Solucionario parte 5 Matemáticas Avanzadas para Ingeniería - 2da Edición - Glyn James

5 The Fourier Transform Exercises 5.2.4 1




0

at −jωt

e e

F (jω) =




−∞

=

e−at e−jωt dt

0

2a + ω2

a2

2

∞

dt +




0

F (jω) =

−jωt

Ae




T

dt +

−T T

0




−Ae−jωt dt

2jA sin ωt dt

= 0

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

=

3

 
T At At −jωt +A e + A e−jωt dt − dt + F (jω) = T T −T 0 
T At + A cos ωt dt − =2 T 0   ωT 2 = AT sinc 2


0



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

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4




2

2Ke−jωt dt = 8K sinc(2ω)

F (jω) = −2 1




Ke−jωt dt = 2K sinc(ω)

G(jω) = −1

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




−1




−jωt

F (jω) =

e

1

−jωt

dt +

−2

e




2

dt +

−1

1

−e−jωt dt

 1  jω 2(e − 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

π −a sin ω πa

(ejat − e−jat )e−jωt dt jat −jωt

e

e

1 dt = 2j




π a

π −a

ej(a−ω)t dt

j(a − ω)

F (jω) = f¯(a) + f¯(−a) =

7


F (jω) = 0

∞

2jω π sin ω ω 2 − a2 a

e−at . sin ω0 t.e−jωt dt

= f¯(ω0 ) − f¯(−ω0 )
∞ 1 ¯ e(−a+j(ω0 −ω)t) dt where f (ω0 ) = 2j 0     1 1 1 1 = = 2j a − j(ω0 − ω) 2j (a + jω) − jω0 ω0 ∴ F (jω) = (a + jω)2 + ω02

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Glyn James: Advanced Modern Engineering Mathematics, Third edition 8

1 Fc (x) = 4
define g(x, b) =


a

a

(ejt + e−jt )(ejxt + e−jxt ) dt

0

ej(b+x)t dt

0

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) =

∞ 0

=

e−at ejxt dt

a + jx a2 + x2

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

a2

Exercises 5.3.6 11

Obvious

12

(jω)2 Y (jω) + 3jωY (jω) + Y (jω) = U (jω) 1 U (jω) (1 − ω 2 ) + 3jω 1 H(jω) = (1 − ω 2 ) + 3jω Y (jω) =

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13 → 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

1 1 T T sin(ω0 − ω) + sin(ω0 + ω) ω0 − ω 2 ω0 + ω 2  T T  sin(ω0 + ω) 2 T sin(ω0 − ω) 2 = + T 2 (ω0 − ω) 2 (ω0 + ω) T2

=

Evaluating at ω = ±ω0 ⇒   T T T sinc(ω0 − ω) + sinc(ω0 + ω) F (jω) = 2 2 2 15


F (jω) = 0

T

cos ω0 t.e−jωt dt

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

T

ej(ω0 −ω)t dt

0

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ω = ±ω0

Glyn James: Advanced Modern Engineering Mathematics, Third edition 1 [ej(ω0 −ω)T − 1] j(ω0 − ω)  1 1 (ej(ω0 −ω)T − 1) F (jω) = 2 j(ω0 − ω)  1 −j(ω0 +ω)T (e − − 1) j(ω0 − ω)  jω0 T /2 T −jωT /2 e sin(ω0 − ω) =e ω0 − ω 2  −jω0 T /2 T e sin(ω0 + ω) + ω0 + ω 2

ω = ω0

=

ω = ±ω0

Checking at ω = ±ω0 gives   T −jωT /2 jω0 T /2 T T −jω0 T /2 e F (jω) = e sinc(ω0 − ω) + e sinc(ω0 + ω) 2 2 2 16




1

F (jω) =

sin 2t.e−jωt dt

−1


1 1 e−j(ω−2)t − e−j(ω+2)t dt = 2j −1
1 ¯ e−j(ω−a)t dt = 2 sinc(ω − a) f (a) = −1

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

Exercises 5.4.3 17 I

H(s) =

s2


H(jω) =

0

∞

h(t) = (e−t − e−2t )ξ(t)

