Fourier transform of rectangular pulse

[nao113]

Homework Statement

Calculate the Fourier transform of rectangular pulse given below. (Height, A; width, 2a)
. I tried to calculate that but I am not sure whether it s correct or not. Do I need to put A there or just omit it so it will just be e^-jwt? thank you

Relevant Equations

available in the question

Here is the question:

Here is my answer

 

[Delta2]

Wolfram says your final result is correct, hold on while I check your in between steps.

 

[nao113]

Wolfram says your final result is correct, hold on while I check your in between steps.

one thing that I am confused, where can I use 2a as width here? I guess width represents time pr frequency, right?

 

[Delta2]

Yes the in between steps seem correct to me.

I don't understand your confusion.
2a is the width of the pulse in the time domain, in the frequency domain it seems to have infinite width,

but

this is typical happening with Fourier transform, a narrow function in time domain has a wide Fourier transform in the frequency domain, and a wide function in the frequency domain has a narrow inverse Fourier transform in the time domain.

 

[nao113]

Yes the in between steps seem correct to me.

I don't understand your confusion.
2a is the width of the pulse in the time domain, in the frequency domain it seems to have infinite width,

but

this is typical happening with Fourier transform, a narrow function in time domain has a wide Fourier transform in the frequency domain, and a wide function in the frequency domain has a narrow inverse Fourier transform in the time domain.

I mean where can I put 2a in my calculation? How can I applied in that math formula? or we don t need to put 2a in that calculation? In my answer, I only used A as height

 

[Delta2]

You have 2a in your final result for F(w) but that doesn't mean that it is the width of F(w) in the frequency domain.
What math formula do you mean? The starting integral of Fourier transform?

 

[DrClaude]

I mean where can I put 2a in my calculation? How can I applied in that math formula? or we don t need to put 2a in that calculation? In my answer, I only used A as height

The width appears through the bounds of the integration. As you can see, your final result depends of the width of the pulse.

 

[Delta2]

Yes well it depends what equivalent expression for F(w) you want to make. If you want to express it in terms of the $sinc x$ function then you must put the 2a there.

 

[nao113]

You have 2a in your final result for F(w) but that doesn't mean that it is the width of F(w) in the frequency domain.
What math formula do you mean? The starting integral of Fourier transform?

yes, that s not 2a for width because I multiplied them with 2 in order to get 2j. Yes, integral of Fourier transform

 

[nao113]

The width appears through the bounds of the integration. As you can see, your final result depends of the width of the pulse.

can you explain to me where the width appears? So rather than using a and -a, I need to use 2a and -2a?

 

[nao113]

Yes well it depends what equivalent expression for F(w) you want to make. If you want to express it in terms of the $sinc x$ function then you must put the 2a there.

where should I put 2a? Did you mean like the picture below?

 

[Delta2]

Well, theoretically speaking you can put 2a inside your calculation steps, in an infinite number of ways, for example you can multiply by the term $\frac{2a\cos(2a)\sin(2a)}{2a\cos(2a)\sin(2a)}$ which is one and nothing changes, but the point is to put it when it has some utility.

Wolfram gives the answer as $$F(w)=2A\frac{\sin(aw)}{w}$$ but if we want to express this in terms of the $sinc(aw)=\frac{\sin(aw)}{aw}$ then we have to multiply and divide by $a$, pretty much like you did.

$SincX$ function is like a basis function for Fourier transforms, many Fourier transforms can be expressed in terms of the sinc function.

 

[Delta2]

where should I put 2a? Did you mean like the picture below?

No you can't put it there like this because you change the function of Fourier transform , you will compute the Fourier transform of the function $g(t)=2aA$ ,$t\in [-a,a]$, but the problem asks for the Fourier transform of $f(t)=A$...

 

[nao113]

Well, theoretically speaking you can put 2a inside your calculation steps, in an infinite number of ways, for example you can multiply by the term $\frac{2a\cos(2a)\sin(2a)}{2a\cos(2a)\sin(2a)}$ which is one and nothing changes, but the point is to put it when it has some utility.

Wolfram gives the answer as $$F(w)=2A\frac{\sin(aw)}{w}$$ but if we want to express this in terms of the $sinc(aw)=\frac{\sin(aw)}{aw}$ then we have to multiply and divide by $a$, pretty much like you did.

$SincX$ function is like a basis function for Fourier transforms, many Fourier transforms can be expressed in terms of the sinc function.

basically, I don't need to multiply a in my calculation based on Wolfram, right? what does a refer to?

 

[Delta2]

basically, I don't need to multiply a in my calculation based on Wolfram, right? what does a refer to?

a is the half width of the pulse, and it is also in the boundaries of the integral of the Fourier transform.

 

[nao113]

a is the half width of the pulse, and it is also in the boundaries of the integral of the Fourier transform.

can I just remove a in my answer? so my answer will be exactly like Wolfram

 

[Delta2]

yes you can but should NOT remove it from inside the $\sin(aw)$.

 

[nao113]

yes you can but should NOT remove it from inside the $\sin(aw)$.

yes, I got it..

No you can't put it there like this because you change the function of Fourier transform , you will compute the Fourier transform of the function $g(t)=2aA$ ,$t\in [-a,a]$, but the problem asks for the Fourier transform of $f(t)=A$...

Thank you, so I don t need to put 2a there. then, why the question provide me with the information of 2a if I will not use them in my answer above?

 

[Delta2]

Well, in the final result $a$ appears inside the $\sin(aw)$ so the final result depends on a, though not exactly on 2a.

 

[nao113]

Well, in the final result $a$ appears inside the $\sin(aw)$ so the final result depends on a, though not exactly on 2a.

Yes, that s what I mean. I also got another question,
What happen if the width 2a is changed?
My answer:
If the width of the rectangular pulse increases then the main lobe becomes narrower. Moreover, the pulse becomes flatter and the magnitude spectrum loops will be thinner and taller. In other words, the zeros (the crossings of the magnitude spectrum with the axis) become closer to the origin.

Is it acceptable?

 

[Delta2]

If the width of the rectangular pulse increases then the main lobe becomes narrower. Moreover, the pulse becomes flatter and the magnitude spectrum loops will be thinner and taller. In other words, the zeros (the crossings of the magnitude spectrum with the axis) become closer to the origin.

Yes this statement looks correct to me, we can infer this statement by examining the behavior of the function $\frac{\sin(aw)}{w}$ as $a$ increases, $a$ is something like the angular frequency of the sin function.

 

[nao113]

Yes this statement looks correct to me, we can infer this statement by examining the behavior of the function $\frac{\sin(aw)}{w}$ as $a$ increases, $a$ is something like the angular frequency of the sin function.

I got it, thank you very much. Then, if I am asked to draw the wave of F(w), can I use Matlab? I tried them. Here they are. Do you think I got it right? I am still not sure how to convert the calculation result in Matlab.

 

[Delta2]

The first screenshot might be the graph of $\frac{\sin(aw)}{w}$ for some $a$, but I don't know about the second screenshot...Calculations for impedance of an RLC circuit? what those have to do here?