Post on 12-Oct-2020
CLTI Impulse Response (4B) 1 Young Won Lim1/20/15
Impulse Matching (N > M)
dN h
d tN + aM +1
dN−M−1 h
d tN−M−1
K1δ(M )(t )
+ aN−1d hd t
+ a1
dN−1 h
d tN−1
+ K 2δ(M−1)
(t)
+⋯
+⋯
+aN h =+ aM
dN−M h
d tN−M
b0
dMδ
d tM
+⋯
+ bM−1d δd t
+ bM δ+⋯
impulse related terms in the left hand side do not have an impulse or its derivatives
yh(t )u( t) + m0δ(t )
yh(t )u( t) + m0δ(t ) + m1˙δ (t ) +⋯+ mM−N δ
(M−N )(t )
yh(t )u( t) (N>M )
(N=M )
(N <M )
h( t ) =
h( t) =
h( t ) =
+ K M+1δ(t ) yh(t )u( t)
h(N−M−1)(t ) = yh(t )u( t)
M+1
Young Won Lim1/20/15
●
CLTI Impulse Response (4B)
Young Won Lim1/20/15
Copyright (c) 2011 - 2015 Young W. Lim.
Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled "GNU Free Documentation License".
Please send corrections (or suggestions) to youngwlim@hotmail.com.
This document was produced by using OpenOffice and Octave.
CLTI Impulse Response (4B) 4 Young Won Lim1/20/15
Solutions of Differential Equations
All the derivatives of h(t) up to N must match a corresponding derivatives of the impulse up to M at time t=0
Case 1 m<n
The linear combination of all the derivatives of h(t) must add to zero for any time t≠0
yh(t)u(t) is such a function
yh(t) is the homogeneous solution of the
differential equation
The derivatives of the yh(t)u(t) provide all
the singularity functions necessary to match the impulse and derivatives of the impulse on the right side and no other terms need to be added
Case 1 m=n
Need to add an impulse term K0δ(t).. and
solve for K0 by matching coefficients of
impulses on both sides
Case 1 m>n
The n-th derivative of the function we add to y
h(t)u(t) must have a term that matches the
m-th derivative of the unit impulse. Must add
Km−n um−n(t ) + K m−n−1 um−n−1(t) + ⋯+ K 0 u0 (t)
d N y ( t )
d tN +a1
d N−1 y (t )
d t N−1 +⋯+aN−1
d y ( t)d t
+aN y ( t ) = b0
dM x (t )
d tM +b1
dM−1 x (t )
d t M−1 +⋯+bM−1
d x ( t)d t
+bM x (t )
= K m−nδ(m−n)
(t) + Km−n−1δ(m−n−1)
( t) +⋯+ K 0δ(t)
CLTI Impulse Response (4B) 5 Young Won Lim1/20/15
Impulse Response Requirements – A
The linear combination of all the derivatives of h(t) must add to zero for any time t≠0
d N h (t )
d tN +a1
dN−1 h( t )
d t N−1 +⋯+aN−1
d h(t )d t
+aN h(t ) = b0
d M δ(t )
d tM +b1
dM−1δ( t )
d tM−1 +⋯+bM− 1
d δ( t)d t
+bM δ( t)
h( t) ⇐ yh(t)u (t ) when t≠0
The requirements of an impulse response h(t)
this is such a function
for t<0, u(t) = 0
for t>0 yh(N )(t) + a1 y h
(N−1)(t )⋯ + aN y h(t ) = 0derivatives of {y h⋅u} producederivatives of δ when t=0
all the derivatives of δ(t) exists only t=0. It is zero for any time t≠0
u(t) = 0 y h(N )(t) + a1 yh
(N−1)(t) ⋯ + aN yh (t) = 0
h(N )(t ) + a1 h(N−1)(t) ⋯ + aN h(t) = 0 (t ≠ 0)
yh(t) homogeneous solution
CLTI Impulse Response (4B) 6 Young Won Lim1/20/15
Impulse Response Requirements – B
All the derivatives of h(t) up to n must match corresponding derivatives of the impulse δ(t) up to m at time t=0
The requirements of an impulse response h(t)
yh(t )u( t) + m0δ(t )
yh(t )u( t) + m0δ(t ) + m1˙δ (t ) +⋯+ mM−N δ
(M−N )(t)
yh(t )u( t) (N >M )
(N=M )
(N <M )
h( t) =
h( t) =
h( t) =
d N h (t )
d t N +a1
dN−1 h( t )
d tN−1 +⋯+aN−1
d h(t )d t
+aN h(t ) = b0
d M δ(t )
d tM +b1
dM−1δ( t )
d t M−1 +⋯+bM− 1
d δ( t)d t
+bM δ(t)
all the derivatives of yh(t)·u(t) may not
include all the required the derivatives of
δ(t) at the time t=0.
