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Laplace Transforms of Periodic Functions

We begin with

Definition 1 LetX be a positive real number. A function f(x) defined on

[0,)isperiodic with periodX if, for allx [0,), f(x + X) = f(x).

The smallest positive numberXfor which this is true, provided such a number

exists, is called theleast period off(x).

We will retain this definition for functions f(x) which may have isolated

singularities, or other isolated points where the function is not defined, pro-

vided f(x+ X) f(x) otherwise. Thus we may speak of f(x) tan x

as being periodic with period X = even though the function has isolated

singularities at odd multiples of /2.

Example 1 It is well known that cos xand sin xare periodic with period

X = 2 and that this is the least period in each case. From this it is clear

that tan x = sinxcosx

must also be periodic with period X = 2 but, due to

cancellation of sign changes in sin xand cos x, tan x is, in fact, also periodic

with period X = , as just indicated, and this is the least period for tan x.

Laplace transforms of periodic functions have a particular structure. Be-

fore developing this structure, however, we need to remind the reader of the

formula for the sum of a geometric progression. Thus if we have the geometricprogression with constant ratio r (real or complex):

Pr : 1, r, r2, , rk, ,

the sum of the terms from k = 0 (i.e., from r0 = 1) to k = n(i.e., torn) is

Sn(Pr) = 1 rn+1

1 r .

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If|r| < 1 the infinite sum S is defined as

S(Pr) = limn

Sn = 1

1 r.

The proof may be found in standard algebra textbooks or in our basic al-

gebrasection.

With this result in hand we suppose that f(x) is a Laplace transformablefunction which is periodic with period X > 0. Then for s > 0 we have

(Lf) (s) =

0esx f(x) dx =

k=0

(k+1)XkX

esx f(x) dx.

For k = 0, 1, 2, we letx = +kX in the k-th integral and obtain, using

the X-periodicity off(x) and the familiar properties ea+b = ea eb, eab =

(ea))b of the exponential,

(Lf) (s) =

k=0

X0

es(+kX) f(+kX) d =

k=0

eskX

X0

es f() d=

k=0

eskX X

0es f() d =

k=0

(esX)k X

0es f() d.

Then, letting r = esX < 1 the first factor in the last expression is seen,

using the formula for S(Pr), to be 11r

= 11 esX

, with the result that we

have

(Lf) (s) =

X0 e

s f() d

1 esX .

Another way to state this is as follows. We let

fX(x) =

f(x), 0 < x < X ;0, x X.

Thus the non-zero values offX(x) encompass just one period off(x); we will

call it thebasic functionassociated with theX-periodic functionf(x). Then

it is clear that

0esx fX(x) dx =

X0

esx f(x) dx

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and therefore

(Lf) (s) = (LfX) (s)

1 esX.

Example 1 Let f(x) = sin ax, a > 0. Thenf(x) has period X = 2a

and we must have

(L sin ax) (s) =2a0 e

sx

sin axdx1 e

2s

a

=

a

s2+a2esx cos ax s

s2+a2esx sin ax

2a0

1 e2s

a

=

as2+a2

1 e

2s

a

1 e2s

a

= a

s2 +a2.

This, of course, is not news. The real interest lies in computing Laplace

transforms of certain piecewise continuous functions which arise as periodicwaveforms in a variety of applications.

Example 2 The square wave of period 2X is the 2X-periodic function

f(x) whose basic function is

f2X(x) =

1, 0 < x < X ;1, X < x < 2X.

Thusf(x) has the value (1)k onkX < x < (k +1)X, k = 0, 1, 2, . From

the formula we have

(Lf) (s) =

X0 e

sx 1 dx 2X

X esx 1 dx

1 e2sX =

1

s

(1 2 esX + e2sX)

1 e2sX

= 1 esX

s(1 + esX).

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Example 3 A periodic function with numerous applications in design of

electronic devices is the so-called saw-tooth function. The basic function is

fX(x) = x, x [0, X) and f(x) is then obtained by periodic continuation

with periodX; on the interval [kX, (k+ 1)X) we havef(x) = x kX. In

this case

(LfX) (s) = X0 e

sx

f(x) dx = X0 e

sx

x dx = d

ds X0 e

sx

dx

= d

ds

1 esX

s =

1 esX

s2

X esX

s .

Then

(Lf) (s) = 1

s2

X esX

s(1 esX).

Example 4 We consider a periodic pulse, i.e., an intense signal of short

duration emitted at periodic intervals. We again let the period be X > 0

and we suppose the basic function, given in terms of an amplitude, A, A > 0

and a pulse duration h, 0 < h < X , is

fX(x) =

A, 0 < x < h;0, h < x < X.

Then X0

esx fX(x) dx = A h0

esx dx = A

s(1 esh).

Taking f(x) to be the X-periodic extension offX(s), our formula gives

(Lf) (s) = A(1 esh)

s(1 esX).

