第一章習題第一章習題第一章習題第一章習題解答解答解答解答

(1)

1. 確認以下的複數值訊號是否具有共軛對稱之特性。

) sin (cos

)

(t A t j t

x = ω + ω

<解答解答解答> 解答

)) sin(

) (cos(

)) sin(

) (cos(

)

( t A t j t A t j t

x − = −ω + −ω = ω − ω

) ( ) sin (cos

)

(t A t j t x t

x^∗ =

ω

−

ω

= −

This signal is conjugate symmetric.

2. 請判斷以下訊號何者為週期性訊號？若是週期性訊號，請找出其週期。

(a) x[n]= Acos(10n) (b) x(t)=cos(

ω

t)u(t)

(c) )

sin(4 3 )

cos(

]

[n n n

x ₌ π ₊ π

(d) x(t)=sin²(ω₀t)

(a) x[n+N]= Acos(10n+10N)= Acos(10n)

? 2

10^N = ^kπ That implies ^N =2^kπ /10. No integer N can be found.

It is a non-periodic signal.

(b) x(t+T)=cos(ω(t+T))u(t+T)=cos(ωt)u(t). No T can satisfy the equation.

It is a non-periodic signal.

(c) )

4 sin(4

3 ) cos(3

)) 4(

sin(

)) 3(

cos(

]

[ π π π π π ^Nπ

N n n N

n N

n

x + = + + + = + + +

Let N =24,

] [ 4 )

sin(

3 ) cos(

) 4 6

sin(

) 3 8

cos(

]

[n N n n n n x n

x + = π + π + π + π = π + π =

It is a periodic signal with period N = 24.

(d) (1 cos(2 ))

2 ) 1 ( sin )

(t ² ₀t ₀t

x = ω = − ω

)) 2 2 cos(

1 2( ))) 1 (

2 cos(

1 2( ) 1

(t T ₀ t T ₀t ₀T

x + = − ω + = − ω + ω

Let ²ω0T =²π,

) ( )) 2 cos(

1 2( ) 1

(t T ₀t x t

x + = − ω =

It is a periodic signal with period T =π/ω0。

(2)

3. 一個訊號的波形描述如下，





 + − ≤ ≤

=

otherwise t t

t x

0

2 ,

cos 1 )

(

ω π

ω π ω

這個訊號稱為 raised-cosine pulse，請計算其全能量。

ω π ω ω

ω ω π

ω π ω

π

ω ω ω

ω π

ω ω

ω ππ ω ω

π ω π

ω π

ω π ω

ππ ω ω π

ω π ω

π ω π

4

| 3 16

) 2 sin(

4 ) 3 2 8 cos(

1 0 4

2

) 2 cos(

1 8 (

| 1 ) 2 sin(

1 2

4

) ( cos ) cos(

2 ) 1

(

/ / /

/

/ /

/ / 2

/

= +

+ +

=

+ +

+

=

+

= +

=

− −

−

∫

dt t t

t dt dt t

t x E

4. 計算以下訊號的平均功率，其中Ω=2π/N ，N 為大於 4 的整數，

x[n]= Asin(nΩ+φ)

2 ) sin(

]

[ ₌ π ₊φ

n N A n x

Ω

=2π/ N

) 2 2 cos(

)) 2 2 cos(

1 ( )

( 1 sin

] 1 [

1

0 2 2

1

0 2 2

2 1

0 2

1

0

φ

φ φ

+ Ω

−

=

+ Ω

−

= + Ω

=

∑

−

=

−

=

−

=

−

=

N n A A

N n n A

N A n N x

P

N

n

N

n N

n

1 0 1 2 1

1 2 1

1 2

) 2 2 2 cos(

/ 4

4 2

/ 4

4 2

2 2 2

) 2 2 ( ) 2 2 1 (

0 1

0

− = + −

−

= −

− + −

−

= −

= + + Ω

−

− Ω

−

− Ω Ω

+ Ω

− +

− Ω

=

−

=

∑

N j

j N

j j j

N j j

N j

n j n

N j

n N

n

e e e

e n e

π π φ

φ φ

φ

φ φ

A2

P=

5. 一個波形x(t)如下圖所示，

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1

−1

t

y(t)=x(2t−5) 請繪出y(t)的波形。

t 1

−1

−6 6

t 1

−1

5 11

t

1

−1

5

6. 一個波形x[n]如下圖所示，

1

−

1

] [n x

n

(4)

] 3 2 [ ]

