Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Integral of Triangle wave

Status
Not open for further replies.

ferrico142

Newbie level 1
Joined
Feb 7, 2011
Messages
1
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,294
Cud anyone tell me...what'll be the resulting waveshape if a triangle waveform is being given as an input to an integrator opamp circuit....

Thnks all....
 

A square law output, I think. Taking the first slope of a triangle, say v=k*t where k is a constant is k*t^2 / 2.

Keith
 

Given that the edges of the triangle wave is perfectly straight, the integral will be square wave.

Remember that interal of a signal represent its slope angle.
 

No. Derivative is the slope and would be a square wave. Integral is the area under the curve.

Keith
 

As I said - derivative is slope, integral is area. The slope of a triangle is a constant because the slope is constant (until the triangle reverses direction). The area under a triangle is its mathematical integral which is a square law. Integral of x is (x^2)/2. Maybe the attached diagram will help.

Keith.
 

Attachments

  • Diff.gif
    Diff.gif
    40.6 KB · Views: 746
Last edited:

That's OK. I had to think very hard to remember what the integral of x was!

Keith.
 

Integral is area under the curve.
During rising slope of triangular wave, the following 2 points may be noted.
1. The area under the curve keeps increasing.
2. As the wave is triangular, the rate at which the area increases, also increases. As captured in the image, this can be seen by seeing the slope of the integral (upper waveform). It may be observed that the slope of the integral is continuously increasing.
During falling slope of triangular wave, following 2 points may be noted.
1. As the value of wave is above zero, the area (i.e. the integral) keeps increasing. From the image, this may be inferred by seeing that the integral waveform is always increasing.
2. Unlike in the rising case, the rate at which the area under the curve increases, actually decreases. It may be observed that the slope of the integral in this region is continuously increasing.
 

That seems a lot of words to say very little.

In response I would say that if the triangle is centred around zero instead of always positive then the integral will decrease and return to zero at the end of a complete cycle.

Keith
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top