half range fourier series solved examples pdf

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Chapter Contents ⊗

• Fourier Series
• Helpful Revision
• 1. Overview of Fourier Series
• 2. Full Range Fourier Series
• 3. Fourier Series of Even and Odd Functions
• Fourier Series Graph Interactive
• 4. Fourier Series of Half Range Functions
• 5. Harmonic Analysis
• 6. Line Spectrum
• 7. Fast Fourier Transform

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4. Half Range Fourier Series

If a function is defined over half the range, say `0` to L , instead of the full range from `-L` to `L`, it may be expanded in a series of sine terms only or of cosine terms only. The series produced is then called a half range Fourier series .

Conversely, the Fourier Series of an even or odd function can be analysed using the half range definition.

Even Function and Half Range Cosine Series

An even function can be expanded using half its range from

• `0` to L or
• `-L` to `0` or
• `L` to `2L`

That is, the range of integration is L . The Fourier series of the half range even function is given by:

`f(t)=a_0/2+sum_(n=1)^oo\ a_n\ cos {:(n pi t)/L:}`

for `n = 1, 2, 3, ...` , where

`a_0=2/Lint_0^Lf(t)dt`

`a_n=2/Lint_0^Lf(t)\ cos {:(n pi t)/L:}dt`

and `b_n= 0`.

Illustration

In the figure below, `f(t) = t` is sketched from `t = 0` to `t = π`.

Graph of `f(t)`.

An even function means that it must be symmetrical about the `f(t)` axis and this is shown in the following figure by the broken line between `t = -π` and `t = 0`.

Graph of `f(t)`, illustrating it's an even function.

It is then assumed that the "triangular wave form" produced is periodic with period `2pi` outside of this range as shown by the dotted lines.

Graph of `f(t)`, a triangular waveform.

We are given that

`f(t)={: {(-t,if, -pi<=t<0),(t,if, 0<=t < pi):}`

and `f(t)` is periodic with period `2π`.

a) Sketch the function for 3 cycles.

b) Find the Fourier trigonometric series for `f(t)`, using half-range series.

(This is the same function we saw in the above Illustration example.)

b) Since the function is even, we have b n = 0 .

In this example, `L=pi`.

`=2/piint_0^pit\ dt`

`=2/pi[t^2/2]_0^pi`

`=2/pi(pi^2)/2`

To find a n , we use a result from before (see Table of Common Integrals ):

`intt\ cos nt\ dt` `=1/n^2(cos nt+nt\ sin nt)`

`a_n=2/Lint_0^Lf(t)cos{:(n pi t)/L:}dt`

`=2/piint_0^pi t\ cos nt\ dt`

`=2/pi[1/n^2(cos nt+nt\ sin nt)]_0^pi`

`=2/(pi n^2)[(cos n pi+0)-(cos 0+0)]`

`=2/(pi n^2)[(cos n pi-1)]`

`=2/(pi n^2)[(-1)^n-1]`

When n is odd, the last line gives us `-4/(pin^2`.

When n is even, the last line equals `0`.

For the series, we need to generate odd values for n . We need to use `(2n - 1)` for `n = 1, 2, 3,...`.

So we have:

`f(t)=a_0/2+sum_(n=1)^oo a_n cos{:(n pi t)/L:}`

`=pi/2-4/pi sum_(n=1)^oo(cos(2n-1)t)/((2n-1)^2)`

`=pi/2` `-4/pi(cos t+1/9cos 3t` `{:+1/25cos 5t+...)`

Check: We graph different terms of the above expression to make sure our answer is correct. Here is the first term involving the cosine expression:

Graph of `f(t)~~pi/2-4/pi cos(t)`.

Now add another term:

Graph of `f(t)` `~~pi/2-4/pi (cos(t)` `{: + 1/9 cos(3t))`.

Graph of `f(t)` `~~pi/2-4/pi (cos(t)` ` + 1/9 cos(3t)` `{:+1/25 cos(5t))`.

We can see our series is rapidly converging to our original triangular function.

Odd Function and Half Range Sine Series

An odd function can be expanded using half its range from `0` to L , i.e. the range of integration has value L . The Fourier series of the odd function is:

Since a o = 0 and a n = 0 , we have:

`f(t)=sum_(n=1)^oo\ b_n\ sin {:(n pi t)/L:}` for n `= 1, 2, 3, ...`

`b_n=2/Lint_0^Lf(t)\ sin {: (n pi t)/L:}dt`

In the figure below, f ( t ) = t is sketched from t = 0 to t = π , as before.

Graph of `f(t)`, for `0 < t < pi`.

An odd function means that it is symmetrical about the origin and this is shown by the red broken lines between t = − π and `t = 0`.

Graph of `f(t)`, showing it's an odd function.

It is then assumed that the waveform produced is periodic of period 2 π outside of this range as shown by the dotted lines.

Graph of `f(t)`, a periodic odd function.

