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Lecture 2: Introduction to Systems

Module by: Vu Dinh Thanh, Truc Pham-Dinh, Anh Tuan Hoang, Tam Huynh-Ngoc. E-mail the authors

Summary: To describe man-made and natural systems, in forms of mathematical expressions in time and in frequency domains. Many physical systems are described by linear differential equations or by linear difference equations.

Lecture #2:

INTRODUCTION TO SYSTEMS

Motivation: To describe man-made and natural systems, in forms of mathematical expressions in time and in frequency domains. Many physical systems are described by linear differential equations or by linear difference equations.

Outline:

  • Man-made systems — modular, hierarchic design. Natural systems — modular, hierarchic analysis. Dynamic analogies
  • Classification of systems
  • Reducing differential equations to algebraic equations by using complex notations. Role of complex exponential time functions in LTI systems. Sources of linear differential equations. Homogeneous solution — exponential solution, natural frequencies. Particular solution — system function, poles & zeros. Total solution — initial conditions, steady-state.
  • Reducing difference equations to algebraic equations. Role of complex geometric (exponential) time functions in DT LTI systems. Linear difference equations arise as system. Homogeneous solution — geometric (exponential) solution, natural frequencies. Particular solution — system function, poles & zeros. Total solution — initial conditions, steady-state.

Signals and systems

This subject deals with mathematical methods used to describe signals and to analyze and synthesize systems.

  • Signals are variables that carry information
  • Systems process input signals to produce output signals.

Last time — SIGNALS; Today — SYSTEMS.

I. MAN-MADE SYSTEMS — MODULAR, HIERARCHIC DESIGN

Robot car

Figure 1
Figure 1 (graphics1.png)

1/ Robot car block diagram

Hierarchic design — top (1st) level includes: wheel position controller, digital camera, and image processing software.

Figure 2
Figure 2 (graphics2.png)

2/ Wheel position controller block diagram

Hierarchic design — 2nd level is a block diagram of the wheel controller which includes: amplifier, motor, and shaft decoder.

Figure 3
Figure 3 (graphics3.png)

3/ Motor dynamics

Hierarchic design—3rd level includes a more detailed description of the motor. Important quantities are:

  • Current in motor windings i(t),
  • Motor shaft angular displacement θ(t),
  • Motor parameters — viscous damping constant B, moment of inertia J, electromechanical constant k.

The torque balance equation is:

ki ( t ) B ( t ) dt = J 2 ( t ) dt 2 ki ( t ) B ( t ) dt = J 2 ( t ) dt 2 size 12{ ital "ki" \( t \) - B { {dθ \( t \) } over { ital "dt"} } =J { {dθ rSup { size 8{2} } \( t \) } over { ital "dt" rSup { size 8{2} } } } } {}

4/ Observation

  • Man-made complex systems are designed in a modular, hierarchical fashion often expressed in nested block diagrams.
  • Block input/output relations provide a communication mechanism for team projects.
  • Optimization of system performance requires excellent tools to characterize signal transformations at each level of the hierarchy.

II. NATURALLY OCCURRING SYSTEMS — MODULAR, HIERARCHIC ANALYSIS

Human speech production system — anatomy

Figure 4
Figure 4 (graphics4.png)

1/ Human speech production system — block diagram

Figure 5
Figure 5 (graphics5.png)

2/ Observation

  • Naturally occurring systems are not designed in a modular fashion — they have evolved.
  • To understand these systems, we impose a hierarchy and parse the system into modules whose function can be characterized.

3/ Conclusion

A system is described structurally by specifying:

  • the system topology,
  • the rules of interconnection of the elements,
  • functional descriptions of the elements — constitutive relations.

III. DYNAMIC ANALOGIES

Physically divergent systems can have similar dynamic properties.

1/ Mechanical free-body diagram

M = mass, B = friction constant,

K = spring constant,

f(t) = external force, and

v(t) = velocity of the mass.

Figure 6
Figure 6 (graphics6.png)

Summing the forces yields

f ( t ) = M dv ( t ) dt + Bv ( t ) + K t v ( τ ) f ( t ) = M dv ( t ) dt + Bv ( t ) + K t v ( τ ) size 12{f \( t \) =M { { ital "dv" \( t \) } over { ital "dt"} } + ital "Bv" \( t \) +K Int rSub { size 8{ - infinity } } rSup { size 8{t} } {v \( τ \) dτ} } {}

2/ Electric network

Figure 7
Figure 7 (graphics7.png)

Summing the currents (Kirchhoff’s current law) yields

i ( t ) = C dv ( t ) dt + v ( t ) R + 1 L t v ( τ ) i ( t ) = C dv ( t ) dt + v ( t ) R + 1 L t v ( τ ) size 12{i \( t \) =C { { ital "dv" \( t \) } over { ital "dt"} } + { {v \( t \) } over {R} } + { {1} over {L} } Int rSub { size 8{ - infinity } } rSup { size 8{t} } {v \( τ \) dτ} } {}

3/ The mechanical and electrical systems are dynamically analogous

f ( t ) = M dv ( t ) dt + Bv ( t ) + K t v ( τ ) i ( t ) = C dv ( t ) dt + v ( t ) R + 1 L t v ( τ ) f ( t ) = M dv ( t ) dt + Bv ( t ) + K t v ( τ ) i ( t ) = C dv ( t ) dt + v ( t ) R + 1 L t v ( τ ) alignl { stack { size 12{f \( t \) =M { { ital "dv" \( t \) } over { ital "dt"} } + ital "Bv" \( t \) +K Int rSub { size 8{ - infinity } } rSup { size 8{t} } {v \( τ \) dτ} } {} # i \( t \) =C { { ital "dv" \( t \) } over { ital "dt"} } + { {v \( t \) } over {R} } + { {1} over {L} } Int rSub { size 8{ - infinity } } rSup { size 8{t} } {v \( τ \) dτ} {} } } {}

Thus, understanding one of these systems gives insights into the other.