(e−t − e−2t )e−jωt dt =

1 1 − 1 + jω 2 + jω

1 as required. 2− + 3jω √ √   √ s+2 3 3 H(s) = 2 t + 3 sin t ξ(t) h(t) = e−1/2t cos s +s+1 2 2 =

II

1 + 3s + 2

ω2

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∞ e−(1/2tjω−jω0 )t dt Consider G(ω0 ) = 0

=

1 2

1 + j(ω − ω0 )

√ √ 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) = 2 4 + 4jω − 4ω 4 + 4jω − 4ω 2 2 + jω = 1 − ω 2 + jω 18

P (jω) = 2AT sinc ωT

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

19

G(s ) =

(s )2 √ (s )2 + 2s + 1

G(jω) =

=

−ω 2 √ 1 − ω 2 + 2jω 1 ω2

1 √ − 1 + 2 ωj

Thus | G(jω) |→ 0 as ω → 0 and | G(jω) |→ 1 as ω → ∞ High-pass filter.

20

g(t) = e−a|t| −→ G(jω) = f (jt) =

a2

2a + ω2

1 G(jt) −→ πg(−ω) = πe−a|ω| 2 c Pearson Education Limited 2004


Glyn James: Advanced Modern Engineering Mathematics, Third edition 21

F{f (t) cos ω0 t} =

1 1 F (j(ω − ω0 )) + F (j(ω + ω0 )) 2 2

F (jω) = 2T sinc ωT ∴ F{PT (t) cos ω0 t}   = T sinc(ω − ω0 )T + sinc(ω + ω0 )T

Exercises 5.5.3 22




1 2π

∞

−∞

πδ(ω − ω0 )e

jωt

1 dω + 2π




∞

−∞

πδ(ω + ω0 )ejωt dω

1 jω0 t + e−jω0 t ) (e 2 = cos ω0 t =

23

24

F{e±jω0 t } = 2πδ(ω ∓ ω0 ) 1 ∴ F{sin ω0 t} = {2πδ(ω − ω0 ) − 2πδ(ω + ω0 )} 2j = jπ[δ(ω + ω0 ) − δ(ω − ω0 )]
∞ 1 jπ[δ(ω + ω0 ) − δ(ω − ω0 )]ejωt dω 2π −∞ j = [e−jω0 t − e+jω0 t ] = sin ω0 t 2 G(jω) =

∞

g(t)e−jωt dt; G(jt) =

−∞


So

∞ −∞

g(ω)e−jωt dω

∞

f (t)G(jt) dt 
∞ 
∞ −jωt f (t) g(ω)e dω dt = −∞ −∞ 
∞ 
∞ −jωt g(ω) f (t)e dt dω = −∞ −∞
∞
∞ g(ω)F (jω)dω = g(t)F (jt) dt = −∞

−∞

−∞

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25

Write result 24 as
∞




∞

f (ω)F{g(t)}dω = −∞ ∞




−∞
∞

f (ω)F{G(jt)}dω =

so −∞

−∞

F{f (t)}g(ω)dω F{f (t)}G(jω)dω

 g(t) → G(jω)  Now G(jt) → 2πg(−ω) symmetry  G(−jt) → 2πg(ω)
∞
∞ Thus f (ω).2πg(ω)dω = F (jω)G(−jω)dω −∞ −∞
∞
∞ 1 or f (t)g(t) dt = F (jω)G(−jω)dω 2π −∞ −∞ 26

F{H(t) sin ω0 t}  
∞   1 1 du πj δ(ω − u + ω0 ) − δ(ω − u − ω0 ) πδ(u) + = 2π −∞ ju    1 1 j 1 = πδ(ω + ω0 ) − πδ(ω − ω0 ) + − 2 2 ω + ω0 ω − ω0  ω0 πj  δ(ω + ω0 ) − δ(ω − ω0 ) − 2 = 2 ω − ω02

27 A an = T




d/2

e−nω0 t dt =

−d/2

f (t) =

F (ω) =

nω0 d Ad sinc , T 2

ω0 = 2π/T

∞ nω0 d nω0 t Ad  e sinc , T n=−∞ 2

∞ 2πAd  nω0 d δ(ω − nω0 ) sinc T n=−∞ 2

Exercises 5.6.6 28 T = 1,

N = 4,

∆ω = 2π/(4 × 1) =

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Glyn James: Advanced Modern Engineering Mathematics, Third edition

G0 =

3 

gn e− ×n×0×π/2 = 2

n=0

G1 =

3 

gn e− ×n×1×π/2 = 0

n=0

G2 =

3 

gn e− ×n×2×π/2 = 2

n=0

G3 =

3 

gn e− ×n×3×π/2 = 0

n=0

G = {2, 0, 2, 0} 29 N = 4, W n = e− nπ/2 

1 0 0 1  gn =  1 0 0 1   1 G00 G  1 G =  10  =  1 G01 0 G11 

Bit reversal gives

    1 0 1 2 0 10 0   =   −1 0 1 0 0 −1 0 0 1 −1 0 0

0 0 1 1

    0 2 2 00 2   =   − 0 0  0 0

  2 0 G=  2 0

30 Computer experiment.

31 Follows by direct substitution. c Pearson Education Limited 2004


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Review exercises 5.9 1


FS (x) =

2




1

t sin xt dt + 0

2

sin xt dt = 1

sin x cos 2x − x2 x

πt π π + H(t − 2) f (t) = − H(−t − 2) + (H(t + 2) − H(t − 2)) 2 4 2 1 + πδ(ω) ω   1 −2jω F{H(t − 2)} = e + πδ(ω) jω     2jω −1 2jω −1 + πδ(−ω) = e + πδ(ω) F{H(−t − 2)} = e jω jω
 π π 2 −jωt π H(t − 2) F{f (t)} = F{− H(−t − 2)} + te dt + F 2 4 −2 2 F{H(t)} =

= 3

−πj sinc 2ω ω

ωT 2 F {cos ω0 t} = π [δ(ω + ω0 ) + δ(ω − ω0 )]

F {H(t + T /2) − H(t − T /2)} = T sinc

Using convolution π F {f (t)} = 2π




∞

T sinc −∞

T (ω − u) (δ(u + ω0 ) + δ(u − ω0 ))du 2

  T T T sinc(ω − ω0 ) + sinc(ω + ω0 ) = 2 2 2 4

  1 1 [πδ(ω − ω0 ) + πδ(ω + ω0 )] ∗ πδ(ω) + F {cos ω0 tH(t)} = 2π jω 1 = 2π

  1 du {πδ(ω − u − ω0 ) + πδ(ω − u + ω0 )} πδ(u) + ju −∞




∞

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277

jω π [δ(ω − ω0 ) + δ(ω + ω0 )] + 2 2 ω0 − ω 2

5 F{f (t) cos ωc t cos ωc t} =

F (jω + jωc ) + F (jω − jωc ) ∗ π [δ(ω − ωc ) + δ(ω + ωc )] 2
1 ∞ = [F (j(u + ωc )) + F (j(u − ωc ))] × 4 −∞ [δ(ω − u − ωc ) + δ(ω − u + ωc )] du 1 1 = F (jω) + [F (jω + 2jωc ) + F (jω − 2jωc )] 2 4

Or write as 1 f (t) (1 + cos 2ωc t) 2 etc.

6 H(t + 1) − H(t − 1) ↔ 2 sinc ω By symmetry 2 sinc t ↔ 2π(H(−ω + 1) − H(−ω − 1)) = 2π(H(ω + 1) − H(ω − 1))

7(a) Simple poles at s = a and s = b . Residue at s = a is eat /(a − b) , at s = b it is ebt /(b − a) , thus

1 at e − ebt H(t) f (t) = a−b 7(b) Double pole at s = 2 , residue is d lim s→2 ds



es t (s − 2) (s − 2)2 2



= te2t

So f (t) = te2t H(t)

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7(c) Simple pole at s = 1 , residue e−t , double pole at s = 0 , residue 

d lim s→0 ds

es t s+1

 = (t − 1)H(t)

Thus f (t) = (t − 1 + e−t )H(t) 8(a)




