A signal x(t) is bandlimited to B Hz

[gabel]

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

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

 

[Dickfore]

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

[tex]
X(f) = c(f) \theta(B - |f|)
[/tex]

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

[tex]
\begin{array}{l}
\mathrm{F.T.}[a(t) b(t)](f) = \int_{-\infty}^{\infty}{a(t) \, b(t) \, e^{-2\pi j f t} \, dt} \\ \\

= \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}}} \\ \\

= \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}}} \\ \\

= \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{df_{1} \, df_{2} \, A(f_{1}) \, B(f_{2}) \, \delta(f_{1} + f_{2} - f)} \\ \\

= \int_{-\infty}^{\infty}{A(f') \, B(f - f') \, df'}
\end{array}
[/tex]

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

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

The integrand is non-zero only when:

[tex]
\left\{\begin{array}{lcl}
B_{1} - |f'| & \ge & 0 \\

B_{2} - |f - f'| & \ge & 0
\end{array}\right.
[/tex]

These conditions limit the domain of integration with respect to [itex]f'[/itex]. But, they also give some necessary conditions on the possible values of [itex]f[/itex] 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 [itex]B_{1}[/itex] and [itex]B_{2}[/itex]?

Then, use Mathematical Induction to prove the band limit of the [itex]n[/itex]-th power.