Physics, asked by omprakash17759, 5 months ago

State and explain Fourier's theorem . Apply this theorem to analyse a square wave??​

Answers

Answered by vandanabajpai1477
5

Explanation:

A square wave is a non-sinusoidal periodic waveform in which the amplitude alternates at a steady frequency between fixed minimum and maximum values, with the same duration at minimum and maximum. In an ideal square wave, the transitions between minimum and maximum are instantaneous.

Sine, square, triangle, and sawtooth waveforms

Square wave sound sample

5 seconds of square wave at 220 Hz

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The square wave is a special case of a pulse wave which allows arbitrary durations at minimum and maximum. The ratio of the high period to the total period of a pulse wave is called the duty cycle. A true square wave has a 50% duty cycle (equal high and low periods).

Square waves are often encountered in electronics and signal processing, particularly digital electronics and digital signal processing. Its stochastic counterpart is a two-state trajectory.

Origin and uses

Definitions

Fourier analysis

Characteristics of imperfect square waves

See also

References

External links

Last edited 22 days ago by LaundryPizza03

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Answered by Karthi2112
5

Answer:

Let f(x) = {1 if -p<x<0, 0 if 0<x<p, and periodic of period 2p}. This function is a square wave; a plot shows the value 1 from x=p to x = 0 followed by the value 0 from x = 0 to x=p and the shape repeats in each interval of length 2p. The problem is to find the coefficients for the Fourier expansion of the square wave.

The Solution:

The Fourier coefficients are found by integration:

a0 = 1,

an>0 = 0, cosine terms are absent,

bn = {0 if n is even and -2/np if n is odd} (only odd sine terms are present).

The Fourier expansion, through n=2m+1 is

F(m,x) = 1/2 - 2/p[sin(x) + 1/3 sin(3x) + 1/5 sin(5x) + + 1/(2m+1) sin((2m+1)x)].

The following plot illustrates the expansion.

Fourier expansion of the square wave. The function f(x) is defined to be f(-p<x<0) = 0 and f(0<x<p)=0 and it is periodic f(x+2p) = f(x). Two Fourier expansions are shown. The two-term expansion, F(1,x) = 1/2 - 0.637 sin(x) is a pure fundamental of frequency 2p shifted upward by the constant 1/2. A much closer approximation is the 7-term series, 1/2 - 0.637sin(x)-0.212sin(3x)-0.127sin(5x)-0.091sin(7x)

-0.071sin(9x)-0.058sin(11x), containing 5 odd harmonics with diminishing amplitudes.

Explanation:

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