Given f(N) as digital signal, find f(2N -1)

[jaus tail]

Homework Statement

f(N) is {3, 4, 5, 6, 7}
It starts at 5 meaning f(o) is 5.
i have to find f{2N-1}

2. Homework Equations [/B]
f(N) is given, so first find f(2N) and then delay by 1.

The Attempt at a Solution

f(N) is {3, 4, 5, 6, 7}
finding f(2N)

f(0) is 5.
f(2 * 0) is 5

f(-1) is 4.
f(2 * -1) is f(-2) is 3.

f(-2) is 3.
f(2*-2) is f(-4) is 0

f(1) is 6
f(2*1) is f(2) is 7.

f(2) is 7
f(2*2) is f(4) is 0

so f(2N) is {0, 3, 5, 7, 0}
Starting value at 5.
So now f(2N-1) is shifting the origin to left by 1 we get f(2N-1) is {3, 5, 7}
starting at 3.

But in book they have done:
f(n) is {3, 4, 5, 6, 7}
starting at 5
then f(n-1) is {3, 4, 5, 6, 7
starting at 4
then f(2n-1) is {4, 6}
how can they multiply 2 only to n and not to (-1).
like shouldn't it be:
f(n)---->f(2n)---->f(2n-2)...

 

[Mark44]

Homework Statement

f(N) is {3, 4, 5, 6, 7}
It starts at 5 meaning f(o) is 5.[/B]

This should be f(0) is 5. o is the letter o, not zero.

i have to find f{2N-1}

2. Homework Equations
f(N) is given, so first find f(2N) and then delay by 1.

The Attempt at a Solution

f(N) is {3, 4, 5, 6, 7}
finding f(2N)

f(0) is 5.
f(2 * 0) is 5

f(-1) is 4.
f(2 * -1) is f(-2) is 3.

f(-2) is 3.
f(2*-2) is f(-4) is 0

f(1) is 6
f(2*1) is f(2) is 7.

f(2) is 7
f(2*2) is f(4) is 0

so f(2N) is {0, 3, 5, 7, 0}
Starting value at 5.
So now f(2N-1) is shifting the origin to left by 1 we get f(2N-1) is {3, 5, 7}

No.
If you know the graph of y = f(n), then the graph of y = f(2n) represents a compression of the graph of y = f(n) toward the vertical axis by a factor of 2.
Then, f(2n - 1) = f(2(n - 1/2)) represents a shift of the graph of y = f(2n) to the right by 1/2 unit.

Original function:
N -2 -1 0 1 2
f(N) 3 4 5 6 7

y = f(2N)
N -1 -1/2 0 1/2 1
f(2N) 3 4 5 6 7

For the graph of y = f(2N - 1) = f(2(N - 1/2)), all the points in the table above are shifted to the right by 1/2 unit.

starting at 3.

But in book they have done:
f(n) is {3, 4, 5, 6, 7}
starting at 5
then f(n-1) is {3, 4, 5, 6, 7
starting at 4
then f(2n-1) is {4, 6}
how can they multiply 2 only to n and not to (-1).
like shouldn't it be:
f(n)---->f(2n)---->f(2n-2)...

 

[jaus tail]

Yes you are right. The f(2n-1) is actually f(2(n-1/2)). I missed that step as it wasn't in the book.
Thanks.