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Fourier Series Examples

• Introduction
• Aperiodicity

This document derives the Fourier Series coefficients for several functions. The functions shown here are fairly simple, but the concepts extend to more complex functions.

Even Pulse Function (Cosine Series)

Consider the periodic pulse function shown below. It is an even function with period T . The function is a pulse function with amplitude A , and pulse width T p . The function can be defined over one period (centered around the origin) as:

Aside: the periodic pulse function

The periodic pulse function can be represented in functional form as Π T (t/T p ) . During one period (centered around the origin)

Π T (t) represents a periodic function with period T and pulse width ½. The pulse is scaled in time by T p in the function Π T (t/T p ) so:

This can be a bit hard to understand at first, but consider the sine function. The function sin(x/2) twice as slow as sin(x) (i.e., each oscillation is twice as wide). In the same way Π T (t/2) is twice as wide (i.e., slow) as Π T (t) .

The Fourier Series representation is

Since the function is even there are only a n terms.

The average is easily found,

The other terms follow from

Any interval of one period is allowed but the interval from -T/2 to T/2 is straightforward in this case.

Since x T (t)=A between -T p /2 to +T p /2 and zero elsewhere the integral simplifies and can be solved

Since sine is an odd function, sin(a)-sin(-a)=2sin(a) , and using the fact that ω 0 =2π/T and

This result is further explored in two examples.

Example 1: Special case, Duty Cycle = 50%

Consider the case when the duty cycle is 50% (this means that the function is high 50% of the time, or T p =T/2 ), A=1 , and T=2 . In this case a 0 =average=0.5 and for n≠0 :

The values for a n are given in the table below. Note: this example was used on the page introducing the Fourier Series. Note also, that in this case a n (except for n=0 ) is zero for even n , and decreases as 1/n as n increases.

The graph shows the function x T (t) (blue) and the partial Fourier Sum (from n=0 to n=N ) (red)

as well as the highest frequency harmonic, $a_Ncos(N\omega_0t)$ (dotted magenta). Lower frequency harmonics in the summation are thin dotted blue lines (but harmonics with $a_n0$ are not shown. You can change n by clicking the buttons. As before , note:

• As you add sine waves of increasingly higher frequency, the approximation improves.
• The addition of higher frequencies better approximates the rapid changes, or details, (i.e., the discontinuity) of the original function (in this case, the square wave).
• Gibb's overshoot exists on either side of the discontinuity.
• Because of the symmetry of the waveform, only odd harmonics (1, 3, 5, ...) are needed to approximate the function.  The reasons for this are discussed below
• The rightmost button shows the sum of all harmonics up to the 21st harmonic, but not all of the individual sinusoids are explicitly shown on the plot.  In particular harmonics between 7 and 21 are not shown.

Example 2: Special case, Duty Cycle = 40%

Now consider the case when the duty cycle is 40%, A=1 , and T=2 . In this case a 0 =average=0.4 and for n≠0 :

The values for a n are given in the table below (note: this example was used on the previous page ).

Note that because this example is similar to the previous one, the coefficients are similar, but they are no longer equal to zero for n even.

Even Square Wave (Exploiting Symmetry)

In problems with even and odd functions, we can exploit the inherent symmetry to simplify the integral. We will exploit other symmetries later . Consider the problem above. We have an expression for a n , n≠0

If x T (t) is even, then the product x T (t)·cos(n·ω 0 t) is even (the product of two even functions is even). We can then use the fact that for an even function, e(t) ,

which generates the same answer as before. This will often be simpler to evaluate than the original integral because one of the limits of integration is zero.

Even Square Wave (Exponential Series)

Consider, again, the pulse function. We can also represent x T (t) by the Exponential Fourier Series

We find the c n

As before the integral is from -T/2 to +T/2 and make use of the facts that the function is constant for |t|<T p /2 and zero elsewhere, and the T·ω 0 =2*·π .

Euler's identities dictate that e +jθ -e -jθ =2jsin(θ) so e -jθ -e +jθ =-2jsin(θ) .

Note that, as expected , c 0 =a 0 and c n =a n /2 , (n≠0) (since this is an even function b n =0 ).

Even Triangle Wave (Cosine Series)

Consider the triangle wave

The average value (i.e., the 0 th Fourier Series Coefficients) is a 0 =0 . For n>0 other coefficients the even symmetry of the function is exploited to give

Between t=0 and t=T/2 the function is defined by x T (t)=A-4At/T so

Perform the integrations (either by hand using integration by parts, or with a table of integrals, or by computer) and use the fact that ω 0 ·T=2·π

Since sin(π·n)=0 this simplifies to

This answer is correct, but noting that

yields an even simpler result

Example 3: Triangle wave

If x T (t) is a triangle wave with A=1 , the values for a n are given in the table below (note: this example was used on the previous page ).

Note: this is similar, but not identical, to the triangle wave seen earlier .

