# Fourier transform of a shifted and time-reversed sign

1. May 27, 2015

### Legend101

1. The problem statement, all variables and given/known data
given a continuous-time signal g(t) . Its fourier transform is G(f) ( see definition in picture / "i" is the imaginary number) . It is required to find the fourier transform of the shifted-time-reversed signal g(a-t) where a is a real constant .
That is , find the fourier transform of g(a-t) based on the knowledge of the fourier transform G(f) of g(t)

2. Relevant equations
The defition of the fourier transform is shown in the attached picture

3. The attempt at a solution
There are 2 properties of the fourier transform : shift property + time scaling.
But i'm not sure how to use them both . I prefer to use the definition of the fourier transform to find the relationship between the fourier transform of g(a-t) and the fourier transform of g(t)

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2. May 27, 2015

### Zondrina

So you know $\mathcal{F} \{g(t) \} = G(f)$.

Try writing this:

$$g(a - t) = g((-1) (t - a))$$

Now you can use both the time scaling and time shifting properties.

3. May 27, 2015

### Legend101

Can you check the workout i did in the picture below ?

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4. May 27, 2015

### Zondrina

The time scaling property states:

$$\mathcal{F} \{g(\alpha x) \} = \frac{1}{|\alpha|} \hat G \left( \frac{f}{\alpha} \right)$$

You know $\mathcal{F} \{g(t) \} = \hat G(f)$ and $g(a - t) = g((-1) (t - a))$.

Let $u = t - a$. Then $g((-1) (t - a)) = g((-1)u) = g(\alpha u)$ with $\alpha = -1$.

Then you know $\mathcal{F} \{g(\alpha u) \} = \frac{1}{|\alpha|} \hat G (\frac{f}{\alpha}) = \hat G (-f)$.

This amounts to saying: Since you know $\mathcal{F} \{g(t) \} = \hat G(f)$, then you can simply apply a negative sign to the argument and get $\mathcal{F} \{ g(a - t) \}$.

5. May 27, 2015

### Legend101

There has to be an exponential factor . Did you check my solution ?

6. May 27, 2015

### Zondrina

The solution is fine, I thought it would be insightful to elaborate on the time-reversal property where $\alpha = -1$.