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I have the following equation:


I tried to apply difference equation on it:

$$y_{n+2}-5y_{n-1}-6y_n = x_{n+1}-x_n$$

I need to find equation expressed as $y_n$

I have the answer (At the end of the chapter) to be: $$y_n = 5y_{n-1}-6y_{n-2}+x_{n-1}-x_{n-2}$$

Can somone please tell me how did in this case the second equation expressd as $y_n$ ? I cant figure it out!

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I see no equation, it's a fraction. – Gigili Jun 26 '12 at 10:49
I don't at all see how the fraction relates to the difference equation, either. You may need to explain what is going on, and what you are doing, a bit more. – Ben Millwood Jun 26 '12 at 12:34
Sorry my mathematical language is not good, I was not sure if this an equation or fraction. This is z transform of a signal and I want to calculate first 5 samples if $x_0=1$ and $x_1$ to $x_4$ are $0$ – Sean87 Jun 26 '12 at 14:51
up vote 1 down vote accepted

The differences between your equation and the book's are:

the index has been shifted by 2

the sign on the 6 has been corrected

two terms have been moved to the other side

To see the index shift, define $m=n+2$ and substitute it in.

Also, the $2$ in the denominator has become $5$ in both your solution and the book's. Did you copy it correctly?

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