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Periodic Solutions of Differential Equations with Periodic Forcing

We consider a differential equation

dny

dtn + a1

dn 1y

dtn 1 + + an1

dy

dt + any = g(t),

whereg(t) is at least piecewise continuous and isperiodicwith periodT > 0

on the interval [0,). Consequently the Laplace transform ofg(t) takes the

form

g(s) = (Lg) (s) = (LgT) (s)

1 esT , (LgT) (s) =

T0

est g(t) dt.

In many applications it is important to be able to find the steady state

periodic solutiony(t) of this differential equation which is also periodic with

period T. Ify(t) is such a solution: y(t+T) y(t), t [0,), then we

also have

y(s) = (Ly) (s) = (LyT) (s)

1 esT , (LyT) (s) =

T0

est y(t) dt.

In the section on Laplace solution of initial value problems we show

that for any solution of this differential equation,

p(s) y(s) q(s) = g(s),

where

q(s) = j,k,, j+k+ = n1

akyjs

andp(s) = sn +a1 sn1 + + an1s+ anis thecharacteristic polynomial

for the differential equation under study. Using the special forms of the

Laplace transforms for periodic y(t) and g(t) this becomes

p(s)(LyT) (s)

1 esT q(s) =

(LgT) (s)

1 esT ,

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Multiplying by 1esT

p(s) we then have

(LyT) (s) =p(s)

.

The left hand side of this differential equation is

y(s) = (LyT) (s) = T

0

est y(t) dt.

This integral clearly has a finite value for every value of s and, in fact,

differentiation under the integral sign is valid and we see that all of the

derivatives of y(s) exist and have finite values for every value ofs. But on

the right hand side the denominator vanishes at each of the roots rm, m =

1, 2, n (counting multiplicity) ofp(s). We can avoid singularities on the

right hand side only if the function (LgT) (s) +

1 esT

q(s) has a root at

each point s = rm of the same multiplicity as p(s) has there.

We consider here only the case where the rm are distinct. Then we need

(LgT) (rm) +

1 ermT

q(rm) = 0, m = 1, 2, , n,

or

(LgT) (rm) +

1 ermT

j,k,, j+k+ = n1

akyjrm

= 0, m = 1, 2, , n.

These constitute n linear algebraic equations for the unknown initial values

y(j)

(0) = yj, j = 0, 1, 2, , n 1, of the periodic solution y(t). Since theyj disappear completely from the m-th equation if 1 ermT = 0, we need

1 ermT = 0, m = 1, 2, , n,

which is the same thing as saying that the characteristic polynomialp(s) must

be such that none of its roots rm have the form of an integer multiple of 2i

T ,

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i.e., the corresponding homogeneous differential equation has no solutions

which are periodic with period T.

Example 5 Let us consider the differential equation

d2y

dt2 + 3

dy

dt + 2 y = f(t),

where f(t) is the saw-tooth function with period T:

f(t) = t kT, kT < t < (k+ 1)T, k = 0, 1, 2, .

We have seen earlier in this section that this function has Laplace transform

(Lf) (s) = (LfT) (s)

1 esT =

1 esT

s2 T e

sT

s

1 esT

= 1

s2

T esT

s(1 esT).

Writing

(Ly) (s) = (LyT) (s)

1 esT

and applying the Laplace transform to both sides of the differential equation

we have, with y (k)(0) yk, k = 0, 1,

s2 + 3 s+ 2

(LyT) (s)1 esT

(s y0+y1) 3 y0 =1 esT

s2 T e

sT

s

1 esT .

Multiplying by 1 esT

s2+3 s+2we have

(LyT) (s) =

1 esT

s2 T e

sT

s +

1 esT

((3 +s)y0+y1)

s2 + 3 s+ 2 .

Since (LyT) (s) must be free of singularities, we need

1 esT

s2

T esT

s +

1 esT

((3 +s)y0+y1) = 0, s = 1, 2,

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since the latter are the points where s2 + 3 s + 2 = 0. This gives the two

equations1 eT +T eT +

1 eT

((2 y0+y1) = 0,

1e2T

4 + T e

2T

2 +

1 e2T

(y0+y1) = 0,

or2 y0 + y1 =

T eT

eT1 1,

y0 + y1 = T e2T

2(e2T1) 1

4.

Solving these equations we have

y0 = T eT

eT1 T e

2T

2(e2T1) 3

4,

y1 = T e2T

e2T1 T eT

eT1 + 1

2.

On the interval [0, T] the periodic solutiony(t) of our differential equation

agrees with the solution of

d2z

dt2 + 3

dz

dt + 2 y = t

with the same initial conditions. The solutionz(t) of this differential equation

with y (0) = y0, y

(0) = y1 can be computed, as in our section on Laplace

solution of initial value problems, to be

z(t) = t

2

3

4 + (y1+ 2 y0+ 1) e

t

y1+y0+1

4

e2t.

Since y(t) is periodic with period Twe have

y(t) = t kT

2

3

4 + (y1+ 2 y0+ 1) e

(tkT)

y1+y0+1

4

e2(tkT), kT < t < (k+ 1)T, k = 0, 1, 2, ,

with y0 and y1 as determined above.

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