Fourier series are essential in chemical engineering for analyzing periodic phenomena such as temperature variations in reactors, concentration oscillations in batch processes, and waveform analysis in instrumentation signals. This chapter introduces the mathematical foundation of Fourier series and provides methods to express periodic functions as sums of sine and cosine terms. Kenyan chemical engineers use these tools to model and solve complex process control and signal processing problems where periodicity is inherent.
Fourier series decompose any periodic function \( f(x) \) defined on an interval into an infinite sum of sines and cosines. The function must satisfy certain conditions (Dirichlet conditions) to be represented by a convergent Fourier series. The general form of the Fourier series for a function with period \( 2\pi \) is
$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx) $$
where the coefficients \( a_0 \), \( a_n \), and \( b_n \) are constants determined by integrals over one period.
The Fourier coefficients for \( f(x) \) defined on \([- \pi, \pi]\) are given by
$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \, dx $$
$$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos nx \, dx, \quad n=1,2,3,... $$
$$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin nx \, dx, \quad n=1,2,3,... $$
Example 1: A temperature profile inside a tubular reactor varies as \( f(x) = x \) for \( -\pi \le x \le \pi \). Find \( a_0 \).
Given: \( f(x) = x \), interval \([- \pi, \pi]\).
$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi x \, dx $$
$$ = \frac{1}{\pi} [ \frac{x^2}{2} ]_{-\pi}^\pi $$
$$ = \frac{1}{\pi} ( \frac{\pi^2}{2} - \frac{\pi^2}{2} ) = 0 $$
Answer: \( a_0 = 0 \)
Example 2: For the same \( f(x) = x \), compute \( a_1 \).
$$ a_1 = \frac{1}{\pi} \int_{-\pi}^\pi x \cos x \, dx $$
Use integration by parts:
Let \( u = x \), \( dv = \cos x \, dx \), then \( du = dx \), \( v = \sin x \).
$$ a_1 = \frac{1}{\pi} [ x \sin x \Big|_{-\pi}^\pi - \int_{-\pi}^\pi \sin x \, dx ] $$
Evaluate boundaries:
$$ x \sin x \Big|_{-\pi}^\pi = \pi \sin \pi - (-\pi) \sin (-\pi) = 0 - 0 = 0 $$
Integral:
$$ \int_{-\pi}^\pi \sin x \, dx = [-\cos x]_{-\pi}^\pi = -\cos \pi + \cos (-\pi) = -(-1) + (-1) = 1 - 1 = 0 $$
$$ a_1 = \frac{1}{\pi} (0 - 0) = 0 $$
Answer: \( a_1 = 0 \)
Example 3: Compute \( b_1 \) for \( f(x) = x \).
$$ b_1 = \frac{1}{\pi} \int_{-\pi}^\pi x \sin x \, dx $$
Using integration by parts:
Let \( u = x \), \( dv = \sin x \, dx \), so \( du = dx \), \( v = -\cos x \).
$$ b_1 = \frac{1}{\pi} [ -x \cos x \Big|_{-\pi}^\pi + \int_{-\pi}^\pi \cos x \, dx ] $$
Evaluate boundaries:
$$ -x \cos x \Big|_{-\pi}^\pi = -\pi \cos \pi + \pi \cos (-\pi) = -\pi (-1) + \pi (-1) = \pi - \pi = 0 $$
Integral:
$$ \int_{-\pi}^\pi \cos x \, dx = [\sin x]_{-\pi}^\pi = \sin \pi - \sin (-\pi) = 0 - 0 = 0 $$
$$ b_1 = \frac{1}{\pi} (0 + 0) = 0 $$
Answer: \( b_1 = 0 \)
A function \( f(x) \) is periodic with period \( T \) if
$$ f(x + T) = f(x) \quad \text{for all } x $$
Common periodic functions include sine and cosine, with period \( 2\pi \). In chemical engineering, periodic behavior arises in cyclic processes and oscillating reaction rates.
Example 1: Verify the period of \( f(t) = \sin (4t) \).
Given: \( f(t) = \sin 4t \)
The general period of \( \sin \omega t \) is \( \frac{2\pi}{\omega} \).
$$ T = \frac{2\pi}{4} = \frac{\pi}{2} $$
Answer: Period \( T = \frac{\pi}{2} \) units of time
Example 2: Check if \( f(x) = \cos (3x) \) is periodic with period \( T = 2\pi \).
Given: \( f(x) = \cos 3x \)
$$ f(x + T) = \cos [3(x + 2\pi)] = \cos (3x + 6\pi) = \cos 3x $$
Since \( \cos \theta \) is \( 2\pi \)-periodic, \( f(x) \) is periodic with period \( \frac{2\pi}{3} \), so \( T = 2\pi \) is not the fundamental period.
Answer: Fundamental period is \( \frac{2\pi}{3} \), not \( 2\pi \)
Example 3: Determine the period of \( f(x) = \sin^2 x \).
Using identity
$$ \sin^2 x = \frac{1 - \cos 2x}{2} $$
Period of \( \cos 2x \) is \( \pi \), so period of \( f(x) \) is \( \pi \).
Answer: Period \( T = \pi \)
The coefficients are obtained by integrating the function multiplied by sine or cosine over one full period.
$$ a_0 = \frac{2}{T} \int_{x_0}^{x_0 + T} f(x) \, dx $$
$$ a_n = \frac{2}{T} \int_{x_0}^{x_0 + T} f(x) \cos \frac{2n\pi x}{T} \, dx $$
$$ b_n = \frac{2}{T} \int_{x_0}^{x_0 + T} f(x) \sin \frac{2n\pi x}{T} \, dx $$
where \( x_0 \) is the start of the period.
Example 1: Find \( a_0 \) for \( f(x) = x \) over \( -\pi \le x \le \pi \).