• As you add sine waves of increasingly higher frequency, the approximation gets better and better, and these higher frequencies better approximate the details, (i.e., the change in slope) in the original function.
• The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as 1/n 2 (which is faster than the 1/n decay seen in the pulse function Fourier Series (above)).   Conceptually, this occurs because the triangle wave looks much more like the 1st harmonic, so the contributions of the higher harmonics are less.  Even with only the 1st few harmonics we have a very good approximation to the original function.
• There is no discontinuity, so no Gibb's overshoot.
• As before, only odd harmonics (1, 3, 5, ...) are needed to approximate the function; this is because of the symmetry of the function .

Odd Function (Sawtooth Wave)

Thus far, the functions considered have all been even. The diagram below shows an odd function.

In this case, a Fourier Sine Series is appropriate

It is easiest to integrate from -T/2 to +T/2 . Over this interval $x_T(t)=2At/T$.

Performing the integration (and using the fact that ω 0 ·T=2·π ) the integral yields

Using two simplification, sin(π·n)=0 and cos(π·n)=(-1) n gives

Aside: using symmetry

In this case since x T (t) is odd and is multiplied by another odd function ( sin(n·ω 0 t) ), their product is even and the integral can be rewritten as:

Example 4: Odd Sawtooth Wave

If x T (t) is a sawtooth wave with A=1 , the values for b n are given in the table below

Note: this is similar, but not identical, to the sawtooth wave seen earlier .

• Since this function doesn't look as much like a sinusoid as the triangle wave , the coefficients decrease less rapidly (as 1/n instead of 1/n 2
• There is Gibb's overshoot caused by the discontinuity.

Functions that are neither even nor odd

So far, all of the functions considered have been either even or odd, but most functions are neither. This presents no conceptual difficult, but may require more integrations. For example if the function x T (t) looks like the one below

Since this has no obvious symmetries, a simple Sine or Cosine Series does not suffice. For the Trigonometric Fourier Series, this requires three integrals

However, an exponential series requires only a single integral

For this reason, among others, the Exponential Fourier Series is often easier to work with, though it lacks the straightforward visualization afforded by the Trigonometric Fourier Series.

Example 5: Neither Even nor Odd

In this case, but not in general, we can easily find the Fourier Series coefficients by realizing that this function is just the sum of the square wave (with 50% duty cycle) and the sawtooth so

From the relationship between the Trigonometric and Exponential Fourier Series

• There is Gibb's overshoot caused by the discontinuities.

Effect of Function Symmetry on Coefficients

If the function x T (t) has certain symmetries, we can simplify the calculation of the coefficients.

Symmetry Trigonometric Series and Symmetry

The first two symmetries are were discussed previously in the discussions of the pulse function ( x T (t) is even) and the sawtooth wave ( x T (t) is odd).

Half-wave symmetry is depicted the diagram below.

The top function, x T1 (t) , is odd ( x T1 (t)=-x T1 (-t) ), but does not have half-wave symmetry. The bottom function, x T2 (t) is nether even nor odd, but since x T2 (t)=-x T2 (t-T/2) , it has halfwave symmetry. To visualize this imagine shifting the function by half a period ( T/2 ); for half-wave symmetry the shifted function should be the mirror image of the original function (about the horizontal axis) as shown below

The reason the coefficients of the even harmonics are zero can be understood in the context of the diagram below. The top graph shows a function, x T (t) with half-wave symmetry along with the first four harmonics of the Fourier Series (only sines are needed because x T (t) is odd). The bottom graph shows the harmonics multiplied by x T (t) .

Now imagine integrating the product terms from -T/2 to +T/2 . The odd terms (from the 1st (red) and 3rd (magenta) harmonics) will have a positive result (because they are above zero more than they are below zero). The even terms (green and cyan) will integrate to zero (because they are equally above and below zero). Though this is a simple example, the concept applies for more complicated functions, and for higher harmonics.

The only funct ion discussed with half-wave symmetry was the triangle wave and indeed the coefficients with even indices are equal to zero (as are all of the b n terms because of the even symmetry). The square wave with 50% duty cycle would have half wave symmetry if it were centered around zero (i.e., centered on the horizontal axis). In that case the a 0 term would be zero and we have already shown that all the terms with even indices are zero, as expected.

Simplifications can also be made based on quarter-wave symmetry , but these are not discussed here.

A periodic function has quarter wave symmetry if it has half wave symmetry and it is either even or odd around its two half-cycles.

Exponential Series and Symmetry

Since the coefficients c n of the Exponential Fourier Series are related to the Trigonometric Series by

(assuming x T (t) is real) we can use the symmetry properties of the Trigonometric Series to find a n and b n and hence c n .

