# RMS of Fullwave rectified sine wave.

• pooface
In summary, the RMS value of a fullwave rectified sine wave can be determined by using the equation RMS = √(1/(b-a) ∫^(b)_(a) (f(x))^2 dx), where a and b are the limits of integration and f(x) is the function representing the wave. For a fullwave rectified sine wave with a period of pi, the RMS value is 1/√2.
pooface

## Homework Statement

Determine the RMS value of fullwave rectified sine wave.

## Homework Equations

RMS = $$\sqrt{({1}/{b-a})\int^{b}_{a}[(fx)]^{2}dx}$$

## The Attempt at a Solution

Notes: The Period of a full wave rectified sine wave is pi.

a=0
b=pi

Let's do square root at the end.

=$$1/pi \int^{pi}_{0}sin^{2}xdx$$

=1/pi [pi/2 - [sin(2pi)]/4] - 1/pi [pi/2 - [sin(2pi)]/4]

=1/2 - 1/2 ? ? ? ? ? ?

Where am i going wrong? Sorry I am not good with latex code, even with the reference.

Last edited:
So you used the double angle identity.

$$sin^2(x) =\frac{1}{2}(1-cos(2x))$$

$$\int_0^{\pi} \frac{1}{2}(1-cos(2x)) = \frac{\pi}{2} - \left[sin(2x)/4 \right]^{x=\pi}_{x=0}$$

$$\int_0^{\pi} \frac{1}{2}(1-cos(2x)) = \frac{\pi}{2}-(0)$$

I'm not sure what you were doing.

I used integral of sin^2(u) du is = u/2 - [sin(2u)]/4 + C

When i sub pi in the term sin2u , then this becomes sin2pi which is 0.

Right so you have

$$\sqrt{\frac{1}{\pi}\frac{\pi}{2}}$$

RMS = 1/√2

## 1. What is the RMS of fullwave rectified sine wave?

The RMS (root mean square) of a fullwave rectified sine wave is the square root of the average of the squared values of the waveform. In simpler terms, it represents the effective value of the waveform and is equal to approximately 0.707 times the peak value of the waveform.

## 2. How is the RMS of fullwave rectified sine wave calculated?

The RMS of a fullwave rectified sine wave can be calculated by taking the square root of the sum of the squares of the positive and negative half-cycles of the waveform, divided by the number of cycles. This can also be expressed as the square root of the area under the waveform, divided by the length of the waveform.

## 3. What is the significance of calculating the RMS of fullwave rectified sine wave?

Calculating the RMS of a fullwave rectified sine wave is important because it gives an accurate representation of the power or energy being delivered by the waveform. It takes into account both the positive and negative values of the waveform, providing a more comprehensive understanding of its characteristics.

## 4. What factors can affect the RMS of fullwave rectified sine wave?

The RMS of a fullwave rectified sine wave can be affected by the amplitude of the waveform, as well as the frequency and duty cycle. The presence of harmonics in the waveform can also impact the RMS value.

## 5. How is the RMS of fullwave rectified sine wave used in practical applications?

The RMS of a fullwave rectified sine wave is commonly used in electrical engineering and power systems to calculate the effective voltage and current of alternating current (AC) power. It is also used in the design and testing of electronic devices to ensure they can handle the peak power delivered by the waveform.

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