From the differentiation-in-frequency property of the DTFT given in Table 3.4 we have

(Using Parseval’s relation with differentiation-in-frequency property)

3.38 (a) ( ) [ ] ⁶ [ ] 10.

3.40 From the differentiation-in-frequency property of the DTFT given in Table 3.4 we have

it follows that ∑^∞ [ ]= ( ⁰). Therefore,

−∞

n=

ej

X n

x .

) ( ⁰

0 ) (

j d

e dX

g X e

j C

j

= ω ω

ω

=

From Table 3.3, ^X(ejω)=F ^{{ [ ]} , 1.}

1

1 α <

= µ

α _{− ω}

α

− e ^j

n n As a result,

2

2 0 (1 )

) 1

0 ( ) (

α

− α

= α ω

− α

=

ω^ω ω = ⁻₋^ω_ω =

j j j

e e d

e

jdX and .

1 0) 1

(e^j = −α

X Hence, .

1−α

= α

Cg

3.41 Let ^G₁^(ejω)= F^{g1^[n^]}.

(b) Noteg₂[n]=g₁[n]+g₁[n−4]. Hence, F^{g₂^[n]}=G₂^(ejω) )

( )

( ⁴ ₁

1 ω + − ω ω

=G e^j e ^j G e^j =

(

¹+e−^j4ω

)

G1⁽e^jω^).

(c) Noteg₃[n]=g₁[−(n−3)]+g₁[n−4]. Now, F ^{g₁^[−n]}=G₁^{(e−jω). Hence,} F^{g₃^[n]}=G₃^{(ejω) =e−j3ωG}₁^{(e−jω)+e−j4ωG}₁^(ejω).

(d) Noteg₄[n]=g₁[n]+g₁[−(n−7)]. Hence, F ^{g4^[n^]}=G4⁽e^jω⁾

).

( )

( ⁷ ₁

1 ω + − ω − ω

=G e^j e ^j G e ^j

3.42 Y(e^jω)=X₁(e^jω)X₂(e^jω), i.e., [ ] ₁[ ] ₂[ ] ⎟⎟.

⎠

⎜⎜ ⎞

⎝

⎛ ∑

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ ∑

∑ = ^∞

−∞

=

ω

∞ −

−∞

=

ω

∞ −

−∞

=

ω

−

n

n j n

n j n

n

j x n e x n e

e n y

(a) Setting ω=0 in the above we get [ ] ₁[ ] ₂[ ]⎟⎟⎠.

⎜⎜ ⎞

⎝

⎛ ∑

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ ∑

∑ = ^∞

−∞

=

∞

−∞

=

∞

−∞

= n n

n

n x n

x n

y

(b) Setting ω=π we get ( 1) [ ] ( 1) ₁[ ] ( 1) ₂[ ]⎟⎟⎠.

⎜⎜ ⎞

⎝

⎛ ∑ −

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ ∑ −

∑ − = ^∞

−∞

=

∞

−∞

=

∞

−∞

= n

n n

n n

ny n x n x n

3.43 x[n]=αⁿµ[n], α <1.From Table 3.3, F _ω

α

− ω =

= j

e

ej

X n x

1

) 1

( ]}

[

{ . The total energy

of ^x[n] is E ^.

3 ) 4

(

2 / 2 1 1

1 0

2 2

1 1 2

1 ∫ ω= ∑ α = =

=

= α α

−

∞

= π

π

− −α − ω

π n

n e j

x d To determine the

80% bandwidth of the signal, we set E ^0.8

2

1 1 2

1 80

, = ^ω∫ ω=

ω

− −α − ω

π d

c

c e j

x E

3

8 4

. 0 ⋅

x=

and solve for ^ω_c, i.e., set E

3 2 . 3 2 / 2 1

sin )

cos 1 (

1 2

1 80

, _{2 2} =

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ ∫ ω

=

= α ω

ω

− −α ω +α ω

π d

c

c

x .

A numerical solution of the above equation yields ω_c =0.5081π.

3.44 Recall x[n]=x_ev[n]+x_od[n],where [ ] ( [ ] [ ])

2

1 x n x n n

x_ev = + − and As is causal, for and

] [n x 0

] [n =

x n<0 x[−n]=0 for Hence, there is no overlap between the nonzero portions of and

.

>0 n ]

[n

x x[ n− except at ] n=0, and we have ]

[ ] 0 [ ] [ ] [ 2 ]

[n x n n x n

x = _ev µ − _ev δ and x[n]=2x_od[n]µ[n]+x_od[0]δ[n]. Moreover, since is real, it follows from Table 3.2 that F ^and

] [n

x {x_ev[n]}= X_re(e^jω)

F{x_ev[n]}= jX_im(e^jω). Taking the DTFT of x[n]=2x_ev[n]µ[n]−x_ev[0]δ[n] we arrive at

= ^π∫

π

−

ν π

ω) ( )

(e^{j 1} X_re e^j

X µ(e^j(ω−ν))dν−x[0].

From Table 3.3 we have

µ^(ejω)= F ⎟+ ∑πδ ω+ π

⎠⎞

⎜⎝

∑ πδ ω+ π = ⎛ − +

=

µ ^∞

−∞

=

∞ ω

−∞

−e^{− ω} k= k

k j

k

n]} j ( 2 ) 1 cot( ) ( 2 ).

[

{ 2 2

1 1

1

Substituting the above in the equation preceding it we get

⎥⎦ ν

⎢⎣ ⎤

⎡ ⎟

⎠

⎜ ⎞

⎝

⎛ − ⎜⎝⎛ ⎟⎠⎞

= ^π∫ ^ω−ν

π

−

ν π

ω X e j d

e

X ^j _re ^j

2 2 1

1 ( ) 1 cot

) (

X (e )

(

( ) 2 k

)

d x[0]

k

re j δ ω−ν + π ν− + ∑ ∫^∞

−∞

= π

π

−

ν

].

0 [ ) ( cot

) 2 (

) 2 (

1

2 d X e x

e j X

d e

X_re ^j ∫ _re ^j ⎜⎝⎛ ⎟⎠⎞ ν+ _re ^j −

− π

∫ ν

= π ^{π ω}

π

−

ν

− ν ω

π π

−

ν

Comparing the last equation with X(e^jω)= X_re(e^jω)+ jX_im(e^jω), we arrive at .

cot ) 2 (

) 1

( ∫ ⎜⎝⎛ 2 ⎟⎠⎞ ν

− π

= ^π

π

−

ν

− ν ω

ω X e d

e

X_im ^j _re ^j

Likewise, taking the DTFT of x[n]=2x_od[n]µ[n]+x_od[0]δ[n] we get

= ^π∫

π

−

ν π

ω) ( )

(e^{j j} X_im e^j

X µ(e^j(ω−ν))dν+x[0].

Substituting the expression for µ(e^jω) given earlier in the above equation we get

⎥⎦ ν

⎢⎣ ⎤

⎡ ⎟

⎠

⎜ ⎞

⎝

⎛ − ⎜⎝⎛ ⎟⎠⎞

= ^π∫ ^ω−ν

π

−

ν π

ω X e j d

e

X ^{j j} _im ^j

2 2

1 1 cot )

( )

(

(

( ) 2

)

[0]

)

(e k d x

X j

k

im j δ ω−ν + π ν+ + ∑ ∫^∞

−∞

= π

π

−

ν

].

0 [ ) ( cot

) ( )

( 2 2

1

2^j X_im e^j d ∫ X_im e^j ⎟dν+ jX_im e^j +x

⎠⎞

⎜⎝ + ⎛

∫ ν

= ^{π ω}

π

−

ν

− ν ω

π π

π

−

ν π

Comparing the last equation with X(e^jω)= X_re(e^jω)+ jX_im(e^jω), we arrive at

. ] 0 [ cot

) 2 (

) 1

( ∫ ⎜⎝⎛ 2 ⎟⎠⎞ ν+

= π ^π

π

−

ν

− ν ω

ω X e d x

e

X_re ^j _im ^j

3.45 If u[n]=zⁿ is the input to the LTI discrete-time system, then its output is given by ).

( ]

[ ]

[ ]

[ ] [ ]

[n h k u n k h k z z hk z z H z

y ⁿ

k

k n

k

k n k

∑ =

∑ =

∑ − =

= ^∞

−∞

=

∞ −

−∞

=

∞ −

−∞

=

where ( )= ∑^∞ [ ] . Hence is an eigenfunction of the system.

−∞

=

− k

z k

k h z

H

If v[n]=zⁿµ[n] is the input to the LTI discrete-time system, then its output is given by .

] [ ]

[ ] [ ]

[ ] [ ]

[ = ∑ − = ∑ µ − = ∑

−∞

=

∞ −

−∞

=

∞ −

−∞

=

n k

k n

k

k n n

k

z k h z z

k n k h z k n v k h n

y

Since in this case the summation depends upon is not an eigenfunction of the system.

3.46 F^{h₁^[n]}=H₁^(ejω)= F^{{ [ ] [ 1]} 1 0.5 ,}

2

1δ − = + − ω

+

δ n n e ^j

F^{h2^[n^]}=H2⁽ej^ω)= F{ [ ] [ 1]} 0.5 0.25 ,

4 1 2

1δ n − δn− = − e−^jω

F^{h₃^[n]}=H₃^(ejω)= F^{{2δ n[ ]}=2,}

F^{h₄^[n]}=H₄^(ejω)= F ^.