CLTI Impulse Response (4B) 7 Young Won Lim1/20/15
Derivatives of yh(t)·u(t)
h(t ) = yh( t)u( t)
h = yh u
h(1) = yh(1)u + yh u(1)
h(2) = yh(2)u + 2 yh
(1)u(1) + yh u(2)
yh(0)δ (t )
2 yh(1)(0)δ(t) + yh(0)δ
(1)(t )
h(3) = yh(3 )u + 3 yh
(2)u(1) + 3 yh(1)u(2) + yh u(3) 3 yh
(2)(0)δ (t ) + 3 yh(1)(0)δ(1)(t) + yh(0) δ
(2)(t )
K1δ(N−1)
( t) + K 2δ(N−2)
(t)
⋯ ⋯ ⋯ ⋯ ⋯ ⋯
+ K Nδ(t )+ K N−1δ(1)(t)+⋯
u(i )(t) = δ(i−1)(t )
f (t )δ(t) = f (0)δ(t )
h(N )(t ) =dN
dt N {y h(t)u(t)}
h(t ) = yh( t)u( t)All the derivatives of h(t) up to N incurs the derivatives of an impulse δ(t) up to N-1
CLTI Impulse Response (4B) 8 Young Won Lim1/20/15
Three different h(t)'s
h(t ) = yh( t)u( t)
h(1)(t ) = yh(t )u( t)
h(2)(t ) = yh(t )u( t )
All the derivatives of h(t) up to N incurs the derivatives of an impulse δ(t) up to N-1
All the derivatives of h(t) up to N incurs the derivatives of an impulse δ(t) up to N-2
All the derivatives of h(t) up to N incurs the derivatives of an impulse δ(t) up to N-3
CLTI Impulse Response (4B) 9 Young Won Lim1/20/15
Derivatives of three different h(t)'s
+ K Nδ(t )+ K N−1δ(1)(t)+⋯
+ K N−1δ( t)+⋯
K1δ(N−1)
( t)
K1δ(N−2)
( t)
+ K 2δ(N−2)
(t)
+ K 2δ(N−3)
(t )
+ K N−2δ( t)+⋯K1δ(N−3)
(t)
h(t ) =∫−∞
t
yh(t )u(t ) dt
h( t ) =∬−∞
t
yh(t )u( t) dt dt
h(t ) = yh( t)u( t) h(N )(t )
h(N )(t )
h(N )(t )
CLTI Impulse Response (4B) 10 Young Won Lim1/20/15
Derivatives & Integrals of yh(t)·u(t)
h(N )( t) +a1h(N−1)(t ) +aN h(0)( t)+aN−1 h(1)(t )+aN−2 h(2)( t)+ ⋯
yh uddt{yh u}
d N
dt N {yh u} d2
dt 2 {yh u}dN−1
dt N−1 {yh u}
∫−∞
t
yh u dtyh udN−1
dt N−1 {yh u}ddt{yh u}d N−2
dt N−2 {yh u}
∫−∞
t
∫−∞
t
yh u dt dt∫−∞
t
yh u dtd N−2
dt N−2 {yh u} yh udN−3
dt N−3 { yhu}
(a)
(b)
(c)
(N = M+1)
(N = M+2)
(N = M+3)
(M−N−1= 0)
(M−N−1=−1)
(M−N−1=−2)
CLTI Impulse Response (4B) 11 Young Won Lim1/20/15
Negative powers denote integration
(a)
(b)
(c)
h( t) = yh( t)u( t)
h(1)(t ) = yh(t )u( t )
h(2)( t) = yh(t )u( t)
h( t) =∫−∞
t
yh(t )u( t) dt
h( t) =∬−∞
t
yh(t )u( t) dt dt
h( t) = yh(t )u( t)
g(−1)( t) ≡∫
−∞
t
g (t) dt
g(−2)( t ) ≡∫
−∞
t
∫−∞
t
g(t ) dt dt
g( t )
CLTI Impulse Response (4B) 12 Young Won Lim1/20/15
Derivatives & Integrals of g(t) = yh(t)·u(t)
h(N )( t) +a1h(N−1)(t ) +aN h(0)( t)+aN−1 h(1)(t )+aN−2 h(2)( t)+ ⋯
g(t )g(1)(t )g(N )( t) g(2)(t)g(N−1)( t)
g(−1)(t)g( t)g(N−1)( t) g(1)(t )g(N−2)(t )
g(−2)(t)g(−1)( t)g(N−2)(t ) g(t )g(N−3)(t )
(a)
(b)
(c)
(N = M+1)
(N = M+2)
(N = M+3)
(M−N−1= 0)
(M−N−1=−1)
(M−N−1=−2)
h(t ) = g(M−N−1)(t ) = g (t)
h( t) = g(M−N−1)(t ) = g(−1)( t)
h(t ) = g(M−N−1)(t ) = g(−2)(t )
CLTI Impulse Response (4B) 13 Young Won Lim1/20/15
Derivatives of h(t) – three cases (2)
h(N )( t) +a1h(N−1)(t ) +aN h(0)( t)+aN−1 h(1)(t )+aN−2 h(2)( t)+ ⋯
yh uddt{yh u}
d N
dt N {yh u} d2
dt 2 {yh u}dN−1
dt N−1 {yh u}
∫−∞
t
yh u dtyh udN−1
dt N−1 {yh u}ddt{yh u}d N−2
dt N−2 {yh u}
∫−∞
t
∫−∞
t
yh u dt dt∫−∞
t
yh u dtd N−2
dt N−2 {yh u} yh udN−3
dt N−3 { yhu}
g(t )g(1)(t )g(N )( t) g(2)(t)g(N−1)( t)
∫−∞
t
g (t ) dtg( t)g(N−1)( t) g(1)(t )g(N−2)(t )
∫−∞
t
∫−∞
t
g ( t) dt dt∫−∞
t
g (t ) dtg(N−2)(t ) g(t )g(N−3)(t )
(a)
(b)
(c)
(a)
(b)
(c)
CLTI Impulse Response (4B) 14 Young Won Lim1/20/15
