[PDF] MATH 231 Laplace transform shift theorems

View - Download MATH 231 Laplace transform shift theorems

Previous PDF

Next PDF

MATH 231

Laplace transform shift theorems

There aretworesults/theorems establishing connections between shifts and exponential factors of

a function and its Laplace transform.Theorem 1:Iff(t) is a function whose Laplace transformLf(t)(s)=F(s), then

A.Lheatf(t)i(s)=F(sa);and

B.LH(ta)f(ta)(s)=easF(s):NeitherofthesetheoremsisstrictlynecessaryforcomputingLaplacetransforms-i.e., whengoing

fromthetimedomainfunctionf(t)toitsfrequencydomaincounterpartLf(t)(s). Suchtransforms can be computed directly from the definition of Laplace transformLf(t)(s)=R1

0estf(t)dt.

Example 1:

We compute

(a)Lhte2ti(s);and (b)LhH(t3)et3i directly from the definition.

For part (a),

Lhte2ti(s)=Z

1 0 estte2tdt=Z 1 0 te(s2)tdt=Z 1 0 testdts7!s2=L[t](s2) 1s 2 s7!s2=1(s2)2:

For part (b),

LhH(t3)et3i=Z

1 0 estH(t3)et3dt=Z 1 3 estet3dt Z 1 0 es(u+3)eudu(by substitution:u=t3) =e3sZ 1 0 esueudu=e3sZ 1 0 estetdt(thenameof the variable of integration is immaterial) =e3sLheti=e3s1s1:Using shift theorems for inverse Laplace transforms It is in findinginverseLaplace transforms where Theorems A and B are indispensible.

Example 2:

Find the inverse Laplace transform for each of the functions (a) se2ss

2+9(b)3(s+1)3(c)2ss

24s+5
Our function in part (a) has an exponential factor, much like in Theorem B. Here, e 2sss

2+9=e2sF(s);whereF(s)=ss

2+9=L[cos(3t)](s):

Thus, L 1 e 2sss 2+9 (t)=H(t2)cos(3(t2)): The functionin part(b) does notlook like anentry in the Laplace transformtable I provide: It is, in fact, a modified version of the table entryn!=sn+1withn=2 but shifted left 1 unit, i.e.,

3(s+1)3=3s

3 s7!s+1=32 2!s 3 s7!s(1): Since L 132
2!s 3 (t)=32 L1 2!s 3 (t)=32 t2; it follows from Theorem A that L

1"3(s+1)3#

(t)=L132 2!s 3 s7!s(1) (t)=32 t2et: The function in part (c) also does not look like an entry in the table of Laplace transforms found at the link above. The denominator is, in fact, anirreducible quadratic (over the reals), havingno realroots. Buta quadratichasaparabolic graph, andanyparabola maybeobtained from the graph ofy=x2via a sequence of shifts and stretches. We can find the shift involved through completing the square: s

24s+5=s24s+4+1=(s2)2+1;

which means the graph ofs24s+5 is the same as the graph ofs2+1 but shifted 2 units to the right. To use Theorem A, we needallinstances ofsto be similarly shifted, so we write 2ss 2+1 s7!s2: [Take a moment to plot, together, the functions 2x=(x24x+5) and (2x+4)=(x2+1). Observe

that the graph of the former is identical to that of the latter, except shifted right 2 units.] Since

L

12s+4s

2+1 =2L1ss 2+1 +4L11s 2+1 =2cost+4sint; it follows from Theorem A that L 12ss 24s+5
=L12s+4s 2+1 s7!s2 =e2t(2cost+4sint): In some cases, we employ partial fraction expansion as part of finding the inverse Laplace trans- form.

