Inverse fourier troubles: e^(-j*infty)

[sam.green]

Hello,

I am working through Signals and Systems Demystified, but I am at an impasse.

I would like to take the inverse Fourier transform of

$$H(f)=\begin{cases}<br /> -j&\text{if } f > 0\\<br /> j&\text{if } f<0\end{cases}$$

So

$$h(t) = \int_{-\infty}^{0} je^{j2\pi f t}df + \int_{0}^{\infty} -je^{j2\pi ft}df <br /> = \int_{-\infty}^{0} je^{j2\pi f t}df + \int_{-\infty}^{0} je^{j2\pi f t}df <br /> = 2\int_{-\infty}^{0} je^{j2\pi f t}df <br /> = \frac{j2e^{j2\pi ft}}{j2\pi t}|_{-\infty}^{0}$$

So

$$h(t) = \frac{1}{\pi t} - \lim_{f \rightarrow -\infty} \frac{e^{j2\pi ft}}{\pi t}$$

The book says that $$h(t) = \frac{1}{\pi t}$$ is the answer, but I don't understand how. Isn't $$\lim_{f \rightarrow -\infty} \frac{e^{j2\pi ft}}{\pi t}$$ periodic and nonconverging?

 

[Dyad]

You have a constant times the sign (not sine) function: sgn(f) = 1 for f >0, and -1 for f<0.

Your problem, as you point out is convergence. Why not try multiplying your function by decaying exponentials, say exp(af) or exp(-af)?

 

[sam.green]

Hi Dyad,

Thanks for your response. Could you please give me more details? Even Mathematica hates the integral.

By the way, I am only confident that the first part is correct.

$$h(t) = \int_{-\infty}^{0} je^{j2\pi f t}df + \int_{0}^{\infty} -je^{j2\pi ft}df <br /> = j(\int_{-\infty}^{0} e^{j2\pi f t}df - \int_{0}^{\infty} e^{j2\pi ft}df)$$

-Sam

 

[Dyad]

What if you had

$$<br /> x(t) = \int_{-\infty}^{\infty} je^{j2\pi f t}e^{-b|f|}df$$

with b real and greater than zero? This is a doable calculation.

Then, after you're done, what happens when you take a certain limit involving a?

 

[sam.green]

Thanks for the help, Dyad!

$$\int_{-\infty}^{\infty}-j Sgn(f)e^{j2\pi ft}e^{-b|f|}df = \\<br /> \int_{-\infty}^{0}je^{j2\pi ft}e^{bf}df - \int_{0}^{\infty}je^{j2\pi ft}e^{bf}df$$

The $$e^{-b|f|}$$ added convergence.

I ended up with

$$j[\frac{1}{j2\pi t+b}+\frac{1}{j2\pi t-b}]$$

But as b approaches 0, we get

$$j\lim_{b \rightarrow 0}[\frac{1}{j2\pi t+b}+\frac{1}{j2\pi t-b}]=\frac{1}{\pi t}$$

 

[Dyad]

Ok, but just be careful; this doesn't always work.

Check the Fourier transform of the unit step function; this method doesn't give you the total answer. Problems arise at f = 0.

(To do this problem, might try writing the step function in terms of the sgn function.)