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

• jaus tail
In summary, when given a function f(N) and asked to find f(2N-1), first find f(2N) by compressing the graph of f(N) by a factor of 2 towards the vertical axis. Then, shift the graph of f(2N) to the right by 1/2 unit to get f(2N-1). This results in a new function with a starting value of 3.
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)...

jaus tail said:

## 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.
jaus tail said:
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.
jaus tail said:
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)...

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.

## 1. What is the meaning of "f(N) as digital signal" in this context?

In this context, "f(N) as digital signal" refers to a function that takes a discrete set of values (usually represented as binary digits) and maps them to another set of values.

## 2. Can you provide an example of a function that would satisfy f(N) as digital signal?

One example could be a digital filter that takes in a set of audio samples and outputs a filtered version of the audio.

## 3. What does finding f(2N - 1) mean and why is it important?

Finding f(2N - 1) means applying the given function to a new input value, which is twice the original input (N) minus one. This is important because it allows us to understand how the function behaves when the input is scaled or shifted.

## 4. How can I determine the value of f(2N - 1) from a given f(N) as digital signal?

To determine the value of f(2N - 1), you would need to know the specific function or algorithm that is being applied to the digital signal. From there, you can plug in the new input value and evaluate the function to get the desired output.

## 5. Are there any limitations to finding f(2N - 1) from a given f(N) as digital signal?

Depending on the complexity of the function and the size of the input, there may be computational limitations to finding f(2N - 1). Additionally, if the function is not well-defined or does not have a clear inverse, it may be difficult to accurately determine the value of f(2N - 1) from a given f(N) as digital signal.

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