5 . 0 1 ]} 2 [ 2

{ 2

1

ω

− −

= −

⎟ µ

⎠⎞

⎜⎝

− ⎛

j n

e n

The overall frequency response of the structure of Figure 2.35 is given by )

( ) ( ) ( ) ( ) ( )

(e^jω =H₁ e^jω +H₂ e^jω H₃ e^jω +H₂ e^jω H₄ e^jω H

^1.

5 . 0 1

) 25 . 0 5 . 0 ( ) 2 25 . 0 5 . 0 ( 2 5

. 0

1 =

−

− −

− +

+

= ^{− ω − ω} _{− ω}^{− ω}

j j j

j

e e e

e

3.46 Denote Hi⁽e^jω)= F{h₁[n]},1≤ i≤5.

(a) The overall frequency of Figure P2.2(a) is then given by

).

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ^jω = ₁ ^jω ₂ ^jω + ₃ ^jω ₄ ^jω + ₁ ^jω ₂ ^jω ₃ ^jω ₅ ^jω

i e H e H e H e H e H e H e H e H e

H

(b) The structure of Figure P2.2(b) can be redrawn as shown below

+

h [n]_o h [n]₃ h [n]₄

where the block with an impulse response represents the part of Figure P2.2(b) with a feedback loop as shown below

h [n]₁ h₂[n]

h [n]5

+

u[n] w[n] v[n]

Let U(e^jω)= F{ nu[ ]}, V(e^jω)= F { nv[ ]}, ^and W(e^jω)= F {w[n]}. Then we have )

( ) ( ) ( )

(e^jω =U e^jω +H₃ e^jω V e^jω

W and

Eliminating from these two equations we get

).

( ) ( ) ( )

(e^jω =H₁ e^jω H₂ e^jω W e^jω V

) (e^jω

(

¹−H1⁽e^jω⁾HW2⁽e^jω⁾H5⁽e^jω⁾

)

V⁽e^jω⁾=H1⁽e^jω⁾H2⁽e^jω⁾U⁽e^jω⁾

which leads to the frequency response of the feedback structure given by

) ( ) ( ) ( 1

) ( ) ( )

( ) ) (

(

5 2

1

2

1ω ω ω

ω ω

ω ω ω

= −

= j j j

j j

j j j

o H e H e H e

e H e H e

U e e V

H

The overall frequency of Figure P2.2(a) is thus given by

. ) ( ) ( ) ( 1

) ( ) ) (

( ) ( ) ( ) ( ) (

5 2

1

2 4 1

3

4 ω ω ω

ω ω ω

ω ω

ω ω

+ −

= +

= j j j

j j j

j o j

j j

e H e H e H

e H e e H

H e

H e H e

H e

H

3.48 F ^{h₁^[n]}=H₁^(ejω)= F^{{2δ[n−2]−3δ[n+1]}=2e−j2ω−3ejω,} F ^{h2^[n^]}=H2⁽ej^ω)= F{δ[n−1]+2δ[n+2]}=e^−jω +2e^j2ω,

F ^{h₃^[n]}=H₃^(ejω)= F^{{5δ[n−5]+7δ[n−3]+2δ[n−1]−δ[n]+3δ[n+1]}}

^{=5e−j5ω+7e−j3ω+2e−jω−1+3ejω.}

The overall frequency of Figure P2.3 is given by

= +

= ^{ω ω ω}

ω) ( ) ( ) ( )

(e^j H₃ e^j H₁ e^j H₂ e^j

H 5e^−j5ω+7e^−j3ω+2e^−jω −1+3e^jω

(

^{2 − 2ω−3 ω}

)(

^{− ω+2 2ω}

)

^{=5 − 5ω +9 − 3ω +2 − ω+3 ω−6 3ω.}

+ e ^j e^j e ^j e^j e ^j e ^j e ^j e^j e^j

3.49 Now, h_ev[n] is the inverse DTFT of H_re(e^jω). Rewriting we get

⎟⎟⎠

⎜⎜ ⎞

⎝ + ⎛

⎟⎟⎠

⎜⎜ ⎞

⎝ + ⎛

⎟⎟⎠

⎜⎜ ⎞

⎝ + ⎛

= ^{ω+ − ω ω+ − ω ω+ − ω}

ω

2 3 3

2 2 2

2 3 4

2 1 )

( ^{j ej e j ej e j ej e j}

re e H

Its inverse DTFT is .

2 2

5 . 1 5

. 1

1+ ^ω + ^{− ω} + ^2ω+ ^{− 2ω}+ ^3ω + ^{− 3ω}

= e^j e ^j e^j e ^j e^j e ^j

].

3 [ 2 ] 3 [ 2 ] 2 [ 5 . 1 ] 2 [ 5 . 1 ] 1 [ ] 1 [ ] [ ]

[n =δ n +δ n+ +δ n− + δ n+ + δn− + δn+ + δn− h_ev

Since is real and causal, and its DTFT exists, it is also absolutely summable. Hence, we can reconstruct from as

] [n

h H(e^jω)

] [n

h h_ev[n] ]

[ ] 0 [ ] [ ] [ 2 ]

[n h n n h n

h = _ev µ − _ev δ =2

(

δ[n]+δ[n−1]+1.5δ[n−2]+2δ[n−3]

)

−δ[n] =δ[n]+2δ[n−1]+3δ[n−2]+4δ[n−3].

3.50 (a) [ ] sin cos( ) sin ( ) sin ( ) .

3 2

1 3

2 1

3 ⎟

⎠⎞

⎜⎝

⎛ ω −

⎟−

⎠⎞

⎜⎝

⎛ ω +

=

⎟ ω

⎠⎞

⎜⎝

= ⎛^π n ^π n ^π n

n

y_a ⁿ _{o o o} Hence, the angular

frequencies present in the output are ( ) .

3 n

o

±π

ω

(b) [ ] cos ( ) cos ( )cos( ) cos(2 ) cos( )

2 1 2 2 1

3 n n n n n

n

y_{b o o o o} ⎥⎦⎤ ω_o

⎢⎣⎡ + ω

= ω ω

= ω

=

[

cos(3 ) cos( )

]

) cos(

) cos(

) 2 cos(

)

cos( 4

1 2

1 2

1 2

1 ω_on + ω_on ω_on = ω_on + ω_on + ω_on

=

).

3 cos(

)

cos( 4

1 4

3 ω_on + ω_on

= Hence, the angular frequencies present in the output are and

ωo

3 ω_o.

(c) y_c[n]=cos(3ω_on). Hence, the angular frequency present in the output is 3ω_o. 3.51 F^{{δ[n]−αδ[n−R]}=H(ejω)=1−α e−jωR. Let α = αejφ.} Then the maximum value

of H(e^jω) is 1+ α and the minimum value is 1− α. There are R peaks of H(e^jω) located at ω=2πk/R, 0≤k≤R−1, and R dips located at

, / ) 1 2

( k+ π R

=

ω 0≤k≤R−1 in the frequency range 0≤ω<2π.

0 0.5 1 1.5 2

0 0.5 1 1.5 2

/

Magnitude

0 0.5 1 1.5 2

-2 -1 0 1 2

/

Phase, in radians

3.52 .

1 ) 1

( ¹

0 − ω

ω ω −

− −

= ω

α

− α

= −

= ∑ α

j M j n M

M j n

n j

e e e

e

G Note G(e^jω)=H(e^jω) for α=1. In order

to have G(e^j0)=1, the impulse response should be multiplied by a factor K, where .

1 1

K M α

− α

= −

3.53 ^H(ejω)=

[

^a₁^cos(2ω)+(a₂ ^+a₃^)cos(ω)+a₃

] [

^{+ j−a}₁^sin(2ω)+(a₂ ^−a₃^)sin(ω)

]

^.

The frequency response will have zero-phase for a₂ =a₃ and a₁=0. 3.54 H(e^jω)=a₁ +a₂e^−jω +a₃e^−j2ω +a₄e^−j3ω+a₅e^−j4ω

( ) ( )

2 3 ^{2 .}

4 2 2

5 2

1 2ω+ − ω − ω + ω + − ω − ω + − ω

= a e^j a e ^j e ^j a e^j a e ^j e ^j a e ^j

If a₁ =a₅ and a₂ =a₄, then we can rewrite the above equation as

The maximum value of H(e^jω) is shown below:

0 0.5 1 1.5 2

Phase, in radians

In this case the maximum value is 5

8

trivial solution.

Solution #2: b₁=±1 and b₀ =sgn(b₁)a₁. In which case .

Solution #1: B₀ =±1,B₁ =sgn(B₀)A₁,φ₁ −φ₀ =θ. In which case

A trivial solution.

0 =±π.

Hence, a non-trivia solution is .

1

j e Therefore, the input-output relation is given by ].

H Therefore, the input-output

relation is given by y[n]+y[n−2]=2x[n−1].

j Therefore, the input-output relation is given by ]. tan sin(

)

3.62 From Eq. (2.20), the input-output relation of a factor-of- L up-sampler is given by The DTFT of is thus given by

where ),

( ]

[ ]

/ [ ]

[ )

( ^{∞ − ω ω}

−∞

= ω

∞ −

=−∞

= ω

∞ −

−∞

=

ω = ∑ = ∑ = ∑ ^{j mL} = ^jL

m n j mL

n n n j n

j y n e x n L e x m e X e

e Y

= F

ω) (e^j

X { nx[ ]}.

3.63 , 1.

1 ) 1

( α <

α

= − _{− ω}

ω

jL j

e e

G Thus, we can write G(e^jω)= X(e^jLω), where .

1 ) 1

( ^ω _{− ω}

α

= −

j j

e e

X From Table 3.3, the inverse DTFT of is

Hence, from the results of Problem 3.62, it follows that

) (e^jω

X x[n]=αⁿµ[n].

