Convolution Help: Understanding Integrals for 0≤t<1

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In summary, the two signals are convolved (integrated) over the time period 0≤t<1 to create a graph with equation y = 1*dτ. The equation of the line for the period of -2 <= t <= 1 is y = m*x + b (where 'b' is the y-intercept when x = 0).
  • #1
unknown7
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Hi there, I'm having trouble convolving two signals, according to this site "http://cnx.org/content/m11541/latest/" " and its example, for the time period 0≤t<1 they've used the integral

yt=∫dτ between 0≤t<1.

My problem is, how did they get this integral?. I get that the height of the two signals is one, hence 1*dτ but I'm still hazy on this.

I'm working on a question where the graph looks like this.
[PLAIN]http://img705.imageshack.us/img705/6834/myconv.png

but i don't know how to get the integral at the position?.
 
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  • #2
Its not the height that they're looking at but rather the equation of the line for that section of time.

Since both of those signals appear as the straight line with equation y = 1, multiplied together (1*1) they give you the 'integral of 1'.

Your lines will have different equations, in different time periods.

Looking at Figure 4 shows an example of the different equations for different lines.
 
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  • #3
I think that was the resulting y(t) graph. Not another input x(t) or h(t). I'm still a little confused of what to do.
 
  • #4
My apologies, you're right there about Figure 4 being the answer and not a 2nd example.

So just to clarify, you had (red) f(t) and made it f(tau) and then flipped (blue) g(t) and made it g(t-tau), correct?

Now you're looking for the integral for the period of time of -2 <= t <= 1 correct?

Whats the equation of the line for red, from -2 to -1, using the dummy variable 'tau' instead of 't'? Do you remember the gradient 'm' being rise/run and the equation of the line being y = m*x + b (where 'b' is the y-intercept when x = 0) ?

Similarly, what's the equation of the line initially for blue? What happens to the equation of the line when you flip and shift it?

Now you should have two equations you can multiply together to get the equation for integration. Do you know the integration bounds for this section by the way?
 
  • #5
do u have soln. to this question?

A quick tip before starting convolution questions.

1) Know the distance of your blue signal. i.e. blue line has a horizontal distance of 1.

this will make it easy for you know how to what's going on when u slide the curve into the other

2) Identify or try and visualise the regions of integration before starting. But how many ROA are here? -1<t<-2, then -1<t-1<-2, then 0<t<1, then 0<t-1<1, then 1<t-1<2, then after these cases the blue line lies outside the red so it is 0

NOTE:I PUT T-1 BUT FOR EASE OF VISUALISATION, HOWEVER WHEN CALCULATING YOU HAVE TO PUT ?<T<?. Don't Leave it as t-1

then proceed with your calculations.
 

1. What is a convolution integral?

A convolution integral is a mathematical operation that combines two functions to create a third function. It is used to represent the way two quantities interact and affect each other. In the context of 0≤t<1, a convolution integral is used to calculate the total effect of two functions over the interval of 0 to 1.

2. Why is understanding integrals important for 0≤t<1?

Integrals are important for understanding the relationship between two functions, particularly in the interval of 0≤t<1. This interval is frequently used in many scientific applications, such as in physics and engineering, making it essential to understand how integrals work in this range.

3. How is a convolution integral calculated?

A convolution integral is calculated by multiplying the two functions, shifting one of them, and then integrating the resulting product over the given interval. This process is repeated for different values of the shift, and the results are added together to produce the final convolution integral.

4. What is the significance of 0≤t<1 in a convolution integral?

The interval of 0≤t<1 is significant in a convolution integral because it represents a specific time period or range. In scientific applications, this range can correspond to a specific process or phenomenon that is being studied, making it a crucial part of the overall calculation.

5. How is a convolution integral used in real-world applications?

Convolution integrals have numerous real-world applications, particularly in fields such as signal processing, image processing, and data analysis. They are used to represent the interaction and effects of different variables in a given system, making them valuable tools for understanding and analyzing complex systems and processes.

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