Vec
Guest
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 Posted:
Sat Jan 01, 2005 7:57 am Post subject:
shift in the delta function "unit impulse" |
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Hello
Could some one please explain to me why the index of the delta function
“unit impulse” is n-8 instead of n+8 in the following example;
consider the signal alfa[n], composed of all zeros except sample number
8, which has a value of –3, the delta function shifted to the right by 8
samples, and multiplied by –3 that is, n-8 is the (total number of
samples – 8) which will not be shifting by 8 to the right but shift by 8
from the end of the samples from the far right of the x axis and thus
will not be the 8th sample, sample 8 should be n+8
alfa[n]=-3 delta[n-8], I thought it should be alfa[n]=-3 delta[n+8],
what am I missing?
Thank |
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Tim Wescott
Guest
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| Posted:
Sat Jan 01, 2005 7:57 am Post subject:
Re: shift in the delta function "unit impulse" |
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Vec wrote:
| Quote: | Hello
Could some one please explain to me why the index of the delta function
“unit impulse” is n-8 instead of n+8 in the following example;
consider the signal alfa[n], composed of all zeros except sample number
8, which has a value of –3, the delta function shifted to the right by 8
samples, and multiplied by –3 that is, n-8 is the (total number of
samples – 8) which will not be shifting by 8 to the right but shift by 8
from the end of the samples from the far right of the x axis and thus
will not be the 8th sample, sample 8 should be n+8
alfa[n]=-3 delta[n-8], I thought it should be alfa[n]=-3 delta[n+8],
what am I missing?
Thank
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The unit impulse is 1 when it's index is 0. If you want a unit impulse
to occur at n = 8 then the index to the unit impulse function must be 0
at that point. If n = 8, then what value does m need to take on for n +
m = 0 to be true?
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com |
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Fred Marshall
Guest
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| Posted:
Sat Jan 01, 2005 7:57 am Post subject:
Re: shift in the delta function "unit impulse" |
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"Vec" <vec2u@valoudari.edu.mx> wrote in message
news:41d5da47$1_1@news.iprimus.com.au...
| Quote: | Hello
Could some one please explain to me why the index of the delta function
“unit impulse” is n-8 instead of n+8 in the following example;
consider the signal alfa[n], composed of all zeros except sample number 8,
which has a value of –3, the delta function shifted to the right by 8
samples, and multiplied by –3 that is, n-8 is the (total number of
samples – 8) which will not be shifting by 8 to the right but shift by 8
from the end of the samples from the far right of the x axis and thus will
not be the 8th sample, sample 8 should be n+8
alfa[n]=-3 delta[n-8], I thought it should be alfa[n]=-3 delta[n+8], what
am I missing?
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Well, first I have to try to write out your descriptions as follows:
"the signal" alpha[n] for n= (what) to (what)???
alpha[x]=0 for x <> 8
alpha[x]=-3 for x=8
(So far, I don't see a delta function)
But, to continue....
I do see a *unit sample* multiplied by -3 for n=x=8
And you say "shifted to the right"
So, if we define the unit sample function as
u[n]=0 n<>0
u[0]=1
And we want this to be shifted to the right (which is generally accepted to
be a delay - it happens later in time).
Then if u[0]=1 by definition it appears we want:
-3*u[t]=1 where t=8; So -3*u[t-8] is what we need for there to be a unit
sample at t=8.
This only shifts the sample from t=zero to t=+8 by subtracting 8 from t.
Now, of course one can select any indexing convention that one likes. So I
wonder if you aren't getting a bit confused about indexing when you say:
"which will not be shifting by 8 to the right but shift by 8
from the end of the samples from the far right of the x axis". Because I
don't get what your meaning is by saying this.... "the far right....."?????
Fred |
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Airy R. Bean
Guest
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| Posted:
Sat Jan 01, 2005 7:03 pm Post subject:
Re: shift in the delta function "unit impulse" |
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I think that the question that you are asking is along the
lines of , "Why is a signal that is delayed by T written
down as f(t -T) and not f(t + T)?".
I will attempt to answer the question above, in the hope
that it is the one that you are asking.
In order to answer it, I'll use a different example, that of t^2.
(t squared)
t^2 has the following shape...... (0,0), (1,1), (2,4), (3,9), (4,16), (5,25)
etc
What you are seeking is for the shape to be delayed by 8, and so to
look like (ignoring the values before 8).....(8,0), (9,1), (10,4), (11,9),
(12,16), (13,25)
How does the function t^2 get to have a value of 0?
Answer - if the value of t is 0 (which is 8 - 8).
How does the function t^2 get to have a value of 1?
Answer - if the value of t is 1. (which is 9 - 8)
How does the function t^2 get to have a value of 4?
Answer - if the value of t is 2. (which is 10 - 8)
How does the function t^2 get to have a value of 9?
Answer - if the value of t is 3. (which is 11 - 8)
How does the function t^2 get to have a value of 16?
Answer - if the value of t is 4. (which is 12 - 8)
How does the function t^2 get to have a value of 25?
Answer - if the value of t is 5. (which is 13 - 8)
So, in order to get the value of the function we expect,
at each point on the curve, we have to _SUBTRACT_
the delay time to get f(t - T).
Or, to put it another way, we want the value at t=8 et seq
to have what it would have been if t = 0 et seq if it had not
been delayed.
By the same arguments, a signal that is _ADVANCED_ in time
by T is written down as f(t + T).
I hope the above is of help to you. It may interest you to hear
that I had the same confusion. It was never explained to me by
my textbooks nor by my lecturers - I had to work it
out for myself. I suggest that you draw a few curves
to evaluate the idea; which is what I did.
*Multitudinous apologies* if the above is not the answer to
the question that you asked.
Happy New Arbitrary-Point-In-The-Earth's-Orbit-Around-The-Sun!!!!!
"Vec" <vec2u@valoudari.edu.mx> wrote in message
news:41d5da47$1_1@news.iprimus.com.au...
| Quote: | Could some one please explain to me why the index of the delta function
“unit impulse” is n-8 instead of n+8 in the following example;
consider the signal alfa[n], composed of all zeros except sample number
8, which has a value of –3, the delta function shifted to the right by 8
samples, and multiplied by –3 that is, n-8 is the (total number of
samples – 8) which will not be shifting by 8 to the right but shift by 8
from the end of the samples from the far right of the x axis and thus
will not be the 8th sample, sample 8 should be n+8
alfa[n]=-3 delta[n-8], I thought it should be alfa[n]=-3 delta[n+8],
what am I missing? |
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