⎩⎨

⎧ = ± ± ±

= 0, otherwise.

, 3 , 2 , , 0 ],

/ ] [

[ x n L n L L LK

n g

3.64 From Table 3.3, .

5 . 0 1 ) 1

( ^ω _{− ω}

= −

j j

e e

H Thus,

) cos(

25 . 1 ) 1

(e^jω = − ω

H and

) . cos(

5 . 0 1

) sin(

5 . tan 0

) ( )}

(

arg{ ¹ ⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

ω

−

ω

= − ω θ

= ⁻

jω

e

H H(e^±jπ/5)=1.3504m j0.6664. Therefore

5059 . 1 ) (e^{± jπ/5} =

H and θ(±π/5)=m0.4585 radians.

Now, for an input x[n]=sin(ω_on)µ[n], the steady-state output is given by

(

^{( )}

)

^.

sin ) ( ]

[n H e^j _on _o

y = ^ωo ω +θ ω For ω_o =π/5, the steady-state output is therefore given by [ ] ( )sin ( ) 1.5059sinsin 0.4585 .

5 5

5 5

/ ⎜⎝⎛ +θ ⎟⎠⎞= ⎜⎝⎛ − ⎟⎠⎞

= H e ^{π π}n ^{π π}n

n

y ^j

3.65 H(e^jω)=h[0]+h[1]e^−jω +h[0]e^−j2ω =h[0](1+e^−j2ω)+h[1]e^−jω We require

(

2h[0]cos( ) h[1]

)

.

e ^j ω +

= ^{− ω} H(e^j0.3) =2h[0]cos(0.3)+h[1]=0 and H(e^j0.6) =2h[0]cos(0.6)+h[1]=1. Solving these two equations we get

and 8461 . 3 ] 0 [ =

h h[1]=−6.3487.

3.66 H(e^jω)=h[0](1+e^−j2ω)+h[1]e^−jω =e^−jω

(

2h[0]cos(ω)+h[1]

)

. We require 1

] 1 [ ) 3 . 0 cos(

] 0 [ 2 )

(e ^0.3 = h +h =

H ^j and H(e^j0.6) =2h[0]cos(0.6)+h[1]=0. Solving these two equations we get h[0]=−3.8461 and h[1]=7.3487.

3.67

3.68 H(e^jω)=h[0](1+e^−j4ω)+h[1](e^−jω +e^−j3ω)+h[2]e^−j2ω

We require

H ^j Solving these three equations we

get h[0]=−13.4866,h[1]=45.228,h[2]=−63.8089, i.e.,

Since the impulse response is real, the value of at equations in matrix form we get

,

Solving these two equations we get .

conditions to be satisfied by the filter are:

.

0 0.2 0.4 0.6 0.8 1 0

0.5 1 1.5 2

ω/pi

Magnitude

0 0.2 0.4 0.6 0.8 1

-2 -1 0 1 2

ω/pi

Phase, radians

3.72 (a) H(e^jω)=h[0]+h[1]e^−jω+h[2]e^−j2ω +h[3]e^−j3ω

(

2 [0]cos(3 /2) 2 [1]cos( /2)

)

. ]

0 [ ]

1 [ ]

1 [ ] 0

[ + + ² + ³ = ^{3 /2} ω + ω

=h h e^−jω h e^{−j ω} h e^{−j ω} e^{−j ω} h h

The two conditions to be satisfied by the filter are:

, 8 . 0 ) 1 . 0 cos(

] 1 [ 2 ) 3 . 0 cos(

] 0 [ 2 )

(e ^0.2π = h π + h π =

H ^j

. 5 . 0 ) 25 . 0 cos(

] 1 [ 2 ) 75 . 0 cos(

] 0 [ 2 )

(e ^0.5π = h π + h π =

H ^j Solving these two equations we

get h[0]=0.0414 and h[1]=0.395.

(b) H(e^jω)=0.0414+0.395e^−jω +0.395e^−j2ω +0.0414e^−j3ω.

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

ω/pi

Magnitude

0 0.2 0.4 0.6 0.8 1

-4 -2 0 2 4

ω/pi

Phase, radians

3.73 (a) H(e^jω)=h[0]+h[1]e^−jω +h[2]e^−j2ω+h[3]e^−j3ω

(

2 [0]sin (3 /2) 2 [1]sin( /2)

)

. ]

0 [ ]

1 [ ]

1 [ ] 0

[ + − ² − ³ = ^{3 /2} ω + ω

=h h e^−jω h e^{−j ω} h e^{−j ω} je^{−j ω} h s h The two conditions to be satisfied by the filter are:

, 2 . 0 ) 25 . 0 ( sin ] 1 [ 2 ) 75 . 0 sin(

] 0 [ 2 )

(e ^0.5π = h π + h s π =

H ^j

. 7 . 0 ) 4 . 0 cos(

] 1 [ 2 ) 2 . 1 sin(

] 0 [ 2 )

(e ^0.8π = h π + h π =

H ^j Solving these two equations we get

and 14 . 0 ] 0 [ =−

h h[1]=0.2815.

(b) H(e^jω)=−0.14+0.2815e^−jω −0.2815e^−j2ω +0.14e^−j3ω.

0 0.2 0.4 0.6 0.8 1 0

0.2 0.4 0.6 0.8 1

ω/π

Magnitude

0 0.2 0.4 0.6 0.8 1

-4 -3 -2 -1 0 1

ω/π

Phase, radians

3.74 (a) H(e^jω)=h[0]+h[1]e^−jω +h[2]e^−j2ω+h[3]e^−j3ω +h[4]e^−j4ω

(

2 [0]sin (2 ) 2 [1]sin( )

)

. ]

0 [ ]

1 [ ]

1 [ ] 0

[ + − ³ − ⁴ = ² ω + ω

=h h e^−jω h e^{−j ω} h e^{−j ω} je^{−j ω} h s h The two conditions to be satisfied by the filter are:

, 8 . 0 ) 4 . 0 ( sin ] 1 [ 2 ) 8 . 0 sin(

] 0 [ 2 )

(e ^0.5π = h π + h s π =

H ^j

. 2 . 0 ) 8 . 0 cos(

] 1 [ 2 ) 6 . 1 sin(

] 0 [ 2 )

(e ^0.8π = h π + h π =

H ^j Solving these two equations we get

and 112 . 0 ] 0 [ =

h h[1]=0.3514.

(b) H(e^jω)=0.112+0.3514e^−jω−0.3514 e^−j3ω−0.112e^−j4ω.

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

ω/π

Magnitude

0 0.2 0.4 0.6 0.8 1

-4 -2 0 2 4

ω/π

Phase, radians

3.75 (a) H_A(e^jω)=0.3−e^−jω+0.3e^−j2ω,H_B(e^jω)=0.3+e^−jω +0.3e^−j2ω.

0 0.2 0.4 0.6 0.8 1 0

0.5 1 1.5 2

ω/π

Magnitude

|H_A(e^jω)|

0 0.2 0.4 0.6 0.8 1

0 0.5 1 1.5 2

ω/π

Magnitude

|H_B(e^jω)|

It can be seen from the above plots that H_A(e^jω) is a highpass filter, whereas is a lowpass filter.

) ( ^jω

B e H

(b) H_C(e^jω)=H_B(e^jω)=H_A(e^j(ω+π)).

3.76 y[n]⁼ y[n⁻1]⁺0.5

(

x[n]⁺x[n⁻1]

)

. Taking the DTFT of both sides we get

(

^{( ) ( )}

)

^.

5 . 0 ) ( )

(e^jω =e^−jωY e^jω + X e^jω +e^−jω X e^jω

Y Hence, the frequency response is

given by .

1 1 2 1 ) (

) ) (

( ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

= +

= − ω

ω

− ω

ω ω

j j j

j j

trap e

e e

X e e Y

H

3.77 [ ] [ 2]

(

[ ] 4 [ 1] [ 2]

)

.

3

1 + − + −

+

−

= y n x n x n x n

n

y Hence,

. 1

4 ) 1

( 2

2 3

1 ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

− +

= + ^{− ω}_{− ω}^{− ω}

ω

j j j j

simpson

e e e e

H

0 0.2 0.4 0.6 0.8 1

0 0.5 1 1.5 2

ω/π

Magnitude

Trapezoidal Simpson

Note: To compare the performances of the Trapezoidal numerical integration formula with that of the Simpson’s formula, we first observe that if the input is then the result of integration is

, )

( ^{j t}

a t e

x = ^ω

. )

( ^{1 j t}

a t j e

y ^ω

= ω Thus, the desired ideal frequency response is

.

H Hence, we take the ratio of the frequency responses of the approximation to the ideal, and plot the two curves as indicated on the previous page. From this plot, it is evident that the Simpson’s formula amplifies high frequencies, whereas, the

trapezoidal formula attenuates them. In the very low frequency range, both formulae yield results close to the ideal. However, Simpson’s formula is reasonably accurate for frequencies close to the midband range.

3.78 G(e^jω)=g₀ +g₁e^−jω+g₂e^−j2ω +g₃e^−j3ω

θ and hence, a constant group delay.

Alternately, if g₀ =−g₃ and g₁=−g₂, then

θ and hence, a constant group delay.

3.79 (a) H_a(e^jω)=a+be^−jω=a+bcosω− jbsinω. Thus, ⎟⎟ tan sin

)

(b) Let From the results of Part (a) we

response of Part (a) and is the frequency response of Part (b). Thus, )

3.80 The group delay of a causal LTI discrete-time system with a frequency response

)

j or, equivalently,

⋅ ω

right-hand side of the above equation is purely imaginary. Hence,

.

