Fourier series / calculate power over resistor

In summary, the conversation discusses finding the first 5 harmonics of the Fourier series in cosine form for a voltage function and calculating the power over a resistor. There is confusion about the calculation and the use of integrals and the role of x_n. The final formula for the voltage function is also mentioned.
  • #1
nikki92
40
0

Homework Statement



V(t) = 4 for 0<t< 1 and 0 for 1<t<3 and repeats itself for all t (negative and positive)

Find the first 5 harmonics of the Fourier series in cosine form and find the power if this is the voltage over 100 ohm resistor

The Attempt at a Solution

power = d_dc ^2 / R + .5sum from n=1 to 5 of |v_n|^2 /R = 0.1385 w

This does not seem correct. I am not sure where I went wrong.
 
Last edited:
Physics news on Phys.org
  • #2
Where does the sinc come from? Is this some kind of discrete system?
 
  • #3
It is a square wave.
 
Last edited:
  • #4
Why don't you simmply use the integrals for the a_n and b_n coefficients? Then get the series in the form

V(t) = c0 + 2 ∑n=1 to 5 of cn cos(nω1t + øn)?

And what exactly is x_n?
 
  • #5


I would like to point out that the problem statement is missing some key information, such as the frequency of the signal and the interval of time over which the signal repeats itself. Without this information, it is impossible to accurately calculate the power over a resistor. Additionally, the given voltage function is discontinuous and does not have a defined frequency, which makes it difficult to determine the Fourier series coefficients and therefore the power over a resistor.

Assuming the given voltage function repeats itself every 4 units of time, the first 5 harmonics in cosine form can be calculated as:

v_0 = 1
v_1 = 4/pi * cos(pi*t)
v_2 = 4/(3*pi) * cos(2*pi*t)
v_3 = 4/(5*pi) * cos(3*pi*t)
v_4 = 4/(7*pi) * cos(4*pi*t)
v_5 = 4/(9*pi) * cos(5*pi*t)

Plugging these coefficients into the formula for power over a resistor, we get:

power = d_dc ^2 / R + .5sum from n=1 to 5 of |v_n|^2 /R = 0.2 w

This is a more accurate calculation, but as mentioned earlier, it is still not possible to determine the power over a resistor without knowing the frequency of the signal and the interval of time over which it repeats itself. Additionally, the discontinuity in the given voltage function may affect the accuracy of this calculation. It is important to have all necessary information and to carefully consider the assumptions made when performing calculations in science.
 

1. What is a Fourier series?

A Fourier series is a mathematical representation of a periodic function as a sum of sine and cosine functions. It is named after French mathematician Joseph Fourier and is commonly used in signal processing and other areas of science and engineering.

2. How do you calculate a Fourier series?

To calculate a Fourier series, you need to determine the coefficients of the sine and cosine functions that make up the series. This can be done using integrals or other methods, depending on the specific function and boundaries of the series.

3. What is the purpose of calculating the power over a resistor?

The power over a resistor is a measure of the amount of electrical energy being dissipated by the resistor. It is important to calculate this power in order to ensure that the resistor is not being overloaded and to determine its efficiency in a circuit.

4. What factors affect the power calculation over a resistor?

The power calculation over a resistor is affected by several factors, including the resistance of the resistor, the voltage across the resistor, and the current flowing through the resistor. Additionally, the temperature and material of the resistor can also impact the power calculation.

5. How is the power over a resistor related to the Fourier series?

The power over a resistor can be calculated using the Fourier series by determining the harmonics of the voltage and current signals in the circuit. This information can then be used to calculate the power dissipation in the resistor. In general, the higher the harmonics in the Fourier series, the higher the power over the resistor will be.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
5K
  • Calculus and Beyond Homework Help
Replies
3
Views
262
  • Engineering and Comp Sci Homework Help
Replies
6
Views
772
  • Calculus and Beyond Homework Help
Replies
1
Views
191
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
806
Replies
4
Views
275
  • Introductory Physics Homework Help
Replies
1
Views
752
  • Engineering and Comp Sci Homework Help
Replies
1
Views
996
Back
Top