Expected power of a square wave

In summary, the expected power of the first three frequency components of a square wave with an amplitude of 0.2V can be found using the equation \frac{1}{T}\int_{0}^{T}|x(t)|^2\,dt, or by using Parseval's Theorem. The latter method requires finding the power contained in the first three frequency components of the square wave.
  • #1
Tjvelcro
18
0

Homework Statement



What is the expected power of the first three frequency components of a square wave with an amplitude of 0.2V?

Homework Equations



None that I know of :(

The Attempt at a Solution



I'm not sure where to start with this since it seems to give me so little information. I think I may have to use V(rms) but I'm not sure. Any help or hints would be appeciated.

Tjvelcro
 
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  • #2
Do you know equation of power for periodic waves?
 
  • #3
No, I am not sure what that equation is... tried to look in my book but cannot find it.
 
  • #4
[tex]
\frac{1}{T}\int_{0}^{T}|x(t)|^2\,dt
[/tex]

Above is the power equation.

However, it seems like you are attempting to finding power contained in first three frequency components of square wave. I only read the title thought you want power of square wave. I would suggest you to look into .. Parseval's Theorem (http://engnet.anu.edu.au/DEcourses/engn2228/Lectures/Lecture15.pdf slide 11).
 
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  • #5
,

I can provide some guidance on how to approach this problem. The first step would be to define the power of a signal. In this case, we are dealing with an electrical signal, so we can use the equation P = V^2/R, where P is power, V is voltage, and R is resistance.

Next, we can look at the formula for a square wave, which is a periodic waveform with a constant amplitude. The square wave can be represented by a series of sine waves with different frequencies. The first three frequency components of a square wave are its fundamental frequency, twice the fundamental frequency, and three times the fundamental frequency.

To calculate the power of each frequency component, we can use the formula P = V^2/R, where V is the amplitude of the square wave (0.2V) and R is the resistance of the circuit. Since we do not have information about the resistance, we can assume a standard value of 1 ohm for simplicity.

So, the expected power of the first three frequency components would be:

- Fundamental frequency: P = (0.2V)^2/1 ohm = 0.04W
- Twice the fundamental frequency: P = (0.2V)^2/1 ohm = 0.04W
- Three times the fundamental frequency: P = (0.2V)^2/1 ohm = 0.04W

Therefore, the total expected power of the first three frequency components of a square wave with an amplitude of 0.2V would be 0.12W.

I hope this helps! Remember, when dealing with signals, it is important to define the parameters and equations that are relevant to the problem. Good luck with your homework!
 

Related to Expected power of a square wave

1. What is the expected power of a square wave?

The expected power of a square wave is equal to the square of its amplitude divided by its period. This can also be expressed as the square of its peak voltage divided by its period.

2. How is the expected power of a square wave calculated?

The expected power of a square wave can be calculated by taking the square of its amplitude and dividing it by its period. This can also be expressed as the square of its peak voltage divided by its period.

3. What is the relationship between the expected power of a square wave and its voltage?

The expected power of a square wave is directly proportional to its voltage. This means that as the voltage of the square wave increases, the expected power also increases.

4. Does the duty cycle affect the expected power of a square wave?

No, the duty cycle does not affect the expected power of a square wave. The expected power is only determined by the amplitude and period of the square wave, not its duty cycle.

5. How does the expected power of a square wave compare to other types of waveforms?

The expected power of a square wave is generally higher than other types of waveforms, such as sine or triangle waves, with the same amplitude and period. This is due to the sharp edges of the square wave, which contain higher frequency components and therefore contribute more to the overall power.

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