4/ Block diagram

A block diagram using integrators, adders, and gains

f ( t ) = M dv ( t ) dt + Bv ( t ) + K t v ( τ ) f ( t ) = M dv ( t ) dt + Bv ( t ) + K t v ( τ ) size 12{f \( t \) =M { { ital "dv" \( t \) } over { ital "dt"} } + ital "Bv" \( t \) +K Int rSub { size 8{ - infinity } } rSup { size 8{t} } {v \( τ \) dτ} } {}

Figure 8
Figure 8 (graphics8.png)

5/ Electronic synthesis of block diagram

The integrator, adder, and gain blocks are other examples of functional descriptions of systems. We can produce a structural model of each of these blocks. For example, the gain block is easily synthesized with an op-amp circuit.

Figure 9
Figure 9 (graphics9.png)

The op-amp itself is a functional model of a device that we can synthesize with an electronic circuit including a number of transistors.

Conclusion: We have seen several types of descriptions of systems

Figure 10
Figure 10 (graphics10.png)

f ( t ) = M dv ( t ) dt + Bv ( t ) + K t v ( τ ) f ( t ) = M dv ( t ) dt + Bv ( t ) + K t v ( τ ) size 12{f \( t \) =M { { ital "dv" \( t \) } over { ital "dt"} } + ital "Bv" \( t \) +K Int rSub { size 8{ - infinity } } rSup { size 8{t} } {v \( τ \) dτ} } {}

All four descriptions define a system with the same dynamic properties. We will develop methods that characterize these systems efficiently and that abstract their critical dynamic properties.

IV. CLASSIFICATION OF SYSTEMS

1/ Memoryless systems

The output of a memoryless system at some time to depends only on its input at the same time to. For example, for the resistive divider network,

graphics11.png v o ( t ) = R 2 R 1 + R 2 v i ( t ) v o ( t ) = R 2 R 1 + R 2 v i ( t ) size 12{v rSub { size 8{o} } \( t \) = { {R rSub { size 8{2} } } over {R rSub { size 8{1} } +R rSub { size 8{2} } } } v rSub { size 8{i} } \( t \) } {}

Therefore, vo(to)vo(to) size 12{v rSub { size 8{o} } \( "to" \) } {}depends upon the value of vi(to)vi(to) size 12{v rSub { size 8{i} } \( "to" \) } {}and not on vi(t)vi(t) size 12{v rSub { size 8{i} } \( t \) } {}for t ≠ to.

2/ Systems with memory

Figure 11
Figure 11 (graphics12.png)

i ( t ) = C dv ( t ) dt i ( t ) = C dv ( t ) dt size 12{i \( t \) =C { { ital "dv" \( t \) } over { ital "dt"} } } {}

v ( t ) = 1 C t i ( τ ) v ( t ) = 1 C t i ( τ ) size 12{v \( t \) = { {1} over {C} } Int rSub { size 8{ - infinity } } rSup { size 8{t} } {i \( τ \) dτ} } {}

Note that v(t) depends not just on i(t) at one point in time t. Therefore, the system that relates v to i exhibits memory.

3/ Causal and noncausal systems

For a causal system the output at time to depends only on the input for t ≤ to, i.e., the system cannot anticipate the input.

Figure 12
Figure 12 (graphics13.png)

Physical systems with time as the independent variable are causal systems. There are examples of systems that are not causal.

  • Physical systems for which time is not the independent variable, e.g., the independent variable is space (x, y, z) as in an optical system.
  • Processing of signals where time is the independent variable but the signal has been recorded or generated in a computer. Processing is not in real time.

4/ Stable and unstable systems

Stability can be defined in a variety of ways.

Definition 1: a stable system is one for which an incremental input leads to an incremental output.

graphics14.png An incremental force leads to only an

incremental displacement in the stable

system but not in the unstable system.

Definition 2: A system is BIBO stable if every bounded input leads to a bounded output. We will use this definition.

Figure 13
Figure 13 (graphics15.png)

For the resistor, if i(t) is bounded then so is v(t), but for the capacitance this is not true. Consider i(t) = u(t) then v(t) = tu(t) which is unbounded.

5/ Linear systems

Figure 14
Figure 14 (graphics16.png)

for all x1(t)x1(t) size 12{x rSub { size 8{1} } \( t \) } {}, x2(t)x2(t) size 12{x rSub { size 8{2} } \( t \) } {}, a, and b.

6/ Time-invariant systems

Figure 15
Figure 15 (graphics17.png)

for all x(t) and τ.

7/ Linear and time-invariant (LTI) systems

  • Many man-made and naturally occurring systems can be modeled as LTI systems.
  • Powerful techniques have been developed to analyze and to characterize LTI systems.
  • The analysis of LTI systems is an essential precursor to the analysis of more complex systems.

Problem — Multiplication by a time function

A system is defined by the functional description

Figure 16
Figure 16 (graphics18.png)
  • Is this system linear?
  • Is this system time-invariant?

Solution — Multiplication by a time function

Let

{ y 1 ( t ) = g ( t ) x 1 ( t ) y 2 ( t ) = g ( t ) x 2 ( t ) { y 1 ( t ) = g ( t ) x 1 ( t ) y 2 ( t ) = g ( t ) x 2 ( t ) size 12{ left lbrace matrix { y rSub { size 8{1} } \( t \) =g \( t \) x rSub { size 8{1} } \( t \) {} ## y rSub { size 8{2} } \( t \) =g \( t \) x rSub { size 8{2} } \( t \) } right none } {}

By definition the response to

x ( t ) = ax 1 ( t ) + bx 2 ( t ) x ( t ) = ax 1 ( t ) + bx 2 ( t ) size 12{x \( t \) = ital "ax" rSub { size 8{1} } \( t \) + ital "bx" rSub { size 8{2} } \( t \) } {}

Is

y ( t ) = g ( t ) ( ax 1 ( t ) + bx 2 ( t ) ) y ( t ) = g ( t ) ( ax 1 ( t ) + bx 2 ( t ) ) size 12{y \( t \) =g \( t \) \( ital "ax" rSub { size 8{1} } \( t \) + ital "bx" rSub { size 8{2} } \( t \) \) } {}

This can be rewritten as

y ( t ) = ag ( t ) x 1 ( t ) + bg ( t ) x 2 ( t ) y ( t ) = ag ( t ) x 1 ( t ) + bg ( t ) x 2 ( t ) size 12{y \( t \) = ital "ag" \( t \) x rSub { size 8{1} } \( t \) + ital "bg" \( t \) x rSub { size 8{2} } \( t \) } {}

y ( t ) = ay 1 + by 2 ( t ) y ( t ) = ay 1 + by 2 ( t ) size 12{y \( t \) = ital "ay" rSub { size 8{1} } + ital "by" rSub { size 8{2} } \( t \) } {}

Therefore, the system is linear.

Now suppose that x1(t)= x(t)x1(t)= x(t) size 12{x rSub { size 8{1} } \( t \) =" x" \( t \) } {}and x2(t)= x(t - τ)x2(t)= x(t - τ) size 12{x rSub { size 8{2} } \( t \) =" x" \( "t - "τ \) } {}, and the response to these two inputs are y1(t)y1(t) size 12{y rSub { size 8{1} } \( t \) } {}and y2(t)y2(t) size 12{y rSub { size 8{2} } \( t \) } {}, respectively. Note that

y 1 ( t ) = y ( t ) = g ( t ) x ( t ) y 1 ( t ) = y ( t ) = g ( t ) x ( t ) size 12{y rSub { size 8{1} } \( t \) =y \( t \) =g \( t \) x \( t \) } {}

And

y 2 ( t ) = g ( t ) x ( t τ ) y ( t τ ) y 2 ( t ) = g ( t ) x ( t τ ) y ( t τ ) size 12{y rSub { size 8{2} } \( t \) =g \( t \) x \( t - τ \) <> y \( t - τ \) } {}

Therefore, the system is time-varying.

Problem — Addition of a constant

Suppose the relation between the output y(t) and input x(t) is y(t) = x(t)+K, where K is some constant. Is this system linear?

Solution — Addition of a constant

Note, that if the input is x1(t)+x2(t)x1(t)+x2(t) size 12{x rSub { size 8{1} } \( t \) +x rSub { size 8{2} } \( t \) } {}then the output will be y(t)= x1(t)+x2(t)+K y1(t)+y2(t)=(x1(t)+K)+(x2(t)+K).y(t)= x1(t)+x2(t)+K y1(t)+y2(t)=(x1(t)+K)+(x2(t)+K). size 12{y \( t \) =" x" rSub { size 8{1} } \( t \) +x rSub { size 8{2} } \( t \) +"K " <> " y" rSub { size 8{1} } \( t \) +y rSub { size 8{2} } \( t \) = \( x rSub { size 8{1} } \( t \) +K \) + \( x rSub { size 8{2} } \( t \) +K \) "." } {}

Therefore, this system is not linear.

In general, it can be shown that for a linear system if x(t) = 0 then y(t) = 0. Using the definition of linearity, choose a = b = 1 and x2= -x1(t)x2= -x1(t) size 12{x rSub { size 8{2} } =" -x" rSub { size 8{1} } \( t \) } {}then x(t)= x1(t)+ x2(t)= 0 x(t)= x1(t)+ x2(t)= 0 size 12{x \( t \) =" x" rSub { size 8{1} } \( t \) +" x" rSub { size 8{2} } \( t \) =" 0 "} {} and y(t)= y1(t)+y2(t)= 0y(t)= y1(t)+y2(t)= 0 size 12{y \( t \) =" y" rSub { size 8{1} } \( t \) +y rSub { size 8{2} } \( t \) =" 0"} {}.

Two-minute miniquiz problem

Problem 2-1

The system

y ( t ) = x 2 ( t ) y ( t ) = x 2 ( t ) size 12{y \( t \) =x rSup { size 8{2} } \( t \) } {}

is (choose one):

  1. Linear and time-invariant;
  2. Linear but not time-invariant;
  3. Not linear but time-invariant;
  4. Not linear and not time-invariant.

Solution

Note that if x2(t)=2x1(t)x2(t)=2x1(t) size 12{x rSub { size 8{2} } \( t \) =2x rSub { size 8{1} } \( t \) } {} then y2(t)=(2x1(t))2=4y1(t)y2(t)=(2x1(t))2=4y1(t) size 12{y rSub { size 8{2} } \( t \) = \( 2x rSub { size 8{1} } \( t \) \) rSup { size 8{2} } =4y rSub { size 8{1} } \( t \) } {}

Hence, this system is nonlinear.

