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

1. Dec 1, 2008

### 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

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

So

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

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?

2. Dec 2, 2008

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)?

3. Dec 8, 2008

### sam.green

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.

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

-Sam

4. Dec 8, 2008

What if you had

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

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?

5. Dec 9, 2008

### sam.green

Thanks for the help, Dyad!

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

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

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

6. Dec 9, 2008