Given: \( T = 2\pi \), \( f(x) = x \)
$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi x \, dx = 0 $$
Answer: \( a_0 = 0 \)
Example 2: Find \( a_1 \) for \( f(x) = x \).
Given: \( a_1 = \frac{1}{\pi} \int_{-\pi}^\pi x \cos x \, dx = 0 \) as shown previously.
Answer: \( a_1 = 0 \)
Example 3: Find \( b_1 \) for \( f(x) = x \).
Given: \( b_1 = \frac{1}{\pi} \int_{-\pi}^\pi x \sin x \, dx = 2 \) (calculate explicitly)
Integration by parts:
\( u = x, dv = \sin x dx \Rightarrow du = dx, v = -\cos x \)
$$ b_1 = \frac{1}{\pi} [ -x \cos x \Big|_{-\pi}^\pi + \int_{-\pi}^\pi \cos x \, dx ] $$
Evaluate boundaries:
$$ -x \cos x \Big|_{-\pi}^\pi = -\pi \cos \pi + \pi \cos (-\pi) = -\pi (-1) + \pi (-1) = \pi - \pi = 0 $$
Integral:
$$ \int_{-\pi}^\pi \cos x \, dx = \sin x \Big|_{-\pi}^\pi = 0 - 0 = 0 $$
$$ b_1 = \frac{1}{\pi} (0 + 0) = 0 $$
Which contradicts the expected result; recheck:
Actually, the function \( f(x) = x \) is an odd function, so \( a_n = 0 \) for all \( n \), but \( b_n \neq 0 \).
Calculate \( b_n \) generally:
$$ b_n = \frac{2}{T} \int_{-T/2}^{T/2} x \sin \frac{2n\pi x}{T} dx $$
For \( T = 2\pi \):
$$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi x \sin nx \, dx $$
Using integration by parts:
\( u = x \), \( dv = \sin nx dx \Rightarrow du = dx \), \( v = -\frac{1}{n} \cos nx \)
$$ b_n = \frac{1}{\pi} [ -\frac{x}{n} \cos nx \Big|_{-\pi}^\pi + \frac{1}{n} \int_{-\pi}^\pi \cos nx \, dx ] $$
Evaluate boundary term:
$$ -\frac{x}{n} \cos nx \Big|_{-\pi}^\pi = -\frac{\pi}{n} \cos n\pi + \frac{\pi}{n} \cos (-n\pi) = -\frac{\pi}{n} (-1)^n + \frac{\pi}{n} (-1)^n = 0 $$
Integral:
$$ \int_{-\pi}^\pi \cos nx \, dx = 0 $$
So \( b_n = 0 \)? This suggests the integral is zero, but \( f(x) = x \) is odd and \( \sin nx \) is odd, so product is even, integral should be nonzero.
Re-express:
The integral of product of two odd functions over symmetric interval is zero, but product of odd and even is odd.
Actually, \( x \) is odd, \( \sin nx \) is odd, product is even (odd × odd = even), so integral over symmetric interval is not zero.
Recalculate \( b_n \):
$$ b_n = \frac{2}{2\pi} \int_{-\pi}^\pi x \sin nx \, dx = \frac{1}{\pi} \int_{-\pi}^\pi x \sin nx \, dx $$
Use integration by parts:
\( u = x \), \( dv = \sin nx dx \Rightarrow du = dx \), \( v = -\frac{1}{n} \cos nx \)
$$ b_n = \frac{1}{\pi} [ -\frac{x}{n} \cos nx \Big|_{-\pi}^\pi + \frac{1}{n} \int_{-\pi}^\pi \cos nx \, dx ] $$
Boundary term:
$$ -\frac{x}{n} \cos nx \Big|_{-\pi}^\pi = -\frac{\pi}{n} \cos n\pi + \frac{\pi}{n} \cos (-n\pi) = -\frac{\pi}{n} (-1)^n + \frac{\pi}{n} (-1)^n = 0 $$
Integral:
$$ \int_{-\pi}^\pi \cos nx \, dx = 0 $$
The mistake is that \( \cos nx \) is even, integral over symmetric interval is nonzero only if \( n=0 \), else zero.
Hence, the integral term is zero.
Wait, but the integral of \( x \sin nx \) over \(-\pi\) to \(\pi\) is:
Since \( x \sin nx \) is even, integrate only from 0 to \(\pi\) and double:
$$ b_n = \frac{2}{\pi} \int_0^\pi x \sin nx \, dx $$
Integration by parts:
\( u = x \), \( dv = \sin nx dx \Rightarrow du = dx \), \( v = -\frac{1}{n} \cos nx \)
$$ \int_0^\pi x \sin nx \, dx = -\frac{x}{n} \cos nx \Big|_0^\pi + \frac{1}{n} \int_0^\pi \cos nx \, dx $$
Evaluate:
$$ -\frac{\pi}{n} \cos n\pi + 0 + \frac{1}{n} [ \frac{\sin nx}{n} \Big|_0^\pi ] = -\frac{\pi}{n} (-1)^n + \frac{1}{n^2} ( \sin n\pi - 0 ) = \frac{\pi}{n} (-1)^{n+1} $$
Since \( \sin n\pi = 0 \),
So,
$$ b_n = \frac{2}{\pi} \times \frac{\pi}{n} (-1)^{n+1} = \frac{2}{n} (-1)^{n+1} $$
Answer: \( b_n = \frac{2}{n} (-1)^{n+1} \)
Functions may be defined over arbitrary periods \( T \). The Fourier series adapts by scaling the argument to fit the standard interval.
The general Fourier series for period \( T \) is
$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty ( a_n \cos \frac{2n\pi x}{T} + b_n \sin \frac{2n\pi x}{T} ) $$
with coefficients
$$ a_0 = \frac{2}{T} \int_{x_0}^{x_0 + T} f(x) \, dx $$
$$ a_n = \frac{2}{T} \int_{x_0}^{x_0 + T} f(x) \cos \frac{2n\pi x}{T} \, dx $$
$$ b_n = \frac{2}{T} \int_{x_0}^{x_0 + T} f(x) \sin \frac{2n\pi x}{T} \, dx $$
Example 1: Find the Fourier series coefficients of \( f(x) = x \) over \( [0, 2\pi] \), period \( T=2\pi \).
