MIT 信号与系统 Lecture 18

麻省理工大学 信号与系统课程 课件MIT Signal & System

6.003: Signals and Systems 6.003: Signals and Systems

Discrete-Time Frequency Representations

Lecture 18

April 13, 2010

April 13, 2010

Historical Perspective

Broad range of CT signal-processing problems: audio

– radio (noise/static reduction, automatic

gain control, etc.) – telephone (equalizers, echo-suppression, etc.) – hi- (bass, treble, loudness, etc.) television (brightness, tint, etc.)

radar and sonar (sensitivity, noise suppression, object detection)

... Increasing important applications of DT signal processing: MP3 JPEG MPEG MRI

...

Signal Processing: Electronic

The development of low-cost electronics enhanced our ability to alter the natural frequency responses of systems.

110 Magnitude (dB) 100 90 80 70

10

10

Frequency (Hz)

10

3

10

4

10

5

Eight drivers faced the wall; one pointed faced the listener. Electronic “equalizer” compensates for limited frequency response.

麻省理工大学 信号与系统课程 课件MIT Signal & System

6.003: Signals and Systems

Lecture 18

April 13, 2010

DT Fourier Series and Frequency Response

Today: frequency representations for DT signals and systems.

Rational System Functions

A system described by a linear di erence equation with constant

coe cients → system function that is a ratio of polynomials in z. Example:

y[n 2]+3y[n 1]+4y[n]=2x[n 2]+7x[n 1]+8x[n] N(z)

=H(z)=≡

2z 2 +7z 1 +8

2+7z +8z2

麻省理工大学 信号与系统课程 课件MIT Signal & System

6.003: Signals and Systems

DT Vector Diagrams

Value of H(z) at z = z0 can be determined by combining the contri­butions of the vectors associated with each of the poles and zeros.

(z0 q0)(z0 q1)(z0 q2) ··· H(z0)=K

0 00 10 2

The magnitude is determined by the product of the magnitudes.

|H(z0)|= |K|

|(z0 q0)||(z0 q1)||(z0 q2)|···0 00 10 2

Lecture 18

DT Frequency Response

Response to eternal sinusoids.

Let x[n] = cosΩ0n (for all time):

1 1 nn

x[n]=ejΩ0n + e jΩ0n = z0+ z1

where z0 = ejΩ0 and z1 = e jΩ0 .

April 13, 2010

The angle is determined by the sum of the angles.

∠H(z0)=∠ K + ∠(z0 q0)+ ∠(z0 q1)+ ··· ∠(z0 p0) ∠(z0 p1) ···

The response to a sum is the sum of the responses:

1 nn

y[n]=H(z0) z0 + H(z1) z1

1

H(ejΩ0 ) ejΩ0n + H(e jΩ0 ) e jΩ0n = Conjugate Symmetry

For physical systems, the complex conjugate of H(ejΩ) is H(e jΩ). The system function is the Z transform of the unit-sample response:

∞

n

H(z)= h[n]z

n= ∞

DT Frequency Response

Response to eternal sinusoids.

Let x[n] = cosΩ0n (for all time), which can be written as

1

x[n]=ejΩ0n + e jΩ0n .

Then

1

jΩ0 )e jΩ0n y[n]=H(ejΩ0 )ejΩ0n + H(e

=Re H(ejΩ0 )ejΩ0n

jΩ0 j∠H(e jΩ0 )jΩ0n

)|ee=Re |H(e

jΩ0

= |H(ejΩ0 )|Re ejΩ0n+j∠H(e )

y[n]= H(ejΩ0 ) cos Ω0n + ∠H(ejΩ0 )

where h[n] is a real-valued function of n for physical systems.

H(e

jΩ

)=

∞

h[n]e jΩn

h[n]ejΩn ≡ H(ejΩ)

H(e jΩ)=

n= ∞

∞

n= ∞

麻省理工大学 信号与系统课程 课件MIT Signal & System

6.003: Signals and Systems

Periodicity of DT Frequency Responses

DT frequency responses are periodic functions of Ω, with period 2π. If Ω2 =Ω1 +2πk where k is an integer then

H(ejΩ2 )=H(ej(Ω1+2πk))=H(ejΩ1 ej2πk)=H(ejΩ1 )

The periodicity of H(ejΩ)results because H(ejΩ)is a function of ejΩ , which is itself periodic in Ω. Thus DT complex exponentials have many “aliases.”

ejΩ2 =ej(Ω1+2πk) =ejΩ1 ej2πk =ejΩ1

Because of this aliasing, there is a “highest” DT frequency: Ω=π.

