Fourier Transform Help: Issues Solving for a & b

[Sharya19]

Hello again. Having some issues on Fourier transform. Can someone please tell me how to proceed? Need to solve this then use some software to check my answer but how to solve for a and b. Plzz help

 

[I like Serena]

Hint for (a): the integral has the form of a convolution $g(u) \star h(u)$. So we can apply the convolution theorem, evaluate $G(f)\cdot H(f)$, and take its inverse Fourier transform.
We can pick for instance $g(u)=\frac{\sin(au)}{au}$ and $h(u)=\frac{\sin(bu)}{bu}$. And look up their Fourier transforms in a table.

 

[Sharya19]

Hint for (a): the integral has the form of a convolution $g(u) \star h(u)$. So we can apply the convolution theorem, evaluate $G(f)\cdot H(f)$, and take its inverse Fourier transform.
We can pick for instance $g(u)=\frac{\sin(au)}{au}$ and $h(u)=\frac{\sin(bu)}{bu}$. And look up their Fourier transforms in a table.

ok thnks. will try

 

[Sharya19]

Hint for (a): the integral has the form of a convolution $g(u) \star h(u)$. So we can apply the convolution theorem, evaluate $G(f)\cdot H(f)$, and take its inverse Fourier transform.
We can pick for instance $g(u)=\frac{\sin(au)}{au}$ and $h(u)=\frac{\sin(bu)}{bu}$. And look up their Fourier transforms in a table.

Hello. i know its simple english u've written but am still not understanding... Sorry... Could u elaborate more? Plzzzz

 

[I like Serena]

The convolution $(g\star h)(x)$ is defined as:
$$(g\star h)(x) = \int_{-\infty}^\infty g(t)h(x-t)\,dt$$
And according to the convolution theorem we have:
$$\mathscr F_x\{(g\star h)(x)\}(f)=G(f)H(f)$$
where $G(f)$ is the Fourier transform of $g(x)$, $H(f)$ is the Fourier transform of $h(x)$, and $\mathscr F_x$ is the Fourier transform with respect to the variable $x$.

Put otherwise, we have:
$$(g\star h)(x) = \mathscr F_f^{-1}\{G(f)H(f)\}(x) \tag 1$$
where $\mathscr F_f^{-1}$ is the inverse Fourier transform with respect to the variable $f$

Now let's define $g(x)=\frac{\sin(ax)}{ax}$ and $h(x)=\frac{\sin(bx)}{bx}$.
Then:
$$\int_{-\infty} \frac{\sin(at)\sin(b(u-t))}{t(u-t)}\,dt = ab \int_{-\infty} \frac{\sin(at)}{at}\cdot\frac{\sin(b(u-t))}{b(u-t)}\,dt = ab \int_{-\infty} g(t)h(u-t)\,dt =ab\cdot (g \star h)(u)
= ab\cdot \mathscr F_f^{-1}\{G(f)H(f)\}(u) \tag 2$$

Before we go any further, can you clarify which version of the Fourier transform you have?
That's because there are many different versions around, and they result in different transformations with different variables.
You should have a definition like:
$$\mathscr F_x\{f(x)\}(f) \overset{\text{def}}= \int_{-\infty}^\infty f(x)e^{2\pi i f x}\,dx$$
Can you quote the definition that you have that is probably slightly different?

Additionally, we need the Fourier transform of $\frac{\sin{x}}{x}$ or something similar.
Do you have a table of Fourier transforms where you can look it up?
If so, what is the transform you have?

 

[Sharya19]

The convolution $(g\star h)(x)$ is defined as:
$$(g\star h)(x) = \int_{-\infty}^\infty g(t)h(x-t)\,dt$$
And according to the convolution theorem we have:
$$\mathscr F_x\{(g\star h)(x)\}(f)=G(f)H(f)$$
where $G(f)$ is the Fourier transform of $g(x)$, $H(f)$ is the Fourier transform of $h(x)$, and $\mathscr F_x$ is the Fourier transform with respect to the variable $x$.