Given: \( f(x) = x \), \( T=2\pi \), \( x_0=0 \)
Calculate \( a_0 \):
$$ a_0 = \frac{1}{\pi} \int_0^{2\pi} x \, dx = \frac{1}{\pi} [ \frac{x^2}{2} ]_0^{2\pi} = \frac{1}{\pi} \times \frac{(2\pi)^2}{2} = \frac{2\pi^2}{\pi} = 2\pi $$
Calculate \( a_n \):
$$ a_n = \frac{1}{\pi} \int_0^{2\pi} x \cos nx \, dx $$
Integration by parts:
\( u = x \), \( dv = \cos nx dx \Rightarrow du = dx \), \( v = \frac{\sin nx}{n} \)
$$ a_n = \frac{1}{\pi} [ x \frac{\sin nx}{n} \Big|_0^{2\pi} - \int_0^{2\pi} \frac{\sin nx}{n} dx ] $$
Boundary term:
$$ 2\pi \frac{\sin 2n\pi}{n} - 0 = 0 $$
Integral:
$$ \int_0^{2\pi} \sin nx \, dx = 0 $$
$$ a_n = 0 $$
Calculate \( b_n \):
$$ b_n = \frac{1}{\pi} \int_0^{2\pi} x \sin nx \, dx $$
Integration by parts:
\( u = x \), \( dv = \sin nx dx \Rightarrow du = dx \), \( v = -\frac{\cos nx}{n} \)
$$ b_n = \frac{1}{\pi} [ -\frac{x}{n} \cos nx \Big|_0^{2\pi} + \frac{1}{n} \int_0^{2\pi} \cos nx \, dx ] $$
Boundary term:
$$ -\frac{2\pi}{n} \cos 2n\pi + 0 = -\frac{2\pi}{n} \times 1 = -\frac{2\pi}{n} $$
Integral:
$$ \int_0^{2\pi} \cos nx \, dx = 0 $$
$$ b_n = \frac{1}{\pi} ( -\frac{2\pi}{n} + 0 ) = -\frac{2}{n} $$
Answer:
$$ a_0 = 2\pi, \quad a_n = 0, \quad b_n = -\frac{2}{n} $$
Example 2: Find the Fourier series coefficients of the function \( f(x) = \sin x \) over the interval \( [0, 4] \), period \( T = 4 \).
Given: \( f(x) = \sin x \), \( T = 4 \), \( x_0 = 0 \)
Calculate \( a_0 \):
$$ a_0 = \frac{2}{4} \int_0^4 \sin x \, dx = \frac{1}{2} [-\cos x]_0^4 = \frac{1}{2} (-\cos 4 + \cos 0) = \frac{1}{2} (-\cos 4 + 1) $$
Calculate \( a_n \):
$$ a_n = \frac{2}{4} \int_0^4 \sin x \cos \frac{2n\pi x}{4} \, dx = \frac{1}{2} \int_0^4 \sin x \cos \frac{n\pi x}{2} \, dx $$
This integral can be evaluated using product-to-sum formulas or numerically.
Calculate \( b_n \):
$$ b_n = \frac{2}{4} \int_0^4 \sin x \sin \frac{n\pi x}{2} \, dx = \frac{1}{2} \int_0^4 \sin x \sin \frac{n\pi x}{2} \, dx $$
Again, this can be evaluated using product-to-sum formulas or numerically.
Answer:
Example 3: Find the Fourier series coefficients for \( f(x) = 3 \) over \( [0, 6] \), period \( T = 6 \).
Given: \( f(x) = 3 \), \( T = 6 \), \( x_0 = 0 \)
Calculate \( a_0 \):
$$ a_0 = \frac{2}{6} \int_0^6 3 \, dx = \frac{1}{3} \times 3 \times 6 = 2 $$
Calculate \( a_n \):
$$ a_n = \frac{2}{6} \int_0^6 3 \cos \frac{2n\pi x}{6} \, dx = \frac{1}{3} \int_0^6 3 \cos \frac{n\pi x}{3} \, dx = \int_0^6 \cos \frac{n\pi x}{3} \, dx $$
$$ \int_0^6 \cos \frac{n\pi x}{3} \, dx = [ \frac{3}{n\pi} \sin \frac{n\pi x}{3} ]_0^6 = \frac{3}{n\pi} (\sin n\pi - 0) = 0 $$
So \( a_n = 0 \) for all \( n \geq 1 \).
Calculate \( b_n \):
$$ b_n = \frac{2}{6} \int_0^6 3 \sin \frac{2n\pi x}{6} \, dx = \frac{1}{3} \int_0^6 3 \sin \frac{n\pi x}{3} \, dx = \int_0^6 \sin \frac{n\pi x}{3} \, dx $$
$$ \int_0^6 \sin \frac{n\pi x}{3} \, dx = [ -\frac{3}{n\pi} \cos \frac{n\pi x}{3} ]_0^6 = -\frac{3}{n\pi} (\cos n\pi - 1) $$
So$$ b_n = -\frac{3}{n\pi} (\cos n\pi - 1) $$
Answer:
A function \( f(x) \) is even if
$$ f(-x) = f(x) $$
and odd if
$$ f(-x) = -f(x) $$
Even functions have only cosine terms in their Fourier series; odd functions have only sine terms.
Example 1: Determine if \( f(x) = x^2 \) is even or odd.
Check:
$$ f(-x) = (-x)^2 = x^2 = f(x) $$
Answer: \( f(x) = x^2 \) is even.
Example 2: Determine if \( f(x) = x^3 \) is even or odd.
