solucionario parte 5 matemáticas avanzadas para ingeniería - 2da edición - glyn james

5

The Fourier Transform

Exercises 5.2.4

1F (jω) =

∫ 0

−∞eate−jωt dt +

∫ ∞

0e−ate−jωt dt

=2a

a2 + ω2

2

F (jω) =∫ 0

−T

Ae−jωt dt +∫ T

0−Ae−jωt dt

=∫ T

02jA sinωt dt

=2jA

ω(1 − cos ωT )

=4jA

ωsin2 ωT

2

= jωAT 2 sinc2(

ωT

2

)

3

F (jω) =∫ 0

−T

(At

T+ A

)e−jωt dt +

∫ T

0

(−At

T+ A

)e−jωt dt

= 2∫ T

0

(−At

T+ A

)cos ωt dt

= AT sinc2(

ωT

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 Glyn James: Advanced Modern Engineering Mathematics, Third edition

4

F (jω) =∫ 2

−22Ke−jωt dt = 8K sinc(2ω)

G(jω) =∫ 1

−1Ke−jωt dt = 2K sinc(ω)

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

5

F (jω) =∫ −1

−2e−jωt dt +

∫ 1

−1e−jωt dt +

∫ 2

1−e−jωt dt

=1jω

[2(ejω − e−jω) − (e2jω − e−2jω)

]= 4 sinc(ω) − 2 sinc(2ω)

6

F (jω) =12j

∫ πa

− πa

(ejat − e−jat)e−jωt dt

f(a) =12j

∫ πa

− πa

ejate−jωt dt =12j

∫ πa

− πa

ej(a−ω)t dt

=sinω π

a

j(a − ω)

F (jω) = f(a) + f(−a) =2jω

ω2 − a2 sinωπ

a

7

F (jω) =∫ ∞

0e−at. sinω0t.e

−jωt dt

= f(ω0) − f(−ω0)

where f(ω0) =12j

∫ ∞

0e(−a+j(ω0−ω)t) dt

=12j

(1

a − j(ω0 − ω)

)=

12j

(1

(a + jω) − jω0

)

∴ F (jω) =ω0

(a + jω)2 + ω20


Glyn James: Advanced Modern Engineering Mathematics, Third edition 269

8

Fc(x) =14

∫ a

0(ejt + e−jt)(ejxt + e−jxt) dt

define g(x, b) =∫ a

0ej(b+x)t dt

=1

j(b + x)[ej(b+x)a − 1]

Fc(x) =14[g(x, 1) + g(x,−1) + g(−x, 1) + g(−x,−1)]

=12

[sin(1 + x)a

1 + x+

sin(1 − x)a1 − x

]

9 Consider F (x) =∫ a

0 1.ejxt dt

=−j

x(cos ax + j sin ax − 1)

Fc(x) = Re F (x) =sin ax

x

Fs(x) = Im F (x) =1 − cos ax

x

10 Consider F (x) =∫∞0 e−atejxt dt

=a + jx

a2 + x2

Fc(x) = Re F (x) =a

a2 + x2

Fs(x) = Im F (x) =x

a2 + x2

Exercises 5.3.6

11 Obvious

12 (jω)2Y (jω) + 3jωY (jω) + Y (jω) = U(jω)

Y (jω) =1

(1 − ω2) + 3jωU(jω)

H(jω) =1

(1 − ω2) + 3jω



13

→ sincω

2

→ (e−iω3/2 + eiω3/2

)sinc

ω

2=

2ω

(sin(2ω) − sin(ω))= 4 sinc(2ω) − 2 sinc(ω)

14

F (jω) =∫ T

2

− T2

cos(ω0t)e−iωt dt

=1

ω0 − ωsin(ω0 − ω)

T

2+

1ω0 + ω

sin(ω0 + ω)T

2ω �= ±ω0

=T

2

[sin(ω0 − ω)T

2

(ω0 − ω)T2

+sin(ω0 + ω)T

2

(ω0 + ω)T2

]

Evaluating at ω = ±ω0 ⇒

F (jω) =T

2

[sinc(ω0 − ω)

T

2+ sinc(ω0 + ω)

T

2

]

15

F (jω) =∫ T

0cos ω0t.e

−jωt dt

=12[f(ω0) + f(−ω0)]

where f(ω0) =∫ T

0ej(ω0−ω)t dt



=1

j(ω0 − ω)[ej(ω0−ω)T − 1] ω �= ω0

F (jω) =12

[1

j(ω0 − ω)(ej(ω0−ω)T − 1)

− 1j(ω0 − ω)

(e−j(ω0+ω)T − 1)]

