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= <math> \sum^{0}_{k = -\infty} 2^{k}u[-n+k]</math> | = <math> \sum^{0}_{k = -\infty} 2^{k}u[-n+k]</math> | ||
| + | |||
| + | other step function changes parameters again | ||
| + | |||
| + | n = k | ||
| + | |||
| + | = <math> sum^{0}_{k = n} 2^{k} </math> | ||
| + | |||
| + | |||
| + | |||
| + | = <math> | ||
Revision as of 10:51, 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} $
=