Check:
$$ f(-x) = (-x)^3 = -x^3 = -f(x) $$
Answer: \( f(x) = x^3 \) is odd.
Example 3: For \( f(x) = \sin x \), check symmetry.
$$ f(-x) = \sin (-x) = -\sin x = -f(x) $$
Answer: \( \sin x \) is odd.
For even functions, the Fourier series contains only cosine terms:
$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos nx $$
with
$$ a_0 = \frac{2}{T} \int_0^{T/2} f(x) \, dx $$
$$ a_n = \frac{2}{T} \int_0^{T/2} f(x) \cos \frac{2n\pi x}{T} \, dx $$
For odd functions, the Fourier series contains only sine terms:
$$ f(x) = \sum_{n=1}^\infty b_n \sin nx $$
with
$$ b_n = \frac{2}{T} \int_0^{T/2} f(x) \sin \frac{2n\pi x}{T} \, dx $$
Example 1: Find the Fourier sine series of the odd function \( f(x) = x \) over \([- \pi, \pi]\).
Given \( f(x) = x \), odd, \( T=2\pi \).
Calculate
$$ b_n = \frac{2}{\pi} \int_0^\pi x \sin nx \, dx $$
Integration by parts as shown before yields
$$ b_n = \frac{2}{n} (-1)^{n+1} $$
Answer:
$$ f(x) = \sum_{n=1}^\infty \frac{2}{n} (-1)^{n+1} \sin nx $$
Example 2: Find the Fourier cosine series of the even function \( f(x) = x^2 \) over \([- \pi, \pi]\).
Calculate
$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi x^2 \, dx = \frac{2}{\pi} \int_0^\pi x^2 \, dx = \frac{2}{\pi} [ \frac{x^3}{3} ]_0^\pi = \frac{2}{\pi} \times \frac{\pi^3}{3} = \frac{2 \pi^2}{3} $$
Calculate
$$ a_n = \frac{2}{\pi} \int_0^\pi x^2 \cos nx \, dx $$
Integration by parts twice (omitted here for brevity) yields
$$ a_n = \frac{4}{n^2} (-1)^n $$
Answer:
$$ f(x) = \frac{\pi^2}{3} + \sum_{n=1}^\infty \frac{4}{n^2} (-1)^n \cos nx $$
Harmonic analysis studies the representation of signals or functions as sums of basic waves (harmonics). In chemical engineering, it is used to analyze oscillations in control loops, vibration analysis in rotating equipment, and signal processing in sensors.
Example 1: A temperature oscillation in a reactor is modeled by \( f(t) = 5 + 2 \cos t + 3 \sin 2t \). Identify the fundamental and harmonics.
Given:
The fundamental frequency corresponds to \( \omega = 1 \).
Harmonics:
Answer:
Fundamental frequency component amplitude: 2
Second harmonic amplitude: 3
Example 2: A pressure signal in a pipeline is given by \( p(t) = 10 + 4 \sin 3t + 2 \cos 6t \). Find amplitudes of harmonics.
Given:
Fundamental frequency \( \omega = 3 \) rad/s.
Harmonics:
Answer:
Amplitudes 4 and 2 for first and second harmonics respectively
When analytical integration is difficult, numerical methods approximate coefficients \( a_n \), \( b_n \) using discrete summation.
Using \( N \) sample points \( x_k = x_0 + k \Delta x \), \( k = 0, 1, ..., N-1 \), and \( \Delta x = \frac{T}{N} \):
$$ a_0 \approx \frac{2}{N} \sum_{k=0}^{N-1} f(x_k) $$
$$ a_n \approx \frac{2}{N} \sum_{k=0}^{N-1} f(x_k) \cos \frac{2n\pi x_k}{T} $$
$$ b_n \approx \frac{2}{N} \sum_{k=0}^{N-1} f(x_k) \sin \frac{2n\pi x_k}{T} $$
Example 1: Approximate \( a_0 \) for \( f(x) = x \) over \( [0, 2\pi] \) using 4 sample points \( x_k = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \).
Calculate \( f(x_k) \):
Sum:
$$ S = 0 + 1.571 + 3.142 + 4.712 = 9.425 $$
Approximate \( a_0 \):
$$ a_0 \approx \frac{2}{4} \times 9.425 = 0.5 \times 9.425 = 4.712 $$
Exact value from integration is \( a_0 = 2\pi = 6.283 \), so approximation has error due to low sample count.
Answer: Approximate \( a_0 = 4.712 \)
Example 2: Approximate \( b_1 \) for \( f(x) = x \) over \( [0, 2\pi] \) with 4 samples.
Calculate \( \sin \frac{2\pi x_k}{T} = \sin n \theta_k \) with \( n=1 \):
Calculate sum:
$$ S = f(0) \times 0 + f(\pi/2) \times 1 + f(\pi) \times 0 + f(3\pi/2) \times (-1) = 0 + 1.571 + 0 - 4.712 = -3.141 $$
Approximate \( b_1 \):
$$ b_1 \approx \frac{2}{4} \times (-3.141) = -1.571 $$
Exact value is \( b_1 = -2 \).
Answer: Approximate \( b_1 = -1.571 \)
Example 3: Estimate \( a_1 \) for \( f(x) = \cos x \) over \( [0, 2\pi] \) using 6 sample points: \( x_k = 0, \frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}, \frac{5\pi}{5} = \pi \).
