Fourier transform of a shifted and time-reversed sign

  • #1
15
0

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]
 

Attachments

Answers and Replies

  • #2
STEMucator
Homework Helper
2,075
140
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
15
0
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 ?
 

Attachments

  • #4
STEMucator
Homework Helper
2,075
140
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
15
0
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 ?
 
  • #6
STEMucator
Homework Helper
2,075
140
The solution is fine, I thought it would be insightful to elaborate on the time-reversal property where ##\alpha = -1##.

shifted-time-reversed signal
 

Related Threads on Fourier transform of a shifted and time-reversed sign

  • Last Post
Replies
4
Views
9K
  • Last Post
Replies
6
Views
225
Replies
6
Views
235
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
9
Views
3K
Replies
7
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
7
Views
2K
Replies
8
Views
2K
Top