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Fourier series: Solved problems pHabala 2012
Solved problems on Fourier series
1. Find the Fourier series for (periodic extension of)
f(t) = 1, t ∈ [0,2);
Determine the sum of this series. −1, t∈[2,4).
2. Find the Fourier series for (periodic extension of)
f(t) = t−1, t ∈ [0,2);
Determine the sum of this series. 3−t, t∈[2,4).
3. Find the sine Fourier series for (periodic extension of)
f(t) = t−1, t ∈ [0,2);
Determine the sum of this series. 3−t, t∈[2,4).
4. Find the cosine Fourier series for (periodic extension of)
f(t) = 1, t ∈ [0,1);
Determine the sum of this series. 0, t ∈ [1,4).
5. Find the Fourier series for (periodic extension of)
2
Determine the sum of this series. f(t) = 1−t , t ∈ [−1,1).
Solutions
1. Parameters: The period length is T = 4, frequency ω = 2π = π.
T 2
T 2 4
2Z 1Z Z
a0 = T f(t)dt = 2 1dt− 1dt =0,
0 0 2
T 2 4
2Z 1Z π Z π
ak = T f(t)cos(kωt)dt = 2 cos k2t dt− cos k2t dt
0 0 2
h i h i
1 2 π 2 1 2 π 4
= 2 kπ sin k2t 0 − 2 kπ sin k2t 2 = 0,
T 2 4
2Z 1Z π Z π
b = f(t)sin(kωt)dt = sin k t dt− sin k t dt
k T 2 2 2
0 0 2
h i h i
1 2 π 2 1 2 π 4 1
= 2 −kπ cos k2t 0− 2 −kπ cos k2t 2 = kπ −cos(kπ)+cos(0)+cos(2kπ)−cos(kπ)
1 k k 2 k 0, k even,
= kπ −(−1) +1+1−(−1) = kπ[1−(−1) ]= 4 , k odd.
kπ
Odd numbers can be expressed as k = 2i + 1, to numbers k = 1,3,5,7,... correspond indeces
i = 0,1,2,3,.... For those we then have a = 4 . We rewrite the resulting series accordingly,
k (2i +1)π
and since the index k is traditional, we pass from i to k at the end.
Thus
∞ ∞ ∞
a0 X X2 k π X 4 π
f ∼ + a cos(kωt)+b sin(kωt) = [1−(−1) ]sin k t = sin (2k+1) t .
2 k k kπ 2 (2k +1)π 2
k=1 k=1 k=0
What is the sum of this series? First we draw a periodic extension of the function f (on the left).
To this we then apply the Jordan criterion. According to it, the resulting series converges to f at all
Fourier series: Solved problems pHabala 2012
points where f (or rather its periodic extension) is continuous. At points of discontinuity of f the
1 + −
series converges to the average 2 f(t ) + f(t ) . Result: On the right is the function to which our
Fourier series converges, i.e. its sum.
1 f(t) 1 F(t)
0 0
−1 2 4 6 8 10 −1 2 4 6 8 10
−2 −2
2. Parameters: The period length is T = 4, frequency ω = 2π = π.
T 2
T 2 4
2Z 1Z Z
a = f(t)dt = t −1dt+ 3−tdt =0.
0 T 2
0 0 2
T 2 4
Z Z Z
2 1 π π
ak = T f(t)cos(kωt)dt = 2 (t − 1)cos k2t dt+ (3−t)cos k2t dt = by parts
0 0 2
2 4
h i Z h i Z
1 2 π 2 1 2 π 1 2 π 4 1 2 π
= 2 (t−1)kπ sin k2t 0 − 2 kπ sin k2t dt+ 2 (3−t)kπ sin k2t 2 + 2 kπ sin k2t dt
h i h 0 i 2
2 π 2 2 π 4
=0+ k2π2 cos k2t 0+0− k2π2 cos k2t 2
= 2 [cos(kπ)−cos(0)]− 2 [cos(2kπ)−cos(kπ)]
k2π2 k2π2
2 k 2 k 4 k 0, k even,
= k2π2[(−1) −1]− k2π2[1−(−1) ] = k2π2[(−1) −1] = − 8 , k odd.
k2π2
T 2 4
Z Z Z
2 1 π π
bk = T f(t)sin(kωt)dt = 2 (t − 1)sin k2t dt+ (3 −t)sin k2t dt = by parts
0 0 2
2
h i2 Z
1 2 π 1 2 π
= 2 −(t−1)kπ cos k2t 0+ 2 kπ cos k2t dt
0
4
h i Z
1 2 π 4 1 2 π
−2 (3−t)kπ cos k2t 2− 2 kπ cos k2t dt
2
h i h i
1 2 π 2 1 2 π 4
=−kπ[cos(kπ)+cos(0)]+ k2π2 sin k2t 0+ kπ[cos(2kπ)+cos(kπ)]− k2π2 sin k2t 2 = 0.
Odd numbers can be expressed as k = 2i + 1, to numbers k = 1,3,5,7,... correspond indeces
i = 0,1,2,3,.... For those we then have a = − 8 . Werewritetheresulting series accordingly,
k (2i +1)π
and since the index k is traditional, we pass from i to k at the end.
Thus
∞ ∞
a0 X X 4 k π
f ∼ 2 + akcos(kωt)+bksin(kωt) = k2π2[(−1) −1]cos k2t
k=1 k=1
∞
X −8 π
= (2k +1)π cos (2k +1)2t .
k=0
Whatis the sum of this series? First we draw a periodic extension of the function f. To this we then
apply the Jordan criterion. According to it, the resulting series converges to f at all points where f
Fourier series: Solved problems pHabala 2012
(or rather its periodic extension) is continuous. Since our extension is continuosu everywhere, this
functions is also the sum of the series.
f(t)
1
0
−2 2 6 10
−4 −1 4 8
Since the extension of f is an even function, we should get a cosine series, which we did indeed.
