1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

12 .7 • SHIFTING THEOREMS AND THE STEP FUNCTION 5 5 7

1 s

We now use Theorem 12.16 and the facts that - and - 2 --are the transforms

s s + 1
of 1 and cost, respectively. We compute the solution, y(t), as

(

e-w•) ( e-"s s)

y (t) = c-^1 -

8

- -c-^1 82 +

1

= U,, (t)- U" (t)cos(t - 11'),

which we then write in the more familiar form

Y (t ) - { - 0 , 1 - cost,

fort< 11';

fort>11'.

• ------.-EXERCISES FOR SECTION 12.7
  
  
  1. Find .C (e' -te').
  
  
  
  

2. Find .C (c^4 'sin3t).

)

s - a

3. Show that .C(e•'cosbt = ( )2 2 ·

s-a +b

4. Show that .C (e•• sinbt) = ( ~ 2 b2.

s-a +

For Exercises 5-8, find c-^1 (F (s) ).

s+2

5. F (s) = z 4 5.

s + s+

8

6. F(s) = s (^2) - (^2) s+ 5
   7.F(s)= s+3
   (s +2)^2 + 1 •
   2 s+ 10
   S. F (s)= s2+6s+25'

For Exercises 9-14, find L (f (t)).

9. f (t) = U2 (t) (t - 2)^2.

10. f (t) = U1 (t) e^1 - •.

11. f (t) = U3,. (t) sin (t - 311').

12. f (t) = 2U1 (t) - U2 (t) - U3 (t).

1 3. Let f (t) be as given in Figure 12 .25.