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Signals can be composed by a superposition of an infinite number of sine and cosine functions. The coefficients of the superpositiondepend on the signal being represented and are equivalent to knowing the function itself.

The classic Fourier series as derived originally expressed a periodic signal (period T ) in terms of harmonically related sines and cosines.

s t a 0 k 1 a k 2 k t T k 1 b k 2 k t T
The complex Fourier series and the sine-cosine series are identical , each representing a signal's spectrum.The Fourier coefficients , a k and b k , express the real and imaginary parts respectively of the spectrum while the coefficients c k of the complex Fourier series express the spectrum as a magnitude and phase. Equating the classic Fourier series to the complex Fourier series , an extra factor of two and complex conjugate become necessary to relate the Fourier coefficients in each. c k 1 2 a k b k

Derive this relationship between the coefficients of the two Fourier series.

Write the coefficients of the complex Fourier series in Cartesian form as c k A k B k and substitute into the expression for the complex Fourier series. k c k 2 k t T k A k B k 2 k t T Simplifying each term in the sum using Euler's formula, A k B k 2 k t T = A k B k 2 k t T 2 k t T = A k 2 k t T B k 2 k t T A k 2 k t T B k 2 k t T We now combine terms that have the same frequency index in magnitude . Because the signal is real-valued, the coefficients of the complex Fourier series have conjugate symmetry: c k c k or A k A k and B k B k . After we add the positive-indexed and negative-indexed terms, each term in the Fourier series becomes 2 A k 2 k t T 2 B k 2 k t T . To obtain the classic Fourier series , we must have 2 A k a k and 2 B k b k .

Just as with the complex Fourier series, we can find the Fourier coefficients using the orthogonality properties of sinusoids. Note that the cosine and sine of harmonically related frequencies, even the same frequency, are orthogonal.

k l k l t 0 T 2 k t T 2 l t T 0
t 0 T 2 k t T 2 l t T T 2 k l k 0 l 0 0 k l k 0 l t 0 T 2 k t T 2 l t T T 2 k l k 0 l 0 T k 0 l 0 k l These orthogonality relations follow from the following important trigonometric identities.
α β 1 2 α β α β α β 1 2 α β α β α β 1 2 α β α β
These identities allow you to substitute a sum of sines and/or cosines for a product of them. Each term in the sum can beintegrated by noticing one of two important properties of sinusoids.
  • The integral of a sinusoid over an integer number of periods equals zero.
  • The integral of the square of a unit-amplitude sinusoid over a period T equals T 2 .

To use these, let's, for example, multiply the Fourier series for a signal by the cosine of the l th harmonic 2 l t T and integrate. The idea is that, because integration is linear, the integration will sift out all but the term involving a l .

t 0 T s t 2 l t T t 0 T a 0 2 l t T k 1 a k t 0 T 2 k t T 2 l t T k 1 b k t 0 T 2 k t T 2 l t T
The first and third terms are zero; in the second, the only non-zero term in the sum results when the indices k and l are equal (but not zero), in which case we obtain a l T 2 . If k 0 l , we obtain a 0 T . Consequently, l l 0 a l 2 T t 0 T s t 2 l t T All of the Fourier coefficients can be found similarly.
a 0 1 T t 0 T s t k k 0 a k 2 T t 0 T s t 2 k t T b k 2 T t 0 T s t 2 k t T

The expression for a 0 is referred to as the average value of s t . Why?

The average of a set of numbers is the sum divided by the number of terms. Viewing signal integration as the limit ofa Riemann sum, the integral corresponds to the average.

What is the Fourier series for a unit-amplitude square wave?

We found that the complex Fourier series coefficients are given by c k 2 k . The coefficients are pure imaginary, which means a k 0 . The coefficients of the sine terms are given by b k 2 c k so that b k 4 k k   odd 0 k   even Thus, the Fourier series for the square wave is

sq t k k 1 3 4 k 2 k t T

Let's find the Fourier series representation for the half-wave rectified sinusoid.

s t 2 t T 0 t T 2 0 T 2 t T
Begin with the sine terms in the series; to find b k we must calculate the integral
b k 2 T t 0 T 2 2 t T 2 k t T
Using our trigonometric identities turns our integral of a product of sinusoids into a sum of integrals of individual sinusoids,which are much easier to evaluate.
t 0 T 2 2 t T 2 k t T 1 2 t 0 T 2 2 k 1 t T 2 k 1 t T 1 2 k 1 0
Thus, b 1 1 2 b 2 b 3 0

On to the cosine terms. The average value, which corresponds to a 0 , equals 1 . The remainder of the cosine coefficients are easy to find, butyield the complicated result

a k 2 1 k 2 1 k 2 4 0 k odd

Thus, the Fourier series for the half-wave rectified sinusoid has non-zero terms for the average, the fundamental, and theeven harmonics.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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