(b) Let 1 0.6 .

0.6969 )

(e^jπ/5 =

H and 0.2063

6821 . 0

1427 . tan 0 ) 5 / ( )}

(

arg{ ^{/5 1} ⎟=

⎠

⎜ ⎞

⎝

= ⎛ π θ

= ⁻

π

ej

H radians.

Since for a frequency response with real coefficient impulse response, H(e^jω) is an even function of ω and θ(ω) is an odd function of ω we have , H(e^{− jπ/5}) =0.6969 and

0.2063.

) 5 / (−π =− θ

Now, for an input x[n]=sin(ω_on)µ[n], the steady-state output is given by

(

^{( )}

)

^.

sin ) ( ]

[n H e^j _on _o

y = ^ωo ω +θ ω Thus, for ω_o =π/5, the steady-state output is given by [ ] ( )sin ( /5) 0.6969sin 0.2063 .

5 5

5

/ ⎜⎝⎛ +θ π ⎟⎠⎞= ⎜⎝⎛ + ⎟⎠⎞

= H e ^{π π}n ^πn

n

y ^j

3.84 ^H(ejω)=h[0]

(

^1+e−j2ω

)

^{+h[1]e−jω =e−jω}

⁽

^{2h[0]cosω+h[1]}

⁾

^. We require and

1 ] 1 [ ) 3 . 0 cos(

] 0 [

2h + h = 2h[0]cos(0.7)+ h[1]=0. Solving these two equations we get h[0]=2.6248 and h[1]=−4.015.

0 0.2 0.4 0.6 0.8 1 1.2

0 0.5 1 1.5 2 2.5

ω

Magnitude

M3.1 r =0.9,θ=0.75

0 0.5 1

-2 0 2 4 6 8

Real part

ω/π

Amplitude

0 0.5 1

-8 -6 -4 -2 0 2

Imaginary part

ω/π

Amplitude

0 0.5 1

0 2 4 6 8

Magnitude Spectrum

ω/π

Magnitude

0 0.5 1

-2 -1.5 -1 -0.5 0 0.5

Phase Spectrum

ω/π

Phase, radians

5 . 0 , 7 . 0 θ=

= r

0 0.5 1

0 1 2 3

4 Real part

ω/π

Amplitude

0 0.5 1

-3 -2 -1

0 Imaginary part

ω/π

Amplitude

0 0.5 1

0 1 2 3 4 5

Magnitude Spectrum

ω/π

Magnitude

0 0.5 1

-1.5 -1 -0.5 0

Phase Spectrum

ω/π

Phase, radians

M3.2 It should be noted that Program 3_1.m uses the function freqz to determine the samples of a DTFT that is rational function in i.e., a ratio of polynomials in

. Their inverse DTFTs are two-sided sequences. However, all sequences of

ω, e− j ω

e− j

Problem 3.19 except that in Part (b) are two-sided finite-length sequences of length , and their DTFTs have both positive and negative powers of As a result, 1

2N+ e^jω.

the frequency sample computed using freqz should be multiplied by the vector evaluated at the frequency points

N j _n

e ^ω ω used in freqz. In Parts (a), (c) and (d), _n

the phase spectra are the plots of the unwrapped phase obtained using the function unwrap.

Moreover, the DTFTs of the sequences in Parts (a), (c) and (d) are real functions of ω and thus have zero phase. More accurate plots of the DTFTs are obtained using the function zerophase.

(a)

⎩⎨

⎧ − ≤ ≤

= 0, otherwise, , 10 10

, ] 1

1[ n

n

y Y₁(e^jω)=

( )

) . 2 / sin(

2 / 21 sin

ω

ω The plots obtained using Program 3_1.m are shown below:

0 0.5 1

-10 0 10 20 30

Real part

ω/π

Amplitude

0 0.5 1

-2 -1 0 1

2x 10^-14 Imaginary part

ω/π

Amplitude

0 0.5 1

0 5 10 15 20 25

Magnitude Spectrum

ω/π

Magnitude

0 0.5 1

-8 -6 -4 -2 0 2

Phase Spectrum

ω/π

Phase, radians

The plot obtained using the function zerophase is shown below:

0 0.2 0.4 0.6 0.8 1

-5 0 5 10 15 20 25

ω/π

Amplitude

(b) Then

⎩⎨

⎧ ≤ ≤

= 0, otherwise.

, 0

, ] 1

2[ n N

n

y Y₂(e^jω) ⎟⎟

⎠

⎜⎜ ⎞

⎝

⎛

ω +

= ^{− ω} ω

) 2 / sin(

) 2 / ] 1 [

2 sin(

/ N

e ^{j N} . The plots

obtained using Program 3_1.m are shown below:

0 0.5 1

-5 0 5 10 15

Real part

ω/π

Amplitude

0 0.5 1

-2 0 2 4 6 8

Imaginary part

ω/π

Amplitude

0 0.5 1

0 5 10 15

Magnitude Spectrum

ω/π

Magnitude

0 0.5 1

-2 -1 0 1 2 3

Phase Spectrum

ω/π

Phase, radians

(c)

⎪⎩

⎪⎨

⎧ − − ≤ ≤

=

otherwise.

, 0

, ,

] 1

3[n N n N

y _N

n

( )

. ) 2 / ( sin

2 / sin ) 1

( 2

2

3 ω

⋅ ω

ω = N

e N

Y ^j The plots obtained

using Program 3_1.m are shown below:

0 0.5 1

-5 0 5

10 Real part

ω/π

Amplitude

0 0.5 1

0 0.5 1

1.5 Imaginary part

ω/π

Amplitude

0 0.5 1

0 2 4 6 8 10

Magnitude Spectrum

ω/π

Magnitude

0 0.5 1

0 1 2 3 4

Phase Spectrum

ω/π

Phase, radians

The plot obtained using the function zerophase is shown below:

0 0.2 0.4 0.6 0.8 1

-5 0 5 10 15 20 25

ω/π

Amplitude

(d)

⎪⎩

⎪⎨

⎧ + − − ≤ ≤

=

otherwise.

, 0

, ,

] 1

4[n N N n N

y _N

n

( )

. ) 2 / ( sin

2 / sin 1 ) 2 / sin(

] [ sin )

( 2

2 2 1

4 ω

⋅ ω ω +

⎟⎠

⎜ ⎞

⎝⎛ω +

⋅

ω = N

N N

N e

Y ^j

0 0.5 1

-10 0 10 20 30 40

Real part

ω^/π

Amplitude

0 0.5 1

-2 -1 0 1

2x 10^-14 Imaginary part

ω^/π

Amplitude

0 0.5 1

0 10 20 30 40

Magnitude Spectrum

ω/π

Magnitude

0 0.5 1

-10 -5 0 5

Phase Spectrum

ω/π

Phase, radians

The plot obtained using the function zerophase is shown below:

0 0.2 0.4 0.6 0.8 1

-5 0 5 10 15 20 25

ω/π

Amplitude

(e)

⎩⎨

⎧ π − ≤ ≤

= 0, otherwise.

, ),

2 / ] cos(

5[

N n N N n n

y

( )

( ) ( )

( )

^.

sin sin 2 1 sin

sin 2 ) 1 (

2 / ) (

)(

( 2

/ ) (

)(

( 5

2 2) 1 2

2 2) 1 2

N N

N

N N N

ej

Y _π

π π

π

+ ω

+ +

ω

− ω

+

−

ω = ⋅ ω + ⋅ The plots obtained using

Program 3_1.m are shown below:

0 0.5 1

-5 0 5 10 15

Real part

ω/π

Amplitude

0 0.5 1

-0.5 0 0.5

x 10^-14 Imaginary part

ω/π

Amplitude

0 0.5 1

0 5 10 15

Magnitude Spectrum

ω/π

Magnitude

0 0.5 1

-5 0 5 10 15

Phase Spectrum

ω/π

Phase, radians

The plot obtained using the function zerophase is shown below:

0 0.2 0.4 0.6 0.8 1

-5 0 5 10 15 20 25

ω/π

Amplitude

M3.3 (a) _{− ω − ω − ω − ω}

ω

− ω

− ω

− ω

ω −

+ +

+ +

+ +

−

= +

4 3

2

4 3

2

4153 . 0 1393

. 0 8258

. 0 2386

. 0 1

) 139

. 0 3519

. 0 139

. 0 1 ( 2418 . ) 0

( j j j j

j j

j j j

e e

e e

e e

e e e

X

The plots obtained using Program 3_1.m are shown below:

0 0.5 1 -0.5

0 0.5

1 Real part

ω/π

Amplitude

0 0.5 1

-1 -0.5 0 0.5

1 Imaginary part

ω/π

Amplitude

0 0.5 1

0 0.2 0.4 0.6 0.8 1

Magnitude Spectrum

ω/π

Magnitude

0 0.5 1

-4 -2 0 2 4

Phase Spectrum

ω/π

Phase, radians

(b) .

