| (8 intermediate revisions by the same user not shown) | |||
| Line 27: | Line 27: | ||
| − | = <math> | + | = <math> \sum^{-n}_{0} \frac{1}{2}^{k} </math> since n <= 0; </math> |
| + | |||
| + | This is then a property that can be exchanged to get rid of the summation signs using a general solution. | ||
| + | |||
| + | { = <math> \frac{1-\frac{1}{2}^{-n+1}}{1-\frac{1}{2}} </math> For all k < n and n <= 0 | ||
| + | |||
| + | |||
| + | { = 0 , n > 0 | ||
| + | |||
| + | The transform of this is then= | ||
| + | <math> (2-2^{n})u[-n]</math> | ||
Latest revision as of 09:57, 12 October 2008
Test Correction of # 3
An LTI system has unit impulse response $ h[n] = u[-n] $
Compute the system's response to the input $ x[n] = 2^{n}u[-n] $
(Simplify answer until all summation signs disappear.)
h[n] = u[-n]
x[n] = $ 2^{n} $u[-n]
y[n] = x[n] * h[n]
= $ \sum^{\infty}_{k = -\infty} 2^{k}u[-k]u[-n--k] $
since -k > 0 and k < 0 the summation parameters change
= $ \sum^{0}_{k = -\infty} 2^{k}u[-n+k] $
other step function changes parameters again
n = k
= $ \sum^{0}_{k = n} 2^{k} $
= $ \sum^{-n}_{0} \frac{1}{2}^{k} $ since n <= 0; </math>
This is then a property that can be exchanged to get rid of the summation signs using a general solution.
{ = $ \frac{1-\frac{1}{2}^{-n+1}}{1-\frac{1}{2}} $ For all k < n and n <= 0
{ = 0 , n > 0
The transform of this is then= $ (2-2^{n})u[-n] $