∞

y(t) = −∞




Thus − sin ω0 t =

∞

h(t − τ )u(τ ) dτ

h(t − τ ) cos ω0 τ dτ = f (t), say

−∞

If u(τ ) = cos ω0 (τ + π/4)


∞

y(t) = −∞




∞

= −∞

h(t − τ ) cos ω0 (τ + π/4) dτ

h(t − (τ − π/4)) cos ω0 τ dτ = f (t + π/4) = − sin ω0 (t + π/4)

8(b) Since sin ω0 t = cos ω0 (t − π/2ω0 )


∞

y(t) = −∞




∞

= −∞




∞

= −∞

h(t − τ ) sin ω0 t dτ

h(t − τ ) cos ω0 (τ − π/2ω0 ) dτ h(t − (τ + π/2ω0 )) cos ω0 τ dτ

= f (t − π/2ω0 ) = − sin(ω0 t − π/2) = cos ω0 t 8(c) ejω0 t = cos ω0 t + j sin ω0 t This is transformed from above to − sin ω0 t + j cos ω0 t = j ejω0 t c Pearson Education Limited 2004


Glyn James: Advanced Modern Engineering Mathematics, Third edition 8(d) Proceed as above using e−jω0 t = cos ω0 t − j sin ω0 t 9 F(sgn(t)) = F(f (t)) = F (jω) =

2 , obvious jω

Symmetry, F (jt) =

2 ↔ 2/πf (−ω) = 2πsgn(−ω) jt

That is

1 ↔ −πsgn(ω) jt 

or −

10

1 1 g(t) = − ∗ f (t) = − πt π

so




1 πt

∞

−∞

1 g(x) = π




 ↔ jsgn(ω)

1 f (τ ) dτ = t−τ π

∞

−∞




∞

−∞

f (τ ) dτ = FHi (t) τ −t

f (t) dt = FHi (x) t−x

So from Exercise 9 FHi (jω) = jsgn(ω) × F (jω) so |FHi (jω)| = |jsgn(ω)| |F (jω)| = |F (jω)| and arg(FHi (jω)) = arg(F (jω)) + π/2, ω ≥ 0 Similarly arg(FHi (jω)) = arg(F (jω)) − π/2, ω < 0 11 First part, elementary algebra. 1 FHi (x) = π




∞

−∞

t (t2

+

a2 )(t

− x)

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dt

279

280

Glyn James: Advanced Modern Engineering Mathematics, Third edition 1 1 = 2 π x + a2




∞



−∞

 a2 t xt dt + − t2 + a2 t − x t2 + a2 a = 2 x + a2

12(a)




∞

f (t) dt = FHi (x) −∞ t − x
1 ∞ f (a + t) dt H{f (a + t)} = π −∞ t − x
1 ∞ f (t) = dt = FHi (a + x) π −∞ t − (a + x) 1 H{f (t)} = π

12(b)


1 ∞ f (at) dt H{f (at)} = π −∞ t − x
1 ∞ f (t) = dt = FHi (ax), a > 0 π −∞ t − ax

12(c)


1 ∞ f (−at) dt H{f (−at)} = π −∞ t − x
1 ∞ f (t) =− dt = −FHi (−ax), a > 0 π −∞ t + ax

12(d)

 H 1 = π



df dt



1 = π




∞

−∞

f  (t) dt t−x

 ∞
∞ f (t)  f (t) + dt 2 t − x −∞ −∞ (t − x)

Provided lim f (t)/t = 0 then |t|→∞

H



df dt



1 = π




∞

−∞

f (t) 1 d dt = 2 (t − x) π dx =




∞

−∞

d FHi (x) dx

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f (t) dt t−x

Glyn James: Advanced Modern Engineering Mathematics, Third edition 12(e) x π




∞

−∞




1 f (t) dt + t−x π

∞

1 f (t) dt = π −∞




∞

−∞

tf (t) dt t−x

= H{tf (t)} 13

From Exercise 10 FHi (t) = −

1 ∗ f (t) πt

So from Exercise 9, F{FHi (t)} = jsgn (ω) × F( ω) so F(jω) = −jsgn (ω) × F{FHi (t)}


