Hahah, yea I just caught this about 30 seconds ago and was coming here to close up the thread. I normally go about everything symbolically but I've had my head up my arse with MathCAD this quarter learning to use it. Thanks so much for your response!
 

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