Problem with substitution in a difference equation

[jwxie]

Homework Statement

Verify that the response of the system governed by the 1st order different equation
$y(k)=bu(k)+ay(k-1)$

is given by this solution
$y(k)=\frac{b}{1+a}\left [ (-1)^{k}+a^{k+1} \right ]$
for u(k) = (-1)^k

The Attempt at a Solution

The solution said we can verify this by substitution. I didn't get how we jump to the second step

(sub)
$\frac{b}{1+a}\left [ (-1)^{k}+a^{k+1} \right ]-\frac{ab}{1+a}\left [ (-1)^{k}+a^{k} \right ]$

2nd step
$\frac{b(-1)^{k}[1+a]}{(1+a)}+\frac{ba^{k}(a-a)}{1+a}$

How did we get the top (1+a) in the first term, and the (a-a) in the second term?
Thank you.

 

[JThompson]

There has to be an error.
Was the first step supposed to be $y(k)-ay(k-1)$? This would either have been
$$\frac{b}{1+a}[(-1)^k+a^{k+1}] + \frac{ab}{1+a}[(-1)^{k}+a^{k}]$$

or

$$\frac{b}{1+a}[(-1)^k+a^{k+1}] - \frac{ab}{1+a}[(-1)^{k-1}+a^{k}]$$.

(Notice the power of the second (-1) term.)

 

[jwxie]

Hi Thompson, yes the first step is indeed to rewrite the difference equation in the standard form, which is y(k) - ay(k-1)
Thanks.

 

[chris99191]

Unsure if this proof is correct

I completed this proof in the attachment but can someone please check my working

The aim was to show that the height of P above floor after the crate is tilted is h(cosb+2sinb)

From the rectangles you can work out
Length of OP=h√5
sin(a)=1/√5
cos(a)=2/√5

therefore
height=h√5.sin(a+b)
=h√5.(sinacosb+cosasinb)
=h√5.(1/√5cosb+2/√5sinb)
=h(cosb+2sinb)

 

[HallsofIvy]

Yes, that's perfectly correct!

But why did you open a new thread for this?
(I'm merging the two threads.)

 

[jwxie]

Hmmm.. excuse me? What is this proof all about?
LOOOL
how is this relevant to my difference equation problem?

 

[jwxie]

anyone? still haven't figure it out yet. Thanks, and sorry for the bump.

 

[JThompson]

Darn it! The first expression I posted had an error! It should be

$$\frac{b}{1+a}[(-1)^k+a^{k+1}] + \frac{ab}{1+a}[(-1)^{k}-a^{k}]$$

Sorry about that. I'm new to doing math via computer, and usually make silly errors.


These two threads were apparently merged by mistake- okay, so let's take the first form of $y(k)-a(yk-1)$ that I posted
$$\frac{b}{1+a}[(-1)^k+a^{k+1}] + \frac{ab}{1+a}[(-1)^{k}-a^{k}]$$

multiply out the terms in the numerators..

$$\frac{b(-1)^k +ba^{k+1}}{1+a} + \frac{ab(-1)^{k}-ba^{k+1}}{1+a}$$

then simply group the terms appropriately. In the reduced form you posted (the second step) they got

$\frac{b(-1)^{k}[1+a]}{(1+a)}+\frac{ba^{k}(a-a)}{1+a}$

from grouping terms immediately instead of expanding the numerators, but you can do it either way.

 

[JThompson]

It's also important for you to see where

$$\frac{b}{1+a}[(-1)^k+a^{k+1}] + \frac{ab}{1+a}[(-1)^{k}-a^{k}]$$

comes from.

$$y(k)-ay(k-1)=\frac{b}{1+a}[(-1)^k+a^{k+1}] - \frac{ab}{1+a}[(-1)^{k-1}+a^{k}]$$

$$= \frac{b}{1+a}[(-1)^k+a^{k+1}] - \frac{ab}{1+a}[\frac{(-1)^{k}}{(-1)}-(-a^{k})]$$

$$= \frac{b}{1+a}[(-1)^k+a^{k+1}] -(-1)\frac{ab}{1+a}[(-1)^{k}+(-a^{k})]$$

Does that make sense?

 

[jwxie]

hey JThompson, thank you very much. Your explanations are very clear. Thanks!