1205 . 0 7275

. 0 1454

. 1 1

) 0911

. 0 0911

. 0 1 ( 1397 . ) 0

( 2 3

3 2

ω

− ω

− ω

−

ω

− ω

− ω

ω −

+ +

+

− +

= −

j j

j

j j

j j

e e

e

e e

e e X

The plots obtained using Program 3_1.m are shown below:

0 0.5 1

-0.2 0 0.2 0.4 0.6 0.8

Real part

ω/π

Amplitude

0 0.5 1

-0.2 0 0.2 0.4 0.6

Imaginary part

ω/π

Amplitude

0 0.5 1

0 0.2 0.4 0.6

0.8 Magnitude Spectrum

ω/π

Magnitude

0 0.5 1

-1 0 1 2

3 Phase Spectrum

ω/π

Phase, radians

M3.4 % Property 1

N = 8; % Number of samples in sequence gamma = 0.5; k = 0:N-1;

x = exp(-j*gamma*k); y = exp(-j*gamma*fliplr(k));

% r = x[-n] then y = r[n-(N-1)]

% so if X1(exp(jw)) is DTFT of x[-n], then

% X1(exp(jw)) = R(exp(jw)) = exp(jw(N-1))Y(exp(jw)) [Y,w] = freqz(y,1,512);

X1 = exp(j*w*(N-1)).*Y;

m = 0:511; w = -pi*m/512;

X = freqz(x,1,w);

% Verify X = X1

% Property 2

k = 0:N-1; y = exp(j*gamma*fliplr(k));

[Y,w] = freqz(y,1,512);

X1 = exp(j*w*(N-1)).*Y;

[X,w] = freqz(x,1,512);

% Verify X1 = conj(X)

% Property 3 y = real(x);

[Y3,w] = freqz(y,1,512);

m = 0:511; w0 = -pi*m/512;

X1 = freqz(x,1,w0);

[X,w] = freqz(x,1,512);

% Verify Y3 = 0.5*(X+conj(X1))

% Property 4

y = j*imag(x); [Y4,w] = freqz(y,1,512);

% Verify Y4 = 0.5*(X-conj(X1))

% Property 5

k = 0:N-1; y = exp(-j*gamma*fliplr(k));

xcs = 0.5*[zeros(1,N-1) x] + 0.5*[conj(y) zeros(1,N-1)];

xacs = 0.5*[zeros(1,N-1) x] - 0.5*[conj(y) zeros(1,N-1)];

[Y5,w] = freqz(xcs,1,512);

[Y6,w] = freqz(xacs,1,512);

Y5 = Y5.*exp(j*w*(N-1));

Y6 = Y6.*exp(j*w*(N-1));

% Verify Y5 = real(X) and Y6 = j*imag(X) M3.5 N = 8; % Number of samples in sequence

gamma = 0.5; k = 0:N-1;

x = exp(gamma*k); y = exp(gamma*fliplr(k));

xev = 0.5*([zeros(1,N-1) x] + [y zeros(1,N-1)]);

xod = 0.5*([zeros(1,N-1) x] - [y zeros(1,N-1)]);

[X,w] = freqz(x,1,512);

[Xev,w] = freqz(xev,1,512);

[Xod,w] = freqz(xod,1,512);

Xev = exp(j*w*(N-1)).*Xev;

Xod = exp(j*w*(N-1)).*Xod;

% Verify real(X) = Xev, and j*imag(X) = Xod M3.5 N = input('The length of the seqeunce = ');

k = 0:N-1; gamma = -0.5;

g = exp(gamma*k);

% g is an exponential seqeunce h = sin(2*pi*k/(N/2));

% h is a sinusoidal sequence with period = N/2 [G,w] = freqz(g,1,512); [H,w] = freqz(h,1,512);

% Property 1

alpha = 0.5; beta = 0.25;

y = alpha*g+beta*h; [Y,w] = freqz(y,1,512);

% Plot Y and alpha*G+beta*H to verify that they are equal

% Property 2

n0 = 12; % Sequence shifted by 12 samples y2 = [zeros(1,n0) g];

[Y2,w] = freqz(y2,1,512);

G0 = exp(-j*w*n0).*G;

% Plot G0 and Y2 to verify they are equal

% Property 3

w0 = pi/2; % the value of omega0 = pi/2

r = 256; % the value of omega0 in terms of number of samples

k = 0:N-1; y3 = g.*exp(j*w0*k);

[Y3,w] = freqz(y3,1,512);

k = 0:511;

w = -w0+pi*k/512; % creating G(exp(w-w0)) G1 = freqz(g,1,w);

% Compare G1 and Y3

% Property 4

k = 0:N-1; y4 = k.*g;

[Y4,w] = freqz(y4,1,512);

% To compute derivative we need sample at pi y0 = ((-1).^k).*g;

G2 = [G(2:512)' sum(y0)]';

delG = (G2-G)*512/pi;

% Compare Y4, delG

% Property 5 y5 = conv(g,h);

[Y5,w] = freqz(y5,1,512);

% Compare Y5 and G.*H

% Property 6

y6 = g.*h;

[Y6,w] = freqz(y6,1,512,'whole');

[G0,w] = freqz(g,1,512,'whole');

[H0,w] = freqz(h,1,512,'whole');

% Evaluate the sample value at w = pi/2

% and verify with Y6 at pi/2

H1 = [fliplr(H0(1:129)') fliplr(H0(130:512)')]';

val = 1/(512)*sum(G0.*H1);

% Compare val with Y6(129) i.e., sample at pi/2

% Can extend this to other points similarly

% Parsevals theorem

val1 m = sum(g.*conj(h)); val2 = sum(G0.*conj(H0))/512;

% Compare val1 with val2 M3.7 The DTFT of nh[n] is ^{( )}.

ω ω d

e dH ^j

j Hence, the group delay τ_g(ω) can be computed at a set of N discrete frequency points ω_k=2πk/N,0≤k≤N−1, as follows:

, Re

)

( { [ ]}

]}

[

{ ⎟

⎠

⎜ ⎞

⎝

= ⎛ ω

τ DFT hn

n h n DFT k

g

where all DFTs are N–points in length with N greater than or equal to the length of ]}.

[ { nh

M3.8 h = [3.8461 -6.3487 3.8461];

[H,w] = freqz(h,1,512);

plot(w/pi,abs(H)); grid

xlabel('\omega/\pi'); ylabel('Magnitude');

0 0.2 0.4 0.6 0.8 1

0 5 10 15

ω/π

Magnitude

M3.9 h = [-13.4866 45.228 -63.8089 45.228 -13.4866];

[H,w] = freqz(h,1,512);

plot(w/pi,abs(H)); grid

xlabel('\omega/\pi'); ylabel(Magnitude);

0 0.2 0.4 0.6 0.8 1 0

50 100 150 200

ω/π

Magnitude

Chapter 4

4.1 Let φ(t) be an arbitrary continuous-time function with a CTFT Φ j( Ω), where Let denote the periodic continuous-time function with a period T obtained by a periodic extension of Note that

is also given by the convolution of .

) ( )

(jΩ = ∫ φ t e ^{j t}dt

Φ ^∞

∞

−

Ω

− φ = ∑^∞φ +

−∞

n=

T(t) (t nT)

~

).

φ(t )

~ (

T t

φ φ(t) with the periodic impulse train

i.e.,

∑δ +

= ^∞

−∞

= n

nT t t

p( ) ( ), ~ ( ) ( ) ( ) . τ τ

∫ τφ −

=

φ ^∞

∞

−

d t p

T t

Tthe CTFT of ~ ( ) is then given by F{ F { }

T t

φ φ~ ( )}=Φ( Ω)⋅

j

T t p(t)

(

^{( )}

)

^{( )}

(

^{( ),}

)

( ^{2 2} _{T T}

n T

n T

T j n jn j n

jΩ ⋅ ∑δ Ω− Ω = ∑ Φ Ω δ Ω− Ω Φ

= ^∞

−∞

=

∞ π

−∞

=

π

)

(4-1)

where 2 . T T

= π

Ω

Now, the Fourier series expansion of is given by A CTFT of both sides of this equation is then

∑φ +

=

φ ^∞

−∞

n=

T(t) (t nT)

~

. )

~ (

= ∑

φ ^∞

−∞

=

Ω n

t n jn

T t a e ^T

F ^{{~ ( )} 2}

(

⁽ _T⁾

)

^. (4-2)

n n

T t = ∑a ⋅ πδ j Ω−nΩ

φ ^∞

−∞

=

Comparing Eqs. (4-1) and (4-2) we arrive at ¹ ( _T).

n T jn

a = Φ Ω Substituting this expression in the Fourier expansion of ~ ( )

T t

φ we therefore arrive at the Poisson’s sum

formula ~ ( ) ( ) 1 ( ) jn t.

n T

n T

T t t nT jn e ^ΩT

∞

−∞

=

∞

−∞

=∑φ + = ∑Φ Ω

= φ

4.2 Consider the continuous-time signal g_a(t)=sin(Ω_mt) which is bandlimited to If we sample at a rate

m.

Ω g_a(t) Ω_T =2Ω_m starting at t=0, then all its samples are zero. Hence, cannot be recovered from its samples obtained by sampling it at the Nyquist rate

) (t g_a

m T = Ω

Ω 2 . As a result, g_a(t)=sin(Ω_mt) must be sampled at a rate Ω_T >2Ω_m to recover it fully from its samples.

4.3 (a) Now, the CTFT of y₁(t) is given by ( ) ( )O* ( )

2 1

1 Ω = Ω Ω

πG j G j

j

Y _{a a} where

denotes the CTFT of and denotes the frequency-domain convolution. The highest frequency present in is therefore twice that of

and hence, the Nyquist frequency of is )

(jΩ

G_a g_a(t) O*

)

1(t y )

(t

g_a y₁(t) 2Ω _m.

(b) The CTFT of y₂(t) is given by Y j g_a ^t e^−jΩtdt

∞

∞

−∫

=

Ω) ( )

( 3

2

The highest frequency present in is therefore one-third of that of and hence, the Nyquist frequency of is

).

