10.5 Periodic and Piecewise Continuous Input Functions
Mathematical models of mechanical or electrical systems often involve functions with discontinuities corresponding to external forces that are turned abruptly on or off. One such simple on–off function is the unit step function that we introduced in Section10.1. Recall that the unit step function at t=at=a is defined by
ua(t)=u(t−a)={0if t<a,1if t≧a.
ua(t)=u(t−a)={01if t<a,if t≧a.
(1)
The notation ua(t)ua(t) indicates succinctly where the unit upward step in value takes place (Fig.10.5.1), whereas u(t−a)u(t−a) connotes the sometimes useful idea of a “time delay” a before the step is made.
Because L{u(t)}=1/s,L{u(t)}=1/s,Eq.(2) implies that multiplication of the transform of u(t) by e−ase−as corresponds to the translation t→t−at→t−a in the original independent variable. Theorem 1 tells us that this fact, when properly interpreted, is a general property of the Laplace transformation.
Note that
u(t−a)f(t−a)={0if t<a,f(t−a)if t≧a.
u(t−a)f(t−a)={0f(t−a)if t<a,if t≧a.
(4)
Thus Theorem1 implies that L−1{e−asF(s)}L−1{e−asF(s)} is the function whose graph for t≧at≧a is the translation by a units to the right of the graph of f(t) for t≧0.t≧0. Note that the part (if any) of the graph of f(t) to the left of t=0t=0 is “cut off” and is not translated (Fig.10.5.2). In some applications the function f(t) describes an incoming signal that starts arriving at time t=0.t=0. Then u(t−a)f(t−a)u(t−a)f(t−a) denotes a signal of the same “shape” but with a time delay of a, so it does not start arriving until time t=at=a.
Before applying Theorem1, we must first write g(t) in the form u(t−3)f(t−3).u(t−3)f(t−3). The function f(t) whose translation 3 units to the right agrees (for t≧3t≧3) with g(t)=t2g(t)=t2 is f(t)=(t+3)2f(t)=(t+3)2 because f(t−3)=t2.f(t−3)=t2. But then
because of the periodicity of the cosine function. Hence Theorem1 gives
L{f(t)}=L{cos2t}−e−2πsL{cos2t}=s(1−e−2πs)s2+4.
L{f(t)}=L{cos2t}−e−2πsL{cos2t}=s(1−e−2πs)s2+4.
Example4
Discontinuous forcing A mass that weighs 32 lb (mass m=1m=1 slug) is attached to the free end of a long light spring that is stretched 1 ft by a force of 4 lb (k=4k=4 lb/ft). The mass is initially at rest in its equilibrium position. Beginning at time t=0t=0 (seconds), an external force F(t)=cos2tF(t)=cos2t is applied to the mass, but at time t=2πt=2π this force is turned off (abruptly discontinued) and the mass is allowed to continue its motion unimpeded. Find the resulting position function x(t) of the mass.
Solution
We need to solve the initial value problem
x″+4x=f(t);x(0)=x′(0)=0,
x′′+4x=f(t);x(0)=x'(0)=0,
where f(t) is the function of Example3. The transformed equation is
If we separate the cases t<2πt<2π and t≧2π,t≧2π, we find that the position function may be written in the form
x(t)={14tsin2tif t<2π,12πsin2tif t≧2π.
x(t)={14tsin2t12πsin2tif t<2π,if t≧2π.
As indicated by the graph of x(t) shown in Fig.10.5.6, the mass oscillates with circular frequency ω=2ω=2 and with linearly increasing amplitude until the force is removed at time t=2π.t=2π. Thereafter, the mass continues to oscillate with the same frequency but with constant amplitude π/2.π/2. The force F(t)=cos2tF(t)=cos2t would produce pure resonance if continued indefinitely, but we see that its effect ceases immediately at the moment it is turned off.
If we were to attack Example4 with the methods of Chapter5, we would need to solve one problem for the interval 0≦t<2π0≦t<2π and then solve a new problem with different initial conditions for the interval t≧2π.t≧2π. In such a situation the Laplace transform method enjoys the distinct advantage of not requiring the solution of different problems on different intervals.
Transforms of Periodic Functions
Periodic forcing functions in practical mechanical or electrical systems often are more complicated than pure sines or cosines. The nonconstant function f(t) defined for t≧0t≧0 is said to be periodic if there is a number p>0p>0 such that
f(t+p)=f(t)
f(t+p)=f(t)
(5)
for all t≧0.t≧0. The least positive value of p (if any) for which Eq.(11) holds is called the period of f. Such a function is shown in Fig.10.5.7. Theorem 2 simplifies the computation of theLaplace transform of a periodic function.
