# Input/output question

1. Aug 12, 2005

### asdf1

if y(x)=e^(-h)*(integral *(e^h)*rdx) + ce^(-h)
then why can you look at that equation from this view:
total output = response to the input+response to the initial data

2. Aug 15, 2005

### SGT

The term $$e^{-h}\int{e^{h}rdx}$$ is the response to the input and the term $$ce^{-h}$$ is the response to the initial conditions.

3. Aug 16, 2005

### asdf1

i know, but why?

4. Aug 16, 2005

### SGT

In the expression, $$r(x)$$ is the input and $$c$$ is the initial state. So, the integral is a function of the input and the second term is a function of the initial state.

5. Aug 17, 2005

### asdf1

@@a
i thought c was a constant? why is that particular one the initial state?
r(x) is the input because usually x is the input, right?

6. Aug 17, 2005

### SGT

$$x$$ is the independent variable. $$r(x)$$ is an arbitrary function of the independent variable and is the input to your system. $$c = y(x_0)$$ is the initial value of the dependent variable $$y$$. $$h = a.(x - x_0)$$.
The integral is in reality a definite one. So:
$$y(x) = e^{-h}\int_{x_0}^{x}e^hr(x')dx' + c e^{-h}$$
At $$x = x_0$$ $$e^{-h} = e^{-a.(x_0 - x_0)} = e^0 = 1$$ and the integral from $$x_0$$ to $$x_0$$ is zero. So,
$$y(x_0) = 0 + c.1 = c$$

7. Aug 18, 2005

### asdf1

thank you very much! :)

8. Aug 18, 2005

### asdf1

by the way, do you mean that a "definite integral" is one that converges?

9. Aug 18, 2005

### SGT

No, a definite integral is one that has specified limits, so its result is a number and not a function of the integration variable. As you can see, in my example, I used x' as integration variable and x as one of the limits. Relative to x', x is a constant.

10. Aug 18, 2005

thanks! :)