Lecture 18

Check Yourself

April 13, 2010

What kind of ltering corresponds to the following?

z-plane

1. high pass 3. band pass

5. none of above

2. low pass

4. band stop (notch)

DT Fourier Series

DT Fourier series represent DT signals in terms of the amplitudes and phases of harmonic components.

x[n]= akejkΩ0n The period N of all harmonic components is the same.

DT Fourier Series

DT Fourier series represent DT signals in terms of the amplitudes and phases of harmonic components.

N 1

akejkΩ0n x[n]=x[n +N]=

k=0

;Ω0 =

2π

N equations (one for each point in time n)inN unknowns (ak).

Example: N =4

j 0·0 x[0]e0·1 x[1] ej = x[2] ej 0·2

x[3] ej 0·3

ej1·0

ej1·1 ej1·2 ej1·3

ej2·0 ej2·1 ej2·2 ej2·3

a0 ej 3·0

3·1 j e a1 3·2 j a2 e3·3 j a3 e

麻省理工大学 信号与系统课程 课件MIT Signal & System

6.003: Signals and Systems

DT Fourier Series

DT Fourier series represent DT signals in terms of the amplitudes and phases of harmonic components.

N 1

akejkΩ0n x[n]=x [n+ N]=

k=0

Lecture 18

DT Fourier Series

April 13, 2010

Solving these equations is simple because these complex exponentials

are orthogonal to each other.

N 1 n=0

;Ω0 =

2π

ejΩ0kn e jΩ0ln =

N 1 n=0

ejΩ0(k l)n

; k = l

=0 ; k =l

N equations (one for each point in time n)inn unknowns (ak). Example: N =4

1 x[0]

x[1] 1

= x[2] 1 x[3] 1

1 1 1 a0

j 1 j a1

1 1 1 a2 j 1 j a3

=

N

jΩ0(k l)N 1 e0= Nδ[k l]

DT Fourier Series

We can use the orthogonality property of these complex exponentials to sift out the Fourier series coe cients, one at a time.

N 1 k=0

DT Fourier Series

Since both x[n] and ak are periodic in N, the sums can be taken over

any N successive indices.

Notation. If f[n] is periodic in N, then

N 1 n=0

N N+1

f[n]= f[n]= f[n]=· ··= f[n]

n=1

n=2

n=<N>

Assume x[n]=

akejkΩ0n

Multiply both sides by the complex conjugate of the lth harmonic,

and sum over time.

N 1 n=0

x[n]e jlΩ0n =

=

N 1 1 N n=0 k=0

N 1 k=0

akejkΩ0n e jlΩ0n =

N 1 k=0

ak

N 1 n=0

ejkΩ0n e jlΩ0n

DT Fourier Series ak = ak+N =

1

x[n]e jΩ0n ;Ω0 =akejkΩ0n

2π

(“analysis” equation) (“synthesis” equation)

akNδ[k l]=N al

n=<N>

x[n]= x[n+ N]=

N 1

1

x[n]e jkΩ0n ak =

n=0

k=<N>

DT Fourier Series

DT Fourier series have simple matrix interpretations.

x[n]=x [n+4] =

akejkΩ0n =

k=<4>

Discrete-Time Frequency Representations

Similarities and di erences between CT and DT.

akjkn

DT frequency response

vector diagrams (similar to CT)

frequency response on unit circle in z-plane (jω axis in CT) DT Fourier series

represent signal as sum of harmonics (similar to CT) nite number of periodic harmonics (unlike CT) nite sum (unlike CT)

x[0] 11 x[1] 1 j =

x[2] 1 1x[3] 1 j ak = ak+4 =

k=<4>

akejkn =

11 a0

1 j a1

1 1 a2

a3 1 j

k=<4>

1 1 jkn 1

x[n]e jkΩ0n =e =x[n]j kn

n=<4>

a0 11 a 1 j 1 =

a2 1 1

1 j a3

x[0] 1 1

1 j x[1]

1 1 x[2]

x[3] 1 j

n=<4>n=<4>

These matrices are inverses of each other.

麻省理工大学 信号与系统课程 课件MIT Signal & System

MIT OpenCourseWare6.003 Signals and Systems

Spring 2010

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