Fourier Representation of Simple Half-Wave Rectifier

[cj]

I, in fact, know the correct Fourier representation
for the following (it was given to me):

$$f(t)=0 \text { if } -\pi \leq \omega t \leq 0$$

and

$$f(t)=sin(\omega t) \text { if } 0 \leq \omega t \leq \pi$$

$$\hrule$$

I'm curious about the derivation that led to it -- specifically how the coefficients were derived.

I know, in general...

$$A_0=\frac{1}{2\pi} \int_{-\pi}^{\pi}f(x)dx$$

$$A_N=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)cos(nx)dx$$

$$B_N=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)sin(nx)dx$$

... but am stuck when it comes to setting-up the
integrals (substitution rules, how integrals might be broken-up into sub-integrals, intervals, etc.)

Comments?

 

[uart]

You understand that multipication by that f(x) is equivalent to simply changing the limits of integration from [-pi, pi] to [0, pi] right.

For the integral of sin(x) sin(nx) and sin(x) cos(nx) just use integration by parts (twice) using sin(x) = d/dx(-cos x) etc, this gets you the original integral on both LHS and RHS which makes it easy to solve.

For example with the sine series integral this gets you something like I = n^2 I, (where I is the integral you're trying to find). This of course tells you that either n=1 or I = 0 and indeed the sine series coefficient are all zero except for the first one (n=1). The exact same procedure will work for the cosine coeficients and this time give a non trivial series.

 

[tacman]

The Fourier representation of a simple half-wave rectifier is a useful tool for understanding the behavior of this circuit. The derivation of the coefficients involves using the properties of trigonometric functions and integrating over the appropriate intervals.

To start, we can rewrite the given function as:

f(t) = \begin{cases} 0, & -\pi \leq \omega t \leq 0 \\ sin(\omega t), & 0 \leq \omega t \leq \pi \end{cases}

We can then use the trigonometric identities:

sin(\omega t) = \frac{1}{2i}(e^{i\omega t} - e^{-i\omega t})

cos(\omega t) = \frac{1}{2}(e^{i\omega t} + e^{-i\omega t})

to rewrite the function as:

f(t) = \frac{1}{2i}(e^{i\omega t} - e^{-i\omega t}), \text{ if } -\pi \leq \omega t \leq 0

and

f(t) = \frac{1}{2}(e^{i\omega t} + e^{-i\omega t}), \text{ if } 0 \leq \omega t \leq \pi

Now, using the given formulas for the coefficients:

A_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) dt

A_N = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) cos(Nt) dt

B_N = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) sin(Nt) dt

We can substitute in the rewritten expressions for f(t) and use the properties of integrals to break up the intervals. For example, we can break up the integral for A_N into two parts, one from -\pi to 0 and one from 0 to \pi:

A_N = \frac{1}{\pi} \left( \int_{-\pi}^{0} \frac{1}{2i}(e^{i\omega t} - e^{-i\omega t}) cos(Nt) dt + \int_{0}^{\pi} \frac{1}{2i}(e^{