Example 3:

Find the inverse Laplace transform for each of the functions (a) 8s

3+4s(b)3s

24s5(c)8e3ss(s2+4)

The denominator of our function in part (a) is a cubic, whose graph cannot be obtained via a shift of anyquadraticfunction. From Calculus, we learn there is a partial fractions expansion of the form

8s(s2+4)=As

+Bs+Cs Equating coecients for the various powers ofs(and using linear algebra?), we discover that

A=2,B=2 andC=0, so

L 18s 3+4s =L12s 2ss 2+4 =2L11s 2L1ss 2+4 =22cos(2t): The demoninator of the function in part (b) is quadratic, but reducible -i.e., it has real roots, exhibited by the fact that it factors s

24s5=(s5)(s+1);

revealingroots(1)and5. (Thequadraticformulawouldalsorevealtheserealroots.) Byusing partial fraction expansion, we can turn function into the sum of functions with denominators which are 1 stdegree polynomials: 3s Equating coecients ofs1ands0, we can solve to findA=1=2;B=1=2. Thus, L 13s 24s5
=L11=2s51=2s+1 =12 L11s5 12

L1"1s(1)#

=12 e5t12 et: The function in part (c) is almost identical to the one in part (a), but for the exponential factore3s. (Think Theorem B!) Piggy-backing on our answer to part (a), we obtain L

1"8e3ss

3+4s# A caution concerning the use of Theorem B to find a Laplace transform We have noted that Theorems A and B are indispensible when finding inverse Laplace transforms (going fromF(s) tof(t)), not for the reverse. That is not the same as saying the theorems are not usefulfor findingF(s) fromf(t). Look back at Example1 , and check that the theorems provide faster ways of obtaining the answers. the Laplace transform of a "switched on" version off(t), but rather a "switched on and shifted" version.

Example 4:

Supposef(t)=t22t+3. Then

L f(t)=Lht2i2L[t]+3Lht0i=2s 32s
2+3s =F(s):1234246810 f(t)=t22t+3t

12345678246810

F(s)=2s

32s
2+3s s Theorem B makes it relatively easy to find the Laplace transform ofH(t1)f(t1)= H (t1)[(t1)22(t1)+3];which has a graph likefbut shifted right 1 unit and shifted on at timet=1. By Theorem B, L

H(t1)f(t1)=es2s

32s
2+3s The graphs of the time and frequency domain functions appear below.123246810

H(t1)[(t1)22(t1)+3]t

12345678246810

e s2s 32s
2+3s s Since (t1)22(t1)+3=t22t+12t+2+3=t24t+4, the graph on the left could have been labeledH(t1)(t24t+4), and the graph on the right isLhH(t1)(t24t+4)i. Now, suppose what we desired was actually the Laplace transform ofH(t1)f(t)= H (t1)(t22t+3), whose graph is depicted at left below. We can only use Theorem B to find it if we find the formula for the functiongfor whichg(t1)=f(t); that is, the function obtained shiftingfone unit to theleft. Since L g(t)=Lht2+2i=Lht2i+2L[1]=2s 3+2s then

LH(t1)f(t)=LH(t1)g(t1)=es2s

3+2s whose graph appears at right below.123246810

H(t1)(t22t+3)t

12345678246810

e s2s 3+2s s

Exercises

1. Graph the function and find its Laplace transform. (a)f(t)=tH(t1)(t1) (c)f(t)=8 >><>>:0;t<3 t

2+3t8;t3

(e)f(t)=e3tsin(4t)(b)f(t)=Ht4 cost4 (d)f(t)=8 >>>><>>>>:0;t< t t<2 0;t2 (f)f(t)=4e2(t5)H(t5)(t5)2 [Note: In the particular case of part (d), you may want to try it both writing it as a series of "shifted, switched-on" functionsanddirectly from the definition of Laplace transform, and decide which you think is easier.] 2. Find the inverse Laplace transform for each function. (a)F(s)=2(s1)s 22s+2
(c)F(s)=4s 24
(e)F(s)=es+e2se3se4ss(b)F(s)=2(s1)e2ss 22s+2
(d)F(s)=4(s2)4+e2ss 2+s2 (f)F(s)=s2s 24s+3
quotesdbs_dbs21.pdfusesText_27

[PDF] inverse laplace transform of 1/s+a

[PDF] inverse matrix 3x3 practice problems

[PDF] inverse matrix bijective

[PDF] inverse matrix calculator 4x4 with steps

[PDF] inverse matrix method

[PDF] inverse of 4x4 matrix example pdf

[PDF] inverse of a 3x3 matrix worksheet

[PDF] inverse of a matrix online calculator with steps

[PDF] inverse of bijective function

[PDF] inverse of linear transformation

[PDF] inverse of matrix product

[PDF] inverse relationship graph

[PDF] inverse relationship science

[PDF] inverseur de source courant continu

[PDF] inverter layout