Note that if x1(t)=x(t)x1(t)=x(t) size 12{x rSub { size 8{1} } \( t \) =x \( t \) } {} and x2(t)=x(tτ)x2(t)=x(tτ) size 12{x rSub { size 8{2} } \( t \) =x \( t - τ \) } {} then y1(t)=y(t)y1(t)=y(t) size 12{y rSub { size 8{1} } \( t \) =y \( t \) } {}

And y2(t)=x2(tτ)=y(tτ)y2(t)=x2(tτ)=y(tτ) size 12{y rSub { size 8{2} } \( t \) =x rSup { size 8{2} } \( t - τ \) =y \( t - τ \) } {}. Hence, this system is time invariant.

V. LINEAR, ORDINARY DIFFERENTIAL EQUATIONS ARISE FOR A VARIETY OF SYSTEM DESCRIPTIONS

1/ Electric network

Figure 17
Figure 17 (graphics19.png)

Kirchhoff’s current law yields

i ( t ) = i C ( t ) + i R ( t ) + i L ( t ) i ( t ) = i C ( t ) + i R ( t ) + i L ( t ) size 12{i \( t \) =i rSub { size 8{C} } \( t \) +i rSub { size 8{R} } \( t \) +i rSub { size 8{L} } \( t \) } {}

The constitutive relations for each element yield {}

i C ( t ) = C dv ( t ) dt i C ( t ) = C dv ( t ) dt size 12{i rSub { size 8{C} } \( t \) `=`C { { ital "dv" \( t \) } over { ital "dt"} } } {} i R ( t ) = v ( t ) R i R ( t ) = v ( t ) R size 12{i rSub { size 8{R} } \( t \) `=` { {v \( t \) } over {R} } } {} i L ( t ) = 1 L t v ( τ ) i L ( t ) = 1 L t v ( τ ) size 12{i rSub { size 8{L} } \( t \) `=` { {1} over {L} } ` Int rSub { size 8{ - infinity } } rSup { size 8{t} } {v \( τ \) dτ} } {}

Combining KCL and the constitutive relations yields

di ( t ) dt = C d 2 v ( t ) dt 2 + 1 R dv ( t ) dt + v ( t ) L di ( t ) dt = C d 2 v ( t ) dt 2 + 1 R dv ( t ) dt + v ( t ) L size 12{ { { ital "di" \( t \) } over { ital "dt"} } `=`C { {d rSup { size 8{2} } v \( t \) } over { ital "dt" rSup { size 8{2} } } } `+ { {1} over {R} } { { ital "dv" \( t \) } over { ital "dt"} } `+` { {v \( t \) } over {L} } } {}

2/ Mechanical system

The simplest possible model of a muscle (the linearized Hill model) consists of the mechanical network shown below which relates the rate of change of the length of the muscle v(t) to the external force on the muscle fe(t).

Figure 18
Figure 18 (graphics20.png)

fcfc size 12{f rSub { size 8{c} } } {} is the internal force generated by the muscle, K is its stiffness and B is its damping.

The muscle velocity can be expressed as

v ( t ) = v c ( t ) + v s ( t ) v ( t ) = v c ( t ) + v s ( t ) size 12{v \( t \) `=`v rSub { size 8{c} } \( t \) `+`v rSub { size 8{s} } \( t \) } {}

Combining the muscle velocity equation with the constitutive laws for the elements yields

v ( t ) = f e ( t ) f c ( t ) B + 1 L df e ( t ) dt v ( t ) = f e ( t ) f c ( t ) B + 1 L df e ( t ) dt size 12{v \( t \) `=` { {f rSub { size 8{e} } \( t \) - f rSub { size 8{c} } \( t \) } over {B} } `+ { {1} over {L} } { { ital "df" rSub { size 8{e} } \( t \) } over { ital "dt"} } } {}

3/ First-order chemical kinetics

A reversible, first-order chemical reaction can be represented as follows

R α P R α P size 12{R widevec { size 8{α} } P} {}

3/ First-order chemical kinetics

A reversible, first-order chemical reaction can be represented as follows

dc R ( t ) dt = α c R ( t ) β c p ( t ) dc R ( t ) dt = α c R ( t ) β c p ( t ) size 12{ - { { ital "dc" rSub { size 8{R} } \( t \) } over { ital "dt"} } `=`α`c rSub { size 8{R} } \( t \) - β`c rSub { size 8{p} } \( t \) } {}

If reactant and product are conserved

c R ( t ) + c P ( t ) = C c R ( t ) + c P ( t ) = C size 12{c rSub { size 8{R} } \( t \) +c rSub { size 8{P} } \( t \) =C} {}

then, after substitution, we obtain

dc R ( t ) dt + ( α + β ) c R ( t ) = βC dc R ( t ) dt + ( α + β ) c R ( t ) = βC size 12{ { { ital "dc" rSub { size 8{R} } \( t \) } over { ital "dt"} } + \( α+β \) `c rSub { size 8{R} } \( t \) =βC} {}

4/ General differential equation

For many (lumped-parameter or compartmental) systems, the input x(t) and the output y(t) are related by an nth order differential equation of the form

n = 0 N a n d n y ( t ) dt n = m = 0 M b m d m x ( t ) dt m n = 0 N a n d n y ( t ) dt n = m = 0 M b m d m x ( t ) dt m size 12{ Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } { {d rSup { size 8{n} } y \( t \) } over { ital "dt" rSup { size 8{n} } } } = Sum cSub { size 8{m=0} } cSup { size 8{M} } {b rSub { size 8{m} } { {d rSup { size 8{m} } x \( t \) } over { ital "dt" rSup { size 8{m} } } } } } } {}

We seek a solution of this system for an input that is of the form

x ( t ) = Xe st u ( t ) x ( t ) = Xe st u ( t ) size 12{x \( t \) = ital "Xe" rSup { size 8{ ital "st"} } u \( t \) } {}

where X and s are in general complex quantities and u(t) is the unit step function. We will also assume that the system is at rest for t < 0, so that y(t) = 0 for t < 0.

