Integrating square to triangle?

In summary, the conversation is discussing the use of an op-amp and the standard formula for determining capacitor and resistor values to achieve a specific output amplitude and frequency. However, when trying to integrate a square wave, the value of the integral always comes out to be 0. The solution is to find the integral from 0 to a time t, as it will result in a function of t that represents a triangle wave. The conversation also mentions the use of arbitrary values for R and C to achieve the desired output.
  • #1
Exulus
50
0
Hi guys, I need a bit of help with this. I've got an op-amp and the standard formula:

[itex]V_{out} = -\frac{1}{RC}\int V_{in} dt[/itex]

And i need to integrate a square wave from it in order to determine some capacitor/resistor values to get an output amplitude of 5V and freq 200Hz (triangle wave). Every time i try to do it i end up getting 0 as an answer (which is kind of logical, considering one half of the cycle is the negative of the other). For example:
[itex] V(t) = V[/itex] for [itex] 0 \leq t \leq 0.0025[/itex]
[itex] V(t) = -V[/itex] for [itex]0.0025 \leq t \leq 0.005[/itex]

[itex]V_{out} = -\frac{1}{RC} [ \int_{0}^{0.0025}Vdt - \int_{0.0025}^{0.005}Vdt][/itex]

[itex]V_{out} = -\frac{1}{RC} (0.0025V - (0.005V - 0.0025V)) = 0[/itex]
How can i get around this problem? Cheers.
 
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  • #2
Exulus said:
Hi guys, I need a bit of help with this. I've got an op-amp and the standard formula:

[itex]V_{out} = -\frac{1}{RC}\int V_{in} dt[/itex]

And i need to integrate a square wave from it in order to determine some capacitor/resistor values to get an output amplitude of 5V and freq 200Hz (triangle wave). Every time i try to do it i end up getting 0 as an answer (which is kind of logical, considering one half of the cycle is the negative of the other). For example:
[itex] V(t) = V[/itex] for [itex] 0 \leq t \leq 0.0025[/itex]
[itex] V(t) = -V[/itex] for [itex]0.0025 \leq t \leq 0.005[/itex]

[itex]V_{out} = -\frac{1}{RC} [ \int_{0}^{0.0025}Vdt - \int_{0.0025}^{0.005}Vdt][/itex]

[itex]V_{out} = -\frac{1}{RC} [ \int_{0}^{0.0025}Vdt - \int_{0.0025}^{0.005}Vdt][/itex]



[itex]V_{out}(t) = -\frac{1}{RC} (0.0025V - (0.005V - 0.0025V)) = 0[/itex]
How can i get around this problem? Cheers.

If you calculate the integral for an entire cycle of the wave you get the area under the curve that is of course 0. What you want is not the integral of one period, but the integral from 0 to a time t, that is a function of t.
[itex]V_{out}(t) = -\frac{1}{RC} [ \int_{0}^{t}V(τ)dτ [/itex]
The function Vout(t) is a triangle wave.
If in the function Vout(t) you replace t by 0.005 you get 0, the value of the integral you have calculated.

edited to inform:
I don't know what happened. Where appears the number 964, there should be the greek letter τ (tau)
 
Last edited by a moderator:
  • #3
SGT, it looks like you used the symbol for tau; however, you have to use \tau in latex. Maybe the latex code on the server messed things up here... 964 is the HTML reference to the greek letter tau.
 
  • #4
SGT said:
If you calculate the integral for an entire cycle of the wave you get the area under the curve that is of course 0. What you want is not the integral of one period, but the integral from 0 to a time t, that is a function of t.
[itex]V_{out}(t) = -\frac{1}{RC} [ \int_{0}^{t}V(\tau)d\tau[/itex]
The function Vout(t) is a triangle wave.
If in the function Vout(t) you replace t by 0.005 you get 0, the value of the integral you have calculated.

Since I cannot anymore edit my previous post, I am posting again with the corrections provided by faust9. Thank you, Faust.
 
  • #5
Cheers, i still really don't get what i need to do though. I've got the function for a square wave as stated above, but how do i go about finding an expression for R and C? I know they can pretty much be unlimited as i will just get a ratio between them but i don't really see how i can go about getting that ratio...im completely lost :(
 
  • #6
You just choose some arbitrary values, J. Doesn't matter as long as the product of the two makes the right number.
 

Related to Integrating square to triangle?

1. What is the concept of integrating square to triangle?

Integrating square to triangle is a mathematical process where the area of a square is converted to the area of a triangle. This is done by dividing the square into smaller triangles and rearranging them to form a triangle.

2. How is integrating square to triangle different from other integration methods?

Unlike other integration methods, integrating square to triangle does not involve finding the antiderivative of a function. Instead, it involves geometric manipulation to find the area of a shape.

3. What are the applications of integrating square to triangle?

Integrating square to triangle is commonly used in geometry and calculus. It can also be used in real-life situations such as calculating the area of a triangular field or determining the volume of a triangular prism.

4. What are the steps involved in integrating square to triangle?

The steps involved in integrating square to triangle are as follows: 1. Divide the square into smaller triangles 2. Rearrange the smaller triangles to form a triangle 3. Use the formula for finding the area of a triangle (base x height / 2) to calculate the area of the new triangle.

5. Can integrating square to triangle be used for shapes other than a square?

Yes, integrating square to triangle can also be used for other shapes such as rectangles, parallelograms, and trapezoids. The process is the same - dividing the shape into smaller triangles and rearranging them to form a triangle.

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