Hello, 
     I am being bugged by this problem for many days. 

     Problem: Is the following system represented by the difference
equation time invariant?
        y(n) = a*y(n-1) + b*x(n)
x(n) is the input, y(n) is the output. a and b are constants. y(-1) =
0

    I proceeded like this
          Solve the difference equation: 
            y(n) = b*x(n) + a*b*x(n-1) +  ... + a^(n-1)*b*x(1) + a^n
*b*x(0)
where a^r, means a raised to r. a^3, for example, means a cubed. 
    Let XI(n) be the input and YI(n) be the corresponding output. 
    Let XI(n) = x(n-n0),n0 is the delay
     YI(n) = b*XI(n) + a*b*XI(n-1) + ... + a^(n-1)*b*XI(1) + a^n
*b*XI(0)
           = b*x(n-n0) + a*b*x(n-n0-1) + ... + a^(n-1)*b*x(1-n0) + a^n
*b*x(-n0)

Calculate y(n-n0). Replace n by n-n0 in y(n)
 y(n-n0) = b*x(n-n0) + a*b*x(n-n0-1)+ ... + a^(n-n0-1)*b*x(1) +
a^(n-n0)*b*x(0)

y(n-n0) is not equal to YI(n-n0). So the system is not time-invariant.
But some websites mentioned it as time invariant. Is my solution
correct? If not, where is the mistake in my solution?
Thanks 

P.S: This was NOT given as an assignment problem

In article 7f6c0a9a.0308041923.473336b7@posting.google.com, Jay at
innocent_802000@yahoo.com wrote on 08/04/2003 23:23:

> Hello, 
> I am being bugged by this problem for many days.
> 
> Problem: Is the following system represented by the difference
> equation time invariant?
> y(n) = a*y(n-1) + b*x(n)
> x(n) is the input, y(n) is the output. a and b are constants. y(-1) =
> 0
> 
> I proceeded like this
> Solve the difference equation:
> y(n) = b*x(n) + a*b*x(n-1) +  ... + a^(n-1)*b*x(1) + a^n
> *b*x(0)
> where a^r, means a raised to r. a^3, for example, means a cubed.
> Let XI(n) be the input and YI(n) be the corresponding output.
> Let XI(n) = x(n-n0),n0 is the delay
> YI(n) = b*XI(n) + a*b*XI(n-1) + ... + a^(n-1)*b*XI(1) + a^n
> *b*XI(0)
> = b*x(n-n0) + a*b*x(n-n0-1) + ... + a^(n-1)*b*x(1-n0) + a^n
> *b*x(-n0)
> 
> Calculate y(n-n0). Replace n by n-n0 in y(n)
> y(n-n0) = b*x(n-n0) + a*b*x(n-n0-1)+ ... + a^(n-n0-1)*b*x(1) +
> a^(n-n0)*b*x(0)
> 
> y(n-n0) is not equal to YI(n-n0).

check your math.  they are equal.  keep in mind that your impulse response
is infinitely long so, to compare apples to apples, you have to make the
same assumptions about x[n] being zero before n=0 in one case, and n=n0 in
the other case.

> So the system is not time-invariant.

it is time-invariant if a and b are constant.

> But some websites mentioned it as time invariant. Is my solution
> correct? If not, where is the mistake in my solution?
> Thanks 
> 
> P.S: This was NOT given as an assignment problem

doesn't matter.

r b-j

robert bristow-johnson wrote:
> 
> In article 7f6c0a9a.0308041923.473336b7@posting.google.com, Jay at
> innocent_802000@yahoo.com wrote on 08/04/2003 23:23:
> 
> > P.S: This was NOT given as an assignment problem
> 
> doesn't matter.
> 
> r b-j

It doesn't matter because you've shown your work and you've asked to
have your mistake pointed out.  comp.dsper's generally don't mind
HELPING with homework - they just don't like to be asked to DO someone
else's homework.

-- 
Jim Thomas            Principal Applications Engineer  Bittware, Inc
jthomas@bittware.com  http://www.bittware.com          (703) 779-7770
Sometimes experience is the only teacher that works - Mike Rosing

robert bristow-johnson wrote:
> 
> In article 7f6c0a9a.0308041923.473336b7@posting.google.com, Jay at
> innocent_802000@yahoo.com wrote on 08/04/2003 23:23:
> 
  ...
> >
> > P.S: This was NOT given as an assignment problem
> 
> doesn't matter.
> 
> r b-j

I speak for myself, and apparently for R.B-j. too. We don't mind helping
with homework at all. We object to being asked to do it when it seems
that the request is only to save work for the requester. "Why doesn't
this work?", "Why does this work?",  and "Please tell me how to get
started" are fine. 

As for Jay's question: The filter is time invariant, but not memoryless.
Those are orthogonal properties.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

robert bristow-johnson wrote:
> 
> In article 7f6c0a9a.0308041923.473336b7@posting.google.com, Jay at
> innocent_802000@yahoo.com wrote on 08/04/2003 23:23:
> 
  ...
> >
> > P.S: This was NOT given as an assignment problem
> 
> doesn't matter.
> 
> r b-j

I speak for myself, and apparently for R.B-J. too. We don't mind helping
with homework at all. We object to being asked to do it when it seems
that the request is only to save work for the requester. "Why doesn't
this work?", "Why does this work?",  and "Please tell me how to get
started" are fine. 

As for Jay's question: The filter is time invariant, but not memoryless.
Those are orthogonal properties.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������