3. Parameters: The length of the given segment is L = 4, after creating an odd function by flipping
the shape about both axes we eventually obtain a function with period T = 8, for sine series we use
the special frequency ω = π = π = π and classical formulas with L in place of T.
2T L 4
Sine series has a0 = ak = 0.
L 2 4
2Z 1Z π Z π
bk = L f(t)sin(kωt)dt = 2 (t − 1)sin k4t dt+ (3 −t)sin k4t dt = by parts
0 0 2
2
h i Z
1 4 π 2 1 4 π
= 2 −(t−1)kπ cos k4t 0+ 2 kπ cos k4t dt
0
4
h i Z
1 4 π 4 1 4 π
−2 (3−t)kπ cos k4t 2− 2 kπ cos k4t dt
2
h i h i
2 π 8 π 2 2 π 8 π 4
=−kπ cos k2 +cos(0) + k2π2 sin k4t 0+ kπ cos(kπ)+cos k2 − k2π2 sin k4t 2
2 k 16 π
= kπ[(−1) −1]+ k2π2 sin k2 .
For k even ve get 0. If k is odd, the first term gives −4, while the second one is (−1)i 16 for
kπ k2π2
−4 i 16
k = 2i+1. Thus for k odd, k = 2i+1 we get bk = (2i+1)π +(−1) (2i+1)2π2. As usual we use k
instead of i.
Thus
∞ ∞h i
a0 X X 2 k 16 π π
f ∼ + a cos(kωt)+b sin(kωt) = [(−1) −1]+ sin k sin k t
2 k k kπ k2π2 2 4
k=1 k=1
∞h i
X −4 k 16 π
= (2k+1)π +(−1) (2k +1)2π2 sin (2k +1)4t .
k=0
What is the sum of this series? First we flip the given shape about both axes, thus creating and odd
function, extending this basic shape we obtain the odd periodic extension of the function f (on the
left). To this we then apply the Jordan criterion. According to it, the resulting series converges to f
at all points where f (or rather its periodic extension) is continuous. At points of discontinuity of f
1 + −
the series converges to the average 2 f(t ) + f(t ) . Result: On the right is the function to which
our Fourier series converges, i.e. its sum.
f(t) F(t)
1 1
0 2 10 0 2 10
−4 −2 −1 4 6 8 −4 −2 −1 4 6 8
Fourier series: Solved problems pHabala 2012
Alternative: It is possible not to memorize the special formula for sine/cosine Fourier, but apply
the usual Fourier series to that extended basic shape of f to an odd function (see picture on the left).
In this way we get T = 8, ω = 2π = π. Then we need to find formulas for the segments that give the
T 4
basic period of odd extension and we can go, for b we get
k
L
b = 2 Z f(t)sin(kωt)dt
k T
−L
−2 0 2 4
1Z π Z π Z π Z π
= 4 (−t−3)sin k4t dt+ (t + 1)sin k4t dt+ (t − 1)sin k4t dt+ (3−t)sin k4t dt .
−4 −2 0 2
This looks tough, perhaps it is better to remember that special formula for sine/cosine series. This
alternative can be made a bit easier by the following reasoning: If f(t) is odd on [−4,4), then
π
f(t)sin k4t is even on [−4,4), thus it is enough to integrate over its right half and take it twice:
L 2 4
2Z 1Z π Z π
b =2· f(t)sin(kωt)dt = (t − 1)sin k t dt+ (3 −t)sin k t dt .
k T 2 4 4
0 0 2
But that’s exactly the formula we got from the special version right away, so it is probably really best
to simply remember the special frequency ω = π for sine/cosine series.
L
4. Parameters: The length of the given part is L = 4, we see that the specification f(t) = 0 on [1,4)
is important since it tells us how long the period is.
For the cosine series we first create by flipping the shape an even function with period T = 8, the we
use the special frequency ω = π = π = π and classical formulas with L in place of T.
2T L 4
L 1
Cosine series has b = 0. a = 2Z f(t)dt = 1Z 1dt = 1.
k 0 L 2 2
0 0
L 1
Z Z h i1
2 1 π 1 4 π 2 π
ak = L f(t)cos(kωt)dt = 2 cos k4t dt = 2 kπ sin k4t = kπ sin k4 .
0
0 0
It is not possible to write this somehow better, since when we try to substitute k = 0, 1, 2, 3, 4, 5,
√ √ √ √
6, 7, we get 0, 2, 1, 2, 0, − 2, −1, − 2, 0, which is too irregular.
2 2 2 2
Thus
∞ ∞
a X 1 X2
f ∼ 0 + a cos(kωt)+b sin(kωt) = + sin kπ cos (2k +1)πt .
2 k k 4 kπ 4 4
k=1 k=1
Whatis the sum of this series? First we flip the given shape about the y-axis, thus obtaining an even
function, by extending it we arrive at the even periodic extension of the function f (on the left). To
this we then apply the Jordan criterion. According to it, the resulting series converges to f at all
points where f (or rather its periodic extension) is continuous. At points of discontinuity of f the
1 + −
series converges to the average 2 f(t ) + f(t ) . Result: On the right is the function to which our
Fourier series converges, i.e. its sum.
1 f(t) 1 F(t)
−4 −10 1 4 7 8 9 −4 −10 1 4 7 8 9
5. Parameters: The period length is T = 2. This function is not given on an interval of the form
[0,T), but somewhere else, however, a shift in an interval is no problem, we find the Fourier series as
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