However, in addition, the coefficients of c n contain some symmetries of their own. In particular,

• The magnitude of the c n terms are even with respect to n: |c -n |=|c n | .
• The angle of the c n terms are odd with respect to n: ∠c -n =-∠c n .
• The real part of c n is even ( Re(c -n ) = Re(c n ) ) and the imaginary part is odd ( Im(c -n ) =-Im(c n ) )
• If x T (t) is even, then b n =0 and c n is even and real.
• If x T (t) is odd, then a n =0 and c n is odd and imaginary.

Some Comments about the Pulse Function

Let's examine the Fourier Series representation of the periodic rectangular pulse function , Π T (t/T p ) , more carefully.

Since the function is even, we expect the coefficients of the Exponential Fourier Series to be real and even(from symmetry properties ). Furthermore, we have already calculated the coefficients of the Trigonometric Series , and could easily calculate those of the Exponential Series . However, let us do it from first principles. The Exponential Fourier Series coefficients are given by

$$\displaylines{ {\Pi _t}\left( {{t \over {{T_p}}}} \right) = \sum\limits_{n = - \infty }^{ + \infty } {{c_n}{e^{jn{\omega _0}t}}} \cr with \cr {c_n} = {1 \over T}\int_T {{\Pi _t}\left( {{t \over {{T_p}}}} \right){e^{ - jn{\omega _0}t}}dt} \cr } $$

We can change the limits of integration to -T p /2 and +T p /2 (since the function is zero elsewhere) and proceed (the function is one in that interval, so we can drop it). We also make use of the fact the ω 0 =2π/T and Euler's identity for sine.

The last step in the derivation is performed so we can use the sinc() function (pronounced like " sink "). This function comes up often in Fourier Analysis.

$${\mathop{\rm sinc}\nolimits} (x) = {{\sin \left( {\pi x} \right)} \over {\pi x}}$$

Note: unfortunately there are two common definitions of the "sinc()" function. Engineers tend to use ${\mathop{\rm sinc}\nolimits} (x) = {{\sin \left( {\pi x} \right)} \over {\pi x}}$; this definition is used in MATLAB. Mathematicians tend to use ${\mathop{\rm sinc}\nolimits} (x) = {{\sin \left( x \right)} \over x}$; this definition is used in Mathematica. We will use the former definition.

Using the "engineering" notation for " sinc() " the coefficients simplify to

Aside: the "sinc()" function

The sinc function has several important features:

• sinc(x)=0 for all integer values of x except at x=0 where sinc(0)=1 . This is because sin(π·n)=0 for all integer values of n. However at n=0 we have sin(π·n)/(π·n) which is zero divided by zero, but by L'Hôpital's rule get a value of 1.
• The first zeros away from the origin occur when x=±1 .
• The function decays with an envelope of 1/(π·x) as x moves from the origin. This is because the sin() function has successive maxima with an amplitude of 1, and the sin function is divided by π·x .

The diagram below shows c n vs n for several values of the duty cycle, T p /T .

The graph on the left shows the time domain function. If you hit the middle button, you will see a square wave with a duty cycle of 0.5 (i.e., it is high 50% of the time). The period of the square wave is T=2·π; . The graph on the right shown the values of c n vs n as red circles vs n (the lower of the two horizontal axes; ignore the top axis for now). The blue line goes through the horizontal axis whenever the argument of the sinc() function, n·T p /T is an integer (except when n=0 .). In particular the first crossing of the horizontal axis is given by n·T p /T=1 or n=T/T p (note this is not an integer values of T p ). There are several important features to note as T p is varied.

• As T p decreases (along with the duty cycle, T p /T ), so does the value of c 0 . This is to be expected because c 0 is just the average value of the function and this will decrease as the pulse width does.
• As T p decreases, the "width" of the sinc() function broadens. This tells us that as the function becomes more localized in time (i.e., narrower) it becomes less localized in frequency (broader). In other words, if a function happens very rapidly in time, the signal must contain high frequency coefficients to enable the rapid change.
• Let us call the "width" of the sinc() function the width of the main lobe (i.e., between the first two zero crossings around ω=0 ), Δn=2·T/T p . If we call the width of the pulse Δt=t p then $$\left( {\Delta n} \right) \cdot \left( {\Delta t} \right) = \left( {2{T \over {{T_p}}}} \right) \cdot \left( {{T_p}} \right) = 2T = {\rm{constant}}$$ This tells us explicitly that the product of width in frequency (i.e., Δn ) multiplied by the width in time ( Δt ) is constant - if one is doubled, the other is halved. Or - as one gets more localized in time, it is less localized in frequency. We will discuss this more later.

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Half-Range Series

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Fourier Series Questions

Fourier series questions with solutions are provided here for students to practice. Fourier series questions are often asked in many competitive exams, and these questions will help and amp up your preparations. Fourier series is a summation that represents periodic functions as waves of simple periodic functions – sine and cosine functions. The main requirement of the Fourier series is to generate infinite periodic sinusoidal waves.