= e−jωT/2[ejω0T/2

ω0 − ωsin(ω0 − ω)

T

2

+e−jω0T/2

ω0 + ωsin(ω0 + ω)

T

2

]ω �= ±ω0

Checking at ω = ±ω0 gives

F (jω) =T

2e−jωT/2

[ejω0T/2 sinc(ω0 − ω)

T

2+ e−jω0T/2 sinc(ω0 + ω)

T

2

]

16

F (jω) =∫ 1

−1sin 2t.e−jωt dt

=12j

∫ 1

−1e−j(ω−2)t − e−j(ω+2)t dt

f(a) =∫ 1

−1e−j(ω−a)t dt = 2 sinc(ω − a)

F (jω) =12j

f(a) − 12j

f(−a), a = 2

= j[sinc(ω + 2) − sinc(ω − 2)]

Exercises 5.4.3

17

I H(s) =1

s2 + 3s + 2h(t) = (e−t − e−2t)ξ(t)

H(jω) =∫ ∞

0(e−t − e−2t)e−jωt dt =

11 + jω

− 12 + jω

=1

2 − ω2 + 3jωas required.

II H(s) =s + 2

s2 + s + 1h(t) = e−1/2t

(cos

√3

2t +

√3 sin

√3

2t

)ξ(t)



Consider G(ω0) =∫ ∞

0e−(1/2tjω−jω0)t dt

=1

12 + j(ω − ω0)

H(jω) =12G(ω0) +

12G(−ω0) +

√3

2j(G(ω0) − G(−ω0)), ω0 =

√3

2

So H(jω) =2 + 4jω

4 + 4jω − 4ω2 +6

4 + 4jω − 4ω2

=2 + jω

1 − ω2 + 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′ + 1G(jω) =

−ω2

1 − ω2 +√

2jω

=1

1ω2 − 1 +

√2 j

ω

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

20 g(t) = e−a|t| −→ G(jω) =2a

a2 + ω2

f(jt) =12G(jt) −→ πg(−ω) = πe−a|ω|



21 F{f(t) cos ω0t} =12F (j(ω − ω0)) +

12F (j(ω + ω0))

F (jω) = 2T sinc ωT

∴ F{PT (t) cos ω0t}= T

[sinc(ω − ω0)T + sinc(ω + ω0)T

]

Exercises 5.5.322

12π

∫ ∞

−∞πδ(ω − ω0)ejωtdω +

12π

∫ ∞

−∞πδ(ω + ω0)ejωtdω

=12(ejω0t + e−jω0t)

= cos ω0t

23F{e±jω0t} = 2πδ(ω ∓ ω0)

∴ F{sinω0t} =12j

{2πδ(ω − ω0) − 2πδ(ω + ω0)}= jπ[δ(ω + ω0) − δ(ω − ω0)]

12π

∫ ∞

−∞jπ[δ(ω + ω0) − δ(ω − ω0)]ejωtdω

=j

2[e−jω0t − e+jω0t] = sinω0t

24 G(jω) =∞∫

−∞g(t)e−jωt dt; G(jt) =

∫∞−∞ g(ω)e−jωtdω

So∫ ∞

−∞f(t)G(jt) dt

=∫ ∞

−∞f(t)

(∫ ∞

−∞g(ω)e−jωtdω

)dt

=∫ ∞

−∞g(ω)

(∫ ∞

−∞f(t)e−jωt dt

)dω

=∫ ∞

−∞g(ω)F (jω)dω =

∫ ∞

−∞g(t)F (jt) dt



25 Write result 24 as∫ ∞

−∞f(ω)F{g(t)}dω =

∫ ∞

−∞F{f(t)}g(ω)dω

so∫ ∞

−∞f(ω)F{G(jt)}dω =

∫ ∞

−∞F{f(t)}G(jω)dω

Nowg(t) → G(jω)

G(jt) → 2πg(−ω)G(−jt) → 2πg(ω)

symmetry

Thus∫ ∞

−∞f(ω).2πg(ω)dω =

∫ ∞

−∞F (jω)G(−jω)dω

or∫ ∞

−∞f(t)g(t) dt =

12π

∫ ∞

−∞F (jω)G(−jω)dω

26 F{H(t) sinω0t}

=12π

∫ ∞

−∞πj[δ(ω − u + ω0) − δ(ω − u − ω0)

][πδ(u) +

1ju

]du

=j

2[πδ(ω + ω0) − πδ(ω − ω0)

]+

12

[1

ω + ω0− 1

ω − ω0

]

=πj

2[δ(ω + ω0) − δ(ω − ω0)

