Triangle Periodic Function: Determining f1 with Heavyside Step Functions

  • Thread starter Thread starter BrianBrian
  • Start date Start date
  • Tags Tags
    Function
BrianBrian
Messages
2
Reaction score
0

Homework Statement


Consider the triangle periodic function where x=0 @ t=0; x=2 @ t=2; x=0 @ t=4; x=2 @ t=6 and x=0 @ t=8. Let f1 be the function that agrees with f on [0,4], and is zero elsewhere. Determine f1 in terms of Heavyside step functions.


Homework Equations


Determine f1 in terms of Heavyside step functions.


The Attempt at a Solution


No idea...please help??!
 
Physics news on Phys.org
If u(t) is the unit step (Heaviside) function and a < b, notice that

u(t-a) - u(t-b) is 1 between a and b and 0 elsewhere. Because of that, this difference is sometimes called a filter function. So if you multiply it by any function f(t), the resulting function will agree with f(t) on (a,b) and be 0 else where. So you can put a few of these together to build sections of your sawtooth function and have 0 outside of the filters.
 
Ok. So you filter between the points where t=0 to t=4?
 
BrianBrian said:
Ok. So you filter between the points where t=0 to t=4?

I would guess that you would be expected to filter each of the 4 pieces on that range and add them up. So you need the equation of each of the four segments for the filters.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top