Put otherwise, we have:
$$(g\star h)(x) = \mathscr F_f^{-1}\{G(f)H(f)\}(x) \tag 1$$
where $\mathscr F_f^{-1}$ is the inverse Fourier transform with respect to the variable $f$

Now let's define $g(x)=\frac{\sin(ax)}{ax}$ and $h(x)=\frac{\sin(bx)}{bx}$.
Then:
$$\int_{-\infty} \frac{\sin(at)\sin(b(u-t))}{t(u-t)}\,dt = ab \int_{-\infty} \frac{\sin(at)}{at}\cdot\frac{\sin(b(u-t))}{b(u-t)}\,dt = ab \int_{-\infty} g(t)h(u-t)\,dt =ab\cdot (g \star h)(u)
= ab\cdot \mathscr F_f^{-1}\{G(f)H(f)\}(u) \tag 2$$

Before we go any further, can you clarify which version of the Fourier transform you have?
That's because there are many different versions around, and they result in different transformations with different variables.
You should have a definition like:
$$\mathscr F_x\{f(x)\}(f) = \int_{-\infty}^\infty f(x)e^{2\pi i f x}\,dx$$
Can you quote the definition that you have that is probably slightly different?

Additionally, we need the Fourier transform of $\frac{\sin{x}}{x}$ or something similar.
Do you have a table of Fourier transforms where you can look it up?
If so, what is the transform you have?

Thnks a lot. as for the Fourier transform table i only have these 2 that my former lecturer left me. As for the definition, can't find any... can any definition be used as I will need to input on Matlab to see if the results corresponds?

 

[I like Serena]

You seem to be missing the actual definition, and you're also missing a table of transforms of common functions.
Anyway, we can refer to the Tables of important Fourier transforms on wiki.
Let's pick the version that transforms $f(x)$ to $\hat f(\nu)$ in those tables.
I expect that is the version that your tables refer to, although it refers to the variable $\nu$ instead of $\omega$.
If you want, you can also refer to these tables that have the same variable convention that you have.

You'll see that the second table on wiki has:
$$\operatorname{sinc}(ax) \longleftrightarrow \frac 1{|a|}\operatorname{rect}\left(\frac{\nu}{2\pi a}\right)$$
where $\operatorname{sinc}(x)\overset{\text{def}}=\frac{\sin(\pi x)}{\pi x}$, and $\operatorname{rect}(x)\overset{\text{def}}=\begin{cases}1&\text{if }-\frac 12< x <\frac 12\\0&\text{otherwise}\end{cases}$.

Can we find the Fourier transform of $\frac{\sin(bx)}{bx}$ with this information?

 

[Sharya19]

You seem to be missing the actual definition, and you're also missing a table of transforms of common functions.
Anyway, we can refer to the Tables of important Fourier transforms on wiki.
I guess we can pick the version that transforms $f(x)$ to $\hat f(\omega)$ in those tables.

You'll see that the second table on wiki has:
$$\operatorname{sinc}(ax) \longleftrightarrow \frac 1{|a|}\operatorname{rect}\frac{\omega}{2\pi a}$$
where $\operatorname{sync}(x)\overset{\text{def}}=\frac{\sin\pi x}{\pi x}$, and $\operatorname{rect}(x)\overset{\text{def}}=\begin{cases}1&\text{if }-\frac 12< x <\frac 12\\0&\text{otherwise}\end{cases}$.

Can we find the Fourier transform of $\frac{\sin(bx)}{bx}$ with this information?

for sin (bx)/bx = 1/|b| rect ω/2πb. is it correct?

 

[I like Serena]

for sin (bx)/bx = 1/|b| rect ω/2πb. is it correct?

I'm afraid not.
Note that the definition of $\operatorname{sinc}$ contains $\pi$, for which we have to compensate.
To be clear, there are also different versions of $\operatorname{sinc}$ around, so we have to be careful. But the version we have here has $\pi$ in it.
We should substitute $a=\frac{b}{\pi}$ so that we get $\operatorname{sinc}(ax)=\frac{\sin(\pi ax)}{\pi ax}=\frac{\sin(\pi \frac b\pi x)}{\pi \frac b\pi x}=\frac{\sin(bx)}{bx}$.Btw, you mentioned Matlab.
Note that Matlab has this definition of the Fourier transform.
Matlab's fourier function matches the version we picked in my previous post.