3 ( 3 )

(

3 ∫ τ ³ τ= Ω

= ^{∞ − Ωτ}

∞

−

j G d e

g_a ^j _a y₂(t)

) (t

g_a y₂(t)

. 3

m / Ω

(c) The CTFT of y₃(t) is given by Y j g_a t e^−jΩtdt

∞

∞

−∫

=

Ω) (3 )

3(

).

( )

( 3 3

3 1 / 3

1 ∞ − Ωτ Ω

∞

−

=

∫ τ τ

= g_a e ^j d G_a j The highest frequency present in y₃(t) is therefore three times of that of and hence, the Nyquist frequency of is

) (t

g_a y₃(t)

. 3Ω_m

(d) The CTFT of y₄(t) is given by

∫ ⎥ τ

⎦

⎢ ⎤

⎣

⎡ ∫ −τ τ

∫ ⎥ =

⎦

⎢ ⎤

⎣

⎡ ∫ −τ τ τ

=

Ω ^∞

∞

−

∞

∞

−

Ω

− Ω

∞ −

∞

−

∞

∞

−

d dt e t g g

dt e d g t g j

Y₄( ) _a( ) _a( ) ^{j t} _a( ) _a( ) ^{j t}

).

( ) ( )

( ) ( )

( )

( Ω τ= Ω ∫ τ τ= Ω Ω

∫ τ

= ^∞

∞

−

τ Ω

∞ −

∞

−

τ Ω

− G j d G j g e d G j G j

e

g_a ^j _{a a a} ^j _{a a} The

highest frequency present in is therefore the same as that of and hence, the Nyquist frequency of is

)

4(t

y g_a(t)

)

4(t

y Ω _m.

(e) Now ( ) ( ) .

2

1 ∫ Ω Ω

= ^{∞ Ω}

∞

π− G j e d

t

g_{a a} ^{j t} Differentiating both sides of this equation

we get ( ) .

2 ) 1

( = ^∞∫ Ω Ω ^Ω Ω

∞

π− j G_a j e^{j t}d

dt t dg_a

Hence, it follows that the CTFT of

dt t dg_a

t

y₅( )= ^{( )} is simply jΩG_a(jΩ). The highest frequency present in is therefore the same as that of and hence, the Nyquist frequency of is

)

5(t y )

(t

g_a y₅(t)

m. Ω

4.4 By Parseval’s relation, the total energy of is given by E

) (t g_a

∫ Ω Ω

∫ =

= ^∞

∞ π−

∞

∞

−

d j G dt

t g

t _{a a}

g_a

2 2

1

2 ( )

) ( )

( ( )² .

2

1 ∫ Ω Ω

= ^Ω

Ω

π− ^m G j d

m

a Likewise, the total energy of g[n] is given by E = ∑ = ^π∫ ω

π

−

ω π

∞

∞

− g n G e^j d

n g

2 2

1 2 ]

[ [ ] ( )

∫ Ω Ω

=

∫ Ω Ω

=

∫ Ω Ω

= ^Ω

Ω π − π

π π − π

π

π− G j d T G j d G j d

m

m T a T

T T a T

T T a

2 2

/ 1 /

2 2

/ 1 /

1 2 2

1 ( ) ( ) ( ) ( )

T

= E1 _{g (t)}.

a

4.5 Sampling period = = 5000

5 .

T 2 sec. Hence, the sampling frequency is

1 =2000

= T

FT Hz. Therefore, the highest frequency component that could be present in the continuous-time signal has a frequency 1000

20000 =2 Hz.

4.6 Since the continuous-time signal is being sampled at kHz rate, the sampled version of its -th sinusoidal component with a frequency will generate discrete-time sinusoidal signals with frequencies

) (t

x_a 2

i F_i

. ,

2000 −∞< <∞

± n n

F_i Hence, the

frequencies generated in the sampled version associated with the sinusoidal components present in are as follows:

Fim

1=300

F Hz ⇒F_1m =300,1700,2300,K Hz

2 =500

F Hz ⇒F_2m =500,1500,2500,K Hz

3 =1200

F Hz ⇒F_3m =1200,800,3200,K Hz

4 =2150

F Hz ⇒F_4m =2150,150,4150,KHz

5 =3500

F Hz ⇒F_5m =3500,1500,5500,500,7500,KHz

After filtering by a lowpass filter with a cutoff at 900 Hz, the frequencies of the sinusoidal components in y_a(t) are 150, 300, 500,800 Hz.

4.7 One possible set of values for the frequencies present in are: Hz, Hz, Hz, and

) (t

y_a F₁=350

2 =575

F F₃ =815 F₄ =9650 Hz. Another possible set of values for the frequencies present in y_a(t) are: F₁=350 Hz, F₂ =575 Hz, Hz, and Hz. Hence, the solution is not unique.

3 =815 F 10575

4 = F

4.8 .

50

nT n

t = = Therefore,

⎟⎠

⎜ ⎞

⎝⎛

⎟+

⎠⎞

⎜⎝ + ⎛

⎟⎠

⎜ ⎞

⎝⎛

⎟−

⎠⎞

⎜⎝

= ⎛ ^{π π π π}

50 176 50

120 50

24 50

20 5cos 3sin 2cos

sin 4 ]

[n ^{n n n n}

x

= ⎜⎝⎛ ^π ⎟⎠⎞− ⎜⎝⎛ ^π ⎟⎠⎞+ ⎜⎝⎛ ^{+ π} ⎟⎠⎞+ ⎜⎝⎛ ^{− π} ⎟⎠⎞

25 ) 12 100 ( 5

) 2 10 ( 25

12 5

2 5cos 3sin 2cos

sin

4 ^{n n n n}

4sin 5cos 3sin 2cos .

25 12 ( 5

2 25

12 5

2 ⎟⎠⎞− ⎜⎝⎛ ⎟⎠⎞+ ⎜⎝⎛ ⎟⎠⎞+ ⎜⎝⎛ ⎟⎠⎞

⎜⎝

= ⎛ ^{πn πn πn πn}

4.9 Both channels are being sampled at kHz. Therefore, there are a total of samples/sec. Each sample is quantized using 12 bits. Hence, the total bit rate of the two channels after sampling and digitization is 108 kpbs.

45 90000

45000

2× =

4.10 .

2 /

) ) sin(

( t

t t h

T

r Ω c

= Ω Therefore, .

2 /

) ) sin(

( nT

nT nT h

T

r Ω c

= Ω Since T =2π/Ω_T, we have

. sin

) (

2

nT n

h ^T

cn

r π

⎟⎠

⎜ ⎞

⎝

⎛

= ^Ω

Ω π

For Ω_c =Ω_T /2, we thus have sin( ) [ ].

)

( n

n nT n

h_r =δ

π

= π 4.11 The spectrum of the sampled signal is as shown below:

0 X (jp Ω)

Ω Ω

_ m Ω__m Ωm

3

Ωm Ωm 2

_2

1/T 1/2T

Now, .

2 2

m m

T Ω

π Ω

π =

= As a result, .

3 3

π Ω

π

Ω =

= ω

m

c m Hence after lowpass filtering the spectrum of the output continuous-time signal y_a(t) will be as shown below:

0 Y (jΩ)a

Ω__m Ω 3 Ω__m

3 _

4.12 (a) Ω₁=100π,Ω₁ =150π. Thus, ∆Ω=Ω₂ −Ω₁=50π. Note is an integer multiple of Hence, we choose the sampling angular frequency as

∆Ω 2.

Ω

, 100

2 ^{2 150}

T M

π

= ×

π

=

∆Ω

=

Ω which is satisfied for M =3. The sampling

frequency is therefore 50 Hz. The CTFT X_p(jΩ)of the sampled sequence and the frequency response H_r(jΩ) of the desired reconstruction filter are shown below.

X (jp Ω)

1/T

0 100π 150π 200π Ω

50π

_ 50π

_200π ^_150π ^_100π

M = 3

Ω T

H (jΩ)r

0 100π 150π

100π 150π _ _

(b) Ω₁=160π,Ω₁=250π. Thus, ∆Ω=Ω₂−Ω₁ =90π. Note is not an integer multiple of Hence, we extend the bandwidth to the left by assuming the lowest frequency to be

∆Ω 2.

Ω

Ω and choose the sampling angular frequency as 0

, )

( 2

2 _{2 0} ^{2 250}

T M

π

= ×

Ω

− Ω

=

∆Ω

=

Ω which is satisfied for Ω₀ =125π and M =2. The sampling frequency is therefore 125 Hz. The CTFT X_p(jΩ)of the sampled sequence and the frequency response H_r(jΩ) of the desired reconstruction filter are shown below.

Ω 1/T

0 160π 250π

160π 250π _

_ ^_90π 90π 340π

340π

_

X (jp Ω)

Ω T

H (jΩ)r

0 160π 250π

160π 250π _ _

(c) Ω₁=110π,Ω₁ =180π. Thus, ∆Ω=Ω₂ −Ω₁=70π. Note ∆Ω is not an integer multiple of Hence, we extend the bandwidth to the left by assuming the lowest frequency to be

2. Ω

Ω and choose the sampling angular frequency as 0

, )

( 2

2 _{2 0} ^{2 180}

T M

π

= ×

Ω

− Ω

=

∆Ω

=

Ω which is satisfied for Ω₀ =90π and

The sampling frequency is therefore Hz. The CTFT

.

=2 M 90 X_p(jΩ)of the sampled sequence and the frequency response H_r(jΩ) of the desired reconstruction filter are shown below.

