TRIGONOMETRIC FOURIER SERIES OF PERIODIC...
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EECS 216 LECTURE NOTES TRIGONOMETRIC FOURIER SERIES OF PERIODIC SIGNALS THEOREM: Let x(t) be a bounded periodic signal with period T . Then x(t) can be expanded as a weighted sum of sinusoids with angular frequencies that are integer multiples of ω 0 = 2π T : x(t)= a 0 + a 1 cos(ω 0 t)+ a 2 cos(2ω 0 t)+ a 3 cos(3ω 0 t)+ ... + b 1 sin(ω 0 t)+ b 2 sin(2ω 0 t)+ b 3 sin(3ω 0 t)+ ... This is the trigonometric Fourier series expansion of x(t). Or, we can use only cosines with phase shifts: x(t)= a 0 + c 1 cos(ω 0 t − φ 1 )+ c 2 cos(2ω 0 t − φ 2 )+ c 3 cos(3ω 0 t − φ 3 )+ ... PROOF: Take a high-level math course to see this done properly. NOTE: A Fourier series is a mathematical version of a prism. COMPUTATION OF FOURIER SERIES COEFFICIENTS: THEOREM: Coefficients a n , b n , c n and φ n can be computed using: a n = 2 T t 0 +T t 0 x(t) cos(nω 0 t)dt and b n = 2 T t 0 +T t 0 x(t) sin(nω 0 t)dt. For n = 0 we have: a 0 = 1 T t 0 +T t 0 x(t)dt=average value of x(t). For cosines with phase shifts: c n = a 2 n + b 2 n ; φ n = tan -1 ( b n a n ),n =0 using a n cos(ωt)+ b n sin(ωt)= a 2 n + b 2 n cos(ωt − tan -1 ( b n a n )) (a n > 0). Derive using phasors: a n +b n e -jπ/2 = a n −jb n = a 2 n + b 2 n e -j tan -1 ( b n a n ) . EXAMPLE OF FOURIER SERIES DECOMPOSITION: x(t)= π/4 for (2kπ) <t< (2k + 1)π Square wave with period −π/4 for (2k − 1)π<t< (2kπ) T =2π → ω o = 2π 2π =1 a n = 2 2π π -π x(t) cos(nω 0 t)dt = 0 by inspection: Set t 0 = − T 2 = −π. b n = 2 2π 2π 0 x(t) sin(nt)dt = 1 π [ π 0 ( π 4 ) sin(nt)dt + 2π π (− π 4 ) sin(nt)dt] = 1 n , if n is odd; 0, if n is even. → x(t) = sin(t)+ 1 3 sin(3t)+ 1 5 sin(5t)+ ... PARSEVAL’S THEORM FOR THIS EXAMPLE Average 1 T T 0 |x(t)| 2 dt = 1 2π 2π 0 |± π 4 | 2 dt = π 2 16 . This agrees with: Power: a 2 0 + 1 2 ∑ ∞ k=1 (a 2 k + b 2 k )= 1 2 ∑ ∞ k odd ( 1 k ) 2 = 1 2 (1 + 1 3 2 + 1 5 2 + ...)= π 2 16 .
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Transcript of TRIGONOMETRIC FOURIER SERIES OF PERIODIC...
EECS 216 LECTURE NOTES
TRIGONOMETRIC FOURIER SERIES OF PERIODIC SIGNALS
THEOREM: Let x(t) be a bounded periodic signal with period T .Then x(t) can be expanded as a weighted sum of sinusoidswith angular frequencies that are integer multiples of ω0 = 2π
T :
x(t) = a0 + a1 cos(ω0t) + a2 cos(2ω0t) + a3 cos(3ω0t) + . . .+ b1 sin(ω0t) + b2 sin(2ω0t) + b3 sin(3ω0t) + . . .
This is the trigonometric Fourier series expansion of x(t).
Or, we can use only cosines with phase shifts:
x(t) = a0 + c1 cos(ω0t−φ1)+ c2 cos(2ω0t−φ2)+ c3 cos(3ω0t−φ3)+ . . .
PROOF: Take a high-level math course to see this done properly.NOTE: A Fourier series is a mathematical version of a prism.
COMPUTATION OF FOURIER SERIES COEFFICIENTS:
THEOREM: Coefficients an, bn, cn and φn can be computed using:
an = 2T
∫ t0+T
t0x(t) cos(nω0t)dt and bn = 2
T
∫ t0+T
t0x(t) sin(nω0t)dt.
For n = 0 we have: a0 = 1T
∫ t0+T
t0x(t)dt=average value of x(t).
For cosines with phase shifts: cn =√
a2n + b2
n;φn = tan−1( bn
an
), n 6= 0
using an cos(ωt)+bn sin(ωt) =√
a2n + b2
n cos(ωt−tan−1( bn
an
)) (an > 0).