]− ω0

ω2 − ω20

27

an =A

T

∫ d/2

−d/2e−nω0t dt =

Ad

Tsinc

nω0d

2, ω0 = 2π/T

f(t) =Ad

T

∞∑n=−∞

sincnω0d

2enω0t,

F (ω) =2πAd

T

∞∑n=−∞

sincnω0d

2δ(ω − nω0)

Exercises 5.6.6

28

T = 1, N = 4, ∆ω = 2π/(4 × 1) =π

2



G0 =3∑

n=0

gne− ×n×0×π/2 = 2

G1 =3∑

n=0

gne− ×n×1×π/2 = 0

G2 =3∑

n=0

gne− ×n×2×π/2 = 2

G3 =3∑

n=0

gne− ×n×3×π/2 = 0

G = {2, 0, 2, 0}

29

N = 4, Wn = e− nπ/2

g′n =

1 0 1 00 1 0 11 0 −1 00 1 0 −1

1010

=

2000

G′ =

G00G10G01G11

=

1 1 0 01 −1 0 01 0 1 −0 0 1

2000

=

2200

Bit reversal gives

G =

2020

30 Computer experiment.

31 Follows by direct substitution.



Review exercises 5.9

1

FS(x) =∫ 1

0t sinxt dt +

∫ 2

1sinxt dt =

sinx

x2 − cos 2x

x

2

f(t) = −π

2H(−t − 2) + (H(t + 2) − H(t − 2))

πt

4+

π

2H(t − 2)

F{H(t)} =1

ω+ πδ(ω)

F{H(t − 2)} = e−2jω[

1jω

+ πδ(ω)]

F{H(−t − 2)} = e2jω[−1

jω+ πδ(−ω)

]= e2jω

[−1jω

+ πδ(ω)]

F{f(t)} = F{−π

2H(−t − 2)} +

π

4

∫ 2

−2te−jωt dt + F

{π

2H(t − 2)

}

=−πjω

sinc 2ω

3

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

2F {cos ω0t} = π [δ(ω + ω0) + δ(ω − ω0)]

Using convolution

F {f(t)} =π

2π

∫ ∞

−∞T sinc

T

2(ω − u) (δ(u + ω0) + δ(u − ω0))du

=T

2

[sinc(ω − ω0)

T

2+ sinc(ω + ω0)

T

2

]

4

F {cos ω0tH(t)} =12π

[πδ(ω − ω0) + πδ(ω + ω0)] ∗[πδ(ω) +

1jω

]

=12π

∫ ∞

−∞{πδ(ω − u − ω0) + πδ(ω − u + ω0)}

[πδ(u) +

1ju

]du



=π

2[δ(ω − ω0) + δ(ω + ω0)] +

jωω2

0 − ω2

5

F{f(t) cos ωct cos ωct}

=F (jω + jωc) + F (jω − jωc)

2∗ π [δ(ω − ωc) + δ(ω + ωc)]

=14

∫ ∞

−∞[F (j(u + ωc)) + F (j(u − ωc))]×

[δ(ω − u − ωc) + δ(ω − u + ωc)] du

=12F (jω) +

14

[F (jω + 2jωc) + F (jω − 2jωc)]

Or write as

f(t)12(1 + cos 2ωct)

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

f(t) =1

a − b

(eat − ebt

)H(t)

7(b) Double pole at s = 2, residue is

lims→2

dds

((s − 2)2

est

(s − 2)2

)= te2t

So f(t) = te2tH(t)



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

lims→0

dds

(est

s + 1

)= (t − 1)H(t)

Thus f(t) = (t − 1 + e−t)H(t)

8(a)

y(t) =∫ ∞

−∞h(t − τ)u(τ) dτ

Thus− sinω0t =

∫ ∞

−∞h(t − τ) cos ω0τdτ = 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 ω0τ dτ = f(t + π/4)

= − sinω0(t + π/4)

8(b) Since sinω0t = cos ω0(t − π/2ω0)

y(t) =∫ ∞

−∞h(t − τ) sinω0t dτ

=∫ ∞

−∞h(t − τ) cos ω0(τ − π/2ω0) dτ

=∫ ∞

−∞h(t − (τ + π/2ω0)) cos ω0τ dτ

= f(t − π/2ω0) = − sin(ω0t − π/2) = cos ω0t

8(c)

ejω0t = cos ω0t + j sinω0t

This is transformed from above to

− sinω0t + j cos ω0t = j ejω0t



8(d) Proceed as above using

e−jω0t = cos ω0t − j sinω0t

9

F(sgn(t)) = F(f(t)) = F (jω) =2jω

, obvious

Symmetry,

F (jt) =2jt

↔ 2/πf(−ω) = 2πsgn(−ω)

That is1jt

↔ −πsgn(ω)

or

−(

1πt

)↔ jsgn(ω)

10

g(t) = − 1πt

∗ f(t) = − 1π

∫ ∞

−∞

f(τ)t − τ

dτ =1π

∫ ∞

−∞

f(τ)τ − t

dτ = FHi(t)

so

g(x) =1π

∫ ∞

−∞

f(t)t − x

dt = FHi(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.