Periodic Functions: For any positive number T, if for a given function F,

F(x + T) = F(x) ∀ x ∈ R

Then F is called a periodic function, and T is called the period of the function. The function repeats its value T, 2T, 3T, … intervals.

For example: Let f = sin x, then sin (x + 2𝜋) = sin x, hence period of sine function is 2n𝜋 for n = 1, 2, 3, …

Fourier Series: For a given function f, we can find the real numbers a 0 , a 1 , a 2 , …a n ,… , b 1 , b 2 , …, b n , … by the Euler-Fourier formulae:

\(\begin{array}{l}a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)cos\:nx\:dx\:\:\:\:(n=0, 1, 2, 3,…) \end{array} \)

\(\begin{array}{l}b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)sin\:nx\:dx\:\:\:\:(n=1, 2, 3,…) \end{array} \)

Then the Fourier series of f is given by:

\(\begin{array}{l}\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}(a_{n}cos\:nx+b_{n}sin\:nx) \end{array} \)

For a function f(x) defined in the interval (–L, L) such that f(x + 2L) = f(x), that is, 2L is the period of f, then the Fourier series of f is given by:

Learn more about Fourier Series .

Fourier Series Questions with Solutions

Now let us solve questions on the Fourier series.

Question 1:

Find the Fourier series of the function f(x) = x 2 , –𝜋 < x < 𝜋.

Let us find the values of the real numbers a 0 , a n , and b n . The period of the given function is 2𝜋, then,

\(\begin{array}{l}a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx =\frac{1}{\pi}\int_{-\pi}^{\pi}x^{2}\:dx \end{array} \)

\(\begin{array}{l}= \frac{1}{\pi}\left [ \frac{x^{3}}{3} \right ]_{-\pi}^{\pi}=\frac{2\pi^{2}}{3}\end{array} \)

⇒ a 0 = 2𝜋 2 /3.

\(\begin{array}{l}a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)cos\:nx\:dx=\frac{2}{\pi}\int_{0}^{\pi}x^{2}cos\:nx\:dx\:\:\:\left [ \int_{-a}^{a} f(x)dx=\left\{\begin{matrix}2\int_{0}^{a} f(x)dx&,when\:f\:is\:an\:even\:function \\ 0&,when\:f\:is\:an\:odd\:function \\\end{matrix}\right.\right ]\end{array} \)

\(\begin{array}{l}=\frac{2}{\pi}\left [ x^{2}\frac{sin\:nx}{n}+2x\left ( \frac{cos\:nx}{n^{2}}\right )+2\left ( -\frac{sin\:nx}{n^{2}} \right ) \right ]_{0}^{\pi}\end{array} \)

\(\begin{array}{l}=\frac{4(-1)^{n}}{n^{2}}\:\:\:\:\:[\because sin\:n\pi = 0\:\: and\:\:cos\:n\pi =(-1)^n]\end{array} \)

\(\begin{array}{l}\Rightarrow a_{n}=\left\{\begin{matrix}-\frac{4}{n^{2}} &,\:when\:n\:is\:odd \\\frac{4}{n^{2}} &,\:when\:n\:is\:even & \\\end{matrix}\right.\end{array} \)

Now x 2 sin nx is an odd function

The Fourier series of x 2 is

\(\begin{array}{l}x^{2}=\frac{\pi^{2}}{3}+\sum_{n=1}^{\infty}\frac{4(-1)^{n}}{n^{2}}cos\:nx+\sum_{n=1}^{\infty}0.sin\:nx\end{array} \)

\(\begin{array}{l}x^{2}=\frac{\pi^{2}}{3}+4\left [ -cos\:x+\frac{1}{4}cos\:2x-\frac{1}{9}cos\:3x+… \right ]\end{array} \)

Question 2:

Find the Fourier series to represent e ax for x ∈ (–𝜋, 𝜋).

Since x ∈ (–𝜋, 𝜋), therefore, the period of the given function is 2𝜋. Then,

\(\begin{array}{l}a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x) dx=\frac{1}{\pi}\int_{-\pi}^{\pi}e^{ax} dx\end{array} \)

\(\begin{array}{l}=\frac{1}{\pi}\left [ \frac{e^{ax}}{a} \right ]_{-\pi}^{\pi} =\frac{1}{\pi}\left [ \frac{e^{a\pi}-e^{-a\pi}}{a} \right ]=\frac{2sinh\:a\pi}{a\pi}\end{array} \)

\(\begin{array}{l}a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)cos\:nx\:dx=\frac{1}{\pi}\int_{-\pi}^{\pi}e^{ax}cos\:nx\:dx\end{array} \)

\(\begin{array}{l}=\frac{1}{\pi}\left [ \frac{e^{ax}}{(a^{2}+n^{2})}(acos\:nx+nsin\:nx) \right ]_{-\pi}^{\pi}=\frac{1}{\pi}\frac{acos\:n\pi}{(a^{2}+n^{2})}(e^{a\pi}-e^{-a\pi})=\frac{1}{\pi}\frac{2a(-1)^{n}.sinh\:a\pi}{(a^{2}+n^{2})}\end{array} \)