Ω 1/T

0 110π 180π

110π 180π _

_ ^_70π 70π 250π

250π

_

X (jp Ω)

Ω T

H (jr Ω)

0 110π 180π

110π 180π _ _

4.13 α_p =−20log₁₀(1−δ_p) dB and α_s =−20log₁₀δ_s dB. Therefore, and

20

10 /

1 ^p

p

α

− −

=

δ δ_s =10^{−αs /20}.

(a) α_p =0.21 dB and α_s =52 dB. Hence, δ_p =0.0239 and δ_s =0.025. (b) α_p =0.03 dB and α_s =69 dB. Hence, δ_p =0.0034 and δ_s =0.00355. (c) α_p =0.33 dB and α_s =57 dB. Hence, δ_p =0.0373 and δ_s =0.0014.

4.14 ( ) .

a s s a H_a

= + Thus, ( ) ,

a j j a H_a

+

= Ω

Ω and hence,

. )

( ) ( )

( 2 2

2 2

a a a j

a a j

a a

a

a j H j H j

H Ω = Ω − Ω = Ω+ ⋅− Ω+ = Ω + As Ω increases from

to ∞ , it can be seen that the square-magnitude function

0 H_a(jΩ)² and hence,

the magnitude function H_a(jΩ)

2 2 a

a + Ω

= decreases monotonically from 1

) 0 ( j =

H_a to H_a(j∞) =0. Let Ω denote the _c 3–dB cutoff frequency. Then

2, 1 2

2 2

2

)

( Ω = =

+

Ω a

a c

a

c

j

H which implies Ω_c =a.

4.15 ( ) .

a s s s G_a

= + Thus, ( ) ,

a j j j G_a

+ Ω

= Ω

Ω and hence,

. )

( ) ( )

( 2 2

2 2

a a j

j a j

j a

a

a j G j G j

G Ω +

Ω +

Ω

− Ω

− + Ω

Ω ⋅ =

= Ω

− Ω

=

Ω As increases from

to ∞ , it can be seen that the square-magnitude function Ω

0 G_a(jΩ)² and hence,

the magnitude function G_a(jΩ)

2 2+a Ω

= Ω increases monotonically from

0 ) 0 (j =

G_a to G_a(j∞) =1. Let Ω denote the _c 3–dB cutoff frequency. Then

2, 1 2

2 2

2

)

( Ω = =

+ Ω

Ω c a

a

c

j c

G which implies Ω_c =a.

4.16

(

^{( ) ( )}

)

2 1 1

2 ) 1

( A₁ s A₂ s

a s

a s a

s s a

H_a ⎟= −

⎠

⎜ ⎞

⎝

⎛ +

− − + =

= and

a s s s G_a

= + ) (

(

^{( ) ( )}

)

^,

2 1 1

2 1

2 1 s A s a A

s a

s ⎟= +

⎠

⎜ ⎞

⎝

⎛ + + −

= where A₁(s)=1 and ₂( ) .

a s

a s s

A +

= − Now,

1 )

1(jΩ =

A and ₂( )² =1

− Ω

+

⋅ Ω + Ω

−

= Ω + Ω

−

− Ω

⋅− + Ω

−

= Ω

Ω j a

a j a j

a j a j

a j a j

a j j

A for all

values of Ω, A₁(s) and A₂(s) are allpass functions.

4.17 ( ) .

2

2 o

a s bs

s bs

H = + +Ω Thus,

2

) 2

( Ω+Ω −Ω

= Ω Ω

o

a jb

j jb

H and hence,

2 2 2

2

2 ( ) ( )

)

( − Ω+Ω −Ω

Ω

⋅ − Ω

− Ω + Ω

= Ω Ω

− Ω

= Ω

o o

a a

a jb

jb jb

j jb H j H j

H

. )

( ^{2 2 2 2 2}

2 2

Ω + Ω

− Ω

= Ω

b b

o

At Ω=0, H_a(j0) =0, at Ω=∞, H_a(j∞) =0, and at

o, Ω

=

Ω H_a(jΩ) has the maximum value of 1. Now,

(

^{( )}

)

^.

) 2 ( )( )

2 2 2 2 2 2

2 2 2 2 2 2

Ω + Ω

− Ω

Ω + Ω Ω

− Ω

= Ω

b b

o

o

( o

Ω Ω d

j H d _a

It therefore follows that in the

frequency range 0≤Ω<Ω_o, ( )² 0, Ω >

Ω d

j H d _a

and in the frequency range

∞,

<

Ω

<

Ω_o ( )² 0.

Ω <

Ω d

j H d _a

Hence, in the frequency range 0≤Ω<Ω_o, )2

(jΩ

H_a is a monotonically increasing function of Ωand in the frequency range

∞,

<

Ω

<

Ω_o H_a(jΩ)² is a monotonically decreasing function of Or in other words, has a bandpass magnitude response. The –dB cutoff frequencies are given by the solution of

Ω. )

(s

H_a 3

2, 1 )

( ^{2 2 2 2 2} 2 2

Ω = + Ω

− Ω

Ω c c

o

c b

b or,

i.e., Substituting

in the last equation we get Let and

be the two roots of this quadratic equation. Then,

and Therefore, From the last two

equations we get

Hence,

, 2 )

(Ω²_o−Ω²_c ² +b²Ω²_c = b²Ω²_c Ω_c⁴ −(b² +2Ω²_o)Ω²_c +Ω_o⁴ =0. 2c

x=Ω x² −(b² +2Ω²_o)x+Ω⁴_o =0. 2 1 1 =Ω x

22 2 =Ω

x x₁x₂ =Ω₁²Ω²₂ =Ω⁴_o

. 2 ²

2 22 12 2

1 x b o

x + =Ω +Ω = + Ω Ω₁Ω₂ =Ω²_o.

. 2

2 )

(

2 _{1 2 2 1} ^{2 2 2 2 2}

22

12 +Ω − Ω Ω = Ω −Ω =b + Ω_o − Ω_o =b Ω

1 .

2−Ω =b Ω

4.18 ( ) .

It therefore follows that in the

frequency range 0≤Ω<Ω_o, ( )² 0,

and in the frequency range

∞,

G_a is a monotonically decreasing function of Ωand in the frequency range Ω_o <Ω<∞, G_a(jΩ)² is a monotonically increasing function of Ω. Or in other words, has a bandstop magnitude response. G_a(s)

The 3 –dB cutoff frequencies are given by the solution of , 2

last equation is exactly the same as in solution of Problem 4.18 from which we get and

4.20 (a) Let ( ) .

function. Since, it is a product of causal, stable allpass functions, and as a result, is also a causal, stable allpass function.

)

To verify using MATLAB, we use the code fragment

[N,Wn]=buttord(2*pi*1500,2*pi*6000,0.25,25,'s');

which yields N = 4 and Wn = 18365.51286.

4.23 The poles are given by p_l = e ^jπ(5+2l)/, 1 ≤ l ≤ 6. Hence,

, 0.7071 +

0.7071

, 9659 . 0 2588 .

0 ₂ ^{(9 /12)}

) 12 / 7

1 e ( j p e j

p = ^{j π} =− + = ^{j π} =−

, 2588 . 0 9659 . 0

0.2588, +

0.9659 ₄ ^{(11 /12)} ₃^*

) 12 / 9

3 e ( j p e p j

p = ^{j π} =− = ^{j π} = =− −

, 7071 . 0 7071 .

* 0

) 2 12 / 13

5 e ( p j

p = ^{j π} = =− − p₆ =e^j(15π/12) = p₁^* =−0.2588− j0.9659. The poles can also be determined in MATLAB using the statement

[z,p,k]=buttap(6) which yields p =

-0.2588 + 0.9659i -0.2588 - 0.9659i -0.7071 + 0.7071i -0.7071 - 0.7071i -0.9659 + 0.2588i -0.9659 - 0.2588i

4.24 From Eq. (4.41) of text, T_N(Ω)=2ΩT_N−1(Ω)−T_N−2(Ω), where is defined in Eq.

(4.40).

(Ω) TN

Case 1: Ω ≤1. Making use of Eq. (4.40) in Eq. (4.41) we get

(

^{− ⋅ Ω}

) (

^{− − ⋅ Ω}

)

Ω

=

Ω) 2 cos( 1) cos⁻¹ cos( 2) cos⁻¹

( N N

T_N

^=2Ωcos

(

^{Ncos−1Ω−cos−1Ω}

) (

^{−cosNcos−1Ω−2cos−1Ω}

)

^=2Ω

[

^{cos(Ncos−1Ω)cos(cos−1Ω)+sin(Ncos−1Ω)sin(cos−1Ω)}

]

⁻

[

^{cos(Ncos−1Ω)cos(2cos−1Ω)+sin(Ncos−1Ω)sin(2cos−1Ω)}

]

=2Ωcos(Ncos⁻¹Ω)cos(cos⁻¹Ω)−cos(Ncos⁻¹Ω)cos(2cos⁻¹Ω) ^{=2Ω2cos(Ncos−1Ω)−cos(Ncos−1Ω)}

[

^{2cos2(cos−1Ω)−1}

]

^{=cos(Ncos−1Ω)}

[

^{2Ω2 −2Ω2 +1}

]

^{=cos(Ncos−1Ω).}

Case 2: Ω >1. Making use of Eq. (4.40) in Eq. (4.41) we get

(

^{( 1) cosh}

)

^cosh

(

^{( 2) cosh}

)

^.

cosh 2 )

(Ω = Ω N− ⋅ ⁻¹Ω − N− ⋅ ⁻¹Ω

T_N Using the trigonometric

identities

), sinh(

) sinh(

) cosh(

) cosh(

)

cosh(A−B = A B − A B sinh(2A)=2sinh(A)cosh(A),and and following a similar algebra as in Case 1, we can show ,

1 ) ( cosh 2 ) 2

cosh( A = ² A − ).

cosh cosh(

)

(Ω = N ⁻¹Ω

T_N

4.25 From the solution of Problem 4.22, we have 1 =4

k and 1 72.9381.