FHi(x) =1π

∫ ∞

−∞

t

(t2 + a2)(t − x)dt



=1π

1x2 + a2

∫ ∞

−∞

[a2

t2 + a2 +t

t − x− xt

t2 + a2

]dt

=a

x2 + a2

12(a)

H{f(t)} =1π

∫ ∞

−∞

f(t)t − x

dt = FHi(x)

H{f(a + t)} =1π

∫ ∞

−∞

f(a + t)t − x

dt

=1π

∫ ∞

−∞

f(t)t − (a + x)

dt = FHi(a + x)

12(b)

H{f(at)} =1π

∫ ∞

−∞

f(at)t − x

dt

=1π

∫ ∞

−∞

f(t)t − ax

dt = FHi(ax), a > 0

12(c)

H{f(−at)} =1π

∫ ∞

−∞

f(−at)t − x

dt

= − 1π

∫ ∞

−∞

f(t)t + ax

dt = −FHi(−ax), a > 0

12(d)

H{

df

dt

}=

1π

∫ ∞

−∞

f ′(t)t − x

dt

=1π

[f(t)t − x

∣∣∣∣∞

−∞+∫ ∞

−∞

f(t)(t − x)2

dt

]

Provided lim|t|→∞

f(t)/t = 0 then

H{

df

dt

}=

1π

∫ ∞

−∞

f(t)(t − x)2

dt =1π

ddx

∫ ∞

−∞

f(t)t − x

dt

=ddx

FHi(x)



12(e)x

π

∫ ∞

−∞

f(t)t − x

dt +1π

∫ ∞

−∞f(t) dt =

1π

∫ ∞

−∞

tf(t)t − x

dt

= H{tf(t)}

13 From Exercise 10

FHi(t) = − 1πt

∗ f(t)

So from Exercise 9,

F{FHi(t)} = jsgn (ω) × F( ω)

so

F(jω) = −jsgn (ω) × F{FHi(t)}

Thus

f(t) =∫ ∞

−∞

1π(t − τ)

FHi(τ)dτ = − 1π

∫ ∞

−∞

1(x − τ)

FHi(x)dx

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 > 00, ω < 0

15

F{H(t)} =1jω

+ πδ(ω) = F (jw)

Symmetry

F ( t) =1jt

+ πδ(t) ↔ 2πH(−ω) = 2π[1 − H(ω)]

= 2π[Fδ(t) − H(ω)]

or

H(ω) ↔ j2πt

+12δ(t)

Thus

F−1{H(ω)} =j

2πt+

12δ(t)



Thenf(t) = 2

[12δ(t) +

j2πt

]∗ f(t) = f(t) − j

(− 1

πt

)∗ f(t)

= f(t) − jFHi(t)

When f(t) = cos ω0t, ω0 > 0, then

F (jω) = π[δ(ω − ω0) + δ(ω + ω0)]

soF{f(t)} = 2πδ(ω − ω0)

whencef(t) = f(t) − jFHi(t) = ejω0t = cos ω0t + j sinω0t

and soFHi(t) = − sinω0t

When g(t) = sinω0t, ω0 > 0

G(jω) = jπ[δ(ω + ω0) − δ(ω − ω0)]

and thusg(t) = −jejω0t = −j(cos ω0t + j sinω0t)

soH{sinω0t} = cos ω0t

16 If h(t) = 0, t < 0, then when t < 0

he(t) =12h(−t), and ho(t) = −1

2h(−t)

i.e. ho(t) = −he(t)

When t > 0 thenhe(t) =

12h(t), and ho(t) =

12h(t)

i.e. ho(t) = he(t)

That isho(t) = sgn (t)he(t) ∀t



Thush(t) = he(t) + sgn (t)he(t)

When h(t) = sin t H(t)

he(t) =

12

sin t, t > 0

−12

sin t, t < 0

and sincesgn (t) he(t) =

12

sin t ∀t

the result is comfirmed.Then taking the FT of the result,

H(jω) = He(jω) + F {sgn (t)he(t)

}

= He(jω) +12π

(2jω

∗ He(jω))

= He(jω) + jH {He(jω)

