Chapter 98 the Laplace Transform of the Heaviside Function

CHAPTER 98 THE LAPLACE TRANSFORM OF THE HEAVISIDE FUNCTION

EXERCISE 357 Page 1042

1. A 6 V source is switched on at time t = 4 s. Write the function in terms of the Heaviside step

function and sketch the waveform.

The function is shown sketched below

The Heaviside step function is: V(t) = 6 H(t – 4)

2 for 0〈〈t 5 2. Write the function Vt()=  in terms of the Heaviside step function and sketch 0 fort 〉 5 the waveform.

The voltage has a value of 2 up until time t = 5; then it is turned off

The function is shown sketched below

The Heaviside step function is: V(t) = 2 H(t) – H(t – 5)

3. Sketch the graph of: f(t) = H(t – 2)

A function H(t – 2) has a maximum value of 1 and starts when t = 2, as shown in the sketch below

4. Sketch the graph of: f(t) = H(t)

A function H(t) has a maximum value of 1 and starts when t = 0, as shown in the sketch below

5. Sketch the graph of: f(t) = 4 H(t – 1)

A function 4H(t – 1) has a maximum value of 4 and starts when t = 1, as shown in the sketch below

6. Sketch the graph of: f(t) = 7H(t – 5)

A function 7H(t – 5) has a maximum value of 7 and starts when t = 5, as shown in the sketch below

π 7. Sketch the graph of: f(t) = Ht− . cos t 4

π Below shows a graph of Ht− . cos t where the graph of cos t does not ‘switch on’ until t = π/4 4

ππ 8. Sketch the graph of: f(t) = 3Ht−− .cos t 26

ππ Below shows a graph of f(t) = 3Ht−− .cos t where the graph of 3 cos(t – π/6) does not 26

‘switch on’ until t = π/2

9. Sketch the graph of: f(t) = Ht( −1.) t2

Below shows a graph of f(t) = Ht( −1.) t2 where the graph of t 2 does not ‘switch on’ until t = 1

t − 10. Sketch the graph of: f(t) = H(t – 2). e 2

t t − − Below shows a graph of f(t) = H(t – 2). e 2 where the graph of e 2 does not ‘switch on’ until t = 2

t − 11. Sketch the graph of: f(t) = [H(t – 2) – H(t – 5)]. e 4

t t − − Below shows the graph of f(t) = [H(t – 2) – H(t – 5)].e 4 where the graph of e 4 does not ‘switch on’ until t = 2, but then ‘switches off’ at t = 5

ππ   12. Sketch the graph of: f(t) = 5Ht−+  .sin t  34  

ππ   Below shows a graph of f(t) = 5Ht−+  .sin t  where the graph of 5 sin(t + π/4) does not 34   ‘switch on’ until t = π/3

EXERCISE 358 Page 1044

1. Determine ℒ{H(t – 1)}

{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = {1} and c = 1 ℒ ℒ 1 Hence, ℒ{H(t – 1)} = e−s  from (i) of Table 95.1, page 1023 s

e−s = s

2. Determine ℒ{7 H(t – 3)}

{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = {7} and c = 3 ℒ ℒ 7 Hence, ℒ{7 H(t – 3) } = e−3s  from (ii) of Table 95.1, page 1023 s

7e−3s = s

3. Determine ℒ{H(t – 2).(t – 2) 2 }

{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = {}t 2 and c = 2 ℒ ℒ 2! Hence, ℒ{H(t – 2).f(t – 2) 2 } = e−2s  from (vii) of Table 95.1, page 1023 s3

2e−2s = s3

4. Determine ℒ{H(t – 3).sin(t – 3)}

ℒ{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = ℒ{sint } and c = 3

1 Hence, ℒ{H(t – 3).sin(t – 3) } = e−3s  from (iv) of Table 95.1, page 1023 s22+1

e−3s = s2 +1

5. Determine ℒ{H(t – 4). et−4 }

ℒ{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = ℒ{et } and c = 4

1 Hence, ℒ{H(t – 4). et−4 } = e−4s  from (iii) of Table 95.1, page 1023 s −1

e−4 s = s −1

6. Determine ℒ{H(t – 5).sin 3(t – 5)}

ℒ{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = ℒ{sin 3t } and c = 5

3 Hence, ℒ{H(t – 5).sin 3(t – 5) } = e− 5s  from (iv) of Table 95.1, page 1023 s22+ 3

3e−5s = s2 + 9

7. Determine ℒ{H(t – 1).(t – 1) 3 }

ℒ{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = ℒ{}t3 and c = 1

3! Hence, ℒ{H(t – 1).(t – 1) 3 } = e− s  from (viii) of Table 95.1, page 1023 s31+

6e−s = s4

8. Determine ℒ{H(t – 6).cos 3(t – 6)}

ℒ{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = ℒ{cos3t } and c = 6

s Hence, ℒ{H(t – 6).cos 3(t – 6)} = e− 6s  from (v) of Table 95.1, page 1023 s22+ 3 s e−6s = s2 + 9