Proof:
The definition of the Laplace transform gives
F(s)=∫∞0e−stf(t)dt=∞∑n=0∫(n+1)pnpe−stf(t)dt.
F(s)=∫∞0e−stf(t)dt=∑n=0∞∫(n+1)pnpe−stf(t)dt.
The substitution t=τ+npt=τ+np in the nth integral following the summation sign yields
with x=e−ps<1x=e−ps<1 (for s>0s>0) to sum the series in the final step. Thus we have derived Eq.(6).
The principal advantage of Theorem 2 is that it enables us to find the Laplace transform of a periodic function without the necessity of an explicit evaluation of an improper integral.
Example5
Figure10.5.8 shows the graph of the square wave function f(t)=(−1)〚t/a〛f(t)=(−1)〚t/a〛 of period p=2a; 〚x〛p=2a; 〚x〛 denotes the greatest integer not exceeding x. By Theorem 2 theLaplace transform of f(t) is
Figure10.5.9 shows the graph of a triangular wave function g(t) of period p=2a.p=2a. Becausethe derivative g′(t)g'(t) is thesquare wave function of Example6, it follows from the formula in (7b) and Theorem 2 of Section10.2 that the transform of this triangular wavefunction is
G(s)=F(s)s=1s2tanhas2.
G(s)=F(s)s=1s2tanhas2.
(8)
Example7
Square wave forcing Consider a mass–spring–dashpot system with m=1,c=4,m=1,c=4, and k=20k=20 in appropriate units. Suppose that the system is initially at rest at equilibrium (x(0)=x′(0)=0)(x(0)=x'(0)=0) and that the mass is acted on beginning at time t=0t=0 by the external force f(t) whose graph is shown in Fig.10.5.10: the square wave with amplitude 20 and period 2π.2π. Find the position function f(t).
Solution
The initial value problem is
x″+4x′+20x=f(t);x(0)=x′(0)=0.
x′′+4x'+20x=f(t);x(0)=x'(0)=0.
The transformed equation is
s2X(s)+4sX(s)+20X(s)=F(s).
s2X(s)+4sX(s)+20X(s)=F(s).
(9)
From Example5 with a=πa=π we see that the transform of f(t) is
which we obtained with the aid of the familiar formula for the sum of a finite geometric progression. A rearrangement of Eq.(16) finally gives, with the aid of Eq.(13),
The last two terms in Eq.(17) give the steady periodic solution xsp.xsp. To investigate it, we write τ=t−nπτ=t−nπ for t in the interval nπ<t<(n+1)π.nπ<t<(n+1)π. Then
Figure10.5.11 shows the graph of xsp(t).xsp(t). Its most interesting feature is the appearance of periodically damped oscillations with a frequency four times that of the imposed force f(t).
10.5 Problems
Find the inverse Laplace transform f(t) of each function given in Problems 1 through 10. Then sketch the graph of f.
F(s)=e−3ss2F(s)=e−3ss2
F(s)=e−s−e−3ss2F(s)=e−s−e−3ss2
F(s)=e−ss+2F(s)=e−ss+2
F(s)=e−s−e2−2ss−1F(s)=e−s−e2−2ss−1
F(s)=e−πss2+1F(s)=e−πss2+1
F(s)=se−ss2+π2F(s)=se−ss2+π2
F(s)=1−e−2πss2+1F(s)=1−e−2πss2+1
F(s)=s(1−e−2s)s2+π2F(s)=s(1−e−2s)s2+π2
F(s)=s(1+e−3s)s2+π2F(s)=s(1+e−3s)s2+π2
F(s)=2s(e−πs−e−2πs)s2+4F(s)=2s(e−πs−e−2πs)s2+4
Find the Laplace transforms of the functions given in Problems 11 through 22.
