How Is Average DC Output Calculated for a Half-Wave Rectified Sine Wave?

[zigg]

Using calculus integrals, evaluate the average dc output of a complete cycle of a half-wave rectified sine wave. Express the average value of the voltage as the area under the voltage vs wt curve, divided by the baseline.

I am completely confused. Should I be using a frequency forumula like:
1/pie (integral of) F(theta) sin [(n)(theta)] d(theta)

 

[Delphi51]

Wouldn't the voltage be V = A*sin(2πft) for the first half period, and zero for the second half? Maybe it isn't necessary to use frequency at all since the frequency should cancel out in the end, but kind of nice to show that.

 

[tomcue]

I would like to clarify that half-wave rectification is a process in which the negative half of an alternating current (AC) signal is removed, leaving only the positive half. This is typically achieved using a diode, which allows current to flow in only one direction.

To evaluate the average direct current (DC) output of a complete cycle of a half-wave rectified sine wave, we can use calculus integrals. The average value of the voltage can be expressed as the area under the voltage vs wt curve, divided by the baseline. This can be written as:

Vavg = (1/T) * ∫0T V(t) dt

Where T is the time period of one complete cycle and V(t) is the voltage at time t.

For a half-wave rectified sine wave, the voltage can be expressed as:

V(t) = V0 * sin(wt) * u(t)

Where V0 is the amplitude of the sine wave, w is the angular frequency (2πf), and u(t) is the unit step function (u(t) = 1 for t > 0, u(t) = 0 for t < 0).

Substituting this into the average voltage equation, we get:

Vavg = (1/T) * ∫0T V0 * sin(wt) * u(t) dt

We can rewrite the unit step function u(t) as a Heaviside function H(t), which is defined as:

H(t) = 0 for t < 0
H(t) = 1 for t > 0

This allows us to rewrite the integral as:

Vavg = (1/T) * ∫0T V0 * sin(wt) * H(t) dt

Now, we can integrate this using the trigonometric identity:

sin(x) * H(x) = (1/2) * [1 - cos(2x)] * H(x)

Applying this to the integral, we get:

Vavg = (1/T) * V0 * (1/2) * ∫0T [1 - cos(2wt)] dt

We can simplify this further by using the fact that the time period T = 2π/w, which means that 2wt = 2π. Therefore, we get:

Vavg = (1/T) * V0 * (1/2) * ∫0T