Derivatives of yh(t)·u(t)
d N h (t )
d t N +a1
dN−1 h( t)
d tN−1 +⋯+aN−1
d h(t )d t
+aN h(t ) = b0
d M δ(t )
d tM +b1
dM−1δ(t )
d tM−1 +⋯+bM−1
d δ( t)d t
+bM δ( t)
g(t ) + m0δ(t )
g(t ) + m0δ(t ) + m1˙δ( t) +⋯+ mM−N δ
(M−N )(t )
g(M−N−1)( t) (N >M )
(N=M )
(N <M )
h( t) =
h( t) =
h( t) =
(N = M+1)
(N = M+2)
(N = M+3)
(M−N−1= 0)
(M−N−1=−1)
(M−N−1=−2)
h(t ) = g(M−N−1)(t ) = g (t)
h( t) = g(M−N−1)(t ) = g(−1)( t)
h(t ) = g(M−N−1)(t ) = g(−2)(t )
g( t) = yh( t)u (t )
CLTI Impulse Response (4B) 15 Young Won Lim1/20/15
Impulse Matching (N > M)
dN h
d tN + aN−M−1
dM+1 h
d t M+1
K1δ(N−1)
( t)
+ aN−1d hd t
+ a1
dN−1 h
d tN−1
+ K 2δ(N−2)
(t)
+⋯
+ K Nδ(t )+ K N−Mδ(M )( t)+⋯
+aN h =+ aN−M
dM h
d t M
b0
dMδ
d tM
+⋯
+ bM−1d δd t
+ bM δ+⋯
+ K N−M+1δ(M−1)
( t) +⋯
impulse related terms in the left hand side must have zero coefficients
yh(t )u( t)
M+1
CLTI Impulse Response (4B) 16 Young Won Lim1/20/15
Impulse Matching (N > M)
dN h
d tN + aM +1
dN−M−1 h
d tN−M−1
K1δ(M )(t )
+ aN−1d hd t
+ a1
dN−1 h
d tN−1
+ K 2δ(M−1)
(t)
+⋯
+⋯
+aN h =+ aM
dN−M h
d tN−M
b0
dMδ
d tM
+⋯
+ bM−1d δd t
+ bM δ+⋯
impulse related terms in the left hand side do not have an impulse or its derivatives
yh(t )u( t) + m0δ(t )
yh(t )u( t) + m0δ(t ) + m1˙δ (t ) +⋯+ mM−N δ
(M−N )(t )
yh(t )u( t) (N>M )
(N=M )
(N <M )
h( t ) =
h( t) =
h( t ) =
+ K M+1δ(t ) yh(t )u( t)
h(N−M−1)(t ) = yh(t )u( t)
M+1
CLTI Impulse Response (4B) 17 Young Won Lim1/20/15
Impulse Matching (N > M)
dN h
d tN + aM +1
dN−M−1 h
d tN−M−1
K1δ(M )(t )
+ aN−1d hd t
+ a1
dN−1 h
d tN−1
+ K 2δ(M−1)
(t)
+⋯
+⋯
+aN h =+ aM
dN−M h
d tN−M
b0
dMδ
d tM
+⋯
+ bM−1d δd t
+ bM δ+⋯
impulse related terms in the left hand side do not have an impulse or its derivatives
yh(t )u( t) + m0δ(t )
yh(t )u( t) + m0δ(t ) + m1˙δ (t ) +⋯+ mM−N δ
(M−N )(t )
yh(t )u( t) (N>M )
(N=M )
(N <M )
h( t ) =
h( t) =
h( t ) =
+ K M+1δ(t ) yh(t )u( t)
h(N−M−1)(t ) = yh(t )u( t)
M+1
h(t ) = ∫−∞
t
∫−∞
t
⋯∫−∞
t
yh(t )u (t )dt⋯dt dt
CLTI Impulse Response (4B) 18 Young Won Lim1/20/15
Impulse Matching (N = M)
b0
dMδ
d tM
dN h
d t N + aN−1d hd t
+⋯ +aN h =+ a1
dN−1 h
d tN−1
+ b1
dM−1δ
d tM−1 + bM−1d δd t
+ bM δ+⋯
+ K Nδ(t )+ K M−2δ(M−2)
(t) +⋯+ K M−1δ(M−1)
(t)K M δ(M )(t)
if h(t) = yh(t)u(t)
if h(t) = yh(t)u(t) + δ(t)
yh(t )u( t) + m0δ(t )
yh(t )u( t) + m0δ(t ) + m1˙δ (t ) +⋯+ mM−N δ
(M−N )(t )
yh(t )u( t) (N>M )
(N=M )
(N <M )
h( t ) =
h( t) =
h( t ) =
yh(t )u( t )
M+1
CLTI Impulse Response (4B) 19 Young Won Lim1/20/15
Impulse Matching (N < M)
b0
dMδ
d tM
dN h
d t N + aN−1d hd t
+⋯ +aN h =+ a1
dN−1 h
d tN−1
+ b1
dM−1δ
d tM−1 + bM−1d δd t
+ bM δ+⋯
+ K Nδ(t )+ ⋯+ K M−1δ(M−1)
( t)K M δ(M )(t)
if h(t) = yh(t)u(t)
if h(t) = yh(t)u(t) + δ(t)+δ'(t)
yh(t )u( t) + m0δ(t )
yh(t )u( t) + m0δ(t ) + m1˙δ (t ) +⋯+ mM−N δ
(M−N )(t )
yh(t )u( t) (N>M )
(N=M )
(N <M )
h( t ) =
h( t) =
h( t ) =
+ b2
dM−2δ
d t M−2
+ K M−2δ(M−2)
(t) yh(t )u( t)
CLTI Impulse Response (4B) 20 Young Won Lim1/20/15
Impulse Response Requirements
All the derivatives of h(t) up to n must match a corresponding derivatives of the impulse δ(t) up to m at time t=0
The linear combination of all the derivatives of h(t) must add to zero for any time t≠0
