How to set up an integral to integrate over a sine wave?

In summary, the goal is to set up an integral to integrate over the equation V(t) = 1/(R*C) integral to t Vin(t) dt, which is the capacitor voltage over time formula. The desired integration is over a sine wave from 9 to 81 degrees with a frequency of 120Hz and amplitude of 120V. The first attempt to integrate using Wolfram Alpha did not produce expected results. In order to properly integrate, the values of t corresponding to the desired phases in the cycle must be used as the limits of the integral.
  • #1
Voltux
30
3
How do I setup an integral to integrate over the following equation:

V(t) = 1/(R*C) integral to t Vin(t) dt

This is the capacitor voltage over time formula.

I want to integrate over a sine wave from 9 to 81 degrees. Frequency of 120Hz, amplitude of 120V.

The formula I used in wolframalpha is not producing results as expected:

integrate 1/(1000*100E-6)*120*sin(2*pi*120*t+x) dx from x=0.15708 to 1.41372

I attempted to integrate by hand by taking the voltage values of a 120V, 120Hz sine wave and applying the capacitor voltage formula to get the following results.

The application of the voltage was calculated at 23.14microseconds*9 degrees (step)

Sine Value Voltage Capacitor Voltage
9 0.1564 18.768 39.04mV
18 0.309 37.08 77.14mV
27 0.4539 54.468 113.31mV
36 0.5877 70.524 146.72mV
45 0.7071 84.852 176.52mV
54 0.809 97.08 201.96mV
63 0.891 106.92 222.43mV
72 0.951 114.12 237.41mV
81 0.9876 118.512 246.55mV
Total: 1.46108V

Thanks kindly!
 
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  • #2
So actually you are trying to integrate the following expression:
$$(\text{some constants})*sin(240 \pi t + x) $$
right ?

Well, remember that ##sin(a + b) = sin(a)cos(b) + sin(b)cos(a)##.
 
  • #3
Wait, I'm confused. The first integral you write is ##dt##. In the second one, you're integrating ##dx## and ##t## is undefined. Is it fixed? What value of ##t## did you use?

It looks like those ##x## values are 9 degrees and 81 degrees, and so ##t## is irrelevant. You're really trying to integrate ##\sin(x) dx## from x = 9 degrees to x = 81 degrees.

But in fact that's not what I think you really want to do. I think you want the time values that correspond to those phases in the cycle of your given frequency. So calculate ##\frac {120} {RC} \int \sin(2\pi * 120t) dt## and use as the limits the values of ##t## such that ##2\pi * 120t## = 9 degrees (expressed in radians) and 81 degrees expressed in radians.

That is, ##2\pi * 120 t_1 = 9 \pi/180 \rightarrow t_1 = 9 / (180 * 240)## and ##2\pi * 120 t_2 = 81 \pi/180 \rightarrow t_2 = 81 / (180 * 240)##
 
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1. How do I determine the limits of integration for a sine wave?

The limits of integration for a sine wave are determined by the range of the independent variable, usually denoted as x. For example, if you are integrating the function sin(x) from 0 to π, your limits of integration would be 0 and π.

2. Can I use any variable for the independent variable in the integral?

Yes, you can use any variable as long as it is consistent throughout the integral. However, it is common practice to use x as the independent variable for sine waves.

3. What is the general formula for setting up an integral to integrate over a sine wave?

The general formula for setting up an integral to integrate over a sine wave is ∫sin(x)dx. This represents the integral of the sine function over the variable x.

4. Do I need to take into account the amplitude and frequency of the sine wave when setting up the integral?

No, the amplitude and frequency do not affect the integral. They only affect the shape and period of the sine wave function.

5. Can I use the same integral to integrate over a cosine wave?

Yes, the same integral can be used to integrate over a cosine wave, as the only difference between the two functions is a phase shift. However, the limits of integration may change depending on the range of the independent variable for the cosine wave.

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