Deduce the solution from weight function

[jwxie]

Homework Statement

Convolution-Summation of a First Order System with Constant Coefficients

[PLAIN]http://dl.dropbox.com/u/14655573/1.PNG

Homework Equations

The Attempt at a Solution

The solution from the lecture
[PLAIN]http://dl.dropbox.com/u/14655573/2.PNG

[PLAIN]http://dl.dropbox.com/u/14655573/3.PNG

What I don't understand is how do we get 4/3 and the minus sign in the solution.

I know I can separate the summation into i = 0 to 0, and i = 1 to k. Thus, the first summation would be 1/3.
Then how did we get that 4/3? Did we just add the 1? Where did that 1 come from? From b_0 ?

How did we get the minus sign in front of the second term in the solution? Was it because we have -1/3 u(k-1) in the original equation?

Can anyone please help me with this? Thank you very much.

 

[LCKurtz]

You have given

$$w_0 = 1,\ w_i = \left(\frac 1 6\right)\left(\frac 1 2\right)^{i-1}$$

So

$$\sum_{i=0}^{k}w_i = 1 + \sum_{i=1}^{k}w_i = 1 + \left(\frac 1 6\right)<br /> \sum_{i=1}^{k}\left(\frac 1 2\right)^{i-1}$$

But

$$\sum_{i=1}^{k}\left(\frac 1 2\right)^{i-1} = 2 - \left(\frac 1 2\right)^{k-1}$$

Put that in for the last sum and simplify it.

 

[jwxie]

hi. i see that since w(0) is 1, then the summation of w(i) when i = 0, we would have 1
that's clear
the 2,... it's the summation of gemoetric series given by (1 - a^(k+1))/ 1-a, am i correct?
i get
(1 - (1/2)^k) / 1-(1/2)
which gives 2* (1- (1/2)^k)
did i do something wrong? when i distribute it the k-power term also gets the 2...

 

[LCKurtz]

hi. i see that since w(0) is 1, then the summation of w(i) when i = 0, we would have 1
that's clear
the 2,... it's the summation of gemoetric series given by (1 - a^(k+1))/ 1-a, am i correct?
i get
(1 - (1/2)^k) / 1-(1/2)
which gives 2* (1- (1/2)^k)
did i do something wrong? when i distribute it the k-power term also gets the 2...

So you distribute the 2 and you get 2 - (1/2)^(k-1) just like I said.

 

[jwxie]

I thought the (1/2)^(k-1) also gets the 2 after the distribution since the top is being divided by the the 1-(1/2)

$(1-\frac{1}{2}^{k-1})*2$

-- edited
okay. i think we are both right.
i think u missed the 2 in the previous post
and doing the simplification
1+ (1/6) * (2) * [1 - (1/2)^(k-1) ] would give the result

thanks.