 

[Sharya19]

I'm afraid not.
Note that the definition of $\operatorname{sync}$ contains $\pi$, for which we have to compensate.
To be clear, there are also different versions of $\operatorname{sync}$ around, so we have to be careful. But the version we have here has $\pi$ in it.
We should substitute $a=\frac{b}{\pi}$ so that we get $\operatorname{sync}(ax)=\frac{\sin(\pi ax)}{\pi ax}=\frac{\sin(\pi \frac b\pi x)}{\pi \frac b\pi x}=\frac{\sin(bx)}{bx}$.Btw, you mentioned Matlab.
Note that Matlab has this definition of the Fourier transform.
Matlab's fourier function matches the version we picked in my previous post.

ok so can we pleasez proceed using the version in your previous post. Am extremely sorry for the wrong answer... am really using my brain(which i think is faulty when it comes to maths)

 

[I like Serena]

ok so can we pleasez proceed using the version in your previous post. Am extremely sorry for the wrong answer... am really using my brain(which i think is faulty when it comes to maths)

No worries. Careful substitution is a skill that is not trivial.

Anyway, substituting $a=\frac b\pi$ or $b=\pi a$, we get:
$$\frac{\sin(bx)}{bx} = \frac{\sin(\pi ax)}{\pi ax} = \operatorname{sinc}(ax) \longleftrightarrow \frac 1{|a|}\operatorname{rect}\left(\frac\omega{2\pi a}\right)
= \frac 1{|\frac b\pi|}\operatorname{rect}\left(\frac\omega{2\pi \frac b\pi}\right) = \frac\pi{|b|}\operatorname{rect}\left(\frac\omega{2 b}\right)$$

To summarize, we have:
$$\begin{cases}g(x)=\frac{\sin(ax)}{ax} \\ h(x)=\frac{\sin(bx)}{bx} \end{cases} \longleftrightarrow
\begin{cases}G(\omega)=\frac\pi{|a|}\operatorname{rect}\left(\frac\omega{2 a}\right) \\ H(\omega)=\frac\pi{|b|}\operatorname{rect}\left(\frac\omega{2 b}\right) \end{cases}\tag 4$$

Can we tell what $G(\omega)H(\omega)$ is?

And can we substitute everything in formula $(2)$ in my previous https://mathhelpboards.com/threads/fourier-transform.28946/post-127174?

 

[Sharya19]

ok so can we pleasez proceed using the version in your previous post. Am extremely sorry for the wrong answer... am really using my brain(which i think is faulty when it comes to maths)

No worries. Careful substitution is a skill that is not trivial.

Anyway, substituting $a=\frac b\pi$ or $b=\pi a$, we get:
$$\frac{\sin(bx)}{bx} = \frac{\sin(\pi ax)}{\pi ax} = \operatorname{sinc}(ax) \longleftrightarrow \frac 1{|a|}\operatorname{rect}\left(\frac\omega{2\pi a}\right)
= \frac 1{|\frac b\pi|}\operatorname{rect}\left(\frac\omega{2\pi \frac b\pi}\right) = \frac\pi{|b|}\operatorname{rect}\left(\frac\omega{2 b}\right)$$

To summarize, we have:
$$\begin{cases}g(x)=\frac{\sin(ax)}{ax} \\ h(x)=\frac{\sin(bx)}{bx} \end{cases} \longleftrightarrow
\begin{cases}G(\omega)=\frac\pi{|a|}\operatorname{rect}\left(\frac\omega{2 a}\right) \\ H(\omega)=\frac\pi{|b|}\operatorname{rect}\left(\frac\omega{2 b}\right) \end{cases}\tag 4$$

Can we tell what $G(\omega)H(\omega)$ is?

And can we substitute everything in formula $(2)$ in my previous https://mathhelpboards.com/threads/fourier-transform.28946/post-127174?

G(ω)H(ω) = [π/|a| rect (ω/2a)]*[ π/|b| rect (ω/2b)]

Formula 2 : ab⋅F−1 f {G(f)H(f)}(u) = ab. F-1 f {[π/|a| rect (ω/2a)]*[ π/|b| rect (ω/2b)]} (u) is it correct?

 

[I like Serena]

G(ω)H(ω) = [π/|a| rect (ω/2a)]*[ π/|b| rect (ω/2b)]

Formula 2 : ab⋅F−1 f {G(f)H(f)}(u) = ab. F-1 f {[π/|a| rect (ω/2a)]*[ π/|b| rect (ω/2b)]} (u) is it correct?

