Fourier transform of a shifted and time-reversed sign

[Legend101]

Homework Statement

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)

Homework Equations

The defition of the Fourier transform is shown in the attached picture

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)[/B]

 

[STEMucator]

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.

 

[Legend101]

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.

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

 

[STEMucator]

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 $\mathcal{F} \{ g(a - t) \}$.

 

[Legend101]

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 $\mathcal{F} \{ g(a - t) \}$.

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

 

[STEMucator]

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

shifted-time-reversed signal