fourier series solved problem

8/9/2019 fourier series solved problem

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Fourier series: Solved problems c pHabala 2012

Solved problems on Fourier series

1. Find the Fourier series for (periodic extension of)

f (t) =

1, t ∈ [0, 2);

−1, t ∈ [2, 4).Determine the sum of this series.

2. Find the Fourier series for (periodic extension of)

f (t) =

t − 1, t ∈ [0, 2);3 − t, t ∈ [2, 4).

Determine the sum of this series.

3. Find the sine Fourier series for (periodic extension of)

f (t) =

t − 1, t ∈ [0, 2);

3 − t, t ∈ [2, 4).Determine the sum of this series.

4. Find the cosine Fourier series for (periodic extension of)

f (t) =

1, t ∈ [0, 1);

0, t ∈ [1, 4).

Determine the sum of this series.5. Find the Fourier series for (periodic extension of)

f (t) = 1 − t2, t ∈ [−1, 1).Determine the sum of this series.

Solutions

1. Parameters: The period length is T = 4, frequency ω = 2πT = π

2.

a0 = 2

T

T

0

f (t) dt = 1

22

0

1 dt −

4

2

1 dt = 0,

ak = 2

T

T 0

f (t) cos(kωt) dt = 1

2

2 0

cos

k π2 t

dt −

4 2

cos

k π2 t

dt

= 1

2

2

kπ sin

k π2 t2

0−

1

2

2

kπ sin

k π2 t4

2= 0,

bk = 2

T

T 0

f (t) sin(kωt) dt = 1

2

2 0

sin

k π2

t

dt −

4 2

sin

k π2

t

dt

= 12

− 2

kπ cos

k π2 t2

0− 1

2

− 2

kπ cos

k π2 t4

2= 1

kπ

− cos(kπ) + cos(0) + cos(2kπ) − cos(kπ)

= 1

kπ

−(−1)k + 1 + 1 − (−1)k

=

2

kπ[1 − (−1)k] =

0, k even,4kπ , k odd.

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 ak = 4

(2i + 1)π. We rewrite the resulting series accordingly,

and since the index k is traditional, we pass from i to k at the end.Thus

f ∼ a02

+ ∞k=1

ak cos(kωt)+ bk sin(kωt)

= ∞

k=1

2kπ

[1−(−1)k]sin

k π2 t

= ∞k=0

4(2k + 1)π

sin

(2k+1) π2 t

.

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




points where f (or rather its periodic extension) is continuous. At points of discontinuity of f theseries converges to the average 1

2

f (t+) + f (t−)

. Result: On the right is the function to which our

Fourier series converges, i.e. its sum.

8 10642−2

1

0

f(t)

8 10642−2

1

0

F(t)

−1−1

2. Parameters: The period length is T = 4, frequency ω = 2πT

= π2 .

a0 = 2

T

T 0

f (t) dt = 1

2

2 0

t − 1 dt +

4 2

3 − t dt

= 0.

ak = 2

T

T 0

f (t)cos(kωt) dt = 1

2

2 0

(t − 1)cos

k π2

t

dt +

4 2

(3 − t)cos

k π2

t

dt

=

by parts

= 1

2

(t − 1)

2

kπ sin

k π2 t2

0−

1

2

2

kπ

2 0

sin

kπ2 t

dt + 1

2

(3 − t)

2

kπ sin

k π2 t4

2+

1

2

2

kπ

4 2

sin

k π2 t

dt

= 0 + 2

k2π2 cos

k π2 t2

0+ 0 −

2

k2π2 cos

k π2 t4

2

= 2

k2π2[cos(kπ) − cos(0)] −

2

k2π2[cos(2kπ) − cos(kπ)]

= 2

k2π2[(−1)k − 1] −

2

k2π2[1 − (−1)k] =

4

k2π2[(−1)k − 1] =

0, k even,

− 8k2π2 , k odd.

bk = 2

T

T 0

f (t)sin(kωt) dt = 1

2

2 0

(t − 1)sin

k π2 t

dt +4

2

(3 − t)sin

k π2 t

dt

=

by parts

= 1

2

−(t − 1)

