Fourier Transform (Triangular Pulse)

[roam]

Homework Statement

What is the Fourier transform of the function graphed below?

According to some textbooks the Fourier transform for this function must be:

$$ab \left( \frac{sin(\omega b/2)}{\omega b /2} \right)^2$$

Homework Equations

The Attempt at a Solution

I believe this triangular pulse is given by:

$$x(t)= \left\{\begin{matrix} a- \frac{|t|}{b} \ \ \ if |t|<b \\ 0 \ \ \ if |t|>b \end{matrix}\right.$$

So we need to find the sum of the two integrals:

$$x_1(\omega)= \int^0_{-b} \left(a+ \frac{|t|}{b} \right) e^{-j \omega t} dt$$

$$x_2(\omega)= \int^b_0 \left(a- \frac{|t|}{b} \right) e^{-j \omega t} dt$$

So using integration by parts we evaluate the two integrals:

$u=a + \frac{t}{b}$, $\frac{du}{dt}=\frac{1}{b}$, $dv=e^{-j\omega t}$, $v=\frac{j}{\omega} e^{-j\omega t}$

$$x_1 (\omega)=[\frac{j}{\omega} (a+ \frac{t}{b} e^{-j\omega t}) ]^0_{-b} - \int^0_{-b} \frac{j}{\omega b} e^{-j \omega t}$$

$\therefore x_1(\omega)= \frac{ja}{\omega}-\frac{ja}{\omega}+\frac{j}{\omega}e^{+j\omega b} + \frac{1}{\omega^2b} - \frac{1}{\omega^2 b} e^{+j \omega b}$ (1)​

Second integral:

$$x_2 (\omega)=[\frac{j}{\omega} (a- \frac{t}{b}) e^{-j \omega t} ]^b_0 - \int^b_0 (-\frac{1}{b}) . \frac{j}{\omega} e^{-j \omega t}$$

$\therefore \ x_2(\omega)= \frac{ja}{\omega} e^{-j\omega b} - \frac{j}{\omega} e^{-j \omega b} - \frac{j}{\omega} a - \frac{1}{\omega^2 b} e^{-j \omega b} + \frac{1}{\omega^2 b}$ (2)​

Combining (1) and (2):

$$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{j}{\omega} (e^{+j \omega b} - e^{-j\omega b}) - \frac{1}{\omega^2 b} (e^{+j \omega b} - e^{-j\omega b}) + \frac{ja}{\omega} e^{-j\omega b}$$

Now writing this in terms of sines and cosines using Euler's formula:

$$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{j}{\omega} (2j sin (\omega b)) - \frac{1}{\omega^2 b} (2 j sin (\omega b)) + \frac{ja}{\omega} (cos (- \omega b)+ j sin (- \omega b))$$

$$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{ja}{\omega} cos (\omega b) + \frac{a}{\omega} sin (\omega b) - \frac{2}{\omega} sin (\omega b) - \frac{2j}{\omega^2 b} sin (\omega b)$$

I'm stuck here. So how can I get from here to $ab (\frac{sin(\omega b/2)}{\omega b /2})^2$?

Where did I go wrong?

Any help would be greatly appreciated.

 

[Samy_A]

Homework Statement

What is the Fourier transform of the function graphed below?

According to some textbooks the Fourier transform for this function must be:

$$ab \left( \frac{sin(\omega b/2)}{\omega b /2} \right)^2$$

Homework Equations

The Attempt at a Solution

I believe this triangular pulse is given by:

$$x(t)= \left\{\begin{matrix} a- \frac{|t|}{b} \ \ \ if |t|<b \\ 0 \ \ \ if |t|>b \end{matrix}\right.$$

So we need to find the sum of the two integrals:

$$x_1(\omega)= \int^0_{-b} \left(a+ \frac{|t|}{b} \right) e^{-j \omega t} dt$$

$$x_2(\omega)= \int^b_0 \left(a- \frac{|t|}{b} \right) e^{-j \omega t} dt$$

So using integration by parts we evaluate the two integrals:

$u=a + \frac{t}{b}$, $\frac{du}{dt}=\frac{1}{b}$, $dv=e^{-j\omega t}$, $v=\frac{j}{\omega} e^{-j\omega t}$

$$x_1 (\omega)=[\frac{j}{\omega} (a+ \frac{t}{b} e^{-j\omega t}) ]^0_{-b} - \int^0_{-b} \frac{j}{\omega b} e^{-j \omega t}$$

$\therefore x_1(\omega)= \frac{ja}{\omega}-\frac{ja}{\omega}+\frac{j}{\omega}e^{+j\omega b} + \frac{1}{\omega^2b} - \frac{1}{\omega^2 b} e^{+j \omega b}$ (1)​

Second integral:

$$x_2 (\omega)=[\frac{j}{\omega} (a- \frac{t}{b}) e^{-j \omega t} ]^b_0 - \int^b_0 (-\frac{1}{b}) . \frac{j}{\omega} e^{-j \omega t}$$

$\therefore \ x_2(\omega)= \frac{ja}{\omega} e^{-j\omega b} - \frac{j}{\omega} e^{-j \omega b} - \frac{j}{\omega} a - \frac{1}{\omega^2 b} e^{-j \omega b} + \frac{1}{\omega^2 b}$ (2)​

Combining (1) and (2):

$$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{j}{\omega} (e^{+j \omega b} - e^{-j\omega b}) - \frac{1}{\omega^2 b} (e^{+j \omega b} - e^{-j\omega b}) + \frac{ja}{\omega} e^{-j\omega b}$$

Now writing this in terms of sines and cosines using Euler's formula:

$$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{j}{\omega} (2j sin (\omega b)) - \frac{1}{\omega^2 b} (2 j sin (\omega b)) + \frac{ja}{\omega} (cos (- \omega b)+ j sin (- \omega b))$$

$$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{ja}{\omega} cos (\omega b) + \frac{a}{\omega} sin (\omega b) - \frac{2}{\omega} sin (\omega b) - \frac{2j}{\omega^2 b} sin (\omega b)$$

I'm stuck here. So how can I get from here to $ab (\frac{sin(\omega b/2)}{\omega b /2})^2$?

Where did I go wrong?

Any help would be greatly appreciated.

Your formula for x(t) is wrong.
I didn't check the details of the rest of your calculation, but you can simplify it by noticing that the function is an even function.

 

[Ray Vickson]

Homework Statement

What is the Fourier transform of the function graphed below?

According to some textbooks the Fourier transform for this function must be:

$$ab \left( \frac{sin(\omega b/2)}{\omega b /2} \right)^2$$

Homework Equations

The Attempt at a Solution

I believe this triangular pulse is given by:

$$x(t)= \left\{\begin{matrix} a- \frac{|t|}{b} \ \ \ if |t|<b \\ 0 \ \ \ if |t|>b \end{matrix}\right.$$

*
*
*

$$x(\omega)=\frac{-ja}{\omega}+\frac{2}{\omega^2 b} + \frac{ja}{\omega} cos (\omega b) + \frac{a}{\omega} sin (\omega b) - \frac{2}{\omega} sin (\omega b) - \frac{2j}{\omega^2 b} sin (\omega b)$$

I'm stuck here. So how can I get from here to $ab (\frac{sin(\omega b/2)}{\omega b /2})^2$?

Where did I go wrong?

Any help would be greatly appreciated.

As 'Sammy_A' pointed out, your expression for $x(t)$ is incorrect, so everything after that is suspect. Again, if you heed the hint given by 'Sammy_A' you should be able to write the FT $X(\omega)$ of $x(t)$ easily in terms of the two integrals $I_1 = \int_0^b 1 \cos(\omega t) \, dt$ and $I_2 = \int_0^b t \cos(\omega t) \, dt$.[Note that I use one letter $x$ for one function and a different letter $X$ for the other. Never, never use the same letter for two different functions in the same problem!]

