Fourier Transform: given X(f), find x(t)

[hastings]

Hi everyone!
I'm not sure if I'm posting this question in the right section. Please don't be mad at me if I'm mistaken.
Can you please help me solve this problem?

Calculate the value of the signal x(t), given its spectrum (see figure in attachment), at the time t=2/W.

Attempted solution:
I don't know how to write the module |X(f)| and phase Φ(f) of X(f). I suppose X(f) should be written in polar form:

X(f)=|X(f)|e^{j\Phi}

Maybe the module could be written like this:

|X(f)|=\frac{1}{16}f \mbox{ rect}_{4W}(f+4W) -\frac{1}{16}f \mbox{ rect}_{4W}(f-4W)

The 4W appearing in subscript of the signal rect, indicates the duration of rect.

The \pm 4W inside the round brackets, tells us that rect is centered in \pm 4W

As to the PHASE, I don't have any idea on how to write it. Does it have anything to do with arctg (...)?

 

[hastings]

Did I get the module correct?
Please help me with the phase, do I have to write a slope's formula?

 

[hastings]

Can someone try and help me? Please :(

 

[Redbelly98]

I would write the amplitude as 2 separate expressions, depending on whether f is between -6W & -2W, or between 2W and 6W:

|X(f)| = ***** , -6W < f < -2W
|X(f)| = ***** , 2W < f < 6W

And similarly for the phase angle. Note that phase is a simple linear function of f.

<br /> X(f)=|X(f)|e^{j\Phi} <br />

That's right. Once you have that expression, do the integrals that give you the inverse transform and then you'll have x(t).

 

[hastings]

That's rather generic. Can you give me another hint for the amplitude?

And similarly for the phase angle. Note that phase is a simple linear function of f.

I suppose it assumes the form of a slope: \phi(f)= m\cdot f

 

[Redbelly98]

That's rather generic. Can you give me another hint for the amplitude?

From the graph, the amplitude is a straight line within each region. You need to use the information in the graph to get the equation of the line in each region.

I.e.,

|X(f)| = (slope · f) + intercept

and you find the values of "slope" and "intercept" from the graph.

I suppose it assumes the form of a slope: \phi(f)= m\cdot f

Yes, and by looking at the numbers given in the graph, you can figure out what the slope "m" is.

 

[hastings]

From the graph, the amplitude is a straight line within each region. You need to use the information in the graph to get the equation of the line in each region.

So you're saying the Amplitude should be something like:

\frac{1}{16}f +\frac{3}{8}W \mbox{, for -6W \leq f \leq -2W}

-\frac{1}{16}f +\frac{3}{8}W \mbox{, for 6W \leq f \leq 2W} \right.

As for the PHASE, it should be:
\phi(f)=-\frac{\pi}{72W}f

Which by the way means that at f=-2W
\phi(-2W)=\frac{\pi}{36}

If I'm right about the phase, how do I express the fact that the phase slope is sort of interrupted between -2W and 2W?

 

[Redbelly98]

Your equations are all correct.

If I'm right about the phase, how do I express the fact that the phase slope is sort of interrupted between -2W and 2W?

Since the amplitude is zero in that region, the phase is not relevant.

 

[hastings]

Sorry I'm late in replying, I was little busy.

So X(f) should be:

<br /> X(f)=<br /> \left\{<br /> \begin{array}{ll}<br /> (\frac{1}{16}f +\frac{3}{8}W)e^{-j(-\frac{2\pi}{144W}f)} &amp;, \text{ for -6W $\leq$ f $\leq$ -2W} \\<br /> (-\frac{1}{16}f +\frac{3}{8}W)e^{-j(-\frac{2\pi}{144W}f)} &amp;,\text{ for 2W $\leq$ f $\leq$ 6W} <br /> \end{array}<br /> \right. <br />

I wrote the exponentials in the form e^{-j2\pi t_0 f} since an exponential in the frequency domain indicates a time shift, might be useful later on.

Ok, I guess I'm ready to antitransform the equations of the system above.
Let's see the first equation:
<br /> \mathcal{F}^{-1}\biggl[ \biggl( \frac{1}{16}f +\frac{3}{8}W \biggr) e^{j(\frac{2\pi}{144W}f)} \bigg|_{-6W}^{-2W} \quad \biggr]= <br /> <br /> \int_{-6w}^{-2W} \biggl[ \biggl(\frac{1}{16}f +\frac{3}{8}W \biggr) e^{j(\frac{2\pi}{144W}f)} \biggr] e^{j2\pi f t}df<br />

Is the integral right? Can I go on?