2

kπ cos

k π2

t2

0+

1

2

2

kπ

2 0

cos

k π2

t

dt

− 1

2

(3 − t)

2

kπ cos

k π2 t4

2−

1

2

2

kπ

4 2

cos

k π2 t

dt

= − 1kπ

[cos(kπ) + cos(0)] +

2k2π2

sin

k π2 t2

0+ 1

kπ[cos(2kπ) + cos(kπ)] −

2

k2π2 sin

k π2 t4

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 ak = − 8

(2i + 1)π. We rewrite the resulting series accordingly,

and since the index k is traditional, we pass from i to k at the end.Thus

f ∼ a0

2 +

∞k=1

ak cos(kωt) + bk sin(kωt)

=

∞k=1

4

k2π2[(−1)k − 1]cos

k π2 t

=∞k=0

−8(2k + 1)π

cos

(2k + 1) π2 t

.

What is the sum of this series? First we draw a periodic extension of the function f . To this we thenapply the Jordan criterion. According to it, the resulting series converges to f at all points where f




(or rather its periodic extension) is continuous. Since our extension is continuosu everywhere, thisfunctions is also the sum of the series.

10

f(t)

−4 −1

1

4 8

62−2

0

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 flippingthe shape about both axes we eventually obtain a function with period T = 8, for sine series we usethe special frequency ω = π

2T = πL = π

4 and classical formulas with L in place of T .Sine series has a0 = ak = 0.

bk = 2

L

L 0


2

2 0

(t − 1) sin

k π4 t

dt +

4 2

(3 − t)sin

kπ4 t

dt

=

by parts

= 1

2

−(t − 1)

4

kπ cos

k π4

t2

0+

1

2

4

kπ

2 0

cos

kπ4

t

dt

− 1

2

(3 − t)

4

kπ cos

k π4 t4

2−

1

2

4

kπ

4 2

cos

k π4 t

dt

= − 2

kπ

cos

k π2

+ cos(0)

+

8k2π2 sin

k π4 t2

0+

2

kπ

cos(kπ) + cos

k π2

− 8k2π2 sin

k π4 t4

2

= 2

kπ[(−1)k − 1] + 16

k2π2 sink π2 .

For k even ve get 0. If k is odd, the first term gives −4kπ , while the second one is (−1)i 16

k2π2 for

k = 2i + 1. Thus for k odd, k = 2i + 1 we get bk = −4

(2i + 1)π + (−1)i

16

(2i + 1)2π2. As usual we use k

instead of i.Thus

f ∼ a0

2 +

∞k=1


=

∞k=1

2

kπ[(−1)k − 1] +

16

k2π2 sin

k π2

sin

k π4 t

=∞k=0

−4(2k+1)π + (−1)k 16

(2k + 1)2π2

sin

(2k + 1) π

4 t

.

What is the sum of this series? First we flip the given shape about both axes, thus creating and oddfunction, extending this basic shape we obtain the odd periodic extension of the function f (on theleft). 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 series converges to the average 12

f (t+) + f (t−)

. Result: On the right is the function to which

our Fourier series converges, i.e. its sum.

0 10

f(t)

−4 −2

1

24 6 8

0 10

F(t)

−4 −22

4 6 8−1

1

−1




Alternative: It is possible not to memorize the special formula for sine/cosine Fourier, but applythe 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

T = π

4. Then we need to find formulas for the segments that give the

basic period of odd extension and we can go, for bk we get

bk = 2

T

L

−L

f (t)sin(kωt) dt

= 1

4

−2 −4

(−t − 3)sin

k π4 t

dt +

0 −2

(t + 1) sin

kπ4 t

dt +

2 0

(t − 1)sin

k π4 t

dt +

4 2

(3 − t)sin

k π4 t

dt

.

This looks tough, perhaps it is better to remember that special formula for sine/cosine series. Thisalternative can be made a bit easier by the following reasoning: If f (t) is odd on [−4, 4), thenf (t)sin

k π4 t

is even on [−4, 4), thus it is enough to integrate over its right half and take it twice:

bk = 2 · 2

T

L

0


2

2

0

(t − 1)sin

k π4 t

dt +

4

2

(3 − t)sin

k π4 t

dt

.

