Frequency of output of full-wave rectifier

Question

Frequency of output voltage of full-wave rectifier is twice of frequency of its input voltage.

This is what I read in a book but don't know how to prove it.Please explain

Answer 1

Frequency is measured by how frequently the period is completed in one second. The output signal completes a period twice as fast as the input frequency, as you can see in the diagram.

This is because the input wave is symmetrical, half positive and half negative. Making all the negative components into positive ones doubles the positive components. The positive component is only half of the original period. Thus the period is halved and the frequency is doubled.

Image from Electrapk.com

Answer 2

The ideal full wave rectifier is effectively a circuit that produces, as output, the absolute value of the input.

Now, elementary algebra tells us that the absolute value is the positive square root of the square:

$$|v| = + \sqrt{v^2}$$

Suppose that the input to a full wave rectifier is a sinusoid of frequency \$ \omega = 2\pi\$:

$$v_i(t) = \cos(2 \pi t)$$

The output of the full wave rectifier is the absolute value which is the square root of the square:

$$v_o(t) = |v_i(t)| = +\sqrt{\cos^2(\omega t)} = +\sqrt{\frac{1}{2}[1 + \cos(2\omega t)]}$$

Note the presence of the sinusoid at twice the frequency of the input under the square root:

Now, for completeness sake, I should mention that, because of the square root, the output of the full-wave rectifier has frequency components (harmonics) that extend to much higher frequencies than twice the input frequency.

So, while we can say that the output has a fundamental frequency that is twice the input, the fact is that the output contains many more frequency components than the fundamental.