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上海交通大学:《数字信号处理 Digital Signal Processing(B)》教学资源_Handouts_ch6_Transfer Functions

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上海交通大学:《数字信号处理 Digital Signal Processing(B)》教学资源_Handouts_ch6_Transfer Functions
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Introduction To Digital Signal Processing 主讲:张君 上克大学

Introduction To Digital Signal Processing 主讲:张君

Equivalent descriptions of Digital Filters Introduction to Digital Signal Processing--Transfer Functions Transfer function Frequency response Block diagram representation Sample processing algorithm Impulse response sequence I/O difference equation Impulse responsen 1/O convolutional equation block processing h(n) I/O difference Transfer function pole/zero Equation(s) H(Z) pattern filter design method frequency response block-diagram filter design +sample processing specifications H() realization 上浒充通大粤

Introduction to Digital Signal Processing—— Transfer Functions Equivalent descriptions of Digital Filters  Transfer function  Frequency response  Block diagram representation  Sample processing algorithm  Impulse response sequence  I/O difference equation Impulse responsen h(n) Transfer function H(z) I/O convolutional equation pole/zero pattern I/O difference Equation(s) frequency response H(ω) block-diagram realization filter design method block processing filter design sample processing specifications

Transfer Function (direct form) Introduction to Digital Signal Processing--Transfer Functions 7.5 H(z)= 5+2z1 1-0.8z=-2.5+ h(n)=-2.5δ(n)+7.5(0.8)"u(n) -0.8z1 x() y(n) H(z)=0.8zH(z)+5+2z1 u Z h(n)=0.8h(n-1)+56(n)+2δ(n-1) w1(m)=x(n-1) v1)=x(-1) 0.8 y(n)=0.8y(n-1)+5x(n)+2x(n-1) y(m)=0.8%(m)+5x(m)+21(m) (n+1)=x(m) M(n+1)=y(m) 上浒充通大粤

Introduction to Digital Signal Processing—— Transfer Functions Transfer Function (direct form) 1 1 1 1 0.8 7.5 2.5 1 0.8 5 2 ( )           z z z H z h(n) 2.5 (n) 7.5(0.8) u(n) n     ( ) 0.8 ( 1) 5 ( ) 2 ( 1) ( ) 0.8 ( ) 5 2 1 1            h n h n n n H z z H z z   y(n)  0.8y(n 1) 5x(n)  2x(n 1)

Transfer Function (frequency response Introduction to Digital Signal Processing--Transfer Functions H(z)= 5+2z15(1+0.4z) 51+0.4eo) 1-0.8z1 Ha)= 1-0.8z 1-0.8ejo 1-ae-ja=1-2acos w+a2 5√1+0.8cos0+0.16 H(O)= 1-1.6c0s0+0.64 H() ∑bWn=la∑ Ho1e-HeL =35 35 ro-s-引 5-235 0 35/21 上降充通大学

Introduction to Digital Signal Processing—— Transfer Functions Transfer Function (frequency response)    j j e e H z z z z H z                 1 0.8 5(1 0.4 ) ( ) 1 0.8 5(1 0.4 ) 1 0.8 5 2 ( ) 1 1 1 1 1 1.6cos 0.64 5 1 0.8cos 0.16 ( )        H  2 2 2 [ ] [ ] n n y n x n        

parallel form Introduction to Digital Signal Processing--Transfer Functions 5+2z1 7.5 H(2)= -0.8z=-25+ 1-0.8z1 wo(=0.8w1(+7.5x(m) 2.5 y(n)=wo(n)-2.5x(n) w(n) x(n) ·y(m wi(n+1)=wo(n) 7.5 wo w(n-1) 上游充通大¥

Introduction to Digital Signal Processing—— Transfer Functions parallel form 1 1 1 1 0.8 7.5 2.5 1 0.8 5 2 ( )           z z z H z

canonical form realization Introduction to Digital Signal Processing--Transfer Functions Y(=H(X()=1-08) W()=1 1-0.82a) W(2)=0.8zw(a)+X(z) Y(z)=(5+2z1)w(z) y(m)=5w(m)+2w(n-1) 1wm)=0.81n-1)+x(n) y(m)=5w(n))+2(n-1) w(n) x(n) y(n) wo(n)=0.8w1(n)+x(n) y()-5wo()+2w1() w1(n+1)=wo(n) w(n-1)w1 0.8N 上游充通大¥

Introduction to Digital Signal Processing—— Transfer Functions canonical form realization

canonical form realization Introduction to Digital Signal Processing--Transfer Functions x(n) y(n) w1(n) 0.8 y(n)=w1(n)+5x(n) w1(n+1)=2x(n)+0.8y(n) 上浒充通大

Introduction to Digital Signal Processing—— Transfer Functions canonical form realization

Transfer Function Introduction to Digital Signal Processing--Transfer Functions H(z)= b0+b121+b222++b221 1+a2+a2z2+.+QMz-M amo会2w yn=-a1yn-1-a2Yn-2-..-aMYn-M+boXn+bixn-1+b2Xn-2+.+bLXn-L 上游充通大学

Introduction to Digital Signal Processing—— Transfer Functions Transfer Function

Example Introduction to Digital Signal Processing--Transfer Functions Example 6.2.1:Determine the transfer function of the following third-order FIR filter with im- pulse response: h=[1,6,11,6] y(n)=x(n)+6x(n-1)+11x(n-2)+6x(n-3) H(z)-ho+h1z1+h2Z2+h3z3=1+6z1+11z2+6z3 H(z)=(1+z1)(1+5z1+6z-2)=(1+z1)(1+2z1)(1+3z1) H(w)=1+6ejw+11e-2jw+6e-3fm=(1+ejm)(1+2ejo)(1+3ejo) O=zeros z-plane (I 24 exact zero 0 -3 0 0 W3 unit circle 上游充通大¥

Example Introduction to Digital Signal Processing—— Transfer Functions

Sinusoidal response-steady state response Introduction to Digital Signal Processing--Transfer Functions y(m)=∑h(m)x(m-m)=∑h(m)eo,a-m)=eo,"Σh(m)ejo,m m m m y(n)=H(@)ej@om X(o)=2πδ(0-o) Y(o)=H(o)X(o)=H(o)2πδ(o-oo)=2πH(oo)δ(o-o0) ym)=2元-Y(do=2元2aHo,8a-o,emdo Ho)=|H()eicHH()otjarge 上游充通大¥

Introduction to Digital Signal Processing—— Transfer Functions Sinusoidal response-steady state response

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