\(\begin{array}{l}b_{n}=\frac{1}{\pi}\frac{2n(-1)^{n}.sinh\:a\pi}{(a^{2}+n^{2})}\end{array} \)

\(\begin{array}{l}\therefore e^{ax}=\frac{sinh\:a\pi}{a\pi}+ \sum_{n=1}^{\infty}\frac{2a(-1)^{n}sinh\:a\pi}{\pi(a^{2}+n^{2})}cos\:nx+\sum_{n=1}^{\infty}\frac{2n(-1)^{n}sinh\:a\pi}{\pi(a^{2}+n^{2})}sin\:nx\end{array} \)

Question 3:

If f(x) = cos x for 0 < x < 𝜋, f(x) = 50 for 𝜋 ≤ x < 2𝜋 and f(x + 2𝜋) = f(x) ∀ x. Find the sum of the Fourier series of f at x = 𝜋.

Sum of the Fourier series at x = 𝜋 is given by

½ [f(𝜋–) + f(𝜋+)] = ½ (cos 𝜋 + 50) = ½ ( –1 + 50) = 24.5.

Question 4:

Find the Fourier series of the periodic function f(x), such that

\(\begin{array}{l}f(x)=\left\{\begin{matrix}-\pi &,when\:-\pi<x<0 \\ x& when\:-0<x<\pi \\\end{matrix}\right.\end{array} \)

Clearly, the period of the function of 2𝜋, then,

\(\begin{array}{l}a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx=\frac{1}{\pi}\int_{-\pi}^{0}-\pi dx+\frac{1}{\pi}\int_{0}^{\pi}xdx=-[x]_{-\pi}^{0}+\frac{1}{\pi}\left [ \frac{x^{2}}{2} \right ]_{0}^{\pi}\end{array} \)

\(\begin{array}{l}=-\pi+\frac{\pi}{2}=-\frac{\pi}{2}\end{array} \)

\(\begin{array}{l}a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)cos \:nx\:dx=\frac{1}{\pi}\int_{-\pi}^{0}(-\pi) cos \:nx\:dx+\frac{1}{\pi}\int_{0}^{\pi}xcos \:nx\:dx\end{array} \)

\(\begin{array}{l}=-\left [ \frac{sin\:nx}{n} \right ]_{-\pi}^{0}+\frac{1}{\pi}\left [ x\frac{sin\:nx}{n}+\frac{cos\:nx}{n^{2}} \right ]_{0}^{\pi}=\frac{cos\:n\pi-1}{\pi n^{2}}\end{array} \)

\(\begin{array}{l}\Rightarrow a_{n}=\left\{\begin{matrix}0 &,when\:n\:is\:even \\-\frac{2}{\pi n^{2}} & ,when\:n\:is\:odd \\\end{matrix}\right. \end{array} \)

\(\begin{array}{l}b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)sin\:nx\:dx=\frac{1}{\pi}\int_{-\pi}^{0}(-\pi)sin\:nx\:dx+\frac{1}{\pi}\int_{0}^{\pi}xsin\:nx\:dx\end{array} \)

\(\begin{array}{l}=\frac{1}{\pi}\left [ -\pi \frac{cos\:nx}{n} \right ]_{-\pi}^{0}+\frac{1}{\pi}\left [ \frac{-xcos\:nx}{n}+\left ( \frac{sin\:nx}{n^{2}} \right ) \right ]_{0}^{\pi}=\frac{1-2cos\:n\pi}{n}\end{array} \)

\(\begin{array}{l}\Rightarrow b_{n}=\left\{\begin{matrix}-\frac{1}{n} & ,when\:n\:is\:even \\\frac{3}{n} & ,when\:n\:is\:odd \\\end{matrix}\right.\end{array} \)

Putting the values of a’s and b’s we get the required fourier series:

f(x) = – 𝜋/4 – 2/𝜋[cos x + 1/3 2 cos 3x + 1/5 2 cos 5x + …] + [3 sin x – ½ sin 2x + 3/3 sin 3x – …]

Question 5:

Prove the following for k = 1, 2, 3, …

\(\begin{array}{l}\int_{-L}^{L}sin\frac{k\pi x}{L}dx=\int_{-L}^{L}cos\frac{k\pi x}{L}=0\end{array} \)

Now, sin (–𝜃) = – sin 𝜃

\(\begin{array}{l}\int_{-a}^{a}f(x)dx = \left\{\begin{matrix}2\int_{0}^{a} f(x)dx & when\:f\:is \:an\:even\:function\\0 & when\:f\:is \:an\:odd\:function \\\end{matrix}\right.\end{array} \)