[n =x n− y

請繪出y[n]的波形。

1

−1

n

1

n

7. 脈衝函數δ(t)^{定義如下，}

))

( 2 2) ( 1( lim )

( 0

−∆

∆ −

∆ +

=∆→ u t u t δ t

請以相似的方式，定義其微分函數。

dt t t d ( )

)

)(

1

( δ

δ = 及

dt t

t d ( )

) (

) 1 ( )

2

(

δ

=

並計算以下的積分結果

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∫

₋^∞_∞ f(t)

^δ

⁽ⁿ⁾(t−t₀)dt

2)) ( 2) ( 1( ) lim ) (

( 0

) 1

( + ∆ − −∆

= ∆

= ∆→ t t

dt t

t dδ δ δ

δ

| 0

) ( 2))

( 2) ( 1 ( lim

2))) (

2) (

1 ( )(

( lim )

( ) (

0 0 0

0 ) 1 (

t

t t

dt f t d

f t

f

t t t

t t

f dt

t t t f

→ =

∆

→

∆

∞

−

∞

−

∆ = +

∆ −

=

− ∆

−

∆ − +

∆ −

=

−

∫

∫ ^δ ^δ ^δ

2)) ( 2)

( 1 ( ) lim ) (

( ⁽¹⁾ ⁽¹⁾

0 )

1 ( )

2

( + ∆ − −∆

= ∆

= ∆→ t t

dt t

t d

δ δ δ

δ

0

0 ( 1) ( ))|

| )) ( (

2)) ( 2)

( 1 (

lim

2))) (

2) (

1( )(

( lim )

( ) (

) 2 ( 2 )

1 ( 0

) 1 ( 0

0 ) 1 ( 0

) 1 ( 0 0

) 2 (

t t t

t f t

t dt f t d

f t

f

t t t

t t

f dt

t t t f

=

→ =

∆

→

∆

∞

−

∞

−

∆ = +

∆ +

−

∆ −

=

− ∆

−

∆ − +

∆ −

=

−

∫

∫ ^δ ^δ ^δ

^∞ f(t) ⁽ⁿ⁾(t t₀)dt ( 1)ⁿ f⁽ⁿ⁾(t))|_t₌_t₀

∞

− − = −

∫ ^δ

8. 以下系統中，判斷其是否 (i)無記憶性，(ii)穩定，(iii)符合因果律，(iv)線性，

及(v)非時變。

(a) y[n]=log|x[n]| (b) ( ) {e x(t)}

dt t d

y = ⁻^at

(c) y[n]=cos(2πx[n+1])+x[n] (d) ^y⁽^t⁾⁼

∫

₋^t_∞^/² ^x⁽

τ

⁾^d

τ

(e) y(t)=x²(t−1) (f) y(t)= x(t+1) (g) y[n]=x[−n]

(h) ) [ 1] 2

(1 ]

[n = ⁺¹x³ n−

y ⁿ

(i) y[n]=x²[n] (j) y(t)=x(2t)

(a) y[n]=log|x[n]|

Memoryless –y[n] depends only on x[n].

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Unstable – example: If |x[n]|=0, y[n]=−∞

Causal – If x[n] starts at n=0, y[n]=0 for n<0 Nonlinear -- y₁[n]=log|x₁[n]| y₂[n]=log|x₂[n]|

] [ ] [

| ] [ ] [

|

log ax₁ n +bx₂ n ≠ay₁ n +by₂ n ^. Time-invariant – log|x[n−m]|= y[n−m]

Conclusion: memoryless, unstable, causal, non-linear, time-invariant

(b) ( ) {e x(t)}

dt t d

y = ⁻^at

dt t e dx t x e t

x dt e t d

y ^t ^t ^t ( )

) ( )}

( { )

( = ⁻ =− ⁻ + ⁻

Memory -- ( ) ( )

) ( )

( c t

dt t e d t e t

h =− ⁻^tδ + ⁻^t δ ≠ δ

Unstable – If x(t)=u(t)

) ( )

) ( ) (

( )

