Monday, February 25, 2013

Back to basics: Urms

Following a question on PhysicsForum concerning the total Urms (Root Mean Square Potential Difference) due to the remaining alternating component in the half wave rectifier.

The half wave rectifier is obtained by using a single diode to let only half of the sine wave go through.


The source can be any signal, but the most frequent would be a sine signal, for instance coming out of a transformer.If we neglect the drop due to the diode, the potential difference on the load will be half of the sine wave.

Now comes a big question: what DC potential difference would be needed to get the same energy dissipated through a purely resistive load? If we call u(t) the varying, periodic signal and U the potential difference needed to generate the same energy, we get


Where T is the signal's period. Remembering that I=U/R (Ohm's law), we get that

R is constant in time, it can be taken out of the integral, and we end with


Let's go through a few examples.

Sine wave of frequency f and peak Up.

Such a signal can be described by the function


Where ω is the pulsation and is



Rectangular wave of amplitude Up and duty cycle ω


The duty cycle ω is the fraction of the period during which the signal will be at Up. Otherwise, the signal is 0.



And thus

And finally ...

Half wave rectified sine of amplitude Up and frequency f.

This is almost the same as for the full sine wave, except that instead of taking the integral from 0 to T, the integral goes only to T/2. This gives the result

Before moving to a different topic, let's establish a property we will use later.

Let's consider the signal u(t) with period T defined by


And


We will prove that


Let's go!


The part labeled A develops like this

The part labeled B is easy to develop and gives

Substituting back, we conclude that


This result will help us later.

{To be continued!}