Yes.
Can we simplify $G(\omega)H(\omega)$ taking into account that $\operatorname{rect}(x)\overset{\text{def}}=\begin{cases}1&\text{if }-\frac 12< x <\frac 12\\0&\text{otherwise}\end{cases}$?

 

[Sharya19]

Yes.
Can we simplify $G(\omega)H(\omega)$ taking into account that $\operatorname{rect}(x)\overset{\text{def}}=\begin{cases}1&\text{if }-\frac 12< x <\frac 12\\0&\text{otherwise}\end{cases}$?

i'm sure we can but how to simplify as there's rect (x) = 1 but we have rect (w/2b), there's no x??

 

[I like Serena]

i'm sure we can but how to simplify as there's rect (x) = 1 but we have rect (w/2b), there's no x??

We need to substitute $x=\frac\omega{2b}$.
Then we find that if $-\frac 12 < \frac\omega{2b} <\frac 12$ then $\operatorname{rect}\left(\frac\omega{2b}\right)=1$, and otherwise it is $0$.
Since we know that $b$ is positive, it means that if $-b < \omega <b$ then $\operatorname{rect}\left(\frac\omega{2b}\right)=1$, and otherwise $0$.

 

[Sharya19]

We need to substitute $x=\frac\omega{2b}$.
Then we find that if $-\frac 12 < \frac\omega{2b} <\frac 12$ then $\operatorname{rect}\left(\frac\omega{2b}\right)=1$, and otherwise it is $0$.
Since we know that $b$ is positive, it means that if $-b < \omega <b$ then $\operatorname{rect}\left(\frac\omega{2b}\right)=1$, and otherwise $0$.

so G(x)H(x) simplified becomes [π/|a|]*[ π/|b|] as rect (a)=rect (b) = 1 for a and b positive

 

[I like Serena]

so G(x)H(x) simplified becomes [π/|a|]*[ π/|b|] as rect (a)=rect (b) = 1 for a and b positive

That parameter is $\omega$ instead of $x$ and the result depends on its value.
Also note that it is given that both $a$ and $b$ are positive.
So more specifically we have:
$$G(\omega)H(\omega) = \begin{cases}\frac \pi a\cdot \frac \pi b &\text{if } -a<\omega<a \land -b<\omega<b \\ 0 &\text{otherwise}\end{cases}$$

This is again a rectangle function.
So we can use the same transformation pair as before to apply the inverse Fourier transform.

 

[Sharya19]

That parameter is $\omega$ instead of $x$ and the result depends on its value.
Also note that it is given that both $a$ and $b$ are positive.
So more specifically we have:
$$G(\omega)H(\omega) = \begin{cases}\frac \pi a\cdot \frac \pi b &\text{if } -a<\omega<a \land -b<\omega<b \\ 0 &\text{otherwise}\end{cases}$$

This is again a rectangle function.
So we can use the same transformation pair as before to apply the inverse Fourier transform.

we use this formula ab⋅F−1 f {G(f)H(f)}(u) = ab. F-1 f {[π/|a|]*[ π/|b|]} (u)... How?

 

[I like Serena]

we use this formula ab⋅F−1 f {G(f)H(f)}(u) = ab. F-1 f {[π/|a|]*[ π/|b|]} (u)... How?

Let $m$ be the minimum of the positive $a$ and $b$.
Then we can write $G(\omega)H(\omega)=\frac\pi a\frac\pi b \operatorname{rect}\left(\frac\omega{2m}\right)$.

We have the transformation pair $\frac{\sin{mx}}{mx} \longleftrightarrow \frac\pi {|m|}\operatorname{rect}\left(\frac\omega{2m}\right)$.
Therefore $\mathscr F_\omega^{-1}\{G(\omega)H(\omega)\}(u) = \frac\pi a\frac\pi b\frac m\pi \frac{\sin mu}{mu}$.

So the original integral is $ab\cdot\frac\pi a\frac\pi b\frac m\pi \frac{\sin mu}{mu}=\frac{\pi\sin mu}{u}$ where $m=\min(a,b)$.

 

[Sharya19]

Let $m$ be the minimum of the positive $a$ and $b$.
Then we can write $G(\omega)H(\omega)=\frac\pi a\frac\pi b \operatorname{rect}\left(\frac\omega{2m}\right)$.

