A signal x(t) is bandlimited to B Hz

[gabel]

A singal x(t) is bandlimited to B Hz. Show that that the signal $$x^{n}(t)$$ is bandlimited to nB Hz.

I have no idee on how to address this problem. Can get some help?

 

[Dickfore]

A band-limited signal has a Fourier component containing a heaviside step function:

$$X(f) = c(f) \theta(B - |f|)$$

The Fourier transform of a product of two functions is their convolution:

$$\begin{array}{l}<br /> \mathrm{F.T.}[a(t) b(t)](f) = \int_{-\infty}^{\infty}{a(t) \, b(t) \, e^{-2\pi j f t} \, dt} \\ \\<br /> <br /> = \int_{-\infty}^{\infty}{dt \, e^{-2\pi j f t} \, \int_{-\infty}^{\infty}{A(f_{1}) \, e^{2\pi j f_{1} t} \, df_{1}} \, \int_{-\infty}^{\infty}{B(f_{2}) \, e^{2\pi j f_{2} t} \, df_{2}}} \\ \\<br /> <br /> = \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{df_{1} \, df_{2} \, A(f_{1}) \, B(f_{2}) \, \int_{-\infty}^{\infty}{dt \, e^{2\pi j (f_{1} + f_{2} - f) t} \, dt}}} \\ \\<br /> <br /> = \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{df_{1} \, df_{2} \, A(f_{1}) \, B(f_{2}) \, \delta(f_{1} + f_{2} - f)} \\ \\<br /> <br /> = \int_{-\infty}^{\infty}{A(f') \, B(f - f') \, df'}<br /> \end{array}$$

Take the signals $A(f)$ and $B(f)$ to be band-limited with band limits $B_{1}$ and $B_{2}$, respectively. Then, their convolution is:

$$A \ast B(f) = \int_{-\infty}^{\infty}{A(f') \, \theta(B_{1} - |f'|) \, B(f - f') \, \theta(B_{2} - |f - f'|) \, df'}$$

The integrand is non-zero only when:

$$\left\{\begin{array}{lcl}<br /> B_{1} - |f'| & \ge & 0 \\<br /> <br /> B_{2} - |f - f'| & \ge & 0<br /> \end{array}\right.$$

These conditions limit the domain of integration with respect to $f'$. But, they also give some necessary conditions on the possible values of $f$ when the above conditions are not contradictory. This makes the convolution also band limited. What is the band limit on the convolution in terms of $B_{1}$ and $B_{2}$?

Then, use Mathematical Induction to prove the band limit of the $n$-th power.