Thus

∞

f (t) = −∞

1 1 FHi (τ )dτ = − π(t − τ ) π




∞

−∞

1 FHi (x)dx (x − τ )

14 fa (t) = f (t) − jFHi (t) F{fa (t)} = F (jω) − j(jsgn (ω))F (jω) = F (jω) + sgn (ω)F (jω)  =

2F (jω), ω > 0 0, ω 0 , then F (jω) = π[δ(ω − ω0 ) + δ(ω + ω0 )] so F{fˆ(t)} = 2πδ(ω − ω0 ) whence fˆ(t) = f (t) − jFHi (t) = ejω0 t = cos ω0 t + j sin ω0 t and so FHi (t) = − sin ω0 t When g(t) = sin ω0 t, ω0 > 0 G(jω) = jπ[δ(ω + ω0 ) − δ(ω − ω0 )] and thus gˆ(t) = −jejω0 t = −j(cos ω0 t + j sin ω0 t) so H{sin ω0 t} = cos ω0 t ¯ = 0, t < 0 , then when t < 0 16 If h(t) ¯ e (t) = 1 h(−t), ¯ ¯ o (t) = − 1 h(−t) ¯ h and h 2 2 ¯ e (t) ¯ o (t) = −h i.e. h When t > 0 then

¯ e (t) = 1 h(t), ¯ ¯ o (t) = 1 h(t) ¯ h and h 2 2 ¯ e (t) ¯ o (t) = h i.e. h

That is ¯ o (t) = sgn (t)h ¯ e (t) ∀t h c Pearson Education Limited 2004


Glyn James: Advanced Modern Engineering Mathematics, Third edition Thus ¯ =h ¯ e (t) + sgn (t)h ¯ e (t) h(t) When h(t) = sin t H(t)

  

1 sin t, t > 0 2 ¯ e (t) = h   − 1 sin t, t < 0 2 and since

¯ e (t) = 1 sin t ∀t sgn (t) h 2

the result is comfirmed. Then taking the FT of the result, ! ¯ e (t) ¯ ¯ e (jω) + F sgn (t)h H(jω) =H 

 2 ¯ e (jω) ∗H jω ! ¯ e (jω) + jH H ¯ e (jω) =H

¯ e (jω) + 1 =H 2π




When ¯ H(jω) =

∞

e−at e−jwt dt =

−∞



then H

a 2 a + ω2



or H Finally

 H

at 2 a + t2



a 2 a + t2

 =−

 H

a2

ω + ω2

a2

x + x2

 =−

x 1 = −x 2 + 2 a +x π

So

a ω − 2 2 +ω a + ω2

a2

t 2 a + t2




∞

−∞

 =

a2

a a2 dt = a2 + t2 a2 + x2

a + x2

17(a)
FH (s) =

∞

e−at (cos 2πst + sin 2πst) dt =

0

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a + 2πs + 4π 2 s2

a2

283

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Glyn James: Advanced Modern Engineering Mathematics, Third edition

17(b)


FH (s) =


18

T

(cos 2πst + sin 2πst) dt = −T




∞

∞

f (t) cos 2πst dt O(s) =

E(s) = −∞

f (t) sin 2πst dt −∞




E(s) −  O(s) =

1 sin 2πst πs

∞

f (t)e−j2πst dt = F (js)

−∞

From 17(a)

whence E(s) =

FH (s) =

1 + πs 2 + 2π 2 s2

1 , 2 + 2π 2 s2

O(s) =

F ( s) =

1 − jπs 2 + 2π 2 s2

so

πs 2 + 2π 2 s2

agreeing with the direct calculation
F (js) =

∞

e−2t e−j2πst dt =

0




19 H{f (t − T )} =


∞

−∞

1 − jπs 2 + 2π 2 s2

f (t − T ) cas 2πst dt

∞

f (τ ) [cos 2πsτ (cos 2πsT + sin 2πsT )+

= −∞

sin 2πsτ (cos 2πsT − sin 2πsT )] dt = cos 2πsT FH (s) + sin 2πsT FH (−s) 20 The Hartley transform follows at once since FH (s) = {F (js)} − {F (js)} =