You may also find it useful to apply the trigonometric identity $\cos(u) = \cos^2(u/2) - \sin^2(u/2) = 1 - 2 \sin^2(u/2)$ at some point.

 

[roam]

Here is another idea:

$$x(t) =\left\{\begin{matrix}a(\frac{b+t}{b}) \ \ (-b<t<0)\\ a(\frac{b-t}{b}) \ \ (0<t<b) \\0 \ \ else\end{matrix}\right.$$

Is this correct now?

I didn't check the details of the rest of your calculation, but you can simplify it by noticing that the function is an even function.

How do I exactly use the property that this is an even function? I am not sure. But I did use the property that cos(-x)=cos(x) and sin(x)=-sin(x) in my calculations...

 

[Samy_A]

Here is another idea:

$$x(t) =\left\{\begin{matrix}a(\frac{b+t}{b}) \ \ (-b<t<0)\\ a(\frac{b-t}{b}) \ \ (0<t<b) \\0 \ \ else\end{matrix}\right.$$

Is this correct now?

Yes.

How do I exactly use the property that this is an even function? I am not sure. But I did use the property that cos(-x)=cos(x) and sin(x)=-sin(x) in my calculations...

As Ray Vickson explained.

In general, the Fourier transform is given by $\displaystyle \int_{- \infty}^{+ \infty} f(t) e^{-i \omega t} dt$
Using the Euler formula you can write this as the sum of two integrals, one with $\cos(\omega t)$, one with $\sin(\omega t)$.

If $f$ is an even function, the integral with the sine is $0$ (as $f.\sin$ is an odd function and the integration range is symmetric around t=0).
You are left with the integral with the cosine. That function ($f.\cos$), is even, so that instead of integrating from -∞ to +∞, you can take the integral from 0 to ∞ twice.

Conclusion: since your function $x$ is even, $\displaystyle X(\omega)=2\int_{0}^{b} x(t) \cos(\omega t)dt$.
(Here I used Ray's notation $X$ for the Fourier transform.)

 

[roam]

Thank you for the hint. I am still having some difficulty getting to $\displaystyle X(\omega)=ab(\sin(\omega b/2)/(\omega b /2))^2$. Here's what I did:

$$\displaystyle X(\omega)=2\int_0^{b} a \left( \frac{b-t}{b} \right) \cos(\omega t)dt =2a \int^b_0 \cos (\omega t) - \frac{t}{b} \cos (\omega t) dt$$

Using integration by parts for the second half:

$$=2a [\frac{t}{\omega b} \sin (\omega t) ]^b_0 - \frac{1}{b \omega} \int^b_0 \sin (\omega t) dt = \frac{1}{\omega} \sin (b \omega) + \frac{1}{b \omega^2} [\cos (\omega t)]^b_0$$

$$\therefore \displaystyle X(\omega)= 2a \left( \frac{1}{\omega} \sin (b \omega) + \frac{1}{b \omega^2} \cos (\omega b) - \frac{1}{b \omega^2} \right)$$

I used Ray's trig identity to further simplify this:

$$\displaystyle X(\omega)= \frac{2a}{\omega} \sin(b \omega) + \frac{2a}{b \omega^2} (1- 2 \sin^2(\frac{\omega b}{2}))- \frac{2a}{b \omega^2} + \frac{2a}{\omega} \sin(b \omega)$$

$$= -2 \sin^2(\frac{\omega b}{2})+ \frac{2a}{\omega} \sin(b \omega)$$

Did I make a mistake somewhere, or do I need to use some other identities to get to that expression for $\displaystyle X(\omega)$?