d N h (t )
d tN +a1
dN−1 h( t )
d t N−1 +⋯+aN−1
d h(t )d t
+aN h(t ) = b0
d M δ(t )
d tM +b1
dM−1δ( t )
d tM−1 +⋯+bM− 1
d δ( t)d t
+bM δ( t)
h( t ) ⇐ yh(t)u (t ) when t≠0
The requirements of an impulse response h(t)
yh(t )u( t ) + m0δ(t )
yh(t )u( t) + m0δ(t ) + m1˙δ (t ) +⋯+ mM−N δ
(M−N )(t)
yh(t )u( t )(N>M )
(N=M )
(N <M )
h( t) =
h(t ) =
h( t) =
this is such a function
u(t) = 0 when t<0y h(t )= 0 when t>0
CLTI Impulse Response (4B) 21 Young Won Lim1/20/15
Case Examples
d N h (t )
d tN +a1
dN−1 h( t )
d t N−1 +⋯+aN−1
d h(t )d t
+aN h(t ) = b0
d M δ(t )
d tM +b1
dM−1δ( t )
d tM−1 +⋯+bM− 1
d δ( t)d t
+bM δ( t)
yh(t )u(t ) + m0δ(t )
yh(t )u( t) + m0δ(t ) + m1˙δ (t ) +⋯+ mM−N δ
(M−N )(t)
yh(t )u( t )(N>M )
(N=M )
(N <M )
h( t) =
h(t ) =
h( t) =
dN h(t)
d tN+a1
dN−1 h(t)
d tN−1+⋯+aN−1
d h(t )d t
+aN h(t) = b0
dN−1δ(t )
d tN−1+b1
dN−2δ(t)
d t N−2+⋯+bN−2
d δ(t)d t
+bN−1δ(t)
m0δ( t)
m0δ( t) + m1 δ (t )
dN h(t)
d tN+a1
dN−1 h(t)
d tN−1+⋯+aN−1
d h(t )d t
+aN h(t) = b0
dN δ(t )
d tN+b1
dN−1δ( t)
d tN−1+⋯+bN−1
d δ( t)d t
+bN δ(t)
dN h(t)
d tN+a1
dN−1 h(t)
d tN−1+⋯+aN−1
d h(t )d t
+aN h(t) = b0
dN +1δ( t)
d tN +1+b1
dN δ(t)
d tN+⋯+bN
dδ(t )d t
+bN+1δ(t)
yh( t)u(t )
These must be added to
M = N−1
M = N
M = N+1
CLTI Impulse Response (4B) 22 Young Won Lim1/20/15
Impulse Response and Other System Responses
t=0−
x( t) = δ(t )
t=0+t=0 t=0− t=0+t=0
h(t )
t=0−
x( t) ≠ δ(t ) y (t )
t=0+t=0 t=0− t=0+t=0
h(t )
y (t) = h(t)
impulse or its derivative functions may exist at t=0
may not be continuous but not contain an impulse function at t=0
CLTI Impulse Response (4B) 23 Young Won Lim1/20/15
Interval of Validity
y (t)
t > 0
x (t )
t ≥ 0
h(t ) y (t)x (t )
d N y ( t)
d tN +a1
d N−1 y (t )
d tN−1 +⋯+aN−1
d y ( t)d t
+aN y ( t) = b0
dM x (t )
d t M +b1
dM−1 x (t )
d t M−1 +⋯+bM−1
d x ( t)d t
+bM x (t )
CLTI Impulse Response (4B) 24 Young Won Lim1/20/15
Initial Conditions & Total Response
zero input response
t=0−
x( t) y( t)
t=0+t=0 t=0− t=0+t=0
zero state response+
y (t ) = y zi(t)
because the input has not started yet
t≤0−
y (0−) = y zi (0−)
y (0−) = y zi (0−)
in general,the total responsey(0−) ≠ y (0+
)
y (0−) ≠ y (0+)
possible discontinuity at t = 0
h(t )
CLTI Impulse Response (4B) 25 Young Won Lim1/20/15
Impulse input creates initial conditions
t=0−
δ( t) h(t )
t=0+t=0 t=0− t=0+t=0
δ(0−) = 0 δ(0+) = 0
● generates energy storage● creates nonzero i. c. at t=0+
h(t )
h(N−1)(0−)
⋮
h(1)(0−)
h(0−)
= 0
= 0
= 0
h(N−1)(0)
⋮
h(1)(0)
h(0)
h(N−1)(0+)
⋮
h(1)(0+)
h(0+)
⋮
h(N−2)(0−) = 0 h(N−2)
(0) h(N−2)(0+)
initially at rest
∃ i , ki ≠ 0
= kN−1
= k1
= k 0
⋮
= kN−2
non-zero initial conditions
= f N−1(ki ,δ(i))
= f 1(ki ,δ( i))
= f 0(k i ,δ(i))
⋮
= f N−2(ki ,δ( i))
assumed finite jumps
Impulse matching
CLTI Impulse Response (4B) 26 Young Won Lim1/20/15
Impulse Response h(t)
t=0−
δ( t) h(t )
t=0+t=0 t=0− t=0+t=0
δ(0−) = 0 δ(0+) = 0
char. mode termst≥0+
(t≠0)
at most an impulse b0δ(t )t=0
h(t) = b0δ(t) + char mode terms t≥0
h(t )
y(N−1)(0+)
⋮
y(1)(0+)
y (0+)
y(N−2)(0+)
∃ i , ki ≠ 0
= kN−1
= k1
= k 0
⋮
= k N−2
non-zero initial conditions
h(t) : ZSR to an impulse input
ZIR with the created i.c.
