Impulse Matching (N > M) · 20/01/2015 · CLTI Impulse Response (4B) 1 Young Won Lim 1/20/15...

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)


Copyright (c) 2011 - 2015 Young W. Lim.

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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)


Impulse Response Requirements – A


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


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


yh(t )u( t) + m0δ(t )


(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.


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


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




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 )


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)


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 )


Derivatives & Integrals of g(t) = yh(t)·u(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 )


Derivatives of h(t) – three cases (2)


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)


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 )



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



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+⋯


yh(t )u( t) + m0δ(t )


(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



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+⋯


yh(t )u( t) + m0δ(t )


(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


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 )


(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


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 )


(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)


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


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


yh(t )u( t ) + m0δ(t )


(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


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 )


(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


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


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 )


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 )


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


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.


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


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 )


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 )


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


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 )


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


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


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)




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



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)


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 )


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)


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


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)


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


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 )


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


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


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


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


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


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


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 )


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)


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

Impulse Matching (N > M) · 20/01/2015 · CLTI Impulse Response (4B) 1 Young Won Lim 1/20/15...

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