Calculate \( f(x_k) \) and \( \cos \frac{2\pi x_k}{T} \) with \( T = 2\pi \), \( n = 1 \):
Now, sum:
$$ S = f(0) \times 1 + f(\frac{\pi}{5}) \times 0.951 + f(\frac{2\pi}{5}) \times 0.809 + f(\frac{3\pi}{5}) \times 0.588 + f(\frac{4\pi}{5}) \times 0.309 + f(\pi) \times (-1) $$
Plug in values:
$$ S = 1 \times 1 + 0.951 \times 0.951 + 0.809 \times 0.809 + 0.588 \times 0.588 + 0.309 \times 0.309 + (-1) \times (-1) $$
$$ S = 1 + 0.904 + 0.655 + 0.346 + 0.095 + 1 = 3.999 $$
Approximate \( a_1 \):
$$ a_1 \approx \frac{2}{6} \times 3.999 = 0.333 \times 3.999 \approx 1.333 $$
Answer: Approximate \( a_1 = 1.333 \)
Find the first three Fourier coefficients \( a_0 \), \( a_1 \), and \( b_1 \) of the function \( f(x) = x \) over the interval \( [-\pi, \pi] \). (6 marks)
Determine the period and write the Fourier series form for \( f(t) = \cos 5t \). (4 marks)
Calculate the Fourier cosine series coefficients \( a_0 \) and \( a_1 \) for the even function \( f(x) = x^2 \) over \( [-\pi, \pi] \). (6 marks)
Using numerical approximation with \( N=6 \) sample points, estimate \( a_0 \) for \( f(x) = \sin x \) over \( [0, 2\pi] \). (6 marks)
For the odd function \( f(x) = x \) defined in \( [0, \pi] \), compute the first two sine Fourier coefficients \( b_1 \) and \( b_2 \). (8 marks)
This chapter introduced the concept of Fourier series as a method to represent periodic functions by decomposing them into sums of sines and cosines. The process of obtaining the Fourier series coefficients \(a_0\), \(a_n\), and \(b_n\) was explained, showing how to analyze functions defined over periods ranging from \(2\pi\) to a general period \(T\). The chapter further explored the classification of functions into odd and even types, demonstrating how these properties simplify the calculation of Fourier series coefficients. Techniques for deriving Fourier series specifically for odd and even functions were covered in detail. The role of harmonic analysis was introduced, highlighting its practical applications in engineering problems involving signal processing and system analysis. Finally, numerical methods for approximating Fourier coefficients were discussed, providing tools for handling functions where analytical integration is difficult. Together, these topics equip students with a thorough understanding of Fourier series and their use in engineering mathematics.
Find the Fourier series coefficients \(a_0\), \(a_n\), and \(b_n\) for the periodic function \(f(x) = x\) defined on the interval \([- \pi, \pi]\). (2 marks)
A temperature variation in a chemical reactor is modeled by the periodic function \(f(t) = 5 + 3 \sin t + 2 \cos 2t\) over the period \(2 \pi\). Calculate the first three non-zero Fourier coefficients. (3 marks)
Determine the Fourier sine series for the function \(f(x) = x\) defined on \([0, \pi]\), assuming it is extended as an odd function. (3 marks)
For a chemical concentration profile defined by \(f(x) = x^2\) on \([- \pi, \pi]\), calculate the first two cosine coefficients \(a_0\) and \(a_2\) of its Fourier series. (3 marks)
A pressure fluctuation in a pipeline is modeled by an even function \(f(x) = |x|\) on \([- \pi, \pi]\). Find the Fourier cosine coefficients \(a_0\) and \(a_1\). (4 marks)
A periodic flow rate \(Q(t)\) in a pump is defined over the period \(T = 4\) seconds by \(Q(t) = t\) for \(0 \le t < 2\) and \(Q(t) = 4 - t\) for \(2 \le t < 4\). Compute the coefficient \(a_0\) and the first sine coefficient \(b_1\). (4 marks)
Using harmonic analysis, estimate the amplitude of the third harmonic component for the function \(f(t) = 10 + 6 \cos(\pi t) + 4 \sin(2 \pi t) + 2 \cos(3 \pi t)\). (3 marks)
A chemical sensor output varies as \(f(t) = t\) for \(-1 \le t \le 1\) and repeats periodically with period 2. Approximate the first two Fourier coefficients \(a_0\) and \(b_1\) using numerical integration by Simpson’s rule with \(n=4\) intervals. (5 marks)
Derive the Fourier series for the function \(f(x) = x^3\) defined on \([- \pi, \pi]\) and calculate the first non-zero sine coefficient \(b_1\). (5 marks)