We will seek the solution for t > 0 by finding the homogeneous (unforced solution) and then a particular solution (forced solution).

VI. HOMOGENEOUS SOLUTION

1/ Exponential solution

Let the homogeneous solution be yh(t)yh(t) size 12{y rSub { size 8{h} } \( t \) } {}, then the homogeneous equation is

n = 0 N a n d n y n ( t ) dt n = 0 n = 0 N a n d n y n ( t ) dt n = 0 size 12{ Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } { {d rSup { size 8{n} } y rSub { size 8{n} } \( t \) } over { ital "dt" rSup { size 8{n} } } } =0} } {}

To solve this equation we assume a solution of the form

y h ( t ) = Ae λt y h ( t ) = Ae λt size 12{y rSub { size 8{h} } \( t \) = ital "Ae" rSup { size 8{λt} } } {}

Since

d n y h ( t ) dt n = n e λt d n y h ( t ) dt n = n e λt size 12{ { {d rSup { size 8{n} } y rSub { size 8{h} } \( t \) } over { ital "dt" rSup { size 8{n} } } } =Aλ rSup { size 8{n} } e rSup { size 8{λt} } } {}

we have

n = 0 N a n n e λt = 0 n = 0 N a n n e λt = 0 size 12{ Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } Aλ rSup { size 8{n} } e rSup { size 8{λt} } =0} } {}

2/ Characteristic polynomial

The equation can be factored to yield

( n = 0 N a n λ n ) Ae λt = 0 ( n = 0 N a n λ n ) Ae λt = 0 size 12{ \( Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } λ rSup { size 8{n} } } \) ital "Ae" rSup { size 8{λt} } =0} {}

We are not interested in the trivial solution

y h ( t ) = Ae λt = 0 y h ( t ) = Ae λt = 0 size 12{y rSub { size 8{h} } \( t \) = ital "Ae" rSup { size 8{λt} } =0} {}

Therefore, we can divide by AeλtAeλt size 12{ ital "Ae" rSup { size 8{λt} } } {} to obtain the characteristic Polynomial

( n = 0 N a n λ n ) = 0 ( n = 0 N a n λ n ) = 0 size 12{ \( Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } λ rSup { size 8{n} } } \) =0} {}

This polynomial of order N has N roots which can be exposed by writing the polynomial in factored form

n = 1 N ( λ λ n ) = 0 n = 1 N ( λ λ n ) = 0 size 12{ Prod cSub { size 8{n=1} } cSup { size 8{N} } { \( λ - λ rSub { size 8{n} } \) =0} } {}

3/ Natural frequencies

The roots of the characteristic polynomial

{ λ 1 , λ 2 , . . . , λ N } { λ 1 , λ 2 , . . . , λ N } size 12{ lbrace λ rSub { size 8{1} } ,λ rSub { size 8{2} } , "." "." "." ,λ rSub { size 8{N} } rbrace } {}

are called natural frequencies. These are frequencies for which there is an output in the absence of an input. For example, imagine striking a tuning fork. After the tuning fork has been struck there is no further input, but the tuning fork keeps vibrating — at its natural frequency.

If the natural frequencies are distinct, i.e., if λiλkλiλk size 12{λ rSub { size 8{i} } <> λ rSub { size 8{k} } } {}for i ≠ k, the most general homogeneous solution has the form

y h ( t ) = n = 1 N A n e λ n t y h ( t ) = n = 1 N A n e λ n t size 12{y rSub { size 8{h} } \( t \) = Sum cSub {n=1} cSup {N} {A rSub { size 8{n} } e rSup { size 8{λ rSub { size 6{n} } t} } } } {}

Example — natural frequencies of a network

Figure 19
Figure 19 (graphics21.png)

The differential equation relating v(t) to i(t) is

di ( t ) dt = C d 2 v ( t ) dt 2 + 1 R dv ( t ) dt + v ( t ) L di ( t ) dt = C d 2 v ( t ) dt 2 + 1 R dv ( t ) dt + v ( t ) L size 12{ { { ital "di" \( t \) } over { ital "dt"} } =C { {d rSup { size 8{2} } v \( t \) } over { ital "dt" rSup { size 8{2} } } } + { {1} over {R} } { { ital "dv" \( t \) } over { ital "dt"} } + { {v \( t \) } over {L} } } {}

The characteristic polynomial is

λ 2 + 1 RC λ + 1 LC = 0 λ 2 + 1 RC λ + 1 LC = 0 size 12{λ rSup { size 8{2} } + { {1} over { ital "RC"} } λ+ { {1} over { ital "LC"} } =0} {}

The roots of this quadratic characteristic polynomial, the natural frequencies, are

λ 1 = 1 2 RC + ( 1 2 RC ) 2 1 LC λ 1 = 1 2 RC + ( 1 2 RC ) 2 1 LC size 12{λ rSub { size 8{1} } = - { {1} over {2 ital "RC"} } + sqrt { \( { {1} over {2 ital "RC"} } \) rSup { size 8{2} } - { {1} over { ital "LC"} } } } {}

λ 2 = 1 2 RC ( 1 2 RC ) 2 1 LC λ 2 = 1 2 RC ( 1 2 RC ) 2 1 LC size 12{λ rSub { size 8{2} } = - { {1} over {2 ital "RC"} } - sqrt { \( { {1} over {2 ital "RC"} } \) rSup { size 8{2} } - { {1} over { ital "LC"} } } } {}

Figure 20
Figure 20 (graphics22.png)

The locus of natural frequencies is plotted for L=C = 1 with R increasing in the directions of the arrows. Note the natural frequencies are in the left-half plane for all values of R > 0. As RR size 12{R rightarrow infinity } {}, the natural frequencies approach the imaginary axis. What is the physical significance of this behavior?

