# Op-Amp Output Units With Oscillatory Input Voltage

1. Jul 18, 2013

### cmmcnamara

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!

2. Jul 18, 2013

### krome

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: Jul 18, 2013
3. Jul 18, 2013

### cmmcnamara

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!