# 数字信号处理 DSP 英文版课件5.1

In This Lecture:
5.4 Frequency Spectrum ·X ( z ) ~ X (? ) (z-domain view of frequency spectrum) 5.5 Inverse z-Transform · Partial fraction expansion method

5.4 Frequency Spectrum
1. Relation of z-transform with spectrum

For a signal x(n) ,
X (? ) ?
n ? ??

x(n)e ? j?n ?
?

?

(DTFT)

(5.4.1)

X ( z) ?

n ? ??

x ( n) z ? n ?

(z-transform)

z ? re j?

?
?n ? j?n

X ( z) |z ?e j? ? ? x(n) z |z ?e j? ? ? x(n)e
n n

? X (?)

------- the evaluation of the z-transform on the unit circle

Illustration in Matlab：

z-plane

X ( z) ? ? x(n)r e
n

? n ? j?n

? DTFT[ x(n)r ]
z 变换扩大了能分析的 信号的范围

?n

X (?) ? X ( z) |z ?e j?
(Definition:

X (? ) ?

n ? ??

x(n)e ? j?n , ?

?

X ( z) ?

n ? ??

x ( n) z ? n ) ?

?

Frequency response of a linear system h(n) :

H (? ) ? DTFT[h(n)] ?

n???

h(n)e ? j?n ?

?

(5.4.3)

H (?) ? H ( z) | z?e j?

( H (z ) : Transfer function)

2. Meaning of the spectrum (a review)

2?f Digital frequency: ? ? fs
f : physical frequency (Hz) Nyquist interval:

(5.4.4)

?

fs

2

? f ?

fs

2

? ?? ? ? ? ? (5.4.5)

Inverse DTFT:

1 x ( n) ? 2?

??
?

?

X (? )e j?n d?

(5.4.6)

It expresses x(n) as a linear combination of discrete-time sinusoids e
j?n

of different frequencies.

The relative amplitudes and phases of these sinusoidal components are given by the DTFT X (? ) .
Sinusoidal frequency component

3. The spectrum of complex sinusoid

Double-sided complex sinusoid (双边复正弦):

x(n) ? e

j?0 n

, ??? n ? ?

?
?

Introducing ? (?) in frequency domain

X (?) ? 2?? (? ? ?0 ) ＋(Nyquist replica)
? 2?

m???

? ? (? ? ?

0

? 2?m)

Assuming

? ? ? ?0 ? ? ,

X (?) ? 2?? (? ? ?0 ) (?? ? ? ? ? )
Verification:

1 2?

2?? (? ? ?0 )e j?n d? ? e j?0n ? x(n) ???

?

or

X ( f ) ? ? ( f ? f0 )

（ ? 0.5 ?

f ? 0.5 ）

For a linear combination of two sinusoids：

x(n) ? A1e j?1n ? A2e j?2n
X (?) ? 2?A1? (? ? ?1 ) ? 2?A2? (? ? ?2 )

(?? ? ? ? ? )

(Assume both ?1 and interval)

?2 lie in the Nyquist

For real-valued cosine and sine signals：

cos(?0 n) ? ?? (? ? ?0 ) ? ?? (? ? ?0 ) sin(?0 n) ? ? j?? (? ? ?0 ) ? j?? (? ? ?0 )
j?0 n

（ cos(?0 n) ?

e

?e 2

? j?0 n

， sin(?0 n) ?

e

j?0 n

?e 2j

? j?0 n

Causal, truncated complex sinusoid（单边复正弦）:

x(n) ? e j?0n u (n)
X ( z) ? ? e
n ?0 ? j?0 n

（p.202, bottom）

z

?n

?

1 1? e
j?0

z

?1

(| z |? 1)

(The spectrum doesn’t exit)

The formal replacement of z by e

j?

yields:

X (? ) ?

1 1? e
j? 0

e

? j?

(

?0 : dominant frequency)

4. Geometric interpretation of frequency spectrum

A system with a zero z ? z1 and a pole

z ? p1 ,

1 ? z1z z ? z1 X ( z) ? ? ?1 z ? p1 1 ? p1z

?1

e j? ? z1 ? X (? ) ? j? e ? p1

?

? | e j? ? z1 | ?| X (? ) |? j? | e ? p1 | ? ?arg( X (? )) ? arg(e j? ? z ) ? arg(e j? ? p ) 1 1 ?

peak

dip

As

e

j?

moves around the unit circle, the denominator(分母)
j?

distance and numerator（分子） distance will vary.

e moves to the point with phase ? ? ?1 ? arg( p1 ) , it will cause a peak in | X (? ) | .
When When

angle

e

j?

moves

to

the

point

with .

phase

angle

? ? ?1 ? arg( z1 ) , it will cause a dip in | X (? ) |

Approximating the “spectrum” of

x ( n) ? e

j?0 n

u ( n) :

Suppose Pole:

x(n) ? r0 e
n

j?0n

u ( n)

（ r0 ? 1 ）

z＝ r0 e

j?0

Spectrum:

1 e X (? ) ? ? j? j?0 ? j? j?0 1 ? r0e e e ? r0e
| X (? ) |? | e j? 1 j?0 ? r0e |

j?