CLTI Impulse Response (4B) 27 Young Won Lim1/20/15
h(t) can have at most a δ(t)
N = M
(DN+a1 DN−1+⋯+aN−1 D+aN )h(t ) = (b0 DN+b1 DN−1+⋯+bN−1 D+bN )δ( t)
δ ' (t ) is included in h(t ) If
then the highest order term
δ(N+ 1)
(t) δ(N )(t) contradiction
h(t) cannot contain δ(i)(t) at all h(t) can contain at most δ(t)
M = N
Q (D) y(t) = P (D) x(t )
dN h
d tN → δ(N+1)(t)
the highest order term
N M (M ≤ N )
only when
CLTI Impulse Response (4B) 28 Young Won Lim1/20/15
Finding h(i)(0)
h(N−1)(0−)
⋮
h(1)(0−)
h(0−)
= 0
= 0
= 0
h(N−1)(0+)
⋮
h(1)(0+)
h(0+)
⋮
h(N−2)(0−) = 0 h(N−2)
(0+)
= kN−1
= k1
= k0
⋮
= k N−2
h(N )(0)
⋮
h(2)(0)
h(1)(0)
h(N−1)(0)
= kN−1δ(t)
= k1δ(t)
= k 0δ( t )
⋮
= kN−2δ(t)
+k 0δ(1)(t )
+ ⋯
+kN−2δ(1)(t)
+k 0δ(N−2)
(t )+kN−3δ(1)(t)
+ ⋯ +k 1δ(N−2)
(t) +k 0δ(N−1)
(t)
h(N )(0)
⋮
h(2)(0)
h(1)(0)
h(N−1)(0)
= kN−1δ(t)
=
⋮
=
+k 0δ(1)(t )
+ ⋯
+kN−2δ(1)(t)
+k 0δ(N−2)
(t )kN−2δ( t)
+ ⋯ +k 1δ(N−2)
(t) +k 0δ(N−1)
(t)
= k 0δ(t )
k 1δ( t )
+k 1δ(N−1)
( t)
⋮
h(N )(0) + a1 h(N−1)(0) +⋯+ aN−1 h(1)(0) + aN h(0) = b0δ(N )(t ) + b1δ
(N−1)( t) +⋯+ bN−1δ(1)(t ) + bN δ(t )
initially at rest assumed finite jumps
Impulse matching
d N y ( t)
d tN +a1
d N−1 y (t )
d tN−1 +⋯+aN−1
d y ( t)d t
+aN y ( t) = b0
dM x (t )
d t M +b1
dM−1 x (t )
d t M−1 +⋯+bM−1
d x ( t)d t
+bM x (t )
CLTI Impulse Response (4B) 29 Young Won Lim1/20/15
h(N−1)(0)
⋮
h(1)(0)
h(0)
h(N−2)(0)
= f N−1(ki ,δ(i))
= f 1(ki ,δ( i))
= f 0(k i ,δ(i))
⋮
= f N−2(ki ,δ( i))
Impulse Matching Method
h(N−1)(0−)
⋮
h(1)(0−)
h(0−)
= 0
= 0
= 0
h(N−1)(0+)
⋮
h(1)(0+)
h(0+)
⋮
h(N−2)(0−) = 0 h(N−2)
(0+)
= kN−1
= k1
= k0
⋮
= k N−2
h(N )(0) + a1 h(N−1)(0) +⋯+ aN−1 h(1)(0) + aN h(0) = b0δ(N )(t ) + b1δ
(N−1)( t) +⋯+ bN−1δ(1)(t ) + bN δ(t )
initially at rest assumed finite jumps
Impulse matching
d N y ( t)
d tN +a1
d N−1 y (t )
d tN−1 +⋯+aN−1
d y ( t)d t
+aN y ( t) = b0
dM x (t )
d t M +b1
dM−1 x (t )
d t M−1 +⋯+bM−1
d x ( t )d t
+bM x (t )
CLTI Impulse Response (4B) 30 Young Won Lim1/20/15
Simplified Impulse Matching
(DN+a1 DN−1+⋯+aN−1 D+aN) ⋅ h(t ) = (bN−M DM+bN− M+1 DM−1+⋯+bN−1 D+bN) ⋅ δ(t )
Q(D)⋅h (t ) = P(D)⋅δ(t )
(DN+a1 DN−1+⋯+aN−1 D+aN) ⋅ h0(t ) = δ(t)
Q(D)⋅h0(t) = δ(t )
h(t) = characteristic mode termst≥0+
A0δ(t)t≥0 h(t) = + characteristic mode terms
Simplified Impulse Matching Method
linear combination of characteristic modeswith the following initial conditions
h0( t)
h(t) = P(D)⋅h0(t)
h(t) = b0δ(t) + [P(D) h0(t)]⋅u( t)
h0(0+ ) = h0
(1)(0+) = h0(2)(0+) ⋯ = h0
(M−2)(0+ ) = 0 h0(M−1)(0+) = 1
CLTI Impulse Response (4B) 31 Young Won Lim1/20/15
Superposition of inputs
h(N )( t) + a1h(N−1)(t ) +⋯+ aN−1 h(1)(t ) + a N h (t ) = b0δ(N )(t ) + b1δ
(N−1)(t ) +⋯+ bN−1δ(1)(t ) + bNδ(t )
h(N )( t) + a1h(N−1)(t ) +⋯+ aN−1 h(1)(t ) + a N h (t ) = δ(N )(t )
h(N )( t) + a1h(N−1)(t ) +⋯+ aN−1 h(1)(t ) + a N h (t ) = δ(N−1)(t )
h(N )( t) + a1h(N−1)(t ) +⋯+ aN−1 h(1)(t ) + a N h (t ) = δ(1)( t)
h(N )( t) + a1h(N−1)(t ) +⋯+ aN−1 h(1)(t ) + a N h (t ) = δ( t)
hN ( t)
hN−1( t)
h0(t)
h1(t )
= h0(N )(t )
= h0(N−1)(t)
= h0(1)(t )
= h0(0 )(t )
⋮⋮⋮⋮⋮⋮⋮
h( t) = b0 hN ( t) + b1 hN−1( t ) + ⋯ + bN−1 h1(t ) + bN h0(t )
= b0 h0(N )( t) + b1 h0
(N−1)(t ) +⋯+ bN−1 h0(1)(t ) + bN h0
(0 )( t )
CLTI Impulse Response (4B) 32 Young Won Lim1/20/15
New Initial Condition created by δ(t)
h0(N )( t) + a1h0
(N−1)(t ) + a2 h0(N−2)( t) + ⋯ + aN−1 h0
(1)(t ) + aN h0( t) = δ(t )
δ(t) u( t)
no jump discontinuity is allowed at t = 0