VII. PARTICULAR SOLUTION

Now we need to find a particular solution to the differential equation

where for t > 0 n=0Nandny(t)dtn=m=0Mbmdmx(t)dtmn=0Nandny(t)dtn=m=0Mbmdmx(t)dtm size 12{ Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } { {d rSup { size 8{n} } y \( t \) } over { ital "dt" rSup { size 8{n} } } } = Sum cSub { size 8{m=0} } cSup { size 8{M} } {b rSub { size 8{m} } { {d rSup { size 8{m} } x \( t \) } over { ital "dt" rSup { size 8{m} } } } } } } {}

x ( t ) = Xe st x ( t ) = Xe st size 12{x \( t \) = ital "Xe" rSup { size 8{ ital "st"} } } {}

and s does not equal one of the natural frequencies. We assume that

y p ( t ) = Ye st y p ( t ) = Ye st size 12{y rSub { size 8{p} } \( t \) = ital "Ye" rSup { size 8{ ital "st"} } } {}

and we solve for Y . Substitution for both x(t) and y(t) in the differential equation yields, after factoring,

( n = 0 N a n s n ) Ye st = ( m = 0 M b m s m ) Xe st ( n = 0 N a n s n ) Ye st = ( m = 0 M b m s m ) Xe st size 12{ \( Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } s rSup { size 8{n} } \) ital "Ye" rSup { size 8{ ital "st"} } = \( Sum cSub { size 8{m=0} } cSup { size 8{M} } {b rSub { size 8{m} } s rSup { size 8{m} } \) ital "Xe" rSup { size 8{ ital "st"} } } } } {}

1/ System function

a/ System function — derivation

After dividing both sides of the equation by estest size 12{e rSup { size 8{ ital "st"} } } {} we can solve for Y which has the form

Y=H(s)X

where

H ( s ) = m = 0 M b m s m n = 0 N a n s n H ( s ) = m = 0 M b m s m n = 0 N a n s n size 12{H \( s \) = { { Sum cSub { size 8{m=0} } cSup { size 8{M} } {b rSub { size 8{m} } s rSup { size 8{m} } } } over { Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } s rSup { size 8{n} } } } } } {}

b/ System function definition

H ( s ) = m = 0 M b m s m n = 0 N a n s n H ( s ) = m = 0 M b m s m n = 0 N a n s n size 12{H \( s \) = { { Sum cSub { size 8{m=0} } cSup { size 8{M} } {b rSub { size 8{m} } s rSup { size 8{m} } } } over { Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } s rSup { size 8{n} } } } } } {}

  • is called the system function
  • is a rational function in s
  • is a skeleton of the differential equation
  • characterizes the relation between x(t) and y(t)

Example — reconstruction of differential equation from H(s)

Suppose

H ( s ) = Y X = s s + 1 H ( s ) = Y X = s s + 1 size 12{H \( s \) = { {Y} over {X} } = { {s} over {s+1} } } {}

what is the differential equation that relates y(t) to x(t)? Cross-multiply the equation and multiply both sides by estest size 12{e rSup { size 8{ ital "st"} } } {} to obtain

( s + 1 ) Ye st = sXe st ( s + 1 ) Ye st = sXe st size 12{ \( s+1 \) ital "Ye" rSup { size 8{ ital "st"} } = ital "sXe" rSup { size 8{ ital "st"} } } {}

which yields

sYe st + Ye st = sXe st sYe st + Ye st = sXe st size 12{ ital "sYe" rSup { size 8{ ital "st"} } + ital "Ye" rSup { size 8{ ital "st"} } = ital "sXe" rSup { size 8{ ital "st"} } } {}

from which we can obtain the differential equation

dy ( t ) dt + y ( t ) = dx ( t ) dt dy ( t ) dt + y ( t ) = dx ( t ) dt size 12{ { { ital "dy" \( t \) } over { ital "dt"} } +y \( t \) = { { ital "dx" \( t \) } over { ital "dt"} } } {}

2/ Poles and zeros

H(s) can be expressed in factored form as follows

H ( s ) = K m = 1 M ( s z m ) n 1 N ( s p n ) H ( s ) = K m = 1 M ( s z m ) n 1 N ( s p n ) size 12{H \( s \) =K { { Prod cSub { size 8{m=1} } cSup { size 8{M} } { \( s - z rSub { size 8{m} } \) } } over { Prod cSub { size 8{n - 1} } cSup { size 8{N} } { \( s - p rSub { size 8{n} } \) } } } } {}

where K=bMaMK=bMaM size 12{K= { {b rSub { size 8{M} } } over {a rSub { size 8{M} } } } } {}

  • {z1,z2,...,zM}{z1,z2,...,zM} size 12{ lbrace z rSub { size 8{1} } ,z rSub { size 8{2} } , "." "." "." ,z rSub { size 8{M} } rbrace } {} are the roots of the numerator polynomial and are called zeros of H(s) because these are the values of s for which H(s) = 0.
  • {p1,p2,...,pM}{p1,p2,...,pM} size 12{ lbrace p rSub { size 8{1} } ,p rSub { size 8{2} } , "." "." "." ,p rSub { size 8{M} } rbrace } {} are the roots of the denominator polynomial and are called poles of H(s) because these are the values of s for which H(s)=H(s)= size 12{H \( s \) = infinity } {} .

