DSP-第6章.ppt

1、Digital Signal Processing,Chapter 6 Structures for Discrete-Time Systems,2,6.0 Introduction,Several ways to describe discrete-time systems: Impulse responses in time domain. Difference equations in time domain. The z transforms in complex frequency domain (transfer functions). Fourier transforms in

2、frequency domain (Frequency response). H(ej) = |H(ej)|ej(),3,6.1 Description of the Digital Filter Structures,Its difference equations in time domain is,因此，网络结构表示一定的运算结构，不同结构所需要的存储单元以及运算次数不同，前者影响结构复杂性，后者影响运算速度。,4,6.1 Description of the Digital Filter Structures,Three basic elements to implement digi

3、tal filters: Delay Multiplier Adder Block diagram（方框图）representation of three basic elements.,5,6.1 Description of the Digital Filter Structures,Signal flowgraph（信号流图）representation of three basic elements.,6,6.1 Description of the Digital Filter Structures,Two classes of digital filters: Finite-dur

4、ation impulse response filters or nonrecursive filters. Its transfer functions are of the polynomial form. Infinite-duration impulse response filters or recursive filters. Its transfer functions are of the rational polynomial form.,7,6.3 Basic structures for IIR digital filters,6.3.1 Direct form I T

5、he transfer function of a recursive filter is given by And the difference equations in time domain is In general, M N.,8,6.3.1 Direct forms I,9,6.3.1 Direct forms I,Direct forms I structure for IIR digital filters,10,6.3.2 Direct forms II,11,6.3.2 Direct forms II,Direct forms II structure for IIR di

6、gital filters,12,Comparison of the two types,Direct forms I,Direct forms II,13,Example 1,Compute H(z) from the following signal flowgraph. Solution:,14,6.3.3 Cascade form,Writing the numerator and denominator polynomials of H(z) as products of second-order factors, respectively, we have that,15,6.3.

7、4 Parallel form,H(z) can also be expressed as an addition of second-order partial-fractions, such that,16,Example 2,Figure the signal flowgraph of the following system by the direct form (type I and II), cascade form and parallel form. Solution:,17,Example 2,18,Example 2,19,Example 3,Determine the t

8、ransfer function of the system below:,20,6.5 Basic structures for FIR digital filters,The difference equation of FIR filters,21,6.5.1 Direct form,22,6.5.1 Direct form,Transposed direct form（直接型结构的转置）,23,Example 4,Compute the transfer function given by the signal flowgraph and the direct form of H(z)

9、.,x(n),y(n),h(0),h(1),h(2),h(3),h(4),h(5),h(6),h(7),h(8),24,Example 4,25,6.5.2 Cascade form,Writing H(z) as a product of second-order factors, we get that,2011.12.7,26,6.5.3 Linear-phase forms（线性相位型）,An important subclass of FIR digital filters is the one that includes linear-phase filters, that is

10、and the frequency response has the following form,27,6.5.3 Linear-phase forms,where b(n) is the inverse Fourier transform of B(), and Since B() is real, So,28,6.5.3 Linear-phase forms,In the common case where all the filter coefficients are real, so If h(n) is causal, that is h(n) = 0, for n 2. So,

11、This equation shows that the h(n) of a linear-phase filter is symmetric or antisymmetric about M/2.,29,6.5.3 Linear-phase forms,30,6.5.3 Linear-phase forms: type I,31,6.5.3 Linear-phase forms: type I,32,6.5.3 Linear-phase forms: type II,33,6.5.3 Linear-phase forms: type II,34,6.5.3 Linear-phase form

12、s: type III,35,6.5.3 Linear-phase forms: type III,36,6.5.3 Linear-phase forms: type IV,37,6.5.3 Linear-phase forms: type IV,38,Example 5,Draw the signal flow-gragh of the direct form and linear-phase form for the FIR system. Solution: direct form,39,Example 5,Linear-phase form,40,6.5.3 Linear-phase

13、forms,Clearly, the linear-phase form structure requires about 50% fewer multiplications than that of the direct forms.,41,Digital network analysis,The analysis of digital networks is realized through the signal flow graph representation. A digital network consists of three devices: delays, multipliers and adders. A signal flow graph is a network composed of directed branches and nodes. A branch delays a signal or multiplies it by a coefficient. The output value of each node is determined by the sum of all other nodes entering the node through branches.,42,Example 6,43,Example 7,Determine