( e u t e t

dt t e du t u dt e

t e dx t x e t

y =− ⁻^t + ⁻^t =− ⁻^t + ⁻^t =− ⁻^t + ⁻^tδ

Causal – If x(t) starts at t =0, y(t)=0, for t <0 Linear – ₁( ) {e x₁(t)}

dt t d

y = ⁻^at ₂( ) {e x₂(t)}

dt t d

y = ⁻^at

) ( ) ( )}

( ( { )}

( ( {

)}

( ( { )}

( ) ( ( {

2 1

t by t ay t

x dt e

b d t x dt e

a d

t bx dt e

t d ax dt e

t d bx t ax dt e

d

at at

at

+

= +

=

+

= +

−

Time-varying – { ( ₀)} ( ₀) {e ⁽ ⁰⁾x(t t₀)}

dt t d t y t

t x dt e

d ⁻at − ≠ − = ⁻at⁻t −

Conclusion: memory, unstable, causal, linear, time-varying

(c) y[n]=cos(2πx[n+1])+x[n] Memory -- ,

Stable -- |x[n]|<M, |y[n]|<1+M Non-causal -- ,

Non-linear -- y₁[n]=cos(2

π

x₁[n+1])+x₁[n] y2^[n^]=^cos(²

π

x2^[n+¹^])+x2^[n^] ]

[ ] [ ]

[n ax₁ n bx₂ n

x = +

] [ ] [ ]) [ ] [ ( ])) 1 [ ] 1 [ ( 2

cos(

π

ax₁ n+ +bx₂ n+ + ax₁ n +bx₂ n ≠ay₁ n +by₂ n Time-invariant -- cos(2πx[n−m])+x[n−m]= y[n−m]

Conclusion: memory, stable, non-causal, non-linear, time-invariant

( ))

( ) (

( ^/² ₁ ^/² ₂ ₁ ₂

2 /

2

) Conclusion: memory, unstable, non-causal, linear, time-varying

(e) y(t)=x²(t−1)

Memory -- y(t) depends on x(t−1) Stable -- |x(t)|<M y(t)<M² Causal --

Non-linear -- y₁(t)=x₁²(t−1), y₂(t)=x₂²(t−1)

) ( ) (

) 1 ( )

1 ( ) 1 ( 2 ) 1 ( ))

1 ( ) 1 ( (

2 1

2 2 2 2

1 2

1 2 2 2

1

t by t ay

t x b t

x t abx t

x a t

bx t

ax

+

≠

− +

−

− +

−

=

− +

−

Time invariant -- Let ^t^→^t⁻ρ x²⁽t−

ρ

−¹⁾= y⁽t−

ρ

⁾ Conclusion: memory, stable, causal, non-linear, time-invariant

(f) y(t)= x(t+1)

Memory -- y(t) depends on x(t+1) Stable -- |x(t)|<M y(t)<M Non-causal --

Linear -- y₁(t)=x₁(t+1), y₂(t)=x₂(t+1) ) ( ) ( ) 1 ( ) 1

( ₂ ₁ ₂

1 t bx t ay t by t

ax + + + = +

Time invariant -- Let t→t−ρ x(t−

ρ

+1)= y(t−

ρ

) Conclusion: memory, stable, non-causal, linear, time-invariant

(g) y[n]=x[−n]

Memory -- y[n] depends on x[−n] Stable -- |x[n]|<M y[n]<M Non-causal --

Linear -- y₁[n]=x₁[−n]^, y₂[n]=x₂[−n] ] [ ] [ ] [ ]

[ ₂ ₁ ₂

1 n bx n ay n by n

ax − + − = +

Time varying -- Let x[−n−m]= y[n+m]≠ y[n−m] Conclusion: memory, stable, non-causal, linear, time-varying

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(h) ) [ 1] 2

(1 ]

[n = ⁺¹x³ n−

y ⁿ

Memory -- y[n] depends on x[n−1] Unstable -- |x[n]|<M ) ¹ ³

2 (1 ]

[n M

y < ⁿ⁺ If n→−∞, y[n]→∞

Causal --

Non-linear -- ) [ 1]

2 (1 ]

[ ¹ ₁³

1 n = ⁺ x n−

y ⁿ , ) [ 1]

2 (1 ]

[ ¹ ₂³

2 n = ⁺ x n−

y ⁿ

] [ ] [ ])

1 [ ] 1 [ ( 2)

(1 ⁿ⁺¹ ax₁ n− +bx₂ n− ³ ≠ay₁ n +by₂ n

Time varying -- Let ) [ 1] [ ] 2

(1 ⁿ⁺¹x³ n−m− ≠ y n−m

Conclusion: memory, unstable, causal, non-linear, time-varying

(i) y[n]=x²[n]

Memoryless -- y[n] depends on x[n] Stable -- |x[n]|<M y[n]<M² Causal --

Non-linear -- y₁[n]=x₁²[n], y₂[n]=x₂²[n] ] [ ] [ ])