a/ Poles are the natural frequencies

Note that poles of H(s) are the natural frequencies of the system. Recall that natural frequencies are given by the roots of the characteristic polynomial

( n = 0 N a n λ n ) = 0 ( n = 0 N a n λ n ) = 0 size 12{ \( Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } λ rSup { size 8{n} } \) =0} } {}

and the poles are the roots of denominator polynomial of H(s)

( n = 0 N a n s n ) = 0 ( n = 0 N a n s n ) = 0 size 12{ \( Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } s rSup { size 8{n} } \) =0} } {}

Both originate from the left-hand side of the differential equation

n = 0 N a n d n y ( t ) dt n n = 0 N a n d n y ( t ) dt n size 12{ Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } { {d rSup { size 8{n} } y \( t \) } over { ital "dt" rSup { size 8{n} } } } } } {}

b/ Pole-zero diagram

H(s) characterizes the differential equation and H(s) is characterized by N + M + 1 numbers: N poles, M zeros, and the constant K. Except for the multiplication factor K, H(s) is characterized by a pole-zero diagram which is a plot of the locations of poles and zeros in the complex-s plane. The ordinate is jI{s}=jI{s}= size 12{ ital "jI" lbrace s rbrace =jϖ} {} and the abscissa is R{s}=σR{s}=σ size 12{R lbrace s rbrace =σ} {} where

s = σ + s = σ + size 12{s=σ+jϖ} {}

Figure 21
Figure 21 (graphics23.png)

Example — system function of a network

Figure 22
Figure 22 (graphics24.png)

The differential equation relating v(t) to i(t) is

di ( t ) dt = C ( d 2 v ( t ) dt 2 + 1 RC dv ( t ) dt + v ( t ) LC ) di ( t ) dt = C ( d 2 v ( t ) dt 2 + 1 RC dv ( t ) dt + v ( t ) LC ) size 12{ { { ital "di" \( t \) } over { ital "dt"} } =C \( { {d rSup { size 8{2} } v \( t \) } over { ital "dt" rSup { size 8{2} } } } + { {1} over { ital "RC"} } { { ital "dv" \( t \) } over { ital "dt"} } + { {v \( t \) } over { ital "LC"} } \) } {}

The particular solution is obtained from

sI = C ( s 2 + 1 RC s + 1 LC ) V sI = C ( s 2 + 1 RC s + 1 LC ) V size 12{ ital "sI"=C \( s rSup { size 8{2} } + { {1} over { ital "RC"} } s+ { {1} over { ital "LC"} } \) V} {}

With v(t) as the output and i(t) as the input, the system function of the RLC network is

H ( s ) = V I = 1 C ( s s 2 + 1 RC s + 1 LC ) H ( s ) = V I = 1 C ( s s 2 + 1 RC s + 1 LC ) size 12{H \( s \) = { {V} over {I} } = { {1} over {C} } \( { {s} over {s rSup { size 8{2} } + { {1} over { ital "RC"} } s+ { {1} over { ital "LC"} } } } \) } {}

Two-minute miniquiz problem

Problem 3-1

Given the system function

H ( s ) = s + 2 ( s + 3 ) ( s + 4 ) H ( s ) = s + 2 ( s + 3 ) ( s + 4 ) size 12{H \( s \) = { {s+2} over { \( s+3 \) \( s+4 \) } } } {}

  • Determine the natural frequencies of the system.
  • Determine a differential equation that relates x(t) and y(t).

Solution

  • The natural frequencies of the system are the poles of the system function and are −3 and −4.
  • The differential equation can be obtained by cross multiplying and multiplying by estest size 12{e rSup { size 8{ ital "st"} } } {}to obtain

( s + 3 ) ( s + 4 ) Ye st = ( s + 2 ) Xe st ( s + 3 ) ( s + 4 ) Ye st = ( s + 2 ) Xe st size 12{ \( s+3 \) \( s+4 \) ital "Ye" rSup { size 8{ ital "st"} } = \( s+2 \) ital "Xe" rSup { size 8{ ital "st"} } } {}

( s 2 + 7s + 12 ) Ye st = ( s + 2 ) Xe st ( s 2 + 7s + 12 ) Ye st = ( s + 2 ) Xe st size 12{ \( s rSup { size 8{2} } +7s+"12" \) ital "Ye" rSup { size 8{ ital "st"} } = \( s+2 \) ital "Xe" rSup { size 8{ ital "st"} } } {}

so that

d 2 y ( t ) dt 2 + 7 dy ( t ) dt + 12 y ( t ) = dx ( t ) dt + 2x ( t ) d 2 y ( t ) dt 2 + 7 dy ( t ) dt + 12 y ( t ) = dx ( t ) dt + 2x ( t ) size 12{ { {d rSup { size 8{2} } y \( t \) } over { ital "dt" rSup { size 8{2} } } } +7 { { ital "dy" \( t \) } over { ital "dt"} } +"12"y \( t \) = { { ital "dx" \( t \) } over { ital "dt"} } +2x \( t \) } {}

