Here we demonstrate that a continuum of waves is required to produce an isolated triangula...

Here we demonstrate that a continuum of waves is required to produce an isolated triangular peak.

Depicted in the accompanying diagram, the function is described by

and its Fourier transform is

We will apply equation (4-21) in steps to sec how well the initial f(x) can be reproduced by a sum. Because A(k) is an even function of k, if we write eitx in equation (4-21) as cos(kx) + i sin (kx), the imaginary part of the integral disappears, for the integrand is an odd function of k. The remaining integrand is an even function, so equation (4-21) becomes:

(a). First, by direct use of equation (4-22), show that the given Fourier transform is indeed correct.

(b). Now, defining H and L to be 1 for convenience, evaluate A (k) at values of k =n(2π/L), where n is I, and 2. Then plot the sum of 2A(k)cos(kx) over these three k values from x = —4 to. v = +4. (You may have to displace your function along the ft-axis a bit, perhaps to 10-6. A(k) is finite there, but your computer may not think so.)

(c). Repeat part (b), but let n run through all integers from 0 to 16.

(d)Repeat part (c), but let n run through all half integers from 0 to 16 (i.e., 0, 0.5, 1, 1.5,...).

(e). Repeat pail (d), but for all quarter integers from O to 16.

(f)Finally, repeat part (e) for sixteenths of an integer,

(g). Your plots may be about right, but something is missing. Multiply the total functions you plotted in parts (e) and (0 by 2π/4 and 2π/16, respectively. How does this fix things, and why?

(h). How do your results show that to produce an iso­lated pulse, a continuum of wave numbers is required.