}When

H(jω) =∫ ∞

−∞e−ate−jwt dt =

a

a2 + ω2 − ω

a2 + ω2

thenH{

a

a2 + ω2

}= − ω

a2 + ω2

orH{

a

a2 + t2

}= − x

a2 + x2

Finally

H{

at

a2 + t2

}= −x

x

a2 + x2 +1π

∫ ∞

−∞

a

a2 + t2dt =

a2

a2 + x2

SoH{

t

a2 + t2

}=

a

a2 + x2

17(a)

FH(s) =∫ ∞

0e−at(cos 2πst + sin 2πst) dt =

a + 2πs

a2 + 4π2s2



17(b)

FH(s) =∫ T

−T

(cos 2πst + sin 2πst) dt =1πs

sin 2πst

18

E(s) =∫ ∞

−∞f(t) cos 2πst dt O(s) =

∫ ∞

−∞f(t) sin 2πst dt

E(s) − O(s) =∫ ∞

−∞f(t)e−j2πst dt = F (js)

From 17(a)

FH(s) =1 + πs

2 + 2π2s2

whenceE(s) =

12 + 2π2s2 , O(s) =

πs

2 + 2π2s2

soF ( s) =

1 − jπs

2 + 2π2s2

agreeing with the direct calculation

F (js) =∫ ∞

0e−2te−j2πst dt =

1 − jπs

2 + 2π2s2

19

H{f(t − T )} =∫ ∞

−∞f(t − T ) cas 2πst dt

=∫ ∞

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

sin 2πsτ(cos 2πsT − sin 2πsT )] dt

= cos 2πsTFH(s) + sin 2πsTFH(−s)

20 The Hartley transform follows at once since

FH(s) = �{F (js)} − {F (js)} =12δ(s) +

1sπ

From time shifting

FH(s) = sinπs

[12δ(−s) − 1

sπ

]+ cos πs

[12δ(s) +

1sπ

]



=12δ(s) +

cos πs − sinπs

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 2πs0t} = δ(s − s0)

2212FH(s − s0) +

12FH(s + s0)

=12

∫ ∞

−∞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 2πs0t [cos 2πst + sin 2πst] dt

= H{f(t) cos 2πs0t}

From Exercise 21, setting f(t) = 1

H{cos 2πs0t} =12(δ(s − s0) + δ(s + s0))



alsoH{sin 2πs0t} = H{cas 2πs0t} − H{cos 2πs0t}

= δ(s − s0) − 12(δ(s − s0) + δ(s + s0)) =

12(δ(s − s0) − δ(s + s0))

23 ∫ t

−∞(1 + τ2)−1 dτ = tan−1 t +

π

2

Thus

F{tan−1 t} = F{∫ t

−∞(1 + τ2)−1 dτ

}− F

{π

2

}

= F{∫ ∞

−∞(1 + τ2)−1H(t − τ)dτ

}− F

{π

2

}

= F{

11 + t2

∗ H(t)}

− F{π

2

}

= F{

11 + t2

}×{

1 ω

+ πδ(ω)}

− π

2× 2πδ(ω)

But from Exercise 1F{

e−|t|}

=2

1 + ω2

and so by symmetry

F{

11 + t2

}= πe−|ω|

whenceF {

tan−1 t}

= πe−|ω| ×{

1jω

+ πδ(ω)}

− π

2× 2πδ(ω)

and so

F {tan−1 t

}=

πe−|ω|

jω

2412

[1 + cos ω0t] ↔ 12

[2πδ(ω) + πδ(ω − ω0) + πδ(ω + ω0)]

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

soF {x(t)} =

∫ ∞

−∞2T sinc (ω − u)



×[πδ(u) +

12

(δ(ω − ω0) + δ(u + ω0))]

du

= T

[sinc ω +

12

sinc (ω − ω0) +12

sinc (ω + ω0)]

25

H(ν) =14

3∑r=0

f(r) cas2πνr

4

H(0) =14

[f(0) + f(1) + f(2) + f(3)]

H(1) =14

[f(0) + f(1) − f(2) − f(3)]

H(0) =14

[f(0) − f(1) + f(2) − f(3)]

H(0) =14

[f(0) − f(1) − f(2) + f(3)]

so

T =14

1 1 1 11 1 −1 −11 −1 1 −11 −1 −1 1

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

T−1H =

1 1 1 11 1 −1 −11 −1 1 −11 −1 −1 1

H(0)H(1)H(2)H(3)

=

f(0)f(1)f(2)f(3)


solucionario parte 5 matemáticas avanzadas para ingeniería - 2da edición - glyn james

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