unit jump discontinuity at t = 0
h0(N )(t) = δ(t )
t u (t) 1(N−2) ! t(N −2) u (t) 1
(N−1)! t(N−1) u(t)
∫⋅d t ∫⋅d t ∫⋅d t ∫⋅d t
yn(N−2)(0+) = ⋯ yn
(2)(0+ ) = yn(1)(0+) = yn(0
+ ) = 0h0(N−1)(0+) = 1
CLTI Impulse Response (4B) 33 Young Won Lim1/20/15
Simplified Impulse Matching Method (1)
(DN+ a1 DN−1
+⋯+ aN−1 D+ aN ) ⋅ y (t) = (bN−M DM+ bN−M+ 1 DM−1
+⋯+ bN−1 D+ bN ) ⋅ x( t)
d N y ( t)
d tN +a1
d N−1 y (t )
d tN−1 +⋯+aN−1
d y ( t)d t
+aN y (t ) = b0
dM x (t )
d tM +b1
dM−1 x (t )
d tM−1 +⋯+bN−1
d x( t)d t
+bN x (t )
(DN+a1 DN−1
+⋯+aN−1 D+aN ) ⋅ y (t ) = x (t )
d N y (t )
d tN +a1
d N−1 y (t )
d t N−1 +⋯+aN−1
d y ( t)d t
+aN y (t ) = x (t )
Derived System S
Base System S0
linear combination of characteristic modeswith the following initial conditions
yn(t )
shares the same characteristic modes
yn(N−2)(0+) = ⋯ = yn ' (0+) = yn(0
+) = 0yn(N−1)(0+) = 1
h( t ) = b0δ( t) + [P(D) yn(t )]u(t)
CLTI Impulse Response (4B) 34 Young Won Lim1/20/15
Simplified Impulse Matching Method (2)
linear combination of characteristic modeswith the following initial conditions
h( t) = b0δ( t) + [P(D) yn(t )]u( t)
yn(t )
Q(D) y (t) = P(D)x (t)
Q(D)w( t) = x ( t)
Q(D) yn(t ) = δ(t)(DN
+a1 DN−1+⋯+aN−1 D+aN )w (t ) = x (t )
w(N )(t ) + a1 w(N−1)(t) +⋯+ aN−1 w(1)(t) + w (t ) = x (t )
Q(D)w( t) = x (t)
Q(D)P(D)w ( t) = P(D)x (t )
Q(D) y (t) = P(D) x (t)y (t ) = P(D )w (t )
h(t) = P(D) yn( t)
(DN+ a1 DN−1
+⋯+ aN−1 D+ a N ) yn(t) = δ(t)
yn(N )(t) + a1 yn
(N−1)(t) + ⋯+ a N−1 yn
(1)(t) + yn( t) = δ(t)
Base System S0
yn(N−2)(0+) = ⋯ = yn ' (0+) = yn(0
+) = 0yn(N−1)(0+) = 1
Derived System S
CLTI Impulse Response (4B) 35 Young Won Lim1/20/15
Simplified Impulse Matching Method (3)
Q (D) yn(t ) = δ(t)
y( t) = P (D)w(t)
h(t ) = P(D) yn(t )
h(t ) = P(D)[ yn(t )u (t )]
h(t ) = boδ(t) + P(D) yn(t) , t≥0
h(t ) = boδ(t) + [P(D) yn(t )]u ( t)
Q (D)w( t) = x (t)
Q (D)P(D)w (t) = P (D)x (t)
Q (D)P(D) yn(t ) = P(D)δ(t )
causal yn( t)u (t )
(DN+a1 DN−1+⋯+aN−1 D+aN)w( t) = x (t)
(DN+a1 DN−1
+⋯+aN−1 D+aN ) y (t) = (b0 DM+⋯+bN−1 D+bN)x ( t)
(DN+a1 DN−1+⋯+aN−1 D+aN) yn(t) = δ(t )
(DN+a1 DN−1
+⋯+aN−1 D+aN )h(t ) = (b0 DM+⋯+bN−1 D+bN) δ(t )
y (t ) = (b0 DM+⋯+bN−1 D+bN)w( t)
h( t) = (b0 DM+⋯+bN−1 D+bN ) yn(t )
No interval restriction
Causality is considered
[P(D) yn( t)]u(t)
CLTI Impulse Response (4B) 36 Young Won Lim1/20/15
New Initial Condition created by δ(t)
dN y (t)
d tN + a1
dN−1 y (t)
d tN−1 + a2
dN−2 y (t )
d tN−2 + ⋯ + aN−1
d y (t)d t
+ aN y(t ) = δ(t)
δ(t) u( t)
no jumpdiscontinuity at t = 0
integrate
yn(N−2)(0+) = yn
(N−1)(0+) = ⋯ yn(1)(0+) = yn(0
+) = 0
unit jump discontinuityat t = 0
yn(N−1)(0+) = 1
t u(t ) 1(N−2 )! t(N−2) u (t ) 1
(N−1) ! t(N−1) u( t )
integrate integrate integrate
(DN+a1 DN−1
+⋯+aN−1 D+aN ) ⋅ y (t ) = x (t )
d N y ( t)
d tN +a1
d N−1 y (t )
d tN−1 +⋯+aN−1
d y ( t)d t
+aN y (t ) = x (t )
CLTI Impulse Response (4B) 37 Young Won Lim1/20/15
CLTI Impulse Response (4B) 38 Young Won Lim1/20/15
Impulse Input
δ( t)
δ(0−) = 0 δ(0+ ) = 0
All initial conditions are zero at t=0−
generates energy storage creates nonzero initial condition at t=0+
y (0−) = 0
d2 y (t )
d t 2 + a1
d y (t )d t
+ a2 y (t ) = b0
d2 x (t )
d t 2 + b1
d x ( t)d t
+ b2 x ( t)
y ' (0−) = 0
h(t )
(1, a1 , a2)
(b0 , b1 , b2)Case N>M (b
0=0)
No delta function
δ( t)
y (0+)= K1
y ' (0+) = K2
Case N=M (b0≠0)
No delta functionh(t )Any two functions that have