\(\begin{array}{l}\therefore \int_{-L}^{L}sin\frac{k\pi x}{L}dx = 0\:\:\:for\:k=1,2,3,…\end{array} \)

\(\begin{array}{l}\int_{-L}^{L}cos\frac{k\pi x}{L}dx = \frac{L}{k\pi x}\left [ sin\frac{k\pi x}{L} \right ]_{-L}^{L}=\frac{L}{k\pi x}(sin\:k\pi + sin\:k\pi)=0\:\:\:\:(\because sin\:n\pi = 0)\end{array} \)

• Linearized Stability
• Existence and Uniqueness Theorems for Initial Value Problem
• Fixed Point Iteration
• Quasi-Linear Partial Differential Equations

Question 6:

Find the Fourier series of |x| where –𝜋 < x < 𝜋.

Let f(x) = |x| and period of f is 2𝜋.

\(\begin{array}{l}a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx=\frac{1}{\pi}\int_{-\pi}^{\pi}|x|dx=\frac{2}{\pi}\int_{0}^{\pi}xdx=\frac{2}{\pi}\left [ \frac{x^{2}}{2} \right ]_{0}^{\pi}dx=\pi\end{array} \)

\(\begin{array}{l}a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)cos\:nx\:dx=\frac{2}{\pi}\int_{0}^{\pi}|x|cos\:nx\:dx\end{array} \)

\(\begin{array}{l}=\frac{2}{\pi}\left [ \frac{xsin\:nx}{n}+\frac{cos\:nx}{n^{2}} \right ]_{0}^{\pi}=\frac{2}{\pi}\left [ \frac{cos\:n\pi}{n^{2}}-\frac{1}{n^{2}} \right ]\end{array} \)

= 0 if n is even

= –4/𝜋n 2 if n is odd.

And b n = 0, since |x| is an even function.

Therefore, the fourier series of |x| is

𝜋/2 + 4/𝜋 [cos x + (cos 3x)/3 2 + (cos 5x)/5 2 + … ]

Question 7:

Find the value of the real number a 0 of the Fourier series if f(x) = x 2 for 0 ≤ x ≤ 2𝜋.

Given f(x) = x 2 , for this also the period of the function is 2𝜋

\(\begin{array}{l}a_{0}=\frac{1}{\pi}\int_{0}^{2\pi}f(x)dx=\frac{1}{\pi}\int_{0}^{2\pi}x^{2}dx=\frac{1}{\pi}\left [ \frac{x^{3}}{3} \right ]_{0}^{2\pi}=\frac{8\pi ^{2}}{3}\end{array} \)

∴ a 0 = 8𝜋 2 /3.

Question 8:

Express f(x) = sin(𝜋x/l) as half range cosine series for 0 ≤ x ≤ l.

Let sin(𝜋x/l) = a 0 /2 + ∑ a n cos(n𝜋x/l) where

\(\begin{array}{l}a_{0}=\frac{2}{l}\int_{0}^{1}f(x)dx=\frac{2}{l}\int_{0}^{1}sin(\frac{\pi x}{l})dx=\frac{2}{l}\left [ \frac{cos\frac{\pi x}{l}}{\pi /l} \right ]_{0}^{l}\end{array} \)

\(\begin{array}{l}=-\frac{2}{l}cos(\pi-1)=\frac{4}{\pi}\end{array} \)

\(\begin{array}{l}a_{n}= \frac{2}{l}\int_{0}^{l}sin\frac{\pi x}{l}cos\frac{n\pi x}{l}dx=\frac{1}{l}\int_{0}^{l}\left [ sin(n+1)\frac{\pi x}{l}-sin(n-1)\frac{\pi x}{l} \right ]dx\end{array} \)

\(\begin{array}{l}=\frac{1}{l}\left [ -\frac{cos(n+1)\frac{\pi x}{l}}{(n+1)\frac{\pi}{l}}+\frac{cos(n-1)\frac{\pi x}{l}}{(n-1)\frac{\pi}{l}} \right ]_{0}^{l}=\frac{1}{\pi}\left [ \left\{ -\frac{cos(n+1)\pi}{(n+1)}\right\} -\left\{ -\frac{1}{n+1}+\frac{1}{n-1}\right\}\right ]\end{array} \)

When n is odd, a n = 0

and when n is even, a n = –4/[𝜋(n + 1)(n – 1)]

∴ the half-range fourier series for sin(𝜋x/l) is given by

\(\begin{array}{l}sin\left ( \frac{\pi x}{l} \right )=\frac{2}{\pi}-\frac{4}{\pi}\left [ \frac{cos\frac{2\pi x}{l}}{1.3}+ \frac{cos\frac{4\pi x}{l}}{3.5}+\frac{cos\frac{6\pi x}{l}}{5.7}+…\right ]\end{array} \)

Question 9:

Express the given function f(x) as Fourier sine half range series.

