# Fourier Transform Properties

Tags:
1. Mar 20, 2016

### roam

1. The problem statement, all variables and given/known data

A certain function $v(x)$ has Fourier transform $V(\nu)$. The plot of the function is shown in the figure attached below.

For each of these functions give their Fourier transform in terms of $V(\nu)$. And also state if the FT is Hermitian/anti-Hermitian, even/odd, imaginary/real.

(a) $v(2x)$

(b) $v(x)-v(-x)$

(c) $v(x+2)+v(x-2)$

2. Relevant equations

Properties of FT.

3. The attempt at a solution

(a)
Using the scaling property of the Fourier transform:

$$FT \Big[ v(2x) \Big] = \frac{1}{|2|} V(\frac{\nu}{2})$$

Now to determine whether this is Hermitian/anti-Hermitian, even/odd, imaginary/real, I have made a table that summarizes the properties (e.g. if a function f if is real F is Hermitian, etc.):

The plot for v(x) looks odd so V(ν) has to be odd. But since we don't have an explicit expression for $v(x)$, how can we determine if $v(2x)$ is even/odd? Likewise how do we know if his function is real?

(b) Using the symmetry property we find:

$$FT \Big[ v(x)-v(-x) \Big] = V(\nu) - V(-\nu)$$

I think there is a property such that that $F(-\nu)=F^*(\nu)$ (* is the complex conjugate). I am not sure if this is applicable in this case, but if so we have:

$$V(\nu) - V^*(\nu) \implies 2 j \ Im[V(\nu)]$$.

The last step comes from the fact that for any complex number $z$ we have $z-z^* = 2j \ Im(z)$. I'm not sure how to further write down the FT for this function.

(c) I think maybe we can use the "Interference" property of the FT:

$$f(t-t_0) + f(t+t_0) \longleftrightarrow 2 \ cos (2 \pi \nu t_0) F(\nu)$$

So we find the equation $2 \ cos (4 \pi \nu) V(\nu).$ Is that okay, or do I need to further develop this equation?

Just as in part (a) and (b), I am not sure how to determine if the function is even/odd, real/imaginary. Any help would be greatly appreciated.

#### Attached Files:

• ###### 2016-03-20_220633.jpg
File size:
3.1 KB
Views:
55
2. Mar 20, 2016

### blue_leaf77

No, it does not. How come you arrive at this conclusion? I think, even if the wording of the question asks you to decided between any a pair of options, there is a chance that the answer does not belong to either options. In this case, one can just say that it's neither odd nor even, or neither real nor imaginary.
Yes, there is such a property, and you are right in using it for this part of the problem. But you have to know why you can use it. Hint: look for the definition of a Hermitian/anti-Hermitian functions.

Last edited: Mar 20, 2016
3. Mar 21, 2016

### roam

My mistake, I think it is neither even nor odd. It is not even as it is not symmetric about the y-axis, and not odd since it doesn't appear to satisfy the definition of odd function ($f(x) = - f(x)$). So, this function's FT is also neither even nor odd?

But (a), (b), and (c) are not $v(x)$, so they could be even/odd. If I want to sketch these three functions where do I start?

Because I believe we need sketches in order to determine if they are even/odd. But the difficulty is that we don't have an explicit expression for the function so I would not be able to plot them using Matlab or Python.

From definition $V(\nu)$ is Hermitian means that $V(\nu) = V^* (\nu)$, and it is anti-Hermitian if $V(\nu)=-V(-\nu)$.

So why are we justified to use this property? (that was my question)

We don't know what $V(\nu)$ or $V^*(\nu)$ are, so I am not sure how to check if the property holds. But if we knew what these two are, we could also determine whether the function is real and even (i.e. $V(\nu) = V^* (\nu)$).

4. Mar 21, 2016

### blue_leaf77

In particular for a), do you think a scaling can alter the parity of a function?
Start from the sketch of $v(x)$ you already have.
Sketches and the table you provided in post #1 is enough.
No, those are not the definitions of Hermitian and anti-Hermitian functions.
What do you think about the reality of $v(x)$? Is it complex, purely real, or purely imaginary?
One more time I will say this, the table you have in your first post and the sketch of the functions in (a), (b), and (c) are all what you need.

5. Mar 24, 2016

### roam

Here are the sketches I made for the different cases:

Hopefully these are correct.

So $v(2x)$ and $v(x+2)+v(x-2)$ are neither even nor odd. But $v(x)-v(-x)$ is odd, and $v(x)+v(-x)$ is even.

No, it can't.

I don't understand this part. We don't have an expression for $v(x)$. So how can we say if it is real or imaginary?

Thank you for the link. The link states that $f$ is Hermitian iff Re(f) is an even function. So basically we need to determine if the functions we sketched in each case are real or imaginary? Therefore we are justified in using $F(- \nu) = F^*(\nu)$ iff the function is Hermitian?

6. Mar 25, 2016

### blue_leaf77

The answer is implied in the fact that you are provided a plot of $v(x)$. Should you be given a plot of a complex function, it must display two components of the function to be able to describe it completely: the real and imaginary parts or the magnitude and phase. But in this problem, you are only given a single plot and you are not told whether it's purely real or purely imaginary. What do you think, given that condition?
Yes.
Yes.

7. Mar 28, 2016

### roam

Thank you very much for the reply. I think we can assume that $v(x)$ and therefore all three functions in question are purely real (since we are provided only with a single plot). So from the table in my post #1, the Fourier transform for all three functions would be Hermitian. Is that correct?

8. Mar 28, 2016

### blue_leaf77

Yes.
Yes, all of the transforms are Hermitian.

9. Mar 28, 2016

### roam

Thank you very much for your help.