f(t)=2f(t)=2 if 0≦t<3;0≦t<3;f(t)=0f(t)=0 if t≧3t≧3
f(t)=1f(t)=1 if 1≦t≦4;1≦t≦4;f(t)=0f(t)=0 if t<1t<1 or if t>4t>4
f(t)=sintf(t)=sint if 0≦t≦2π;0≦t≦2π;f(t)=0f(t)=0 if t>2πt>2π
f(t)=cosπtf(t)=cosπt if 0≦t≦2;0≦t≦2;f(t)=0f(t)=0 if t>2t>2
f(t)=sintf(t)=sint if 0≦t≦3π;0≦t≦3π;f(t)=0f(t)=0 if t>3πt>3π
f(t)=sin2tf(t)=sin2t if π≦t≦2π;π≦t≦2π;f(t)=0f(t)=0 if t<πt<π or if t>2πt>2π
f(t)=sinπtf(t)=sinπt if 2≦t≦3;2≦t≦3;f(t)=0f(t)=0 if t<2t<2 or if t>3t>3
f(t)=cos12πtf(t)=cos12πt if 3≦t≦53≦t≦5; f(t)=0f(t)=0 if t<3t<3 or if t>5t>5
f(t)=0f(t)=0 if t<1;t<1;f(t)=tf(t)=t if t≧1t≧1
f(t)=tf(t)=t if t≦1;t≦1;f(t)=1f(t)=1 if t>1t>1
f(t)=tf(t)=t if t≦1;t≦1;f(t)=2−tf(t)=2−t if 1≦t≦2;1≦t≦2;f(t)=0f(t)=0 if t>2t>2
f(t)=t3f(t)=t3 if 1≦t≦2;1≦t≦2;f(t)=0f(t)=0 if t<1t<1 or if t>2t>2
Apply Theorem 2 with p=1p=1 to verify that L{1}=1/s.L{1}=1/s.
Apply Theorem 2 to verify that L{coskt}=s/(s2+k2)L{coskt}=s/(s2+k2).
Apply Theorem 2 to show that the Laplace transform of the square wave function of Fig.10.5.12 is
L{f(t)}=1s(1+e−as).
L{f(t)}=1s(1+e−as).
Apply Theorem 2 to show that the Laplace transform of the sawtooth function f(t) of Fig.10.5.13 is
F(s)=1as2−e−ass(1−e−as).
F(s)=1as2−e−ass(1−e−as).
Let g(t) be the staircase function of Fig.10.5.14. Show that g(t)=(t/a)−f(t),g(t)=(t/a)−f(t), where f is the sawtooth function of Fig.10.5.14, and hence deduce that
L{g(t)}=e−ass(1−e−as).
L{g(t)}=e−ass(1−e−as).
Suppose that f(t) is a periodic function of period 2a with f(t)=tf(t)=t if 0≦t<a0≦t<a and f(t)=0f(t)=0 if a≦t<2a.a≦t<2a. Find L{f(t)}L{f(t)}.
Suppose that f(t) is the half-wave rectification of sinkt,sinkt, shown in Fig.10.5.15. Show that
L{f(t)}=k(s2+k2)(1−e−πs/k).
L{f(t)}=k(s2+k2)(1−e−πs/k).
Let g(t)=u(t−π/k)f(t−π/k),g(t)=u(t−π/k)f(t−π/k), where f(t) is the function of Problem29 and k>0.k>0. Note that h(t)=f(t)+g(t)h(t)=f(t)+g(t) is the full-wave rectification of sinktsinkt shown in Fig.10.5.16. Hence deduce from Problem29 that
L{h(t)}=ks2+k2cothπs2k.
L{h(t)}=ks2+k2cothπs2k.
Discontinuous Forcing
In Problems 31 through 35, the values of mass m, spring constant k, dashpot resistance c, and force f(t) are given for a mass–spring–dashpot system with external forcing function. Solve the initial value problem
mx″+cx′+kx=f(t);x(0)=x′(0)=0
mx′′+cx'+kx=f(t);x(0)=x'(0)=0
and construct the graph of the position function x(t).
m=1,k=4,c=0;m=1,k=4,c=0;f(t)=1f(t)=1 if 0≦t<π,0≦t<π,f(t)=0f(t)=0 if t≧πt≧π
m=1,k=4,c=5;m=1,k=4,c=5;f(t)=1f(t)=1 if 0≦t<2,0≦t<2,f(t)=0f(t)=0 if t≧2t≧2
m=1,k=9,c=0;m=1,k=9,c=0;f(t)=sintf(t)=sint if 0≦t≦2π,0≦t≦2π,f(t)=0f(t)=0 if t>2πt>2π
m=1,k=1,c=0;m=1,k=1,c=0;f(t)=tf(t)=t if 0≦t<1,0≦t<1,f(t)=0f(t)=0 if t≧1t≧1
m=1,k=4,c=4;m=1,k=4,c=4;f(t)=tf(t)=t if 0≦t≦2,0≦t≦2,f(t)=0f(t)=0 if t>2t>2
Transient and Steady Periodic Motions
In Problems 36 and 37, a mass–spring–dashpot system with external force f(t) is described. Under the assumption that x(0)=x′(0)=0,x(0)=x'(0)=0, use the method of Example7 to find the transient and steady periodic motions of the mass. Then construct the graph of the position function x(t). If you would like to check your graph using a numerical DE solver, it may be useful to note that the function
has the value +A+A if 0<t<π,0<t<π, the value −A−A if π<t<2π,π<t<2π, and so forth, and hence agrees on the interval [0,6π][0,6π] with the square wave function that has amplitude A and period 2π.2π. (See also the definition of a square wave function interms of sawtooth and triangular wave functions in the application material for this section.)
m=1,k=4,c=0;m=1,k=4,c=0;f(t) is a square wave function with amplitude 4 and period 2π2π.
m=1,k=10,c=2;m=1,k=10,c=2;f(t) is a square wave function with amplitude 10 andperiod 2π2π.