finite values everywhere and differ in value only at a finite number of points are equivalent in the system response or transform
y (0+)
m = y '(0+)
m = y '(0+)y (0+)
IVP (Initial Value Problem)
CLTI Impulse Response (4B) 39 Young Won Lim1/20/15
Derivatives of a delta function at the output
if contains
d2 h(t)
d t2+ a1
d h( t)d t
+ a2 h (t ) = b0
d2δ( t )
d t 2 + b1
d δ(t )d t
+ b2δ(t )
δ(1)(t )
δ(3)( t) + ⋯ b0δ(2)(t ) + b1δ
(1)( t) + b2δ(t )≠
h( t)
h( t) can contain at most δ( t)
h(t )δ( t)
h( t) cannot contain any derivatives of δ( t )
(b0 ≠ 0)
δ(t) , δ( t) , δ(t) , ⋯
b0δ(t ) + char modes
(b0 ≠ 0)the impulse response
the impulse response
CLTI Impulse Response (4B) 40 Young Won Lim1/20/15
Impulse Response h(t)
h(t) = characteristic mode terms only
t≥0+
(t≠0)
h(t) can have at most an impulse or finite jumpb0δ(t )
t=0
h(t) = b0δ(t ) + char mode terms t≥0
d2 h(t)
d t2+ a1
d h( t)d t
+ a2 h (t ) = b0
d2δ( t )
d t 2 + b1
d δ(t )d t
+ b2δ(t )
h(t ) = b0δ(t) + (∑i
ci e−λ i t
)u(t )
c0, c1
y (0+) , y ' (0+)
Impulse Matching
-then determine
-determine h(t )
(1, a1 , a2)
(b0 , b1 , b2)
c0, c1
y (0+)= 0, y ' (0+) = 1
Simplified Impulse Matching
-then determine
-use h0(t)
(1, a1 , a2)
(0 , 0 , 1)
h(t ) = b0δ(t) + [(b0 D2+ b1 D + b2)(∑i
ci e−λ i t )]u(t )
Case N>M (b0=0)
No delta function
δ( t )Case N=M (b
0≠0)
No delta functionh(t )
y (0+)
m = y '(0+)
m = y '(0+)
y (0+)
CLTI Impulse Response (4B) 41
Base System Impulse Matching
yn(N )(t ) + a1 yn
(N−1)(t ) +⋯+ aN−1 yn(1)(t) + aN yn(t ) = δ( t)
x (t) w (t )δ(t ) yn(t)(1, a1 , ⋯, aN)
must continuousyn(t )
If yn(t) is discontinuous, then
there should exist derivatives of delta : δ(i)(t)
If yn
(1)(t) is discontinuous,
then there should exist derivatives of delta : δ(i)(t)
must continuousyn(1)( t)
must be discontinuousyn(N−1)(t )
yn(0+) = yn(0
−) = 0
yn(1)(0+) = yn
(1)(0−) = 0
yn(0)(N−2)= ⋯ = yn(0) = 0yn
(N−1)(0) = 1
yn(t ) contains no impulse, butcharacteristic modes only
yn(N−1)(0+) = 1, yn
(N−1)(0−) = 0N > M (=0)
h0(t)
(0 , ⋯, 0 , 1)
CLTI Impulse Response (4B) 42
General System Impulse Matching
y (N )+ a1y
(N−1)+ ⋯ + aN−1 y
(1)+ aN y = b0x
(N )+ b1x
(N−1)+ ⋯ + bN−1x
(1)+ bNx
h(N)+ a1h
(N−1)+ ⋯ + aN−1h
(1)+ aNh = b0δ
(N)+ b1δ
(N−1)+ ⋯ + bN−1δ
(1)+ bN δ
h=C δ + ⋯
Cδ(N )
h=C δ(1 ) + ⋯
Cδ(N+1)
h=C δ(N ) + ⋯
C δ(2N)
h(t) can have at most an impulse no derivatives at all
b0δ(t )t=0
if
if
if
b0δ(N )
+ b1δ(N−1)
+ ⋯ + bN−1δ(1)
+ bN δ
b0δ(N )
+ b1δ(N−1)
+ ⋯ + bN−1δ(1)
+ bN δ
b0δ(N)
+ b1δ(N−1)
+ ⋯ + bN−1δ(1)
+ bN δ
CLTI Impulse Response (4B) 43
General System's Zero Input Response
x (t) y (t)h(t )
x (t) w (t )δ(t ) h0(t)
δ(t ) h(t )(1, a1 , ⋯, aN)
(b0 , b1 , ⋯, bN)
(1, a1 , ⋯, aN)
General System S
Base System S0
h(t ) : the impulse response of S
h0(t) : the impulse response of S0
M < N : include no impulse h0(t ) δ(t)
h0(t ) include characteristic modes only
System S and S0 have the same characteristic modes
yn(t) : viewed as the zero input response of S0
natural response homogeneous response
M : the highest order of input derivatives (rhs)N: the highest order of output derivatives (lhs)
h0(t)
(0 , ⋯, 0 , 1)
all the lumped char modescf)
generates energy storage creates nonzero initial condition at t=0+
h0(t) = yn(t )u( t) yn(t )
CLTI Impulse Response (4B) 44
Causal Systems
h(t ) = b0 y n(N )(t) + b1 yn
(N−1)(t ) + ⋯ + bN yn (t)x (t) y (t)h(t )
δ(t ) h(t )(1, a1 , ⋯, aN)
(b0 , b1 , ⋯, bN)
