Fourier

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5 The Fourier Transform Exercises 5.2.4
1
0

F (jω) =
−∞

eat e−jωt dt +
0



e−at e−jωt dt

2a = 2 a + ω2

2 F (jω) =

0 −T T

Ae−jωt dt +
0

T

−Ae−jωt dt

=
0

2jA sin ωt dt

=

2jA (1 − cos ωT ) ω 4jA ωT sin2 = ω 2 ωT = jωAT 2 sinc2 2

3
0

F (jω) =
−T T

At + A e−jωt dt + T −

T 0



At + A e−jωt dt T

=2

At + A cos ωt dt T 0 ωT = ATsinc2 2

Exercise 2 is T × derivative of Exercise 3, so result 2 follows as (jω × T ) × result 3. Sketch is readily drawn.

c Pearson Education Limited 2004

268 4

Glyn James: Advanced Modern Engineering Mathematics, Third edition

2

F (jω) =
−2 1

2Ke−jωt dt = 8K sinc(2ω) Ke−jωt dt = 2K sinc(ω)

G(jω) =
−1

H(jω) = F (jω) − G(jω) = 2K(4 sinc(2ω) − sinc(ω)) 5 F (jω) =
−2

−1e−jωt dt +

1 −1

e−jωt dt +
1

2

−e−jωt dt

=

1 2(ejω − e−jω ) − (e2jω − e−2jω ) jω = 4 sinc(ω) − 2 sinc(2ω)

6 1 F (jω) = 2j 1 ¯ f (a) = 2j =

π a π −a π a

(ejat − e−jat )e−jωt dt ejat e−jωt 1 dt = 2j
π a π −a

j(a − ω) ω2 2jω π sin ω 2 −a a

π −a sin ω π a

ej(a−ω)t dt

¯ ¯ F (jω) = f (a) + f (−a) =

7 F (jω) =
0



e−at . sin ω0 t.e−jωt dt

¯ ¯ = f(ω0 ) − f (−ω0 ) ∞ ¯(ω0 ) = 1 e(−a+j(ω0 −ω)t) dt where f 2j 0 1 1 1 = = 2j a − j(ω0 − ω) 2j ω0 ∴ F (jω) = 2 (a + jω)2 + ω0

1 (a + jω) − jω0

c Pearson Education Limited 2004

Glyn James: Advanced Modern Engineering Mathematics, Third edition 8 1 Fc (x) = 4
a a 0

269

(ejt + e−jt )(ejxt + e−jxt ) dt

define g(x, b) =
0

ej(b+x)t dt

1 [ej(b+x)a − 1] j(b + x) 1 Fc (x) = [g(x, 1)+ g(x, −1) + g(−x, 1) + g(−x, −1)] 4 1 sin(1 + x)a sin(1 − x)a + = 2 1+x 1−x = 9 Consider F (x) = =
a 0

1.ejxt dt

−j (cos ax + j sin ax − 1) x sin ax Fc (x) = Re F (x) = x 1 − cos ax Fs (x) = Im F (x) = x 10 Consider F (x) =
∞ −at jxt e e 0

dt

=

a + jx a2 + x2 a2

a + x2 x Fs (x) = Im F (x) = 2 a + x2 Fc (x) = Re F (x) =

Exercises 5.3.6
11 12 Obvious (jω)2 Y (jω) + 3jωY (jω)+ Y (jω) = U (jω) Y (jω) = H(jω) = 1 (1 − (1 − ω2 ) 1 ω2 ) + 3jω + 3jω U (jω)

c Pearson Education Limited 2004

270 13

Glyn James: Advanced Modern Engineering Mathematics, Third edition

→ sinc

ω 2

ω → e−iω3/2 + eiω3/2 sinc 2 2 = (sin(2ω) − sin(ω)) ω = 4 sinc(2ω) − 2 sinc(ω) 14 F (jω) = =
T 2 T −2

cos(ω0 t)e−iωt dt ω = ±ω0

1 1 T T sin(ω0 − ω) + sin(ω0 + ω) ω0 − ω 2 ω0 + ω2 T T sin(ω0 + ω) 2 T sin(ω0 − ω) 2 = + 2 (ω0 − ω) T (ω0 + ω) T 2 2

Evaluating at ω = ±ω0 ⇒ F (jω) = T T T sinc(ω0 − ω) + sinc(ω0 + ω) 2 2 2

15 F (jω) =
0

T

cos ω0 t.e−jωt dt

1 ¯ ¯ = [f (ω0 ) + f (−ω0 )] 2 ¯ where f (ω0 ) =
T 0

ej(ω0 −ω)t dt

c Pearson Education Limited 2004

Glyn James: Advanced Modern Engineering Mathematics, Third edition 1 [ej(ω0 −ω)T − 1] j(ω0 − ω) 11 (ej(ω0 −ω)T − 1) F (jω) = 2 j(ω0 − ω) 1 (e−j(ω0 +ω)T − 1) − j(ω0 − ω) = =e
−jωT /2

271

ω = ω0

ejω0 T /2 T sin(ω0 − ω) ω0 − ω 2 ω = ±ω0

+ Checking at ω = ±ω0 gives F (jω) =

T e−jω0 T /2 sin(ω0 + ω) ω0 + ω 2

T −jωT /2 jω0 T /2 T T e sinc(ω0 − ω) + e−jω0 T /2 sinc(ω0 + ω) e 2 2 2

16 F (jω) =

1 −1

sin 2t.e−jωt dt
1 −1

= ¯ f (a) = F (jω) =

1 2j
1 −1

e−j(ω−2)t −e−j(ω+2)t dt

e−j(ω−a)t dt = 2 sinc(ω − a)

1 ¯ 1 ¯ f (a) − f (−a), a = 2 2j 2j = j[sinc(ω + 2) − sinc(ω − 2)]

Exercises 5.4.3
17 I H(s) = H(jω) =
0

1 s2 + 3s + 2


h(t) = (e−t − e−2t )ξ(t) 1 1 − 1 + jω 2 + jω √ √ 3 3 t + 3 sin t ξ(t) cos 2 2 √

(e−t − e−2t )e−jωt dt = ω2 1 as required. + 3jω h(t) = e
−1/2t

= II

2−

s+2 H(s) = 2 s +s+1

c Pearson Education Limited 2004 272

Glyn James: Advanced Modern Engineering Mathematics, Third edition


Consider G(ω0 ) = =

e−(1/2tjω−jω0 )t dt

0 1 2

√ √ 3 3 1 1 (G(ω0 ) − G(−ω0 )), ω0 = H(jω) = G(ω0 ) + G(−ω0 ) + 2 2 2j 2 6 2 + 4jω + So H(jω) = 4 + 4jω − 4ω 2 4 + 4jω − 4ω 2 2 + jω = 1 − ω 2 + jω 18

1 + j(ω − ω0 )

P (jω) = 2AT sinc ωT

So F (jω) = (e−jωτ + eiωτ )P (jω) = 4AT cos ωτ sinc ωT (s )2 √...
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