A temperature profile in a chemical reactor is given by a triangular wave with period \(T = 6\), defined by \(f(t) = 2t\) for \(0 \le t < 3\) and \(f(t) = 6 - 2t\) for \(3 \le t < 6\). Calculate the coefficient \(a_0\) and the first two cosine coefficients \(a_1\) and \(a_2\). (6 marks)
1. Fourier series coefficients for \(f(x) = x\), \([- \pi, \pi]\)
Given: \(f(x) = x\), period \(2 \pi\)
Formulae:
$$a_0 = \frac{1}{2 \pi} \int_{- \pi}^{\pi} f(x) \, dx$$$$a_n = \frac{1}{\pi} \int_{- \pi}^{\pi} f(x) \cos n x \, dx$$$$b_n = \frac{1}{\pi} \int_{- \pi}^{\pi} f(x) \sin n x \, dx$$
Calculate \(a_0\):
$$a_0 = \frac{1}{2 \pi} \int_{- \pi}^{\pi} x \, dx = \frac{1}{2 \pi} [ \frac{x^2}{2} ]_{- \pi}^{\pi} = \frac{1}{2 \pi} ( \frac{\pi^2}{2} - \frac{\pi^2}{2} ) = 0$$
Calculate \(a_n\):
Since \(x\) is odd and \(\cos n x\) is even, the product \(x \cos n x\) is odd:
$$a_n = \frac{1}{\pi} \int_{- \pi}^{\pi} x \cos n x \, dx = 0$$
Calculate \(b_n\):
Since both \(x\) and \(\sin n x\) are odd, their product is even:
$$b_n = \frac{1}{\pi} \int_{- \pi}^{\pi} x \sin n x \, dx = \frac{2}{\pi} \int_0^{\pi} x \sin n x \, dx$$
Integrate by parts:
Let \(u = x\), \(dv = \sin n x \, dx\), then \(du = dx\), \(v = - \frac{\cos n x}{n}\)
$$\int x \sin n x \, dx = - \frac{x \cos n x}{n} + \frac{1}{n} \int \cos n x \, dx = - \frac{x \cos n x}{n} + \frac{1}{n^2} \sin n x + C$$
Evaluate definite integral:
$$\int_0^{\pi} x \sin n x \, dx = [ - \frac{x \cos n x}{n} + \frac{1}{n^2} \sin n x ]_0^{\pi} = - \frac{\pi \cos n \pi}{n} + \frac{1}{n^2} \sin n \pi - (0 + 0)$$
Since \(\sin n \pi = 0\),
$$= - \frac{\pi \cos n \pi}{n} = - \frac{\pi (-1)^n}{n}$$
$$b_n = \frac{2}{\pi} \times ( - \frac{\pi (-1)^n}{n} ) = \frac{2 (-1)^{n+1}}{n}$$
Final answer:
$$a_0 = 0, \quad a_n = 0, \quad b_n = \frac{2 (-1)^{n+1}}{n}$$
2. Fourier coefficients for \(f(t) = 5 + 3 \sin t + 2 \cos 2t\), period \(2 \pi\)
Given function is already in Fourier form:
The first three non-zero coefficients are:
$$a_0 = 5, \quad b_1 = 3, \quad a_2 = 2$$
3. Fourier sine series for \(f(x) = x\) on \([0, \pi]\), odd extension
Formula for sine coefficients:
$$b_n = \frac{2}{\pi} \int_0^{\pi} x \sin n x \, dx$$
Integrate by parts:
Let \(u = x\), \(dv = \sin n x \, dx\), so \(du = dx\), \(v = - \frac{\cos n x}{n}\)
$$\int_0^{\pi} x \sin n x \, dx = [ - \frac{x \cos n x}{n} ]_0^{\pi} + \frac{1}{n} \int_0^{\pi} \cos n x \, dx$$
Calculate:
$$- \frac{\pi \cos n \pi}{n} + \frac{1}{n} [ \frac{\sin n x}{n} ]_0^{\pi} = - \frac{\pi (-1)^n}{n} + 0 = - \frac{\pi (-1)^n}{n}$$
$$b_n = \frac{2}{\pi} \times ( - \frac{\pi (-1)^n}{n} ) = \frac{2 (-1)^{n+1}}{n}$$
Final answer:
Sine series:
$$f(x) = \sum_{n=1}^\infty \frac{2 (-1)^{n+1}}{n} \sin n x$$
4. Fourier coefficients for \(f(x) = x^2\), \([- \pi, \pi]\)
Calculate:
$$a_0 = \frac{1}{2 \pi} \int_{- \pi}^{\pi} x^2 \, dx = \frac{1}{2 \pi} \times 2 \int_0^{\pi} x^2 \, dx = \frac{1}{\pi} [ \frac{x^3}{3} ]_0^{\pi} = \frac{1}{\pi} \times \frac{\pi^3}{3} = \frac{\pi^2}{3}$$
Calculate \(a_2\):
$$a_n = \frac{1}{\pi} \int_{- \pi}^{\pi} x^2 \cos n x \, dx = \frac{2}{\pi} \int_0^{\pi} x^2 \cos n x \, dx$$
Use integration by parts twice:
Let \(I = \int_0^{\pi} x^2 \cos n x \, dx\).
First integration by parts:
\(u = x^2\), \(dv = \cos n x \, dx\), so \(du = 2x dx\), \(v = \frac{\sin n x}{n}\)
$$I = [ x^2 \frac{\sin n x}{n} ]_0^{\pi} - \frac{2}{n} \int_0^{\pi} x \sin n x \, dx$$
Since \(\sin n \pi = 0\) and \(\sin 0 = 0\), first term is zero.
Now, evaluate \(\int_0^{\pi} x \sin n x \, dx\) (from problem 1):
$$\int_0^{\pi} x \sin n x \, dx = - \frac{\pi (-1)^n}{n}$$
$$I = - \frac{2}{n} \times ( - \frac{\pi (-1)^n}{n} ) = \frac{2 \pi (-1)^n}{n^2}$$
Finally,
$$a_n = \frac{2}{\pi} \times I = \frac{2}{\pi} \times \frac{2 \pi (-1)^n}{n^2} = \frac{4 (-1)^n}{n^2}$$
For \(n=2\):
$$a_2 = \frac{4 (-1)^2}{2^2} = \frac{4 \times 1}{4} = 1$$
Final answers:
$$a_0 = \frac{\pi^2}{3}, \quad a_2 = 1$$
5. Fourier cosine coefficients for \(f(x) = |x|\), \([- \pi, \pi]\)
Since \(f\) is even:
$$a_0 = \frac{1}{\pi} \int_{- \pi}^{\pi} |x| \, dx = \frac{2}{\pi} \int_0^{\pi} x \, dx = \frac{2}{\pi} \times \frac{\pi^2}{2} = \pi$$
Calculate \(a_1\):
$$a_1 = \frac{1}{\pi} \int_{- \pi}^{\pi} |x| \cos x \, dx = \frac{2}{\pi} \int_0^{\pi} x \cos x \, dx$$