[ ] [

(ax₁ n +bx₂ n ² ≠ay₁ n +by₂ n Time invariant -- x²[n−m]= y[n−m]

Conclusion: memoryless, stable, causal, non-linear, time-invariant

(j) y(t)=x(2t)

Memory -- y(t) depends on x(2t) Stable -- |x(t)|<M y(t)<M Non-causal --

Linear -- y₁(t)=x₁(2t)^, y₂(t)=x₂(2t) ) ( ) ( ) 2 ( ) 2 (

(ax₁ t +bx₂ t =ay₁ t +by₂ t

Time varying -- x(2t−t₀)≠ y(t−t₀)=x(2t−2t₀)

Conclusion: memory, stable, non-causal, linear, time-varying

9. 一個 RC 電路如下，v(t)為輸入，v_c(t)為輸出。

(9)

如果電阻是時間的函數R(t)，請問這系統是否是線性系統？是否是非時變系統？

) ) (

) ( ( )

( v t

dt t Cdv t R t

v = ^C + _C ( ) ( )

) ( )

( y t

dt t Cdy t R t

x = +

)) ( )) (

( ) (

( )]

) ( ) (

( [ )

( ¹ ₁ ¹ ₁

1 ay t

dt t ay Cd t R t dt y

t Cdy t R a t

ax = + = +

)) ( )) (

( ) (

( )]

) ( ) (

( [ )

( ² ₂ ² ₂

2 by t

dt t by Cd t R t dt y

t Cdy t R b t

bx = + = +

))]

( )) (

( ) (

( [ ))]

( )) (

( ) (

( [ ) ( ) ( )

( ₁ ₂ ¹ ₁ ² by₂ t

dt t by Cd t R t

dt ay t ay Cd t R t bx t ax t

x = + = + + +

)], ( ) ( )] [

( ) ( ) [

( )

( ¹ ² ay₁ t by₂ t

dt t by t ay Cd t R t

x = + + + y(t)=ay₁(t)+by₂(t)

linear

The system is time-varying, but is linear.

10. 一個非線性元件具有平方的功能，

y(t)=x²(t) 如果輸入訊號為

x(t)= A₁cos(

ω

₁t+

φ

₁)+A₂cos(

ω

₂t+

φ

₂)

請計算它的輸出訊號y(t)，並說明這個輸出訊號含有那些頻率成分。

) )

(cos((

) )

(cos((

) 2 2

cos(

1 2 ( ) 2 2 cos(

1 2 (

)) cos(

) cos(

( ) ( ) (

2 1 2 1 2

1 2 1 2 1 2

1

2 2 2

2 1 1 2

1

2 2 2 2

1 1 1

2

φ φ ω ω φ

φ ω ω

φ ω φ

ω

φ ω φ

ω

+ + + +

− +

+ +

+ + +

=

+ +

+

=

t A

A t

A A

A t A t

t A

t x t y

The frequency components are at

2 1 2

1 2

1, 2 , ,

2 ,

0

ω ω ω ω ω ω ω ω ω ω

ω

= = = = − = +

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11. 判斷以下系統是否可以有逆向系統，若有，請寫出其逆向系統；

(a) y[n]=nx[n] (b) y(t)= x(3t)

(a) y[n]=nx[n] n n y n

x[ ]= [ ]/

(b) y(t)= x(3t) ) 3 / ( ) (t y t

x =

第一章習題第一章習題第一章習題第一章習題解答解答解答解答

ω

ω

ω

ω π

ω π ω

ω π ω ω

ω ω π

ω π ω

π

ω ω ω

ω π

ω ω

∫

∫

∫

∫

φ

φ φ

∑

∑

∑

∑

∑

∑

t

−

n

n

δ

δ

∫

δ

∫

∫ δ δ δ

δ δ δ

δ

∫

∫ δ δ δ

∫ δ

∫

τ

τ

π

π

π

∫

τ

τ

τ

∫

τ

τ

∫

τ

τ

∫

∫

∫

τ τ τ

τ τ

τ τ

∫

τ

ρ

τ

∫

τ

τ

ρ

ρ

ρ

ρ

ρ

ω

φ

ω

φ

φ φ ω ω φ

φ ω ω

^δ

∫ ^δ ^δ ^δ

∫ ^δ ^δ ^δ

∫ ^δ

^τ

^τ

^τ

^τ

^τ ^τ ^τ

^τ ^τ

^τ ^τ