9. Determine ℒ{5 H(t – 5).sinh 2(t – 5)}

ℒ{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = ℒ{sinh 2t } and c = 5

2 Hence, ℒ{5 H(t – 5).sinh 2(t – 5) } = 5 e− 5s  from (x) of Table 95.1, page 1023 s22− 2

10e−5s = s2 − 4

ππ  10. Determine ℒ{ Ht−−.cos 2  t } 33 

π ℒ{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = ℒ{cos 2t } and c = 3

π ππ  − s s Hence, ℒ{tt−−. cos 2 } = e 3  from (v) of Table 95.1, page 1023 33  s22+ 2

π − s s e 3 = s2 + 4

11. Determine ℒ{2 H(t – 3). et −3 }

ℒ{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = ℒ{et } and c = 3

1 Hence, ℒ{2 H(t – 3). et−3 } = 2e−3s  from (iii) of Table 95.1, page 1023 s −1

2e−3 s = s −1

12. Determine ℒ{3 H(t – 2).cosh(t – 2)}

ℒ{H(t – c).f(t – c)} = e−cs F(s) where in this case, F(s) = ℒ{cosht } and c = 2

s Hence, ℒ{3 H(t – 2).cosh(t – 2) } = 3 e− 2s  from (ix) of Table 95.1, page 1023 s22−1

3es −2s = s2 −1

EXERCISE 359 Page 1045

e−9 s 1. Determine ℒ −1  s

Part of the numerator corresponds to e−cs where c = 9. This indicates H(t – 9) 1 Then = F(s) = ℒ{1} from (i) of Table 97.1, page 1033 s

e−9 s Hence, ℒ −1 = H(t – 9) s

4e−3 s 2. Determine ℒ −1  s

Part of the numerator corresponds to e−cs where c = 3. This indicates H(t – 3) 4 Then = F(s) = ℒ{4} from (ii) of Table 97.1, page 1033 s

4e−3 s Hence, ℒ −1 = 4 H(t – 3) s

2e−2 s 3. Determine ℒ −1  s2

The numerator corresponds to e−cs where c = 2. This indicates H(t – 2) 1 = F(s) = ℒ{t} from (vii) of Table 97.1, page 1033 s2

2e−2 s Then ℒ −1  = 2 H(t – 2).(t – 2) s2

5e−2 s 4. Determine ℒ −1  s2 +1

Part of the numerator corresponds to e−cs where c = 2. This indicates H(t – 2)

5 1 may be written as: 5 s2 +1 s22+1 1 Then 5 = F(s) = ℒ{5 sin t} from (iv) of Table 97.1, page 1033 s22+1

5e−2 s Hence, ℒ −1 = H(t – 2).5 sin(t – 2) = 5 H(t – 2).sin(t – 2) s2 +1

3es −4 s 5. Determine ℒ −1  s2 +16

Part of the numerator corresponds to e−cs where c = 4. This indicates H(t – 4)

3s s may be written as: 3 s2 +16 s22+ 4

s Then 3 = F(s) = ℒ{3 cos 4t} from (v) of Table 97.1, page 1033 s22+ 4

3es −4 s Hence, ℒ −1 =H(t – 4).3 cos 4(t – 4) = 3 H(t – 4).cos 4(t – 4) s22+ 4

6e−2 s 6. Determine ℒ −1  s2 −1

Part of the numerator corresponds to e−cs where c = 2. This indicates H(t – 2)

6 1 may be written as: 6 s2 −1 s22−1

1 Then 6 = F(s) = ℒ{6 sinh t} from (x) of Table 97.1, page 1033 s22−1

6e−2 s Hence, ℒ −1 =H(t – 2).6 sinh (t – 2) = 6 H(t – 2).sinh (t – 2) s2 −1

3e−6 s 7. Determine ℒ −1  s3

The numerator corresponds to e−cs where c = 6. This indicates H(t – 6)

1 1 = F(s) = ℒ t 2 from (viii) of Table 97.1, page 1033 s3 2

3e−6 s 1 Then ℒ −1  = 3 H(t – 6). (t – 6) 2 = 1.5 H(t – 6). (t – 6) 2 s2 2

2es −4 s 8. Determine ℒ −1  s2 −16

Part of the numerator corresponds to e−cs where c = 4. This indicates H(t – 4)

2 s s may be written as: 2 s2 −16 s22− 4 s Then 2 = F(s) = ℒ{2 cosh 4t} from (ix) of Table 97.1, page 1033 s22− 4

2es −4 s Hence, ℒ −1 =H(t – 4).2 cosh 4(t – 4) = 2 H(t – 4).cosh 4(t – 4) s22− 4

1 − s 2es 2 9. Determine ℒ −1  s2 + 5 

1 1 Part of the numerator corresponds to e−cs where c = . This indicates Ht − 2 2  2 s s may be written as: 2 2 2 s + 5 2 + s ( 5)

 s Then 2 = F(s) = from (v) of Table 97.1, page 1033 2 ℒ{2cos 5 t} 2 + s ( 5)

1 − s 2 −1 2es 1  11   1 Hence, ℒ 2 = Ht−.2cos 5 t −= 2 Ht  − .cos 5 t − 2 + 2  22   2 s ( 5)

4e− 7 s 10. Determine ℒ −1  s −1

Part of the numerator corresponds to e−cs where c = 7. This indicates H(t – 7)

1 Then = F(s) = ℒ{ et } from (iii) of Table 97.1, page 1033 s −1

4e−7 s Hence, ℒ −1 = 4 H(t – 7). et−7 s −1