By John M.H. Olmsted

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Therefore its area is 1/6 times as great or 1/6. 1 bx What is the area of the function, δ (bx ) = lim tri for b positive and a→0 a a for b negative ? for “b” positive and for “b” negative? It is simply “1/|b|”. 39. Using a change of variable and the definition of the unit impulse, prove that Solutions 2-35 M. J. Roberts - 7/12/03 δ ( a(t − t0 )) = 1 δ (t − t0 ) . a δ ( x) = 0 , x ≠ 0 , [ ∞ ∫ δ ( x )dx = 1 −∞ ] δ a(t − t0 ) = 0 , where a(t − t0 ) ≠ 0 or t ≠ t0 ∞ [ ] Strength = ∫ δ a(t − t0 ) dt −∞ Let a(t − t0 ) = λ and ∴ adt = dλ Then, for a > 0, ∞ Strength = ∫ δ (λ ) −∞ and for a < 0, −∞ ∞ 1 1 dλ 1 = ∫ δ (λ )dλ = = a a −∞ a a −∞ ∞ 1 1 1 dλ 1 Strength = ∫ δ (λ ) = ∫ δ (λ )dλ = − ∫ δ (λ )dλ = − = a a∞ a −∞ a a ∞ Therefore for a > 0 and a < 0, 1 1 Strength = and δ a(t − t0 ) = δ (t − t0 ) .

1 ... -2 ... 2 4 6 8 10 12 t 1 ... -2 1 4 -2 ... 2 4 6 8 10 12 ... 2 4 6 8 10 12 t 1 ... -2 t 1 2 ... -2 ... 2 4 6 8 10 12 t g'3 (t) g'2 (t) g'1 (t) ... g3(t) g2(t) g1(t) ... 2 4 6 8 10 12 t 1 ... -2 ... 2 4 6 8 10 12 t Average derivative is zero in each case. 47. A function, g( t) , has this description: It is zero for t < −5. It has a slope of –2 in the range, −5 < t < −2 . It has 1 Hz plus a constant in the shape of a sine wave of unit amplitude and with a frequency of 4 the range, −2 < t < 2 .

In both cases that strength is − . 2 d (g(t)) dt 3π 2 t 3 2 Alternate solution: 1 πt 1 g( t) = 3 sin u t + − u t − 2 2 2 d πt 1 1 πt 1 1 3π g( t)) = 3 sin δ t + − δ t − + cos u t + − u t − ( 2 2 2 2 2 2 2 dt Solutions 2-37 M. J. Roberts - 7/12/03 d πt π 1 π 1 3π g( t)) = 3 sin − δ t + − 3 sin δ t − + cos rect ( t) ( 2 dt 4 2 4 2 2 2 1 d πt 1 3π g( t)) = −3 cos rect ( t) δ t + + δ t − + ( 2 2 2 2 2 dt 42.