Chapter 6: The Laplace Transform 6.3 Step Functions and Dirac δ 2 April 2018
Step Function Definition: Suppose c is a fixed real number. The unit step function u c is defined as follows: u c (t) = { 0 if t < c 1 if c t (1) If follows immediately, 1 u c (t) = { 1 if t < c 0 if c t
The Graph of u = u c (t)
It also follows: For two real numbers c < d, we have 0 if t < c u c u d (t) = 1 if c t < d 0 if d t (2)
Impulse Function Dirac δ(t) In physics, to represent a unit impulse, the Dirac Delta Function is defined, by the following two conditions: { δ(t) = 0 if t 0 δ(t)dt = 1 (3) This may appear to be an unusual way to define a function, because δ(0) is not given in any form. For of this reason, in Algebra, this will not qualify as a function,.
Approximation to Dirac δ(t) For τ > 0, define a function d τ, as follows: d τ (t) = 1 2τ (u τ u τ )(t) = Refer to the graph in the next frame. 0 if t < τ 1 2τ if τ t < τ 0 if τ t
Continued Graph of y = d τ (t):
Continued We have the following: d τ(t)dt = 1 for all τ 0. And, lim d τ (t)dt = 1 τ 0 { 0 if t 0 lim τ 0 d τ (t) = if t = 0 So, one can view δ = lim d τ τ 0
Laplace Transform of u c For c 0, the Laplace Transform L{u c (t)} = e cs s s > 0. Proof. By definition L{u c (t)} = 0 e st u c (t)dt = = c 0 c 0 0dt + e st u c (t)dt + c c e st dt = e cs s e st u c (t)dt
Translation of a Function Given a function f (t) on the domain t 0, define { 0 if t < c g(t) = f (t c) if t c Write g(t) = u c (t)f (t c). g(t) is translation of f to the right by c.
Laplace Transform and Translation Theorem 6.3.1 Suppose c > 0 and F (s) = L{f (t)}. Then, L{u c (t)f (t c)} = e cs F (s) = Therefore, Proof. L{u c (t)f (t c)} = f (t c)e st dt = L 1 {e cs L{f (t)}} = u c (t)f (t c). c c 0 u c (t)f (t c)e st dt f (x)e s(x+c) dx = e cs F (s)
Laplace Transform and Translation Theorem 6.3.2 Suppose c > 0 and F (s) = L{f (t)}. Then, Therefore, Proof. Similar to the above. L{e ct f (t c)} = F (s c) L 1 {F (s c)} = e ct f (t c). Remark. The formulas in these two theorems are in the charts.
Example 1 6.5 Dirac Delta Function Example 1 Example 2 Example 3 Express the following function in terms of step functions: 2 if 0 t < 1 2 if 1 t < 2 f (t) = 2 if 2 t < 3 2 if 3 t < 4 0 if t 4 Use the (2) for u c u d. The first line is given by (u 0 u 1 ), the second line is given by (u 1 u 2 ), the third line is given by (u 2 u 3 ), the fourth line is given by (u 3 u 4 ).
Example 1 Example 2 Example 3 So, f (t) = 2(u 0 u 1 ) 2(u 1 u 2 ) + 2(u 2 u 3 ) 2(u 3 u 4 ) = 2u 0 4u 1 + 4u 2 4u 3 + 2u 4
Example 2 6.5 Dirac Delta Function Example 1 Example 2 Example 3 Compute the Laplace transform of the function: { 0 if t < 2 f (t) = t 2 4t + 5 if t 2 To use theorem 6.3.1 (or the Charts), rewrite f (t): { 0 if t < 2 f (t) = (t 2) 2 + 1 if t 2
Example 1 Example 2 Example 3 With g(t) = t 2 + 1, we have f (t) = u 2 (t)g(t 2) By Theorem 6.3.1 (Use Charts), L{u 2 (t)g(t 2)} = e 2s L{t 2 + 1} ( 2 = e 2s s + 1 ) 3 s
Example 1 Example 2 Example 3 Example 3 Compute the inverse Laplace transform of the function: F (s) = (s 2)e 2s s 2 4s + 5 Write F (s) = e 2s H(s) where H(s) = Also write L 1 {H(s)} = h(t) By Theorem 6.3.1 (Use Charts) (s 2) s 2 4s+5. L 1 {F (s)} = u 2 (t)h(t 2) (4)
Example 1 Example 2 Example 3 { } (s 2) h(t) = L 1 {H(s)} = L 1 s 2 4s + 5 { } { } (s 2) s = L 1 = e 2t L 1 (Use Chart) (s 2) 2 + 1 s 2 + 1 = e 2t cos t
Example 1 Example 2 Example 3 By (4) L 1 {F (s)} = u 2 (t)h(t 2)} = u 2 (t)e 2(t 2) cos(t 2)
Unit Impulse δ t 0 (t) time t = t 0 Laplace Transform of d τ (t t 0 ) Laplace Transform of Dirac δ t 0 (t) The Dirac Delta δ(t), represents Unit Impulse, at time t = 0. So, the Unit Impulse at time t = t 0, is represented, by the translation δ(t t 0 ), which we denote by δ t 0 (t) := δ(t t 0 ). More directly, define { δ t 0 (t) = 0 if t t 0 δt 0 (t)dt = 1
Laplace Transform of d τ (t t 0 ) Unit Impulse δ t 0 (t) time t = t 0 Laplace Transform of d τ (t t 0 ) Laplace Transform of Dirac δ t 0 (t) Fix, t 0 > 0 and τ > 0 such that t 0 > τ. Define, 0 if t < t 0 τ d t 0 1 τ (t) := d τ (t t 0 ) = if t 2τ 0 τ t < t 0 + τ 0 if t 0 + τ t So, d t 0 τ = 1 2τ (u t 0 τ u t0 +τ). So, the Laplace Transform = 1 2τ L{d t 0 τ }(s) = 1 2τ (L{u t 0 τ}(s) L{u t0 +τ(s)}) ( e (t 0 τ)s s e (t 0+τ)s s ) ( ) = e t 0s e sτ e sτ 2s τ
Laplace Transform of Dirac δ t 0(t) Unit Impulse δ t 0 (t) time t = t 0 Laplace Transform of d τ (t t 0 ) Laplace Transform of Dirac δ t 0 (t) Theorem For time t 0 > 0, the Laplace Transform of Dirac δ t 0 (t) is: L{δ t 0 (t)} = e t 0s Outline of the Proof. We can view lim τ 0 d t 0 τ = δ t 0. So, ( L{δ t 0 } = lim L{d t 0 e t 0 ( s e sτ e sτ τ } = lim τ 0 τ 0 2s τ To compute this limit, one can use L Hospital s Rule. )) = e t 0s