x (t) w (t )yn(t)δ(t ) yn(t)(1, a1 , ⋯, aN)
(0 , ⋯, 0 , 1)
yn =1
aN
(δ − y n(N)− a1 yn
(N−1)−⋯− aN−1 yn
(1))
yn(t )u(t )causal
causal
h(t ) = b0δ(t) + [P(D) yn(t)]u(t)
h(t ) = b0 [ yn (t)u(t )](N ) + b1[ y n(t)u(t)]
(N−1) + ⋯ + bN [ yn(t )u(t )]
causal(M ≤ N)
yn(0)(N−2)= ⋯ = yn(0) = 0yn
(N−1)(0) = 1
CLTI Impulse Response (4B) 45
Zero State Response
x (t) y (t)h(t )
δ(t ) h(t )
(1, a1 , ⋯, aN)
(b0 , b1 , ⋯, bN)
x ( t) = limΔ τ→0
∑τ
x (nΔ τ) p (t−nΔ τ) = limΔ τ→0
∑τ
x (nΔ τ)p (t−nΔ τ)
Δ τΔ τ = lim
Δ τ→0∑τ
x (nΔ τ)δ(t−nΔ τ)Δ τ
δ(t−nΔ τ) h(t−nΔ τ)
x (nΔ τ)δ(t−nΔ τ)Δ τ x (nΔ τ)h(t−nΔ τ)Δ τ
limΔ τ→0
x (nΔ τ)δ(t−nΔ τ)Δ τ limΔ τ→0
x (nΔ τ)h(t−nΔ τ)Δ τ
x (t) y (t)
y (t) = limΔ τ→0
x (nΔ τ)h(t−nΔ τ)Δ τ = ∫−∞
+∞
x (τ)h(t−τ) d τ
Δ τ
1Δτ
1Δτ
CLTI Impulse Response (4B) 46
Causality
x (t) y (t)h(t )
δ(t ) h(t )
(1, a1 , ⋯, aN)
(b0 , b1 , ⋯, bN)
Causal system: the response cannot begin before the input
Causal signals: the input signals starts at time t=0
Causal system: h(t) cannot begin before t=0
h(t) : causal signal
x (t) Causal signal:x(t)=0 (t<0)
h(t)=0 (t<0)
y (t) = ∫−∞
+∞
x( τ)h(t−τ) d τx( τ)=0 ( τ<0)
h(t−τ)=0 (t−τ<0)
y (t) = x(t) ∗ h(t) = ∫0−
tx( τ)h(t−τ) d τ
CLTI Impulse Response (4B) 47 Young Won Lim1/20/15
CLTI Impulse Response (4B) 48 Young Won Lim1/20/15
x(t) = 1
1
y (0) = 2
h(t )
y ' ( t) =−e−ty (t ) = 1 + e−t
y ' ( t ) + y (t ) = 1
y (0) = 1 y (t ) = 0.5 (1 + e−t) y ' (t) =−0.5 e−t
y (0) = 0 y (t ) = 1− e−t y ' ( t) = +e−t
CLTI Impulse Response (4B) 49 Young Won Lim1/20/15
x(t) = u(t)
1
y (0) = 0h(t )
y ' ( t) = [e−t u( t) + (1− e−t)δ(t )]
y (t ) = (1 − e−t)u (t)y ' ( t) + y ( t ) = 1
y ' ( t) = e−t u( t)
Step Response
u(t ) s(t)h(t )
δ(t ) h(t )h(t )
dd t
dd t
CLTI Impulse Response (4B) 50 Young Won Lim1/20/15
x(t) = δ(t)
t=0−
δ( t)
t=0+t=0 t=0− t=0+t=0
All initial conditions are zero at t=0−
Generates energy storage creates nonzero initial condition at t=0+
y (0−) = 0 y (0) = 1
h(t )
h' (t ) =−a e−at u( t) + e−atδ( t)
h ' (t ) =−a e−at u(t) + δ( t)
h( t) = e−at u (t )y ' ( t) + a y (t ) = x (t )
Any two functions that have finite values everywhere and differ in value only at a finite number of points are equivalent in the system response or transform
CLTI Impulse Response (4B) 51 Young Won Lim1/20/15
x(t) & x'(t) forcing functions
δ(t ) h(t )h(t )
δ ' (t) h ' (t)h(t )
dd t
dd t
y (0) = 1
h' (t ) =−a e−at u( t) + e−atδ( t)
h ' (t ) =−a e−at u(t) + δ( t)
h( t) = e−at u (t )y ' ( t) + a y (t ) = x (t )
y ' ( t) + a y (t ) = x ' (t )
CLTI Impulse Response (4B) 52 Young Won Lim1/20/15
Unit Step Function
g1(t ) =∫0−
0+
g i(t )dt = 0
The area under a single point is zero, regardless of the point's value , if it is finite
y1
y2
(t < 0)
(t > 0)
(t = 0)undefined
g2(t ) =
y1
y2
g3(t) =
y1
y2
g4 (t ) =
y1
y2
( y1+ y2)/2
y1
y2
(t < 0)
(t > 0)
(t = 0)
(t < 0)
(t > 0)
(t = 0)
(t < 0)
(t > 0)
(t = 0)
The system response is the same
The transform is the same also
(i = 1,2,3,4)
Any two functions that have finite values everywhere and differ in value only at a finite number of points are equivalent in the system response or transform
∫α
β
g i(t )dt = ∫α
β
g j (t )dt (i ≠ j)
Young Won Lim1/20/15
References
[1] http://en.wikipedia.org/[2] J.H. McClellan, et al., Signal Processing First, Pearson Prentice Hall, 2003[3] B.P. Lathi, Linear Systems and Signals (2nd Ed)[4] M.J. Roberts, Fundamentals of Signals and Systems