 

[Redbelly98]

Looks right so far. I forget how the +/- signs work in the exponential terms, but trust that you have looked up the formulas and done that part correctly.

 

[hastings]

I thought of breaking the equation in 2 pieces:
\underbrace{\mathcal{F}^{-1}[ \frac{1}{16}f e^{j2\pi \frac{1}{144W}f }]}_{\text{1st addend}}+ \underbrace{\mathcal{F}^{-1}[\frac{3}{8}We^{j2\pi\frac{1}{144W}f } ]}_{\text{2nd addend}}

I know that \mathcal{F}[\delta(t-t_0) ]=e^{j2\pi t_0 f} therefore the 2nd addend should be

\mathcal{F}^{-1}[\frac{3}{8}We^{j2\pi\frac{1}{144W}f } ] =\frac{3W}{8}\delta(t+\frac{1}{144W})
However I don't know how to evaluate this last expression for -6W \leq f \leq -2W

As to the 1st addend \mathcal{F}^{-1}[ \frac{1}{16}f e^{j2\pi \frac{1}{144W}f }] = \frac{1}{16}\int_{-6W}^{-2W} f e^{j2\pi \frac{1}{144W}f } e^{j2\pi t f} df= \frac{1}{16}\int_{-6W}^{-2W} f e^{j2\pi (\frac{1}{144W}+t) f }df

I looked up the integration by part formula:
\int f(x)g&#039;(x) dx= f(x)g(x) -\int f&#039;(x) g(x) dx

To simplify notations I called a=j2\pi(\frac{1}{144W}+t) since I'm integrating in f.

Here's the work out:
\int f \cdot e^{af}df= f \frac{e^{af}}{a} -\int \frac{e^{af}}{a}df = f \frac{e^{af}}{a} - \frac{e^{af}}{a^2}

where f(x)=f \text{, } g&#039;(x)= e^{af} \text{ and } g(x)= \frac{e^{af}}{a}

Now, introducing the extremes of integration, the solution should be:
\frac{1}{16}\bigg[ f \frac{e^{af}}{a} - \frac{e^{af}}{a^2} \bigg]_{-6W}^{-2W} = \frac{1}{16} \bigg\{ e^{-6Wa} \bigg[\frac{6W}{a}+\frac{1}{a^2}\bigg] -e^{-2Wa}\bigg[\frac{2W}{a} + \frac{1}{a^2}\bigg] \bigg\}

I'm sure there's some mistake somewhere, please correct me if I'm wrong.

 

[Redbelly98]

I thought of breaking the equation in 2 pieces:
\underbrace{\mathcal{F}^{-1}[ \frac{1}{16}f e^{j2\pi \frac{1}{144W}f }]}_{\text{1st addend}}+ \underbrace{\mathcal{F}^{-1}[\frac{3}{8}We^{j2\pi\frac{1}{144W}f } ]}_{\text{2nd addend}}

I know that \mathcal{F}[\delta(t-t_0) ]=e^{j2\pi t_0 f}

... for -∞ < f < ∞.

But that is not what we have, since your expression holds only for

-6W < f < -2W​

and is zero outside that range. So it is not a δ-function in the time domain.

Can you do the integral for the inverse transform? Your expressions are of the forms

A ∫ e^cx dx​

and

A ∫ x e^cx dx​

These are standard integrals, which you can look up here:
http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions

 

[hastings]

Regarding the 1st addend, is it correct?

Here's the work out:
<br /> \int f \cdot e^{af}df= f \frac{e^{af}}{a} -\int \frac{e^{af}}{a}df = f \frac{e^{af}}{a} - \frac{e^{af}}{a^2}<br />

\frac{1}{16}\bigg[ f \frac{e^{af}}{a} - \frac{e^{af}}{a^2} \bigg]_{-6W}^{-2W}

where

To simplify notations I called
<br /> a=j2\pi(\frac{1}{144W}+t) <br />

Meanwhile I'll work out the 2nd addend.

 

[Redbelly98]

Looks good so far.