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Lets say we have

$h[n] = ((1/2)^n )(u[n])$

now if we are ask, find h[k-n], then isn't it we should just swapped every 'n' with 'k-n'. So it turns out

$h[k-n] = ((1/2)^{k-n})(u[k-n])$

But why here in letter benter image description here

h[n] there is $((1/2)^n)(u[n])$ and if we find h[m], we simply change every 'n' with m so it yiels

$h[m] = ((1/2)^m)(u[m])$

now, if we find h[n-m], it should become

$h[n-m] = ((1/2)^{n-m})(u[n-m])$

but h[n-m] according here in solution manual enter image description here is

$h[n-m] = ((1/2)^m)(u[n-m])$.

Why is that the exponent of '1/2' is only 'm', not 'n-m'?

Wolphram jonny
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vvavepacket
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  • Please use $\LaTeX$; It makes everything much more understandable as opposed to not using it! – JohnWO May 16 '13 at 00:34
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    I agree with your result, it could be a typo (not that uncommon), however you showed y[n], not h[n] in the scanned solution, are you sure you read it well? – Wolphram jonny May 16 '13 at 01:10
  • well actually y[n] consists of h[n](which i defined) and x[n], the book defines what it is. hmmm. this is from MIT http://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/assignments/ Chapter 4, convolution P4.4b – vvavepacket May 16 '13 at 01:17
  • btw @julianfernandez thanks for LATEXing it. i appreciate it – vvavepacket May 16 '13 at 01:17
  • hmmm, if its a typo, then that solution in MIT is wrong at all :( – vvavepacket May 16 '13 at 01:33
  • as far as I could see (I might be wrong) nowhere it says that $h[n-m] = ((1/2)^m)(u[n-m])$. – Wolphram jonny May 16 '13 at 02:01

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