We have the transformation pair $\frac{\sin{mx}}{mx} \longleftrightarrow \frac\pi {|m|}\operatorname{rect}\left(\frac\omega{2m}\right)$.
Therefore $\mathscr F_\omega^{-1}\{G(\omega)H(\omega)\}(u) = \frac\pi a\frac\pi b\frac m\pi \frac{\sin mu}{mu}$.

So the original integral is $ab\cdot\frac\pi a\frac\pi b\frac m\pi \frac{\sin mu}{mu}=\frac{\pi\sin mu}{u}$ where $m=\min(a,b)$.

is this πsinmu/u the answer?
Are there more steps? Need to attend work(will finish at 0500am, won't be able to clarify/answer u if i have any issues concerning the next step if any. Sorry for taking up ur time on a saturday but when it comes to maths well i try i really really try but ...Plzz be online tomorrow to help me again...

 

[I like Serena]

is this πsinmu/u the answer?
Are there more steps? Need to attend work(will finish at 0500am, won't be able to clarify/answer u if i have any issues concerning the next step if any. Sorry for taking up ur time on a saturday but when it comes to maths well i try i really really try but ...Plzz be online tomorrow to help me again...

Yes. That is the answer for (a). There are no more steps.

As for (b), hint: substitute $u=0$ in the result of (a).

 

[Sharya19]

Yes. That is the answer for (a). There are no more steps.

As for (b), hint: substitute $u=0$ in the result of (a).

Thanks a lot for the help... Am having an issue on Matlab can u help?? The answer am getting is cos at - bt... can u help please?

 

[I like Serena]

Thanks a lot for the help... Am having an issue on Matlab can u help?? The answer am getting is cos at - bt... can u help please?

What did you evaluate?

 

[Sharya19]

What did you evaluate?

checking some vids on u tube
syms t v w...: Fw=(sin... The enter the f(t). Not an expert on matlab..Sorry...

 

[I like Serena]

Note that the result should be a function of $u$.

Btw, I don't have Matlab myself.
Instead I have access to the opensource Octave, which is identical, except that it doesn't have the symbolic version of the Fourier function that Matlab has...
It does look as if you can use MatLab to do the Fourier transforms that we did by hand.
That is, following the examples given, something like:

syms a b u t
assume(a > 0)
f = sin(a*t)/t
f_FT = fourier(f)
f_FT =

will likely work and should show the rectangle function we found.

Moreover, I see MatLab also has the sinc function, that has the same definition as the one we used. That is, it includes $\pi$.
So we should also be able to do:

g = sinc(a*t/pi)
g_FT = fourier(g)
g_FT =

If the previous version did not work to find a "clean" rectangle function, this one might.

 

[Sharya19]

Note that the result should be a function of $u$.

Btw, I don't have Matlab myself.
Instead I have access to the opensource Octave, which is identical, except that it doesn't have the symbolic version of the Fourier function that Matlab has...
It does look as if you can use MatLab to do the Fourier transforms that we did by hand.
That is, following the examples given, something like:

syms a b u t
assume(a > 0)
f = sin(a*t)/t
f_FT = fourier(f)
f_FT =

will likely work and should show the rectangle function we found.

Moreover, I see MatLab also has the sinc function, that has the same definition as the one we used. That is, it includes $\pi$.
So we should also be able to do:

g = sinc(a*t/pi)
g_FT = fourier(g)
g_FT =

If the previous version did not work to find a "clean" rectangle function, this one might.

ok thnks will re try it. Thanks a lot for the help. God Bless you

 

[Sharya19]

Note that the result should be a function of $u$.

Btw, I don't have Matlab myself.
Instead I have access to the opensource Octave, which is identical, except that it doesn't have the symbolic version of the Fourier function that Matlab has...
It does look as if you can use MatLab to do the Fourier transforms that we did by hand.
That is, following the examples given, something like:

syms a b u t
assume(a > 0)
f = sin(a*t)/t
f_FT = fourier(f)
f_FT =

will likely work and should show the rectangle function we found.

Moreover, I see MatLab also has the sinc function, that has the same definition as the one we used. That is, it includes $\pi$.
So we should also be able to do:

g = sinc(a*t/pi)
g_FT = fourier(g)
g_FT =

If the previous version did not work to find a "clean" rectangle function, this one might.

Thanks a lot for ur help. Wll try it again. Really Really thankful for ur help. God Bless YOU, UR PARENTS AND UR FAMILY. Thanks a lot lot lot lot lot lot