Suppose the function x(t) satisfies the initial value problem
mx″+cx′+kx=F(t),x(a)=b0,x′(a)=b1
mx′′+cx'+kx=F(t),x(a)=b0,x'(a)=b1
for t≧at≧a and x(t)=0x(t)=0 for t<a.t<a. Then show that X(s)=L{x(t)}X(s)=L{x(t)} satisfies the equation
This is an alternate approach to Example7 that was suggested by Keng C. Wu of Lockheed Martin (Maritime Systems Sensors). Let x(t) denote the steady periodic solution of the given differential equation x″+4x′+20x=F(t).x′′+4x'+20x=F(t). Suppose we write v(t) for the restriction of x(t) to the first half [0,π][0,π] of the fundamental interval [0,2π],[0,2π], and w(t) for its restriction to the second half-interval [π,2π].[π,2π]. We may regard v(t) as the solution of the initial value problem
v″+4v′+20v=+20,v(0)=b0,v′(0)=b1
v′′+4v'+20v=+20,v(0)=b0,v'(0)=b1
and w(t) as the solution of the initial value problem
w″+4w′+20w=−20,w(π)=c0,w′(π)=c1,
w′′+4w'+20w=−20,w(π)=c0,w'(π)=c1,
where the initial values b0,b1b0,b1 and c0,c1c0,c1 are to be determined so that x(t) and x′(t)x'(t) are continuous.
Transform the first of these initial value problems to show that V(s)=A(s)b0+B(s)b1+C(s),V(s)=A(s)b0+B(s)b1+C(s), where
Apply the result of Problem38 to the second initial value problem above to show that W(s)=e−πs(A(s)c0+B(s)c1−C(s)),W(s)=e−πs(A(s)c0+B(s)c1−C(s)), where the coefficient functions A(s),B(s),C(s)A(s),B(s),C(s) are as defined in part (a).
solve the four continuity equations v(π)=c0,v′(π)=c1,w(2π)=b0,w′(2π)=b1v(π)=c0,v'(π)=c1,w(2π)=b0,w'(2π)=b1 to find the values of the previously undetermined initial values. Conclude that
Finally, use these expressions to verify that the graph of the steady periodic solution x(t) looks as indicated in Fig.10.5.11.
10.5 Application Engineering Functions
Periodic piecewise linear functions occur so frequently as input functions in engineering applications that they are sometimes called engineering functions. Computations with such functions are readily handled by computer algebra systems. In Mathematica, for instance, the SawToothWave, TriangleWave, and Square-Wave functions can be used to create the corresponding inputs with specified range, period, etc. Alternatively, we can define our own engineering functions using elementary functions available in any computer algebra system:
Plot each of the functions to verify that it has period 2 and that its name is aptly chosen. For instance, the result of
Plot[squarewave[t], {t, 0, 6}]
should look like Fig.10.5.8. If f(t) is one of these engineering functions and p>0,p>0, then the function f(2t/p)f(2t/p) will have period p. To illustrate this, try
Plot[triangularwave[ 2 t/p ], {t, 0, 3 p}]
with various values of p. Now let’s consider the mass–spring–dashpot equation
diffEq = m x″ [t] + c x′ [t] + k x[t] == input
with selected parameter values and an input forcing function with period p and amplitude F0F0.
m = 4; c = 8; k = 5; p = 1; F0 = 4;
input = F0 squarewave[2 t/p];
You can plot this input function to verify that it has period 1:
Plot[input, {t, 0, 2}]
Finally, let’s suppose that the mass is initially at rest in its equilibrium position and solve numerically the resulting initial value problem.
In the resulting Fig.10.5.17 we see that after an initial transient dies out, the response function x(t) settles down (as expected?) to a periodic oscillation with the same period as the input.
Investigate this initial value problem with several mass–spring–dashpot para–meters–for instance, selected digits of your student ID number—and with input engineering functions having various amplitudes and periods.