But that’s exactly the formula we got from the special version right away, so it is probably really bestto simply remember the special frequency ω = π

L for sine/cosine series.

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 weuse the special frequency ω = π

2T = πL = π

4 and classical formulas with L in place of T .

Cosine series has bk = 0. a0 = 2

L

L

0

f (t) dt = 1

2

1

0

1 dt = 1

2 .

ak = 2

L

L 0

f (t)cos(kωt) dt = 1

2

1 0

cos

k π4 t

dt = 1

2

4

kπ sin

k π4 t1

0=

2

kπ sin

k π4

.

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,√ 22 , 1,

√ 22 , 0, −

√ 22 , −1, −

√ 22 , 0, which is too irregular.

Thus

f ∼ a0

2 +

∞

k=1ak cos(kωt) + bk sin(kωt)

=

1

4 +

∞

k=1

2

kπ sin

k π4

cos

(2k + 1) π

4 t

.

What is the sum of this series? First we flip the given shape about the y-axis, thus obtaining an evenfunction, by extending it we arrive at the even periodic extension of the function f (on the left). Tothis we then apply the Jordan criterion. According to it, the resulting series converges to f at allpoints where f (or rather its periodic extension) is continuous. At points of discontinuity of f theseries converges to the average 1

2

f (t+) + f (t−)

. Result: On the right is the function to which our

Fourier series converges, i.e. its sum.

f(t)

−1−4 97 8410

1 F(t)

−1−4 97 8410

1

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




usual. Frequency is ω = 2πT = π.

a0 = 2

T

T/2 −T/2

f (t) dt =

1 −1

1 − t2dt = 4

3.

ak = 2T

T/2 −T/2

f (t)cos(kωt) dt =

1 −1

(1 − t2)cos(kπt) dt =

by parts

= 1

kπ(1 − t2) sin(kπt)

1−1

+ 2

kπ

1 −1

t sin(kπt) dt = 0 +−

2

k2π2t cos(kπt)

1−1

+ 2

k2π2

1 −1

cos(kπt) dt

= − 2

k2π2[cos(kπ) + cos(−kπ)] +

2

k3π3 sin(kπt)

1−1

= − 2

k2π2[cos(kπ) + cos(kπ)] + 0

= −(−1)k 4

k2π2.

bk = 2

T

T/2 −T/2

f (t)sin(kωt) dt =

1 −1

(1 − t2)sin(kπt) dt =

by parts

=−

1

kπ(1 − t2)cos(kπt)

1−1

− 2

kπ

1 −1

t cos(kπt) dt = 0 − 2

k2π2t sin(kπt)

1−1

+ 2

k2π2

1 −1

sin(kπt) dt

= − 2

k2π2[sin(kπ) + sin(−kπ)] +

−

2

k3π3 cos(kπt)

1−1

= 0 − 2

k3π3[cos(kπ) − cos(−kπ)]

= − 2

k3π3[cos(kπ) − cos(kπ)] = 0.

Thus

f ∼ a0

2 +

∞k=1


=

2

3 +

∞k=1

(−1)k+1 4

k2π2 cos(kπt).

What is the sum of this series? First we draw a periodic extension of the function f . To this we thenapply the Jordan criterion. According to it, the resulting series converges to f at all points where f

(or rather its periodic extension) is continuous. Since our extension is continuosu everywhere, thisfunctions is also the sum of the series.

31−1 5 70

1

2

f(t)

Notice that the given function is even after we extended it, so the resulting series is naturally a cosineseries. If we drew the extension first, we need not have calculated bk, we could have just writtenbk = 0.Since the interval [−1, 1) extends to both sides from the origin, the symmetry is decided right fromthe start. We see that the cosine series is possible, since the function f (t) = 1 − t2 is even on theinterval [−1, 1), but we cannot make an odd function out of f and thus sine series is not possible.

fourier series solved problem

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