Integrate by parts:
\(u = x\), \(dv = \cos x \, dx\), so \(du = dx\), \(v = \sin x\)
$$\int_0^{\pi} x \cos x \, dx = [ x \sin x ]_0^{\pi} - \int_0^{\pi} \sin x \, dx = \pi \sin \pi - ( - \cos x)_0^{\pi} = 0 - ( - \cos \pi + \cos 0) = 1 + 1 = 2$$
$$a_1 = \frac{2}{\pi} \times 2 = \frac{4}{\pi}$$
Final answers:
$$a_0 = \pi, \quad a_1 = \frac{4}{\pi}$$
6. Fourier coefficients for periodic flow \(Q(t)\), period \(T=4\)
Given:
\[ Q(t) = \begin{cases} t, & 0 \le t < 2 \\ 4 - t, & 2 \le t < 4 \end{cases} \]
Calculate \(a_0\):
$$a_0 = \frac{1}{T} \int_0^T Q(t) \, dt = \frac{1}{4} ( \int_0^2 t \, dt + \int_2^4 (4 - t) \, dt )$$
Calculate integrals:
$$\int_0^2 t \, dt = \frac{2^2}{2} = 2$$
$$\int_2^4 (4 - t) \, dt = [ 4t - \frac{t^2}{2} ]_2^4 = (16 - 8) - (8 - 2) = 8 - 6 = 2$$
$$a_0 = \frac{1}{4} (2 + 2) = 1$$
Calculate \(b_1\):
$$b_1 = \frac{2}{T} \int_0^T Q(t) \sin ( \frac{2 \pi t}{T} ) dt = \frac{1}{2} ( \int_0^2 t \sin \frac{\pi t}{2} dt + \int_2^4 (4 - t) \sin \frac{\pi t}{2} dt )$$
First integral:
Use integration by parts:
Let \(u = t\), \(dv = \sin \frac{\pi t}{2} dt\), so \(du = dt\), \(v = - \frac{2}{\pi} \cos \frac{\pi t}{2}\)
$$\int_0^2 t \sin \frac{\pi t}{2} dt = [ - \frac{2t}{\pi} \cos \frac{\pi t}{2} ]_0^2 + \frac{2}{\pi} \int_0^2 \cos \frac{\pi t}{2} dt$$
Evaluate first term:
At \(t=2\):
$$- \frac{4}{\pi} \cos \pi = - \frac{4}{\pi} (-1) = \frac{4}{\pi}$$
At \(t=0\):
$$0$$
Integral:
$$\int_0^2 \cos \frac{\pi t}{2} dt = [ \frac{2}{\pi} \sin \frac{\pi t}{2} ]_0^2 = \frac{2}{\pi} ( \sin \pi - 0) = 0$$
So,
$$\int_0^2 t \sin \frac{\pi t}{2} dt = \frac{4}{\pi} + 0 = \frac{4}{\pi}$$
Second integral:
\[ \int_2^4 (4 - t) \sin \frac{\pi t}{2} dt = 4 \int_2^4 \sin \frac{\pi t}{2} dt - \int_2^4 t \sin \frac{\pi t}{2} dt \]
Calculate each term:
\[ 4 \int_2^4 \sin \frac{\pi t}{2} dt = 4 [ - \frac{2}{\pi} \cos \frac{\pi t}{2} ]_2^4 = - \frac{8}{\pi} ( \cos 2\pi - \cos \pi ) = - \frac{8}{\pi} (1 - (-1)) = - \frac{8}{\pi} \times 2 = - \frac{16}{\pi} \]
Calculate \(\int_2^4 t \sin \frac{\pi t}{2} dt\) by integration by parts:
Let \(u = t\), \(dv = \sin \frac{\pi t}{2} dt\), \(du = dt\), \(v = - \frac{2}{\pi} \cos \frac{\pi t}{2}\)
$$\int_2^4 t \sin \frac{\pi t}{2} dt = [ - \frac{2t}{\pi} \cos \frac{\pi t}{2} ]_2^4 + \frac{2}{\pi} \int_2^4 \cos \frac{\pi t}{2} dt$$
Evaluate first term:
At \(t=4\):
$$- \frac{8}{\pi} \cos 2\pi = - \frac{8}{\pi} \times 1 = - \frac{8}{\pi}
The Fourier series representation of \( f(t) \) with period \( T \) is$$ f(t) = a_0 + \sum_{n=1}^{\infty} [ a_n \cos(\frac{2n\pi t}{T}) + b_n \sin(\frac{2n\pi t}{T}) ] $$ where$$ a_0 = \frac{1}{T} \int_0^T f(t) dt, \quad a_n = \frac{2}{T} \int_0^T f(t) \cos(\frac{2n\pi t}{T}) dt, \quad b_n = \frac{2}{T} \int_0^T f(t) \sin(\frac{2n\pi t}{T}) dt. $$
Since \( f(-x) = -x = -f(x) \), \( f(x) = x \) is an odd function.
Given \( f(x) = 3 + 2\cos(x) \),$$ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} (3 + 2\cos x) dx = \frac{1}{2\pi} ( 3 \times 2\pi + 2 \times 0 ) = 3. $$
For \( f(x) = x^2 \),$$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos(nx) dx = \frac{2}{\pi} \int_0^{\pi} x^2 \cos(nx) dx $$ (The integral evaluates to)$$ a_n = \frac{4(-1)^n}{n^2}. $$
Harmonic components represent frequencies at multiples of the fundamental frequency. They help identify periodic disturbances or oscillations in the pressure signal, indicating potential issues in the distillation process.
For an odd function,$$ b_n = \frac{2}{L} \int_0^L f(x) \sin(\frac{n\pi x}{L}) dx, \quad a_n = 0, \quad a_0 = 0. $$
The Trapezoidal rule is suitable due to its balance between accuracy and computational simplicity for approximating integrals of measured data.
$$ a_n = \frac{2}{T} \int_0^T f(t) \cos(\frac{2n\pi t}{T}) dt, \quad b_n = \frac{2}{T} \int_0^T f(t) \sin(\frac{2n\pi t}{T}) dt. $$
Even function: \( f(-x) = f(x) \), example: temperature distribution along a symmetric reactor vessel.
Odd function: \( f(-x) = -f(x) \), example: flow velocity profile with direction reversal.
Harmonic analysis isolates frequency components of signals from sensors; abnormal harmonics indicate faults like valve oscillations or sensor malfunctions.
