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Chapter 98 the Laplace Transform of the Heaviside Function

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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 1475 © 2014, John Bird 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 1476 © 2014, John Bird 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 1477 © 2014, John Bird 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 1478 © 2014, John Bird 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 1479 © 2014, John Bird 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)} 1480 © 2014, John Bird 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 1481 © 2014, John Bird 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) 1482 © 2014, John Bird 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.
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