Integrating square to triangle?

[Exulus]

Hi guys, I need a bit of help with this. I've got an op-amp and the standard formula:

$V_{out} = -\frac{1}{RC}\int V_{in} dt$

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:
$V(t) = V$ for $0 \leq t \leq 0.0025$
$V(t) = -V$ for $0.0025 \leq t \leq 0.005$

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

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

 

[SGT]

Hi guys, I need a bit of help with this. I've got an op-amp and the standard formula:

$V_{out} = -\frac{1}{RC}\int V_{in} dt$

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:
$V(t) = V$ for $0 \leq t \leq 0.0025$
$V(t) = -V$ for $0.0025 \leq t \leq 0.005$

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

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



$V_{out}(t) = -\frac{1}{RC} (0.0025V - (0.005V - 0.0025V)) = 0$
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.
$V_{out}(t) = -\frac{1}{RC} [ \int_{0}^{t}V(τ)dτ$
The function V_out(t) is a triangle wave.
If in the function V_out(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)

 

[faust9]

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.

 

[SGT]

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.
$V_{out}(t) = -\frac{1}{RC} [ \int_{0}^{t}V(\tau)d\tau$
The function V_out(t) is a triangle wave.
If in the function V_out(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.

 

[Exulus]

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 :(

 

[Sojourner01]

You just choose some arbitrary values, J. Doesn't matter as long as the product of the two makes the right number.