Fourier transform of rect function (1 Viewer)

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

1. The problem statement, all variables and given/known data
From the definition of the Fourier transform, find the Fourier transform of rect(t-5).


2. Relevant equations
[tex]G(w) = \int^{\infty}_{-\infty}g(t)e^{jwt}dt[/tex]


3. The attempt at a solution

So, I sketched the function which has area 1 and centre at 5, with its lower bound @ 4.5 and upper at 5.5. SOOOO cannot I not just write


[tex]G(w) = \int^{5.5}_{4.5}1e^{jwt}dt[/tex]

for it's Fourier transform? Is that allowed. I know the rect function some how turns into sinc, but in this case how?

I can prove with limits -a and +a but when you have 2 positive or 2 negative limits you don't get the sin(-ax) = -sin(ax) which keeps the sine and gets rid of the cosine!
 
Last edited:

lanedance

Homework Helper
3,305
2
why not try shifting the integal by using the transform s = t-5
 
I got the answer in the end by a clever little exponent trick. However the very next question on my question sheet is "Using the time shift property, compute again the Fourier transform of rect(t-5) and compare the 2 results"

The results being the one I compute for this one and the previous one (which I got sinc(w/2)e^(-jw5))

So basically I'm shifting the rect back around 0, so from -0.5 to 0.5. I believe the shift theorem says that a shift in the time domain translate to a shift in the phase... so a shift of 5 means I have to multiply the exponent value by 5.

Therefore I'm integrating (wrt t) between -0.5 and 0.5 e^(-5jwt)

which gives [tex]\frac{1}{-5jw}[e^{-5jwt}]^{0.5}_{-0.5}[/tex]

Using euler to expand and sticking in the limits I get sinc(5/2 jwt)

Does that sound correct? I don't see how the 2 answers compare. Have I done it wrong? Can someone shed some light?

Thanks
Thomas
 
Ok so I have to multiply by e^-j2pi(a) where a is the distance of the rect function (centre) to the origin? I was using omega. What's going on?

Euuughhhh what does w (omega) equal again? Isn't it 2pi*f? I don't have an f? Maybe I'm just confusing myself...


After an hours break....

I think i have it:
Can I confirm that all you do is perform the fourier transform THEN multiply by e^-jwa

and w = 2pi/t (which is 2pi f)

Right?
 

vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,281
1,029
After an hours break....

I think i have it:
Can I confirm that all you do is perform the fourier transform THEN multiply by e^-jwa
Yes, that's right. Note you get this result simply by making the substitution t' = t-a and then integrating.
and w = 2pi/t (which is 2pi f)

Right?
Not quite or you're being sloppy. [itex]\omega=2\pi f[/itex] is correct, but [itex]\omega=2\pi/t[/itex] isn't (t is your integration variable).
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top