Derive using phasors: an+bne−jπ/2 = an−jbn =√
a2n + b2
ne−j tan−1( bn
an).
EXAMPLE OF FOURIER SERIES DECOMPOSITION:
x(t) =
{
π/4 for (2kπ) < t < (2k + 1)π Squarewavewith period−π/4 for (2k − 1)π < t < (2kπ) T = 2π → ωo = 2π
2π= 1
an = 22π
∫ π
−πx(t) cos(nω0t)dt = 0 by inspection: Set t0 = −T
2 = −π.
bn = 22π
∫ 2π
0x(t) sin(nt)dt = 1
π [∫ π
0(π
4 ) sin(nt)dt +∫ 2π
π(−π
4 ) sin(nt)dt]
=
{
1n , if n is odd;0, if n is even.
→ x(t) = sin(t) + 13 sin(3t) + 1
5 sin(5t) + . . .
PARSEVAL’S THEORM FOR THIS EXAMPLE
Average 1T
∫ T
0|x(t)|2dt = 1
2π
∫ 2π
0| ± π
4 |2dt = π2
16 . This agrees with:
Power: a20 + 1
2
∑
∞
k=1(a2k + b2
k) = 12
∑
∞
k odd(1k )2 = 1
2 (1 + 132 + 1
52 + . . .) = π2
16 .
EECS 216 LECTURE NOTES
PROOF OF THEOREM: First we need the following lemma:
LEMMA: The sine and cosine functions are orthogonal functions:∫ t0+T
t0cos(iω0t) cos(jω0t)dt =
∫ t0+T
t0sin(iω0t) sin(jω0t)dt =
{
T/2, if i = j;0, if i 6= j.
∫ t0+T
t0cos(iω0t) sin(jω0t)dt = 0 (even if i = j). These assume i, j > 0.
PROOF OF LEMMA: Adding and subtracting the cosine additionformula cos(x ± y) = cos(x) cos(y) ∓ sin(x) sin(y) gives the formulae
2 cos(x) cos(y) = cos(x − y) + cos(x + y)2 sin(x) sin(y) = cos(x − y) − cos(x + y).
Setting x = iω0t and y = jω0t and∫ t0+T
t0dt gives
2∫ t0+T
t0cos(iω0t) cos(jω0t)dt =
∫ t0+T
t0[cos((i + j)ω0t) + cos((i − j)ω0t)]dt.
Since the integral of a sinusoid with nonzero frequency over an integernumber i ± j of periods is zero, the first part of the lemma follows.The other two parts follow similarly. QED.
PROOF: Multiply the Fourier series by cos(nω0t) and∫ t0+T
t0dt:
∫ t0+T
t0x(t) cos(nω0t)dt =
∫ t0+T
t0a0 cos(nω0t)dt
+∫ t0+T
t0a1 cos(nω0t) cos(ω0t)dt +
∫ t0+T
t0a2 cos(nω0t) cos(2ω0t)dt + . . .
+∫ t0+T
t0b1 cos(nω0t) sin(ω0t)dt +
∫ t0+T
t0b2 cos(nω0t) sin(2ω0t)dt + . . .
= 0 + 0 + . . . + 0 + anT2
+ 0 + . . . from which the an formula follows.
The formulae for a0 and bn follow similarly. QED.
COMMENTS:
1. t0 is arbitrary; all integrals are over one period.2. ω0 = 2π
T is angular frequency in RADIANSECOND ; this is 1
T Hertz.3. The sinusoid at frequency ω0 is called the fundamental;
the sinusoid at frequency nω0 is the nth harmonic.Harmonics are also called overtones (not used much anymore).Some say the nth harmonic is at frequency (n + 1)ωo.
4. The more terms (i.e., harmonics) we keep in the Fourier series,the better the approximation the truncated series is to x(t).
5. If x(t) has a discontinuity at t = t1, the Fourier series converges to12(x(t−1 ) + x(t+1 )), where x(t−) and x(t+) are the values of x(t) on
either side of the discontinuity.
0 2 4 6 8 10−1
−0.5
0
0.5
1sin(t)
0 2 4 6 8 10−1
−0.5
0
0.5
1sin(t)+sin(3t)/3+sin(5t)/5
0 2 4 6 8 10−1
−0.5
0
0.5
1sin(t)+...+sin(15t)/15
0 2 4 6 8 10−1
−0.5
0
0.5
1sin(t)+...+sin(25t)/25
0 2 4 6 8 10−2
−1
0
1
2cos(t)+cos(3t)/9+cos(5t)/25
0 2 4 6 8 10−2
−1
0
1
2cos(t)+...+cos(9t)/81