1 1 δ(s) + 2 sπ

From time shifting    1 1 1 1 + cos πs δ(s) + FH (s) = sin πs δ(−s) − 2 sπ 2 sπ 

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Glyn James: Advanced Modern Engineering Mathematics, Third edition

285

cos πs − sin πs 1 δ(s) + 2 πs

=




21

∞

H{δ(t)} =

δ(t) cas 2πst dt = 1 −∞

From Exercise 18 it follows that the inversion integral for the Hartley transform is


∞

f (t) = −∞

FH (s) cas 2πstds

and so the symmetry property is simply f (t) ↔ FH (s) =⇒ FH (t) ↔ f (s) Thus H{1} = δ(s)


At once H{δ(t − t0 )}

∞

−∞

δ(t − t0 ) cas 2πst dt = cas 2πst0

By symmetry H{cas 2πs0 t} = δ(s − s0 )

22

1 = 2




1 1 FH (s − s0 ) + FH (s + s0 ) 2 2 ∞

−∞

f (t) {cos 2π(s − s0 )t + sin 2π(s − s0 )t

+ cos 2π(s + s0 )t + sin 2π(s + s0 )t} dt
∞ = f (t) cos 2πs0 t [cos 2πst + sin 2πst] dt −∞

= H{f (t) cos 2πs0 t} From Exercise 21, setting f (t) = 1 H{cos 2πs0 t} =

1 (δ(s − s0 ) + δ(s + s0 )) 2

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Glyn James: Advanced Modern Engineering Mathematics, Third edition

also H{sin 2πs0 t} = H{cas 2πs0 t} − H{cos 2πs0 t} 1 1 = δ(s − s0 ) − (δ(s − s0 ) + δ(s + s0 )) = (δ(s − s0 ) − δ(s + s0 )) 2 2 23




t

π 2

(1 + τ 2 )−1 dτ = tan−1 t +

−∞




Thus −1

F{tan

t} = F


=F

∞

−∞

t

2 −1

(1 + τ )

 dτ

−∞

(1 + τ 2 )−1 H(t − τ )dτ

−F

 −F

π 

2 π 



2

 π  1 =F ∗ H(t) − F 1 + t2 2     1 π 1 =F + πδ(ω) − × 2πδ(ω) × 2 1+t ω 2 But from Exercise 1

  F e−|t| =

and so by symmetry

 F

whence −1

F tan and so

24

!

1 1 + t2

−|ω|

t = πe

 ×



2 1 + ω2

= πe−|ω|

 π 1 + πδ(ω) − × 2πδ(ω) jω 2

! πe−|ω| F tan−1 t = jω

1 1 [1 + cos ω0 t] ↔ [2πδ(ω) + πδ(ω − ω0 ) + πδ(ω + ω0 )] 2 2

and H(t + T /2) − H(t − T /2) ↔ 2T sinc ω


so F {x(t)} =

∞

−∞

2T sinc (ω − u)

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Glyn James: Advanced Modern Engineering Mathematics, Third edition   1 × πδ(u) + (δ(ω − ω0 ) + δ(u + ω0 )) du 2   1 1 = T sinc ω + sinc (ω − ω0 ) + sinc (ω + ω0 ) 2 2 25

287

3

1 2πνr H(ν) = f (r) cas 4 r=0 4 1 [f (0) + f (1) + f (2) + f (3)] 4 1 H(1) = [f (0) + f (1) − f (2) − f (3)] 4 1 H(0) = [f (0) − f (1) + f (2) − f (3)] 4 1 H(0) = [f (0) − f (1) − f (2) + f (3)] 4

H(0) =

so



1 1 1 T=  4 1 1

 1 1 1 1 −1 −1   −1 1 −1 −1 −1 1

By elementary calculation T2 = 1/4T and if T−1 exists, T−1 = 4T . Since T−1 T = I , it does. Then 

1 1 T−1 H =  1 1

    1 1 1 H(0) f (0) 1 −1 −1   H(1)   f (1)   =  −1 1 −1 H(2) f (2) −1 −1 1 H(3) f (3)

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