Magnitude spectrum:

When If

? ? ?0 ,

| e j? ? r0 e j?0 | decreases , | X (? ) | increases.

r0 －>1, then | X (?0 ) |? ?

5. Properties of DTFT
◆ For real-valued signals x(n)，

X (? ) ? X (?? )
*

(5.4.8)

?| X (? ) |?| X (??) | ?? ?arg X (?) ? ? arg X (??)

(5.4.9)

The magnitude response is even (偶) in response odd（奇）

? and the phase

For example, x =[1 1 1 1 1 1 1 1 1 1]

◆ y (n) ?

x ( n ) * h ( n ) ――> Y (? ) ? X (? ) H (? )
（5.4.10）

◆ Parseval’s equation

1 ? | x(n) | ? 2? n ? ??
2

?

??? | X (? ) | d?
2

?

(5.4.7)

5.5 Inverse z-Transform

Partial fraction expansion method（部分分式展开法）

N ( z) N ( z) X ( z) ? ? D( z ) (1 ? p1 z ?1 )(1 ? p2 z ?1 )?(1 ? p M z ?1 )
A1 A2 AM ? ? ?? ?1 ?1 1 ? p1 z 1 ? p2 z 1 ? pM z ?1
（5.5.1）

pi (i ? 1 ~ M ) ：zeros of the denominator polynomial D(z) ，
or the poles of X (z )

According to the degrees of N(z) and D(z), the representation differs slightly:
（1）The degree of N(z) is strictly less than M，

A1 A2 AM X ( z) ? ? ?? ?1 ?1 ?1 1 ? p1z 1 ? p2 z 1 ? pM z
Ai ? [(1 ? pi z ?1 ) X ( z )] | z ? pi （5.5.2） the coefficients

（2）The degree of N(z) is exactly equal to M，

A1 A2 AM X ( z ) ? A0 ? ? ?? ?1 ?1 1 ? p1 z 1 ? p2 z 1 ? pM z
（5.5.3）

A0 ? X ( z ) | z ?0
Ai (i ? 0)

（5.5.4）

are computed in the same way as (5.5.2)

（3）The degree of N(z) is strictly greater than M，

R( z ) X ( z ) ? Q( z ) ? D( z )
（ ， ） Q(z) ：quotient（商式） R(z) ：remainder（余式）

R( z ) Technique 1： ordinary partial fraction expansion for D( z)
Technique 2： Remove / Restore

1 X ( z) ? N ( z) ? ? N ( z )W ( z ) D( z )

Example 5.5.6： Determine the causal inverse z-transform

6? z X ( z) ? ?2 1 ? 0.25 z

?5

Solution: Technique 1: divide the denominator into the numerator

6 ? 16 z ?1 X ( z ) ? ?16 z ?1 ? 4 z ?3 ? 1 ? 0.25 z ?2 19 13 ?1 ?3 ? ?16 z ? 4 z ? ? ?1 1 ? 0.5 z 1 ? 0.5 z ?1

? x(n) ? ?16? (n ?1) ? 4? (n ? 3)
? 19 ? 0.5 u(n) ? 13(?0.5) u(n)
n n

Technique 2：Remove / Restore

6 ? z ?5 X ( z) ? 1 ? 0.25 z ?2

1 0.5 0.5 W ( z) ? ? ? ?2 ?1 1 ? 0.25 z 1 ? 0.5 z 1 ? 0.5 z ?1

w(n) ? 0.5 ? 0.5 u (n) ? 0.5(?0.5) u (n)
n n

X ( z ) ? (6 ? z ?5 )W ( z ) ? 6W ( z ) ? z ?5W ( z )

? x(n) ? 6w(n) ? w(n ? 5)
? 3(0.5) n u(n) ? 3(?0.5) n u(n) ? 0.5(0.5) n?5 u(n ? 5) ? 0.5(?0.5) n?5 u(n ? 5)

For X(z) with real-valued coefficients, the poles are real or complex conjugate pairs（复共轭对）.
* A1 A1 A2 X ( z) ? ? ? ?? ?1 * ?1 ?1 1 ? p1z 1 ? p1 z 1 ? p2 z

（p.211,bottom）

x(n) ? A p u (n) ? A p u (n) ? A2 p u (n) ? ...
n 1 1 * *n 1 1 n 2 n ? 2 Re[ A1 p1n ]u (n) ? A2 p2 u (n) ? ...

Problems： 5-8 (b) (c) (e)