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