Simplifying the Discrete Time Signal Summation

[Chandasouk]

It's been a while since I took Calc 2 and I am in a Linear Systems and Signals class right now. I'm looking at a solution on how to obtain a zero state response of a discrete time signal, but performing the summation confuses me. Can someone explain the steps they did?

This is part A

http://imageshack.us/photo/my-images/684/partal.jpg/

This is part B

http://imageshack.us/photo/my-images/11/part2sy.jpg/

In Part A, where did the +1 come from?
Also, how would you simplify the first fraction appearing after the equal sign of Ystep?

In Part B, where did the -1 's come from? Originally, h[m] was just (6/5)$\sum$(-1/2)^m + (24/5)Ʃ(1/3)^m

I have a Final next week on this stuff so, any help would be appreciated.

 

[SammyS]

It's been a while since I took Calc 2 and I am in a Linear Systems and Signals class right now. I'm looking at a solution on how to obtain a zero state response of a discrete time signal, but performing the summation confuses me. Can someone explain the steps they did?

This is part A

http://imageshack.us/photo/my-images/684/partal.jpg/

This is part B

http://imageshack.us/photo/my-images/11/part2sy.jpg/

In Part A, where did the +1 come from? Also, how would you simplify the first fraction appearing after the equal sign of Ystep?

In Part B, where did the -1 's come from?

I have a Final next week on this stuff so, any help would be appreciated.

Part A: Where did the 1 come from?

$\displaystyle \sum_{m=0}^{k}(-2)^m=(-2)^0+\sum_{m=1}^{k}(-2)^m$

and of course, (-2)⁰=1​

As for simplifying the fraction, they have done it quite well.

$\displaystyle \frac{1-(-2)^{k+1}}{1-(-2)}=\frac{1-(-2)(-2)^{k}}{3}$

$\displaystyle =\frac{1}{3}+\frac{2}{3}(-2)^{k}$​

 

[SammyS]

...

In Part B, where did the -1 's come from? Originally, h[m] was just (6/5)$\sum$(-1/2)^m + (24/5)Ʃ(1/3)^m

I have a Final next week on this stuff so, any help would be appreciated.

I assume the sum in the original form of h[m] started with m=1. The -1s come from the fact that in Part B, the sums start with m=0.

BTW: You appear to still be very rusty regarding working with summations, considering that your final is next week.

Good Luck !

 

[Chandasouk]

For Part A, the index for the final answer is m = 0 though. You have it at m=1 for (-2)^m

 

[SammyS]

For Part A, the index for the final answer is m = 0 though. You have it at m=1 for (-2)^m

Yes. I left it for you to identify what is what & for you to do a little algebra.

$\displaystyle \underbrace{\sum_{m=0}^{k}(-2)^m}_{\text{from 2nd line}}=(-2)^0+\underbrace{\sum_{m=1}^{k}(-2)^m)}_{\text{from 1st line}}$