 

[Samy_A]

Thank you for the hint. I am still having some difficulty getting to $\displaystyle X(\omega)=ab(\sin(\omega b/2)/(\omega b /2))^2$. Here's what I did:

$$\displaystyle X(\omega)=2\int_0^{b} a \left( \frac{b-t}{b} \right) \cos(\omega t)dt =2a \int^b_0 \cos (\omega t) - \frac{t}{b} \cos (\omega t) dt$$

Using integration by parts for the second half:

$$=2a [\frac{t}{\omega b} \sin (\omega t) ]^b_0 - \frac{1}{b \omega} \int^b_0 \sin (\omega t) dt = \frac{1}{\omega} \sin (b \omega) + \frac{1}{b \omega^2} [\cos (\omega t)]^b_0$$

$$\therefore \displaystyle X(\omega)= 2a \left( \frac{1}{\omega} \sin (b \omega) + \frac{1}{b \omega^2} \cos (\omega b) - \frac{1}{b \omega^2} \right)$$

I used Ray's trig identity to further simplify this:

$$\displaystyle X(\omega)= \frac{2a}{\omega} \sin(b \omega) + \frac{2a}{b \omega^2} (1- 2 \sin^2(\frac{\omega b}{2}))- \frac{2a}{b \omega^2} + \frac{2a}{\omega} \sin(b \omega)$$

$$= -2 \sin^2(\frac{\omega b}{2})+ \frac{2a}{\omega} \sin(b \omega)$$

Did I make a mistake somewhere, or do I need to use some other identities to get to that expression for $\displaystyle X(\omega)$?

You seem to have lost the first integral:
$\displaystyle 2a \int^b_0 \cos (\omega t) dt$

There is also something wrong with the sign (and the $a$).

 

[roam]

You seem to have lost the first integral:
$\displaystyle 2a \int^b_0 \cos (\omega t) dt$

There is also something wrong with the sign (and the $a$).

Sorry, that was a typo. I didn't miss the first integral, in fact I evaluated both integrals separately:

$$X(\omega) = 2a \left( \underbrace{ \int^b_0 \cos(\omega t) dt}_{{\sin(\omega b)}} + \underbrace{\int^b_0 \frac{t}{b} \cos(\omega t) dt}_{{\frac{1}{\omega} \sin(\omega b) + \frac{1}{\omega^2 b} \cos(\omega b) - \frac{1}{\omega^2 b}}} \right)$$

$$\therefore \ X(\omega) =2a \left( \sin(\omega b) + \frac{1}{\omega} \sin(\omega b) + \frac{1}{\omega^2 b} \cos(\omega b) - \frac{1}{\omega^2 b} \right)$$

What is wrong with the sign?

Using Ray's trig identity the above becomes:

$$X(\omega) =2a \left( \sin(\omega b) + \frac{1}{\omega} \sin(\omega b) - \frac{2}{\omega^2 b} \sin^2(\frac{\omega b}{2}) \right)$$

So, how do I get from here to $ab \left( \frac{\sin(\omega b/2)}{\omega b/2} \right)$?

 

[Samy_A]

Sorry, that was a typo. I didn't miss the first integral, in fact I evaluated both integrals separately:

$$X(\omega) = 2a \left( \underbrace{ \int^b_0 \cos(\omega t) dt}_{{\sin(\omega b)}} + \underbrace{\int^b_0 \frac{t}{b} \cos(\omega t) dt}_{{\frac{1}{\omega} \sin(\omega b) + \frac{1}{\omega^2 b} \cos(\omega b) - \frac{1}{\omega^2 b}}} \right)$$

$$\therefore \ X(\omega) =2a \left( \sin(\omega b) + \frac{1}{\omega} \sin(\omega b) + \frac{1}{\omega^2 b} \cos(\omega b) - \frac{1}{\omega^2 b} \right)$$

What is wrong with the sign?

A number of small errors, adding up to a wrong result.
1) The second integral with the $t\cos(\omega t)$ misses a minus sign (look at the formula for $x(t)$).
2) In the result of the first integral, you miss an $\omega$ in the denominator.

Once you have fixed the errors, you will see that Ray's identity leads right to the desired result.

 

[roam]

Thank you so much, I finally got it. I really thank you for your time.