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I'm not sure if I've done this correctly in the first place. I would like your input before I actually attempt to inverse it. | I'm not sure if I've done this correctly in the first place. I would like your input before I actually attempt to inverse it. | ||
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| − | * | + | *http://img.photobucket.com/albums/v89/weirdly_cool/Untitled-2.jpg |
| − | + | ||
| − | + | *Thank you.'''I tried the LaTeX but it failed miserably. ''' | |
| − | + | *Maybe it would be easier if you wrote it out while you were trying to help me, otherwise I can come in and see you on Monday. | |
| − | Answer from Prof. MImi | + | x[n] = a^n(u[n-2]+u[n]) |
| − | *In the step where you replaced -n by k, you forgot to replace the n inside the summation. Also, the first sum should then go from -2 to -infinity, instead of infinity. | + | |
| − | *Actually, I do not see why you replaced -n by k in both sums. IN the first sum, you should have set k=n-2. In the second sum, you did not need to make any change of variable. | + | *X(z) |
| − | *The arrow in the middle of your computations, and the one towards the end should both be replaced by equal signs. | + | *= sum[n=-inf,inf] of x[n]z^-n |
| + | |||
| + | *= sum[n=-inf,fin] of a^n(u[n-2]+u[n]) | ||
| + | *= sum[n=2,inf] of a^n(z^-n) + sum[n=0,inf] of a^n(z^-n) | ||
| + | *Let k = -n | ||
| + | *Therefore, sum[k=-2,inf] of (a/z)^n + sum[k=0,inf] of (a/z)^n | ||
| + | *= sum[k=0,inf] of [(a/z)^n + 2] + sum[k=0,inf] of (a/z)^n | ||
| + | *= (1/1-a/z)+2 + (1/1-a/z) | ||
| + | *= (z/z-a)+2 + (z/z-a) | ||
| + | *= (z/z-a)+(2(z-a)/z-a) + (z/z-a) | ||
| + | *= (4z-2a)/(z-a) | ||
| + | *= (4-2a/z)/(1-a/z) for |z|>a | ||
| + | **EDIT: so if I end up with something that says 1/1-(1/z), I am confused. does it converge when |z|<a or when |z|>a?? | ||
| + | *:~ksoong | ||
| + | |||
| + | Answer from Prof. MImi | ||
| + | |||
| + | *In the step where you replaced -n by k, you forgot to replace the n inside the summation. Also, the first sum should then go from -2 to -infinity, instead of infinity. | ||
| + | *Actually, I do not see why you replaced -n by k in both sums. IN the first sum, you should have set k=n-2. In the second sum, you did not need to make any change of variable. | ||
| + | *The arrow in the middle of your computations, and the one towards the end should both be replaced by equal signs. | ||
*The simplification of the first summation following the arrow is incorrect: you would need to add two terms instead of just one. | *The simplification of the first summation following the arrow is incorrect: you would need to add two terms instead of just one. | ||
| − | *The equality following the arrow is only valid when |z| | + | *The equality following the arrow is only valid when |z|>|a|.You must write this next to the equality! |
*This explanation would be much clearer if you had typed in your answer: this way I could make notes directly inside the computations and cross-out and replace stuff using different colors. | *This explanation would be much clearer if you had typed in your answer: this way I could make notes directly inside the computations and cross-out and replace stuff using different colors. | ||
| + | |||
---- | ---- | ||
| − | Anybody sees anything else? Do you have more questions? | + | |
| + | Anybody sees anything else? Do you have more questions? | ||
| + | |||
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Revision as of 22:03, 11 September 2010
I'm not sure if I've done this correctly in the first place. I would like your input before I actually attempt to inverse it.
- Thank you.I tried the LaTeX but it failed miserably.
- Maybe it would be easier if you wrote it out while you were trying to help me, otherwise I can come in and see you on Monday.
x[n] = a^n(u[n-2]+u[n])
- X(z)
- = sum[n=-inf,inf] of x[n]z^-n
- = sum[n=-inf,fin] of a^n(u[n-2]+u[n])
- = sum[n=2,inf] of a^n(z^-n) + sum[n=0,inf] of a^n(z^-n)
- Let k = -n
- Therefore, sum[k=-2,inf] of (a/z)^n + sum[k=0,inf] of (a/z)^n
- = sum[k=0,inf] of [(a/z)^n + 2] + sum[k=0,inf] of (a/z)^n
- = (1/1-a/z)+2 + (1/1-a/z)
- = (z/z-a)+2 + (z/z-a)
- = (z/z-a)+(2(z-a)/z-a) + (z/z-a)
- = (4z-2a)/(z-a)
- = (4-2a/z)/(1-a/z) for |z|>a
- EDIT: so if I end up with something that says 1/1-(1/z), I am confused. does it converge when |z|<a or when |z|>a??
- ~ksoong
Answer from Prof. MImi
- In the step where you replaced -n by k, you forgot to replace the n inside the summation. Also, the first sum should then go from -2 to -infinity, instead of infinity.
- Actually, I do not see why you replaced -n by k in both sums. IN the first sum, you should have set k=n-2. In the second sum, you did not need to make any change of variable.
- The arrow in the middle of your computations, and the one towards the end should both be replaced by equal signs.
- The simplification of the first summation following the arrow is incorrect: you would need to add two terms instead of just one.
- The equality following the arrow is only valid when |z|>|a|.You must write this next to the equality!
- This explanation would be much clearer if you had typed in your answer: this way I could make notes directly inside the computations and cross-out and replace stuff using different colors.
Anybody sees anything else? Do you have more questions?
