Question

Determine transform of the waveforms ?(see attachment)

Answer 1

If the graph shown in the first exercise of the attachment consists of parabolas and line segments, then its equation over [0,1] is y(t) = 8(t - t^2). If f(t) is defined over the interval
[0, 2], then f(t) = y(t) for t in [0,1] and 0 for t in [1, 2]. Integrating f(t) over [0,2] sets a(0) = 4/3, integrating
f(t) cos(n*pi*t) over [0, 2] sets
a(n) = 16[1 - (-1)^n] / (n*pi)^3, and integrating
f(t) sin(n*pi*t) over[0, 2] sets b(n) = -8[1 - (-1)^n]/(n*pi)^2.

Answer 2

Restricting n to even integers simplifies a(2n) to 4/(n*pi)^3 and b(2n) to -4/(n*pi)^2. All odd-numbered terms vanish, and the Fourier series expansion is done. The next step is to take the Laplace transforms of the terms individually, beginning with a(0). Because it is a constant, its transform is 4/(3s). For n > 0,

a(2n) cos(2*n*pi*t) transforms to
[4/((2*n*pi^3)] s/[s^2 + (2*n*pi)^2]
= [1/(2n^3*pi^3}s/[s^2 + (2*n*pi)^2], and

b(2n) sin(2*n*pi*t) transforms to
-[4/(2*n*pi)^2]*(2*n*pi)/[s^2 + (2*n*pi)^2]
= -[2/(n*pi)]/[s^2 + (2*n*pi)^2]

The final step is to sum up all the transformed terms.
to get 4/(3s)
+ sum(n, 1, infty, [1/(2n^3*pi^3)]s/[s^2 + (2*n*pi)^2]
- sum(n, 1, infty, [2/(n pi)]*{1/[s^2 + (2*n*pi)^2].
The answer here is of a different form from that of any of the four choices shown in the attachment. The point I'm making here is that different methods can, and often do, give different-looking forms; nevertheless they are mathematically equivalent.

If my answer is erroneous, then I made a wrong assumption about the graph or I introduced errors into my calculations.

Answer 3

I now try the problem using the Laplace transform of a periodic function.

Answer 4

Thank you for your time !!!!

Answer 5

You are right the Laplace transform should be used.

Answer 6

If f(t) is a periodic function, then its Laplace transform
is integral (exp(-s*t)f(t), t, 0, T)/(1 - e^(s*T)), where T is the period of f(t). The function for the first exercise in the attachment has a period T = 2. I will assume that the graph represents a half-wave rectification of 2*sin(pi*t).

Answer 7

Over the interval [0, 2], the function integrates to 4/pi., so its Laplace transform is (4/pi)/[1 - e^(-2s)].

Does not look like any of the four choices shown in the attachment.

Answer 8

In order to attempt bringing it to the desired form, we can express 1/[1 - e^(-2s)] as a geometric series.
1/(1 - r) = sum(r^n, n, infty). In this case, r = e^(-2s).
The nth term, then, is e^(-2n*s). The half-wave rectiffication of 2 sin(pi*t) is(4/pi) transforms to
(4/pi) sum(e^(-2*n*s), n, 0, infty).

I forgot to multiply the function by e^(-s*t) before integrating
over [0, 2].

Answer 9

The integral of e^(-s*t)f(t) over [0, 2] is
2*pi(1 + e^(-s))/(s^2 + pi^2), so f(t) transfotms to
2*pi(1 + e^(-s))/((s^2 + pi^2)(1 - e^(-2s)). The geometric series expansion of 1/(1 - e^(-2s) is
sum(e^(-2*n*s), n , 0, infty).

Answer 10

The Laplace transform of the half-wave rectification of 2sin(pi*t) is
[2*pi(1 + e^(-s))/(s^2 + pi^2)}sum(e^(-2*n*s), n, 0, infty].

It's getting closer to the desired form. Looks like manipulating the expression the right way will get it exactly right.

Answer 11

Thank you,for your time !!!!!!