\(\begin{array}{l}f(x)=\left\{\begin{matrix}x &,0\leq x \leq \frac{\pi}{2} \\\pi -x &,\frac{\pi}{2}< x \leq \pi & \\\end{matrix}\right.\end{array} \)

Let the Fourier sine hailf-range series is given by

f(x) = ∑ n b n sin nx where

\(\begin{array}{l}b_{n}=\frac{2}{\pi}\int_{0}^{\pi}f(x)sin\:nx\:dx=\frac{2}{\pi}\int_{0}^{\pi /2}xsin\:nx\:dx+\frac{2}{\pi}\int_{\pi /2}^{\pi}(\pi – x)sin\:nx\:dx\end{array} \)

\(\begin{array}{l}=\frac{2}{\pi}\left [ x.\left ( -\frac{cos\:nx}{n} \right )+\left ( \frac{sin\:nx}{n^{2}} \right ) \right ]_{0}^{\pi /2}+\frac{2}{\pi}\left [ (\pi-x).\left ( -\frac{cos\:nx}{n} \right )-\left ( \frac{sin\:nx}{n^{2}} \right ) \right ]_{\pi /2}^{\pi }\end{array} \)

\(\begin{array}{l}=\frac{2}{\pi}\left [ -\frac{\pi}{2n}cos\frac{n\pi}{2}+\frac{1}{n^{2}}sin\frac{n\pi}{2} \right ]+\frac{2}{\pi}\left [\frac{\pi}{2n}cos\frac{n\pi}{2}+\frac{1}{n^{2}} sin\frac{n\pi}{2} \right ]=\frac{4}{\pi n^{2}}sin\frac{n\pi}{2}\end{array} \)

When n is even, b n = 0

When n is odd, b n = 4/n 2 𝜋

Hence, the Fourier sine half-range series is given by

\(\begin{array}{l}f(x)= \frac{4}{\pi}\left [ \frac{1}{1^{2}}sin\:x-\frac{1}{3^{2}}sin\:3x+\frac{1}{5^{2}}sin\:5x-… \right ]\end{array} \)

Question 10:

Deduce the half-range cosine series for f(x) = kx when 0 ≤ x ≤ l/2 and f(x) = k(l – x) when l/2 < x ≤ l.

Let the Fourier cosine half-range series of f is

f(x) = a 0 /2 + ∑ n a n cos (n𝜋x/l)

\(\begin{array}{l}a_{0}= \frac{2}{l}\int_{0}^{l}f(x)dx=\frac{2}{l}\int_{0}^{l/2}kx\:dx+\frac{2}{l}\int_{l/2}^{l}k(l-x)dx\end{array} \)

\(\begin{array}{l}=\frac{2}{l}\left [ \frac{kx^{2}}{2} \right ]_{0}^{l/2}+\frac{2}{l}\left [ klx-\frac{klx^{2}}{2} \right ]_{l/2}^{l}=\frac{kl}{2}\end{array} \)

And for n = 1, 2, 3, …

\(\begin{array}{l}a_{n}=\frac{2}{l}\int_{0}^{l}f(x)cos\frac{n\pi x}{l}dx=\frac{2}{l}\int_{0}^{l2}kx\:cos\frac{n\pi x}{l}dx+\frac{2}{l}\int_{l/2}^{l}k(l-x)cos\frac{n\pi x}{l}dx\end{array} \)

\(\begin{array}{l}=\frac{2}{l}\left [ \frac{2kl^{2}}{n^{2}\pi ^{2}}cos \frac{n \pi}{2}-\frac{kl^{2}}{n^{2}\pi ^{2}}-\frac{kl^{2}}{n^{2}\pi ^{2}}cos\:n\pi \right ]=\frac{2kl}{n^{2}\pi^{2}}\left [ 2cos\frac{n\pi}{2}-1-cos\:n\pi \right ]\end{array} \)

When n is odd cos (n𝜋/2) = 0 and cos n𝜋 = –1 ∴ a n = 0

When n is even,

a 2 = 8kl/2 2 𝜋 2 , a 4 = 0, a 6 = 8kl/6 2 𝜋 2 and so on.

Hence, the Fourier half-range cosine series of f is

\(\begin{array}{l}f(x)=\frac{kl}{4}-\frac{8k}{\pi ^{2}}\left ( \frac{1}{2^{2}}cos \frac{2\pi x}{l}+\frac{1}{6^{2}}cos \frac{6\pi x}{l} +…\right )\end{array} \)

Practice Questions on Fourier Series

1. Find the Fourier series of the following:

(i) x – x 2 for –𝜋 ≤ x ≤ 𝜋

(ii) f(x) = –1 for –𝜋 < x < 0 and f(x) = 1 for 0 ≤ x ≤ 𝜋.

2. For f(x) = x 2 (1 – x 4 ) find the b n of the Fourier series x ∈ (–l,l).

3. Write the Fourier sine series for f(x) = k in (0, 𝜋).

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