1

k = Hence,

. 2.4151 )

/ 1 ( cosh

) / 1 ( cosh

1 1 1

=

= ⁻₋ k

N k We choose the filter order as N =3. The filter order obtained using the MATLAB statement

[N,Wn]=cheb1ord(2*pi*1500,2*pi*6000,0.25, 25, 's') results in N=3.

4.26 0.25,

1 log 1

10 10 2⎟⎟⎠=−

⎜⎜ ⎞

⎝

⎛ ε

+ which yields ε=0.2434. 1 25,

log

10 10 2⎟⎟=−

⎠

⎜⎜ ⎞

⎝

⎛ A

which yields A² =316.2278. Now, 0.25

1500 =6000 Ω =

=Ω

s

k p and = =

−

= ε

2278 . 315

2434 . 0

2 1

1

A k

0.0137.

= Substituting the value of in Eq. (4..55a) we get Then from Eq.

(4.55b) we get Substituting the value

k k^' =0.9682.

. 0.004

0 =

ρ ρ in Eq. (4.55c) we get ₀ ρ=0.004.

Finally, from Eq. (4.54) we arrive at N =2.0591. We choose the next higher integer as the filter order N =3.

The filter order obtained using the MATLAB statement

[N,Wn]=ellipord(2*pi*1500,2*pi*6000,0.25, 25, 's') results in N=3.

4.27 B_N(s)=(2N−1)B_N−1(s)+s²B_N−2(s), where B₁(s)= s+1 and B₂(s)=s² +3s+3. (a) Thus, B₃(s)=5B₂(s)+s²B₁(s)=5(s² +3s+3)+s²(s+1)=s³ +6s² +15s+15,

) 3 3 ( ) 15 15 6

( 7 ) ( )

( 7 )

( ₃ ² ₂ ^{3 2 2 2}

4 s = B s +s B s = s + s + s+ +s s + s+ B

=s⁴ +10s³ +45s² +105s+105.

(b) B₅(s)=9B₄(s)+s²B₃(s)=9(s⁴ +10s³ +45s² +105s+105)+s²(s³ +6s² +15s+15) =s⁵ +15s⁴ +105s³ +420s² +945s+945.

4.28 Ω_p =2π×0.24 and Ωˆ _p =2π×3. The mapping is thus . ˆ

72 . 0 4 ˆ

ˆ 2

s s Ω^psΩ^p = π ×

=

Denote Hence, the desired highpass transfer function is given by

28.4245.

72 . 0 4π² × =

= K

10 2835

. 9 309

. 4 ) 10

( ˆ)

(

ˆ 2

ˆ 3

ˆ /ˆ

+

⎟⎠

⎜ ⎞

⎝⎛ +

⎟⎠

⎜ ⎞

⎝⎛ +

⎟⎠

⎜ ⎞

⎝⎛

=

= ₌

s K s

K s

s K K LP s

HP s H s

H

3 2

2 3

3

10ˆ 2835 ˆ

. ˆ 9 309 . 4

10ˆ

s s

K s

K K

s

+ +

= +

22966 sˆ

3481.5 sˆ

263.8785 sˆ

10

sˆ 10

2 3

3

+ +

= + . 2296.6 sˆ

348.15 sˆ

26.38785 sˆ

sˆ

2 3

3

+ +

= +

4.29 Ω_p =2π×0.9 and Ωˆ _p =2π×3. The mapping is thus . ˆ

7 . 2 4 ˆ

ˆ 2

s s Ω^psΩ^p = π ×

= Denote

Hence, the desired lowpass transfer function is given by 106.5917.

7 . 2 4π²× =

= K

100 prototype lowpass filter is thus given by 3.8579.

)

The order of the prototype lowpass filter obtained using the MATLAB statement [N,Wn]=buttord(1,13/3,0.5, 40, 's') results in N=4.

The order of the desired highpass filter is also 4.

4.32 ˆ 20 10 , ˆ 45 10 , ˆ 10 10³, and Thus,

2 8

1ˆ 9 10 ˆ_p F_p = ×

F and Since we can either

increase left stopband edge or decrease the left passband edge to make

We choose to increase to a new value given by in

which case The center angular frequency of the

bandpass filter is therefore The passband width is .

10 5 . ˆ 7

ˆ ⁸

2

1 _s = ×

s F

F ˆ ˆ ˆ ˆ ,

2 1 2

1 p s s

p F F F

F >

ˆ 1

Fs F^ˆ_p1

ˆ . ˆ ˆ

ˆp1Fp2 Fs1Fs2

F = ˆ ₁

Fs ˆ 18 10³,

1 = × Fs

. 10 ˆ 9

ˆ ˆ ˆ

ˆ ^{2 8}

2 1 2

1 _p = _{s s} = _o = ×

p F F F F

F

. 10 30 ˆ =2π× × ³ Ω_o

. 10 25 ˆ 2

ˆ ³

2

1−Ω = π× × Ω

= _{p p}

Bw

To determine the bandedges of the prototype lowpass filter we set Ω_p =1 and thus, .

28 . 25 1 18

18 30 ˆ

ˆ

ˆ ^{2 2}

1 21

2 =

×

= − Ω

Ω

− Ω Ω

= Ω

w s

s p o

s B

Now, 0.78125.

28 . 1

1 = Ω =

= Ω

s

k p Hence, k'= 1−k² =0.62421826.

Next, 0.25

1 log 1

10 10 2⎟⎟=−

⎠

⎜⎜ ⎞

⎝

⎛ ε

+ or equivalently, log₁₀(1+ε²)=0.025 which yields 5

0.05925372 1

10^0.025

2 = − =

ε or ε=0.243421. Likewise, 1 50

log

10 10 2⎟⎟=−

⎠

⎜⎜ ⎞

⎝

⎛ A or, equivalently, log₁₀(A²)=5 which yields A² =10⁵ =100000. Therefore,

. 058635856 .

0 ) ' 1 ( 2

' , 1

10 69768 . 7 1

4 0

1 2 =

+

= − ρ

×

=

−

= ε ⁻

k k A

k As a result,

. 058637246 .

0 ) ( 150 )

( 15 ) (

2 ₀ ⁵ ₀ ⁹ ₀ ¹³

0 + ρ + ρ + ρ =

ρ

=

ρ Hence,

. 0328 . ) 6 / 1 ( log

) / 4 ( log 2

10 1

10 =

= ρk

N We choose N =7 as the order of the prototype lowpass filter.

Note that the order can also estimated using the specifications of the bandpass filter. To this end, the statement to use is

[N,Wn]=ellipord([20 45],[15 50],0.25,50,'s') which also yields N=7 as the order of the prototype lowpass filter. The order of the desired bandpass filter is

therefore 7×2=14.

4.33 ˆ 10 10 , ˆ 70 10 , ˆ 20 10⁶, and Thus,

6 1 6 2

1 = × _p = × _s = ×

p F F

F ˆ 45 10⁶.

2 = × Fs

and Since we can either

increase left passband edge or decrease the left stopband edge to make We choose to increase to a new value given by

2 13

1ˆ 70 10

ˆ_p F_p = ×

F ˆ ˆ 90 10¹³.

2

1 _s = ×

s F

F ˆ ˆ ˆ ˆ ,

2 1 2

1 p s s

p F F F

F <

ˆ 1

Fp ˆ₁

Fs ˆ .

ˆ ˆ

ˆp1Fp2 Fs1Fs2

F = ˆ ₁

Fp

, 10 8571 . ˆ 12

ˆ ˆ ˆ

ˆ ⁶

2 2 2 1

1 = = ×

p s p s

p F

F F F

F in which case ˆ ˆ ˆ ˆ ² 700 10¹².

2 1 2

1 _p = _{s s} = _o = ×

p F F F F

F

The width of the stopband is and the center angular frequency of the stopband is

1 6

2 ˆ 2 25 10

ˆ −Ω = π× × Ω

= _{s s}

Bw

. 10 700

4 ^{2 12}

2 = π × ×

Ω_o

To determine the bandedges of the prototype lowpass filter we set resulting in its passband edge

=1 Ω_s .

4375 . ˆ 0

ˆ ˆ

21 2

1 =

Ω

− Ω Ω Ω

= Ω

p o

w p p p

B

Now, 0.5

1 log 1

10 10 2⎟⎟=−

⎠

⎜⎜ ⎞

⎝

⎛ ε

+ or equivalently, log₁₀(1+ε²)=0.05 which yields or

1220184543 .

0 1 10^0.05

2 = − =

ε ε=0.349114. Likewise, 1 30

log

10 10 2⎟⎟=−

⎠

⎜⎜ ⎞

⎝

⎛ A or, equivalently, log₁₀(A²)=3 which yields A² =10³ =1000. Therefore,

2.2857 4375

. 0

1

1 = =

Ω

= Ω

p s

k and 90.4836236.

349114 .

0 999 1

1 ²

1

= ε =

= A − k

Substituting the values of k 1 and

1

1

k in Eq. (4.43) we get .

5408 . 3 ) 2857 . 2 ( cosh

) 4836236 .

90 ( cosh

1

1 =

= ⁻ ₋

N We therefore choose N =4 as the order of the prototype lowpass filter. The order of the desired bandstop filter is thus 8.

N We therefore choose N =4 as the order of the prototype lowpass filter. The order of the desired bandstop filter is thus 8.