数字信号处理 DSP 英文版课件4.0

w1(n), w2(n): contents of the two registers
At time n:
Example 4.1.5： An IIR filter h(n) 0.75 u (n) . Using convolution, derive
n
closed-form expressions(闭合形式表示) for the output signal y(n) when the input is: （a）A unit step(单位阶跃) x(n)=u(n) （b）An alternating step x(n) (1) u (n) (正负交替的阶跃)
Algorithmic form:
for each input sample x，do：
y：= w1
w1：＝x
The delay register is typically initialized (初始化) to zero： w1(0)=0.
double delay：
I/O equation: y(n)=x(n-2)
State-space representation：
y(n)=w1(n) (output equation) w1 (n+1)=x (n) (state updating equation) w1(n)： content of the register or internal state at time n
4.1.6 Flip-and-Slide Form
Flip-and-Slide Form of convolution:
h(n) is flipped(反转) around and slid (slide平移) over the input sequence x(n). y(n) is obtained by computing the dot product(点积) of flipped h with the input samples aligned(对准) below it.
ydc h(m)
m
(4.1.24)
Example 4.1.7：An integrator（积分器）-like FIR filter：
y(n) G[ x(n) x(n 1) ... x(n 14)] ，
G, or h(n) 0, 0 n 14 otherwise
Direct form
y ( n)
min( n , M )
m max( 0,n L 1)
h( m) x ( n m)
n
(4.1.16)
可分阶段写为： （p.136）
●
0 n M （ input-on） y (n)
m 0
h ( m) x ( n m )
M
●
n
（c）A square pulse of duration L=25, x(n)=u(n)-u(n-25) (持续时 长 25 点的方脉冲) In each case, determine the steady-state response of the filter.
Solution: （a） x(n) u (n)
M n L 1 （steady state）
y ( n ) h ( m) x ( n m )
m 0
（4.1.23）
●
L 1 n L 1 M (input-off) y (n)
m n L 1
h ( m) x ( n m )
M
(4.1.23) can be considered as the generic I/O equation for FIR filters. But, for programming purposes, one must work with (4.1.16) which does not let the indices exceed the array bounds. （p.137） 稳态阶段的 y (n) h(m) x(n m) (4.1.23)可作为通用式子，
y(n) 4
（b） x(n) (1) u (n)
n
做法与（a）类似 （c） x(n) u(n) u(n 25) 响应中只有 input-on 和 input-off 暂态阶段。 当 n=0~24, y (n)
m 0 n
x(n m)h(m) (0.75) m
steady-state (稳态）:
yn h0 xn h1 xn 1 ... hM xn M
M+1 items
input-off transients:
The last M outputs after the input has been turned off
4.1.7 Transient and Steady State Behavior
Three subranges of y(n):
0 n M （ input-on transients）
M n L 1 （steady state） L 1 n L 1 M (input-off transients)
If Lx<Lh, the steady-state range does not exist－ the input is too short to exhibit steady behavior .
Basic building blocks: adder, multiplier, delay
4.2.1 Pure Delays singlt each time instant n, two steps are carried out：
(a) the current content w1(n) = x(n-1) is clocked out to the output，y(n)=w1(n)； (b) the current input x(n) gets stored in the register, where it will be held for one sampling instant and become the output at the next time (n+1)： w1 (n+1)=x (n).
m 0 M
但编程时应用 y ( n)
min( n , M )
m max( 0 ,n L 1)
h( m) x ( n m)
(4.1.16)避免下标出错。
4.1.8 Convolution of Infinite Sequences x 和 h 中有一个或均为无限长时，无限长 y(n) (0 n ) 的 暂稳态情况： M= , L< , input-on transients and input-off transients (Lx<Lh, the input is too short to exhibit steady behavior) M< , L= , input-on transients and steady state M= , L= , input-on transients. The steady state response is the limit of y(n) for n―>
input-on transients（暂态）of y(n): The first M output samples when the input is turned on. Less than M+1 input samples are used for obtaining y(n), or the input sequence is assumed to be padding zeros (补零) to compute (M+1)-dimensional dot product.
G=0.1时，稳态输出值＝DC gain
ydc h(m) G 1.5
m 0 m 0
14
14
Example 4.1.8：x(n) 同上例
ban a) h( n) 0
14
for
0 n 14
,
M=14
otherwise
ydc h(m) 1 a M 1 0.987
h(n) 0.75 n u (n)
x is flipped
y(n) h(m) x(n m) 0.75m u(m)u(n m)
m
n
m
1 0.75 n 1 0.75 m 4 3(0.75) n 1 0.75 m 0
(n 0)
因 h(n)和 x(n)无限长，y(n)的稳态阶段出现在 n ，即
m 0
b) differentiator （微分器） h=[0.2, -1, 2, -2, 1, -0.2], M=5
ydc
m0
h ( m) 0
5
Example 4.1.8 (a):
Example 4.1.8 (b):
4.1.10 Overlap-Add Block Convolution Method
m 0
n
n
当 n 25 y (n)
m n 24
(0.75) m
4.1.9 Programming Considerations
DC gain: The steady-state value of the output of a filter when the input remains constant for a long period of time. For a unity input, the DC gain is: