Lecture #4:

THE SUPERPOSITION SUM (DT SYSTEM) AND THE SUPERPOSITION INTEGRAL (CT SYSTEM) OF LTI SYSTEMS

Motivation:

Outline:

I. DERIVATION OF SUPERPOSITION SUM/INTEGRAL

We will show that for a DT system

{} y [ n ] = ∑ m = − ∞ ∞ x [ m ] h [ n − m ] y [ n ] = ∑ m = − ∞ ∞ x [ m ] h [ n − m ] size 12{y \[ n \] = Sum cSub { size 8{m= - infinity } } cSup { size 8{ infinity } } {x \[ m \] h \[ n - m \] } } {}

where x[n] is an arbitrary input, h[n] is the unit sample response, y[n] is the output, and the above relation is called the superposition sum.

We will show that for a CT system

y ( t ) = ∫ − ∞ ∞ x ( τ ) h ( t − τ ) dτ y ( t ) = ∫ − ∞ ∞ x ( τ ) h ( t − τ ) dτ size 12{y \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {x \( τ \) h \( t - τ \) dτ} } {}

where x(t) is an arbitrary input, h(t) is the unit impulse response, y(t) is the output, and the above relation is called the superposition integral.

II. THE SUPERPOSITION SUM FOR DT SYSTEMS

1/ Graphic view of superposition sum

2/ Derivation of superposition sum

3/ Convolution sum

The relation

x [ n ] = ∑ m x [ m ] δ [ n − m ] x [ n ] = ∑ m x [ m ] δ [ n − m ] size 12{x \[ n \] = Sum cSub { size 8{m} } {x \[ m \] δ \[ n - m \] } } {}

expresses the sifting property of the unit sample. Note that the only non-zero term in this sum occurs when m = n, hence demonstrating the validity of the equation. The major conclusion of the derivation is that for an arbitrary input x[n], the output is

y [ n ] = ∑ m x [ m ] h [ n − m ] y [ n ] = ∑ m x [ m ] h [ n − m ] size 12{y \[ n \] = Sum cSub { size 8{m} } {x \[ m \] h \[ n - m \] } } {}

which is called the superposition sum. Such a relation is called a convolution sum when it involves arbitrary functions, i.e.,

z [ n ] = ∑ m x 1 [ m ] x 2 [ n − m ] z [ n ] = ∑ m x 1 [ m ] x 2 [ n − m ] size 12{z \[ n \] = Sum cSub { size 8{m} } {x rSub { size 8{1} } } \[ m \] x rSub { size 8{2} } \[ n - m \] } {}

Thus, the superposition sum is a special case of the convolution sum.

4/ Notation

We shall write the convolution sum of two DT signals as

z [ n ] = x 1 [ n ] ∗ x 2 [ n ] = ∑ m x 1 [ m ] x 2 [ n − m ] z [ n ] = x 1 [ n ] ∗ x 2 [ n ] = ∑ m x 1 [ m ] x 2 [ n − m ] size 12{z \[ n \] =x rSub { size 8{1} } \[ n \] * x rSub { size 8{2} } \[ n \] = Sum cSub { size 8{m} } {x rSub { size 8{1} } \[ m \] x rSub { size 8{2} } \[ n - m \] } } {}

The symbol for convolution in various textbooks includes

x 1 [ n ] ∗ x 2 [ n ] , x 1 [ n ] ∗ x 2 [ n ] and x 1 [ n ] ⊗ x 2 [ n ] x 1 [ n ] ∗ x 2 [ n ] , x 1 [ n ] ∗ x 2 [ n ] and x 1 [ n ] ⊗ x 2 [ n ] size 12{x rSub { size 8{1} } \[ n \] * x rSub { size 8{2} } \[ n \] ,x rSub { size 8{1} } \[ n \] * x rSub { size 8{2} } \[ n \] matrix { {} # {} } ital "and" matrix { {} # {} } x rSub { size 8{1} } \[ n \] ⊗x rSub { size 8{2} } \[ n \] } {}

5/ Mechanics

To compute the superposition sum

y [ n ] = x [ n ] ∗ h [ n ] = ∑ m x [ m ] h [ n − m ] y [ n ] = x [ n ] ∗ h [ n ] = ∑ m x [ m ] h [ n − m ] size 12{y \[ n \] =x \[ n \] * h \[ n \] = Sum cSub { size 8{m} } {x \[ m \] h \[ n - m \] } } {}

Step 1 Plot x and h vs m since the convolution sum is on m.

Step 2 Flip h[m] around the vertical axis to obtain h[−m].

Step 3 Shift h[−m] by n to obtain h[n − m].

{}Step 4 Multiply to obtain x[m]h[n − m].

Step 5 Sum on m to compute. ∑mx[m]h[n−m]∑mx[m]h[n−m] size 12{ Sum cSub { size 8{m} } {x \[ m \] h \[ n - m \] } } {}

Step 6 Index n and repeat Steps 3-6.

Demo of DT convolution

6/ DT convolution properties — commutative property

x[n]*h[n]=h[n]*x[n]

Proof:

y [ n ] = x [ n ] ∗ h [ n ] = ∑ m x [ m ] h [ n − m ] y [ n ] = x [ n ] ∗ h [ n ] = ∑ m x [ m ] h [ n − m ] size 12{y \[ n \] =x \[ n \] * h \[ n \] = Sum cSub { size 8{m} } {x \[ m \] h \[ n - m \] } } {}

Let n − m = l then

y [ n ] = x [ n ] ∗ h [ n ] = ∑ 1 x [ n − 1 ] h [ 1 ] = ∑ 1 h [ 1 ] x [ n − 1 ] = h [ n ] ∗ x [ n ] y [ n ] = x [ n ] ∗ h [ n ] = ∑ 1 x [ n − 1 ] h [ 1 ] = ∑ 1 h [ 1 ] x [ n − 1 ] = h [ n ] ∗ x [ n ] size 12{y \[ n \] =x \[ n \] * h \[ n \] = Sum cSub { size 8{1} } {x \[ n - 1 \] h \[ 1 \] = Sum cSub { size 8{1} } {h \[ 1 \] x \[ n - 1 \] } =h \[ n \] * x \[ n \] } } {}

7/ DT convolution properties — associative property

{}

y [ n ] = ( x [ n ] ∗ h 1 [ n ] ) ∗ h 2 [ n ] = x [ n ] ∗ ( h 1 [ n ] ∗ h 2 [ n ] ) y [ n ] = ( x [ n ] ∗ h 1 [ n ] ) ∗ h 2 [ n ] = x [ n ] ∗ ( h 1 [ n ] ∗ h 2 [ n ] ) size 12{y \[ n \] = \( x \[ n \] *h rSub { size 8{1} } \[ n \] \) *h rSub { size 8{2} } \[ n \] =x \[ n \] * \( h rSub { size 8{1} } \[ n \] *h rSub { size 8{2} } \[ n \] \) } {}

Proof:

y [ n ] = ∑ 1 ( ∑ m x [ m ] h 1 [ 1 − m ] ) h 2 [ n − 1 ] = ∑ m x [ m ] ( ∑ 1 h 1 [ 1 − m ] h 2 [ n − 1 ] ) y [ n ] = ∑ 1 ( ∑ m x [ m ] h 1 [ 1 − m ] ) h 2 [ n − 1 ] = ∑ m x [ m ] ( ∑ 1 h 1 [ 1 − m ] h 2 [ n − 1 ] ) size 12{y \[ n \] = Sum cSub { size 8{1} } { \( Sum cSub { size 8{m} } {x \[ m \] h rSub { size 8{1} } \[ 1 - m \] } \) h rSub { size 8{2} } \[ n - 1 \] = Sum cSub { size 8{m} } {x \[ m \] \( Sum cSub { size 8{1} } {h rSub { size 8{1} } } \[ 1 - m \] h rSub { size 8{2} } \[ n - 1 \] \) } } } {}

Let k = l − m to obtain

y [ n ] = ∑ m x [ m ] ( ∑ k h 1 [ k ] h 2 [ n − m − k ] ) = ∑ m x [ m ] h [ n − m ] y [ n ] = ∑ m x [ m ] ( ∑ k h 1 [ k ] h 2 [ n − m − k ] ) = ∑ m x [ m ] h [ n − m ] size 12{y \[ n \] = Sum cSub { size 8{m} } {x \[ m \] \( Sum cSub { size 8{k} } {h rSub { size 8{1} } } \[ k \] h rSub { size 8{2} } \[ n - m - k \] \) = Sum cSub { size 8{m} } {x \[ m \] h \[ n - m \] } } } {}

where

h [ n ] = h 1 [ n ] ∗ h 2 [ n ] h [ n ] = h 1 [ n ] ∗ h 2 [ n ] size 12{h \[ n \] =h rSub { size 8{1} } \[ n \] *h rSub { size 8{2} } \[ n \] } {}

8/ DT convolution properties — distributive property

y [ n ] = x [ n ] ∗ ( h 1 [ n ] + h 2 [ n ] ) = x [ n ] ∗ h 1 [ n ] + x [ n ] ∗ h 2 [ n ] y [ n ] = x [ n ] ∗ ( h 1 [ n ] + h 2 [ n ] ) = x [ n ] ∗ h 1 [ n ] + x [ n ] ∗ h 2 [ n ] size 12{y \[ n \] =x \[ n \] * \( h rSub { size 8{1} } \[ n \] +h rSub { size 8{2} } \[ n \] \) =x \[ n \] *h rSub { size 8{1} } \[ n \] +x \[ n \] *h rSub { size 8{2} } \[ n \] } {}

Proof:

y [ n ] = ∑ m x [ m ] ( h 1 [ n − m ] + h 2 [ n − m ] ) = ∑ m x [ n ] h 1 [ n − m ] + ∑ m x [ m ] h 2 [ n − m ] y [ n ] = ∑ m x [ m ] ( h 1 [ n − m ] + h 2 [ n − m ] ) = ∑ m x [ n ] h 1 [ n − m ] + ∑ m x [ m ] h 2 [ n − m ] alignl { stack { size 12{y \[ n \] = Sum cSub { size 8{m} } {x \[ m \] \( h rSub { size 8{1} } \[ n - m \] +h rSub { size 8{2} } \[ n - m \] \) } } {} # matrix { {} # {} } = Sum cSub { size 8{m} } {x \[ n \] h rSub { size 8{1} } \[ n - m \] + Sum cSub { size 8{m} } {x \[ m \] h rSub { size 8{2} } \[ n - m \] } } {} } } {}

9/ DT convolution properties — delay accumulation

If

y[n] = x[n]∗h[n]

then

x[n − j]∗h[n − k] = y[n − k − j]

Proof:

∑ m x [ m − j ] h [ n − k − m ] = ∑ 1 x [ 1 ] h [ n − k − j − 1 ] = y [ n − k − j ] ∑ m x [ m − j ] h [ n − k − m ] = ∑ 1 x [ 1 ] h [ n − k − j − 1 ] = y [ n − k − j ] size 12{ Sum cSub { size 8{m} } {x \[ m - j \] h \[ n - k - m \] = Sum cSub { size 8{1} } {x \[ 1 \] h \[ n - k - j - 1 \] =y \[ n - k - j \] } } } {}

III. THE SUPERPOSITION INTEGRAL FOR CT SYSTEMS

1/ Graphic view of staircase approximation to the superposition integral

2/ Definition of a tall and narrow pulse

To represent a continuous time function with a staircase approximation, it is useful to define a tall and narrow pulse.

As Δ is decreased, the width of the pulse decreases and the amplitude increases while the area remains 1.

3/ Derivation of staircase approximation to superposition integral

4/ The superposition integral

The derivation shows that a staircase approximation to the input x(t)

x Δ ( t ) = ∑ k x ( kΔ ) δ Δ ( t − kΔ ) Δ x Δ ( t ) = ∑ k x ( kΔ ) δ Δ ( t − kΔ ) Δ size 12{x rSub { size 8{Δ} } \( t \) = Sum cSub { size 8{k} } {x \( kΔ \) δ rSub { size 8{Δ} } \( t - kΔ \) Δ} } {}

yields a staircase approximation to the output

y Δ ( t ) = ∑ k x ( kΔ ) h Δ ( t − kΔ ) Δ y Δ ( t ) = ∑ k x ( kΔ ) h Δ ( t − kΔ ) Δ size 12{y rSub { size 8{Δ} } \( t \) = Sum cSub { size 8{k} } {x \( kΔ \) h rSub { size 8{Δ} } \( t - kΔ \) Δ} } {}

Now we take the limit as Δ→0, k→∞, kΔ=τ, xΔ(t) → x(t), hΔ(t−kΔ) → h(t−τ), and δΔ(t−kΔ) → δ(t−τ) in a generalized function sense. Then the sums approach the integrals

x ( t ) = ∫ − ∞ ∞ x ( τ ) δ ( t − τ ) dτ x ( t ) = ∫ − ∞ ∞ x ( τ ) δ ( t − τ ) dτ size 12{x \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {x \( τ \) δ \( t - τ \) dτ} } {}

y ( t ) = ∫ − ∞ ∞ x ( τ ) h ( t − τ ) dτ y ( t ) = ∫ − ∞ ∞ x ( τ ) h ( t − τ ) dτ size 12{y \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {x \( τ \) h \( t - τ \) dτ} } {}

5/ Sifting property of the unit impulse

We examine the sifting property of the unit impulse which is inherent in the expression

x ( t ) = ∫ − ∞ ∞ x ( τ ) δ ( t − τ ) dτ x ( t ) = ∫ − ∞ ∞ x ( τ ) δ ( t − τ ) dτ size 12{x \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {x \( τ \) δ \( t - τ \) dτ} } {}

To see what this means we approximate the impulse with a tall, narrow pulse.

6/ Mechanics

To compute the superposition integral

y ( t ) = x ( t ) ∗ h ( t ) = ∫ − ∞ ∞ x ( τ ) h ( t − τ ) dτ y ( t ) = x ( t ) ∗ h ( t ) = ∫ − ∞ ∞ x ( τ ) h ( t − τ ) dτ size 12{y \( t \) =x \( t \) *h \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {x \( τ \) h \( t - τ \) dτ} } {}

Step 1 Plot x and h vs τ since the convolution integral is on τ.

Step 2 Flip h(τ) around the vertical axis to obtain h(−τ).

Step 3 Shift h(τ) by t to obtain h(t − τ).

Step 4 Multiply to obtain x(τ)h(t − τ).

Step 5 Integrate on τ to compute ∫−∞∞x(τ)h(t-τ) dτ∫−∞∞x(τ)h(t-τ) dτ size 12{ Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {x \( τ \) h \( "t-"τ \) " d"τ} } {}.

Step 6 Increase t and repeat Steps 3-6.

Demo of CT convolution

7/ CT convolution properties

The convolution integral has properties similar to those of the convolution sum.

Commutative y(t)=x(t)∗h(t)=h(t)∗x(t)y(t)=x(t)∗h(t)=h(t)∗x(t) size 12{y \( t \) =x \( t \) *h \( t \) =h \( t \) *x \( t \) } {}

Associative y(t)=(x(t)∗h1(t))∗h2(t)y(t)=(x(t)∗h1(t))∗h2(t) size 12{y \( t \) = \( x \( t \) *h rSub { size 8{1} } \( t \) \) *h rSub { size 8{2} } \( t \) } {}

y ( t ) = x ( t ) ∗ ( h 1 ( t ) ∗ h 2 ( t ) ) y ( t ) = x ( t ) ∗ ( h 1 ( t ) ∗ h 2 ( t ) ) size 12{y \( t \) =x \( t \) * \( h rSub { size 8{1} } \( t \) *h rSub { size 8{2} } \( t \) \) } {}

Distributive y(t)=x(t)∗(h1(t)+h2(t))y(t)=x(t)∗(h1(t)+h2(t)) size 12{y \( t \) =x \( t \) * \( h rSub { size 8{1} } \( t \) +h rSub { size 8{2} } \( t \) \) } {}

y ( t ) = x ( t ) ∗ h 1 ( t ) + x ( t ) ∗ h 2 ( t ) y ( t ) = x ( t ) ∗ h 1 ( t ) + x ( t ) ∗ h 2 ( t ) size 12{y \( t \) =x \( t \) *h rSub { size 8{1} } \( t \) +x \( t \) *h rSub { size 8{2} } \( t \) } {}

Delay accomutation y(t−τ1−τ2)=x(t−τ1)∗h(t−τ2)y(t−τ1−τ2)=x(t−τ1)∗h(t−τ2) size 12{y \( t - τ rSub { size 8{1} } - τ rSub { size 8{2} } \) =x \( t - τ rSub { size 8{1} } \) *h \( t - τ rSub { size 8{2} } \) } {}

Derivative accomutation y[n+m](t)=x[n](t)∗h[m](t)y[n+m](t)=x[n](t)∗h[m](t) size 12{y rSup { size 8{ \[ n+m \] } } \( t \) =x rSup { size 8{ \[ n \] } } \( t \) *h rSup { size 8{ \[ m \] } } \( t \) } {}

The last property implies that differentiating the input n times and the impulse response m times results in an output that is differentiated n+m times.

IV. RELATION OF TIME DOMAIN AND TRANSFORM DOMAIN CHARACTERIZATION OF LTI SYSTEMS

1/ Two methods for characterizing LTI systems

2/ Convolution in the time domain corresponds to multiplication in the transform domain

x [ n ] ∗ h [ m ] ⇔ Z X ~ ( z ) . H ~ ( z ) and x ( t ) ∗ h ( t ) ⇔ L X ( s ) H ( s ) x [ n ] ∗ h [ m ] ⇔ Z X ~ ( z ) . H ~ ( z ) and x ( t ) ∗ h ( t ) ⇔ L X ( s ) H ( s ) size 12{x \[ n \] *h \[ m \] { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) "." {H} cSup { size 8{ "~" } } \( z \) matrix { {} # {} } ital "and" matrix { {} # {} } x \( t \) *h \( t \) { dlrarrow } cSup { size 8{L} } X \( s \) H \( s \) } {}

{}If the input is a unit sample/impulse, x[n] = δ[n] and x(t) = δ(t) then since δ[n] ⇔Z⇔Z size 12{ { dlrarrow } cSup { size 8{Z} } } {}1 and δ(t) ⇔L⇔L size 12{ { dlrarrow } cSup { size 8{L} } } {}1, we have

h [ n ] ⇔ Z H ~ ( z ) and h ( t ) ⇔ L H ( s ) h [ n ] ⇔ Z H ~ ( z ) and h ( t ) ⇔ L H ( s ) size 12{h \[ n \] { dlrarrow } cSup { size 8{Z} } {H} cSup { size 8{ "~" } } \( z \) matrix { {} # {} } ital "and" matrix { {} # {} } h \( t \) { dlrarrow } cSup { size 8{L} } H \( s \) } {}

V. THE UNIT SAMPLE/IMPULSE RESPONSE

The unit sample/impulse response of an LTI system characterizes that system. How can we measure this response on a real system? To indicate how this might be done we will determine the impulse response of a CT LTI system that is a lowpass filter.

To find the impulse response we will determine the response to a tall, narrow, rectangular pulse.

Two-minute miniquiz problem

Problem 14-1 — Rectangular pulse response of a LPF

Determine vo(t) in response to the pulse shown.

[Hint: use superposition.]

Solution

Note that vi(t) can be written as the difference of a step and a step delayed that is scaled as follows

v i ( t ) = 1 Δ ( u ( t ) − u ( t − Δ ) ) v i ( t ) = 1 Δ ( u ( t ) − u ( t − Δ ) ) size 12{v rSub { size 8{i} } \( t \) = { {1} over {Δ} } \( u \( t \) - u \( t - Δ \) \) } {} {}

Therefore, we need only find the response of the network to a unit step vi(t)=u(t). We denote the step response as vo(t)=s(t) which is

s ( t ) = ( 1 − e − αt ) u ( t ) s ( t ) = ( 1 − e − αt ) u ( t ) size 12{s \( t \) = \( 1 - e rSup { size 8{ - αt} } \) u \( t \) } {}

where α = 1/(RC). Therefore, the response to the pulse is

v ο ( t ) = 1 Δ ( ( 1 − e − αt ) u ( t ) − ( 1 − e − α ( t − Δ ) ) u ( t − Δ ) ) v ο ( t ) = 1 Δ ( ( 1 − e − αt ) u ( t ) − ( 1 − e − α ( t − Δ ) ) u ( t − Δ ) ) size 12{v rSub { size 8{ο} } \( t \) = { {1} over {Δ} } \( \( 1 - e rSup { size 8{ - αt} } \) u \( t \) - \( 1 - e rSup { size 8{ - α \( t - Δ \) } } \) u \( t - Δ \) \) } {}

1/ Rectangular pulse response of LPF

The pulse response can be written as

{} v 0 ( 0 ) = { 0 ( t < 0 ) 1 Δ ( 1 − e − αt ) ( 0 ≤ t < Δ ) 1 Δ ( e αΔ − 1 ) e − αt ( t ≥ Δ ) v 0 ( 0 ) = { 0 ( t < 0 ) 1 Δ ( 1 − e − αt ) ( 0 ≤ t < Δ ) 1 Δ ( e αΔ − 1 ) e − αt ( t ≥ Δ ) size 12{v rSub { size 8{0} } \( 0 \) = left lbrace matrix { 0`` matrix { {} # {} } ` matrix { {} # {} } ` matrix { {} # {} } \( t<0 \) {} ## { {1} over {Δ} } \( 1 - e rSup { size 8{ - αt} } \) matrix { {} # {} } `` \( 0 <= t<Δ \) {} ## { {1} over {Δ} } \( e rSup { size 8{αΔ} } - 1 \) e rSup { size 8{ - αt} } matrix { {} # {} } \( t >= Δ \) {} } right none } {}

2/ Convergence of rectangular pulse response to impulse response

As αΔ → 0

v 0 ( t ) → { 0 ( t < 0 ) α Δ t ( 0 ≤ t < Δ ) αe − αt ( t ≥ Δ ) v 0 ( t ) → { 0 ( t < 0 ) α Δ t ( 0 ≤ t < Δ ) αe − αt ( t ≥ Δ ) size 12{v rSub { size 8{0} } \( t \) rightarrow left lbrace matrix { 0``` matrix { {} # {} } ` matrix { {} # {} } \( t<0 \) {} ## { {α} over {Δ} } t``` matrix { {} # {} } ` \( 0 <= t<Δ \) {} ## αe rSup { size 8{ - αt} } ``` matrix { {} # {} } ` \( t >= Δ \) {} } right none } {}

Hence, the impulse response of the LPF is

h ( t ) = αe − αt u ( t ) h ( t ) = αe − αt u ( t ) size 12{h \( t \) =αe rSup { size 8{ - αt} } u \( t \) } {}

Demo showing that the shape of the pulse does not matter

3/ Alternative method for finding the impulse response

To determine the impulse response, we need to evaluate the derivative which we do by parts

h ( t ) = d dt ( ( 1 − e − αΔ ) u ( t ) ) = αe − αt u ( t ) + ( 1 − e − αt ) δ ( t ) h ( t ) = d dt ( ( 1 − e − αΔ ) u ( t ) ) = αe − αt u ( t ) + ( 1 − e − αt ) δ ( t ) size 12{h \( t \) = { {d} over { ital "dt"} } \( \( 1 - e rSup { size 8{ - αΔ} } \) u \( t \) \) =αe rSup { size 8{ - αt} } u \( t \) + \( 1 - e rSup { size 8{ - αt} } \) δ \( t \) } {}

To simplify the impulse response we need to interpret the term (1−e−αt)δ(t)(1−e−αt)δ(t) size 12{ \( 1 - e rSup { size 8{ - αt} } \) δ \( t \) } {} which we do by placing that term in an integral and noting that

∫ − ∞ ∞ ( 1 − e − αt ) δ ( t ) dt = ( 1 − e − αt ) t = 0 = 0 ∫ − ∞ ∞ ( 1 − e − αt ) δ ( t ) dt = ( 1 − e − αt ) t = 0 = 0 size 12{ Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } { \( 1 - e rSup { size 8{ - αt} } \) δ \( t \) ital "dt"= \( 1 - e rSup { size 8{ - αt} } \) rSub { size 8{t=0} } } =0} {} {}

Therefore, the second term is an impulse of area 0 which equals 0, so that the impulse response is h(t)=αe−αtu(t)h(t)=αe−αtu(t) size 12{h \( t \) =αe rSup { size 8{ - αt} } u \( t \) } {}

This agrees with the result obtained by finding the response to a tall, narrow pulse.

So how do you “measure” the impulse response of an LTI system?

Apply a brief pulse (the waveshape does not matter) and measure the response. Repeat the measurement with a pulse of briefer duration but the same area.

Response of the LPF to pulses of different durations.

Repeat the process with briefer pulses until the changes in the pulse responses do not matter to you. The pulse response to the briefest of these pulses is an estimate of the impulse response.

4/ Unit impulse/sample responses of different classes of LTI systems

The condition for the unit impulse/sample response for a BIBO stable system requires some justification.

5/ BIBO stable systems

We examine the condition for a DT LTI system, the proof is similar for a CT LTI system. For a DT LTI system the output y[n] can be expressed in terms of the input x[n] as

y [ n ] = ∑ m = − ∞ ∞ h [ m ] x [ n − m ] y [ n ] = ∑ m = − ∞ ∞ h [ m ] x [ n − m ] size 12{y \[ n \] = Sum cSub { size 8{m= - infinity } } cSup { size 8{ infinity } } {h \[ m \] x \[ n - m \] } } {}

For a BIBO stable system, if x[n] < ∞ then y[n] < ∞. Therefore,

∣ ∑ m = − ∞ ∞ h [ m ] x [ n − m ] ∣ = ∑ m = − ∞ ∞ ∣ h [ m ] x [ n − m ] ∣ < x max ∑ m = − ∞ ∞ ∣ h [ m ] ∣ ∣ ∑ m = − ∞ ∞ h [ m ] x [ n − m ] ∣ = ∑ m = − ∞ ∞ ∣ h [ m ] x [ n − m ] ∣ < x max ∑ m = − ∞ ∞ ∣ h [ m ] ∣ size 12{ lline Sum cSub { size 8{m= - infinity } } cSup { size 8{ infinity } } {h \[ m \] x \[ n - m \] } rline = Sum cSub { size 8{m= - infinity } } cSup { size 8{ infinity } } { lline h \[ m \] x \[ n - m \] rline } ~<~x rSub { size 8{"max"} } Sum cSub { size 8{m= - infinity } } cSup { size 8{ infinity } } { lline h \[ m \] rline } } {}

where xmax is the maximum value of |x[m]|. Therefore, if xmax <∞, and if

∑ m = − ∞ ∞ ∣ h [ m ] ∣ < ∞ ∑ m = − ∞ ∞ ∣ h [ m ] ∣ < ∞ size 12{ Sum cSub { size 8{m= - infinity } } cSup { size 8{ infinity } } { lline h \[ m \] rline } < infinity } {}

then y[n] < ∞. This condition is both necessary and sufficient.

VI. CONCLUSIONS

x [ n ] = ∑ m x [ m ] δ [ n − m ] and x ( t ) = ∫ − ∞ ∞ x ( τ ) δ ( t − τ ) dτ x [ n ] = ∑ m x [ m ] δ [ n − m ] and x ( t ) = ∫ − ∞ ∞ x ( τ ) δ ( t − τ ) dτ size 12{x \[ n \] = Sum cSub { size 8{m} } {x \[ m \] δ \[ } n - m \] ~ ital "and"~x \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {x \( τ \) δ \( t - τ \) dτ} } {}

y [ n ] = ∑ m x [ m ] h [ n − m ] and y ( t ) = ∫ − ∞ ∞ x ( τ ) h ( t − τ ) dτ y [ n ] = ∑ m x [ m ] h [ n − m ] and y ( t ) = ∫ − ∞ ∞ x ( τ ) h ( t − τ ) dτ size 12{y \[ n \] = Sum cSub { size 8{m} } {x \[ m \] h \[ n - m \] ~ ital "and"~y \( t \) = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {x \( τ \) h \( t - τ \) dτ} } } {}

Exercises .

Solution of exercises.

Content actions

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This work is licensed by Truc Pham-Dinh, Tam Huynh-Ngoc, Anh Tuan Hoang, and Thanh Dinh Vu under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.

Last edited by vocw on Jul 7, 2009 3:42 am -0500.