Question 11 (Compulsory - 20 marks)
Kenya Pipeline Company (KPC) monitors pressure fluctuations in its pipelines using sensors producing periodic signals.
a) Given a pressure signal \( p(t) \) with period \( T=6 \) seconds, derive the general Fourier series representation for \( p(t) \). (10 marks)
b) Using numerical integration with the Trapezoidal rule, outline the steps to approximate the first three Fourier coefficients \( a_0, a_1, b_1 \) from sampled sensor data. (10 marks)
Question 12 (20 marks)
A temperature profile in a chemical reactor is modelled by the function \( f(x) = x \) for \( -\pi \leq x \leq \pi \) and extended periodically with period \( 2\pi \).
a) Determine the Fourier sine series representation of \( f(x) \).
b) Calculate the first four nonzero Fourier coefficients \( b_n \).
Question 13 (20 marks)
In a chemical plant, vibration signals from rotating equipment are periodic and analyzed to detect faults.
a) Explain how the Fourier series helps in harmonic analysis of vibration signals.
b) For a function defined over \( [0, T] \), derive expressions for \( a_n \) and \( b_n \) when the function is even or odd, respectively.
Question 14 (20 marks)
A flow rate function \( f(t) \) given over period \( T = 4 \) seconds is sampled at 5 equally spaced points: \( t = 0, 1, 2, 3, 4 \) seconds with values \( f = [2, 3, 5, 3, 2] \) m\(^3\)/s.
a) Use the Trapezoidal rule to approximate \( a_0 \).
b) Approximate \( a_1 \) and \( b_1 \) using numerical integration.
Question 11
a) The Fourier series of \( p(t) \) with period \( T=6 \) s is:$$ p(t) = a_0 + \sum_{n=1}^\infty [ a_n \cos( \frac{2n\pi t}{6} ) + b_n \sin( \frac{2n\pi t}{6} ) ] $$
where$$ a_0 = \frac{1}{6} \int_0^6 p(t) dt, \quad a_n = \frac{2}{6} \int_0^6 p(t) \cos( \frac{2n\pi t}{6} ) dt, \quad b_n = \frac{2}{6} \int_0^6 p(t) \sin( \frac{2n\pi t}{6} ) dt. $$
b) Steps for numerical approximation using Trapezoidal rule:
Compute integrals for \( a_0, a_1, b_1 \) by approximating integrals as sums:$$ a_0 \approx \frac{1}{6} \sum_{i=0}^{N} p(t_i) \Delta t_i $$$$ a_1 \approx \frac{2}{6} \sum_{i=0}^N p(t_i) \cos( \frac{2\pi t_i}{6} ) \Delta t_i $$$$ b_1 \approx \frac{2}{6} \sum_{i=0}^N p(t_i) \sin( \frac{2\pi t_i}{6} ) \Delta t_i $$
Use Trapezoidal weights: average adjacent function values multiplied by interval width.
Question 12
a) Since \( f(x) = x \) is an odd function, its Fourier series contains only sine terms:$$ f(x) = \sum_{n=1}^\infty b_n \sin(nx), \quad b_n = \frac{2}{\pi} \int_0^\pi x \sin(nx) dx. $$
b) Calculation of \( b_n \):
Using integration by parts,$$ b_n = \frac{2}{\pi} [ \frac{\pi (-1)^{n+1}}{n} + \frac{1 - (-1)^n}{n^2} ] = \frac{2(-1)^{n+1}}{n}. $$
First four nonzero coefficients:
\( b_1 = 2 \), \( b_3 = \frac{2}{3} \), \( b_5 = \frac{2}{5} \), \( b_7 = \frac{2}{7} \) (with alternating signs).
Question 13
a) Fourier series decomposes vibration signals into harmonic components with frequencies corresponding to multiples of the fundamental frequency, enabling identification of abnormal vibrations indicating faults.
b) For even functions:$$ a_n = \frac{2}{T} \int_0^{T} f(t) \cos( \frac{2n\pi t}{T} ) dt, \quad b_n = 0. $$ For odd functions:$$ b_n = \frac{2}{T} \int_0^{T} f(t) \sin( \frac{2n\pi t}{T} ) dt, \quad a_n = 0, \quad a_0 = 0. $$
Question 14
Given \( T=4 \) s, \( N=4 \) intervals with \( \Delta t = 1 \) s.
a) Approximate \( a_0 \):$$ a_0 = \frac{1}{4} \int_0^4 f(t) dt \approx \frac{1}{4} \times \text{Trapezoidal sum} $$ Trapezoidal sum:$$ \frac{1}{2}(2 + 3) + \frac{1}{2}(3 + 5) + \frac{1}{2}(5 + 3) + \frac{1}{2}(3 + 2) = (2.5 + 4 + 4 + 2.5) = 13 $$ Integral approx:$$ \int_0^4 f(t) dt \approx 13 \times 1 = 13 $$$$ a_0 \approx \frac{13}{4} = 3.25 \text{ m}^3/\text{s} $$
b) Approximate \( a_1 \) and \( b_1 \):
Calculate$$ a_1 = \frac{2}{4} \int_0^4 f(t) \cos( \frac{2\pi t}{4} ) dt = \frac{1}{2} \int_0^4 f(t) \cos( \frac{\pi t}{2} ) dt $$
and$$ b_1 = \frac{1}{2} \int_0^4 f(t) \sin( \frac{\pi t}{2} ) dt. $$
Using Trapezoidal rule with sample points:
Calculate \( f(t_i) \cos(\frac{\pi t_i}{2}) \) and \( f(t_i) \sin(\frac{\pi t_i}{2}) \) at \( t_i = 0,1,2,3,4 \), then sum with trapezoidal weights and multiply by \( \frac{1}{2} \).
Final numerical values:
\( a_1 \approx -0.75 \) m\(^3\)/s,
\( b_1 \approx 0.5 \) m\(^3\)/s.