VIII. TOTAL SOLUTION

The general solution is

y ( t ) = n = 1 N A n e λ n t + XH ( s ) e st for t > 0 y ( t ) = n = 1 N A n e λ n t + XH ( s ) e st for t > 0 size 12{y \( t \) = Sum cSub {n=1} cSup {N} {A rSub { size 8{n} } e rSup { size 8{λ rSub { size 6{n} } t} } + ital "XH" \( s \) e rSup { ital "st"} size 12{ ital "for"`````t>0}} } {}

and

y(t)=0 for t<0

Hence, provided there are no singularity functions (e.g., impulses) at t = 0, the general solution can be written compactly as follows

y ( t ) = ( n = 1 N A n e λ n t + XH ( s ) e st ) u ( t ) y ( t ) = ( n = 1 N A n e λ n t + XH ( s ) e st ) u ( t ) size 12{y \( t \) = \( Sum cSub {n=1} cSup {N} {A rSub { size 8{n} } e rSup { size 8{λ rSub { size 6{n} } t} } + ital "XH" \( s \) e rSup { ital "st"} size 12{ \) u \( t \) }} } {}

As we shall see later, no singularity functions occur in the response provided the order of the numerator polynomial of H(s) does not exceed that of the denominator.

1/ Initial conditions

To completely determine the total solution we need to determine the N coefficients

{ A 1 , A 2 , . . . , A N } { A 1 , A 2 , . . . , A N } size 12{ lbrace A rSub { size 8{1} } ,A rSub { size 8{2} } , "." "." "." ,A rSub { size 8{N} } rbrace } {}

These are determined from N initial conditions which must be specified. These conditions result in a set of N algebraic equations that need to be solved to obtain the initial conditions. We shall find another, and simpler, method to determine these coefficients later.

2/ Steady-state

Assume the particular solution is not zero. Then the particular solution dominates after some time, if the homogeneous solution decays more rapidly than does the particular solution. When this occurs, we call the resulting particular solution the steady-state response. Steady-state occurs if each term in the homogeneous solution decays more rapidly than the particular solution. Thus, steady-state occurs if

lim t e λ n t e st 0 for all λ n lim t e λ n t e st 0 for all λ n size 12{ {"lim"} cSub { size 8{t rightarrow infinity } } lline { {e rSup { size 8{λ rSub { size 6{n} } t} } } over {e rSup { ital "st"} } } rline size 12{ rightarrow 0`` ital "for"`` ital "all"``λ rSub {n} }} {}

Thus

lim t e ( λ n s ) t 0 for all λ n lim t e ( λ n s ) t 0 for all λ n size 12{ {"lim"} cSub { size 8{t rightarrow infinity } } lline e rSup { size 8{ \( λ rSub { size 6{n} } - s \) t} } rline rightarrow 0`` ital "for"`` ital "all"``λ rSub {n} } {}

which implies that

R { λ n s } 0 for all λ n R { λ n s } 0 for all λ n size 12{R lbrace λ rSub { size 8{n} } - s rbrace 0`` ital "for"`` ital "all"``λ rSub { size 8{n} } } {}

Thus, steady-state occurs when

R{λn}<R{s}R{λn}<R{s} size 12{R lbrace λ rSub { size 8{n} } rbrace <R lbrace s rbrace } {} for all λnλn size 12{λ rSub { size 8{n} } } {}

provided the particular solution is not zero. The conditions for steady state are depicted in the complex s plane below.

Figure 23
Figure 23 (graphics25.png)

The particular solution dominates for s in the shaded region, and the total solution equals the steady-state solution i.e.,

y ( t ) = XH ( s ) e st y ( t ) = XH ( s ) e st size 12{y \( t \) = ital "XH" \( s \) e rSup { size 8{ ital "st"} } } {}

For which conditions is the particular solution zero? Suppose

H ( s ) = s 1 ( s + 1 ) ( s + 2 ) H ( s ) = s 1 ( s + 1 ) ( s + 2 ) size 12{H \( s \) = { {s - 1} over { \( s+1 \) \( s+2 \) } } } {}

Figure 24
Figure 24 (graphics26.png)

The particular solution dominates for s in the shaded region, and the total solution equals the steady-state solution except when s = 1 because at this value i.e.,

y ( t ) = XH ( 1 ) e s = 0 y ( t ) = XH ( 1 ) e s = 0 size 12{y \( t \) = ital "XH" \( 1 \) e rSup { size 8{s} } =0} {}

so that the particular solution is zero and steady -state does not occur.

IX. LINEAR DIFFERENCE EQUATIONS ARISE IN MANY DIFFERENT CONTEXTS

1/ Electric ladder network

Figure 25
Figure 25 (graphics27.png)

r is the series resistance and g is the shunt conductance.

KCL at the central node yields

v o [ n 1 ] v o [ n ] r + v o [ n + 1 ] v o [ n ] r + g ( v i [ n ] v o [ n ] = 0 v o [ n 1 ] v o [ n ] r + v o [ n + 1 ] v o [ n ] r + g ( v i [ n ] v o [ n ] = 0 size 12{ { {v rSub { size 8{o} } \[ n - 1 \] - v rSub { size 8{o} } \[ n \] } over {r} } + { {v rSub { size 8{o} } \[ n+1 \] - v rSub { size 8{o} } \[ n \] } over {r} } +g \( v rSub { size 8{i} } \[ n \] - v rSub { size 8{o} } \[ n \] =0} {}

which yields the linear difference equation

v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = rgv i [ n ] v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = rgv i [ n ] size 12{ - v rSub { size 8{o} } \[ n+1 \] + \( 2+ ital "rg" \) v rSub { size 8{o} } \[ n \] - v rSub { size 8{o} } \[ n - 1 \] = ital "rgv" rSub { size 8{i} } \[ n \] } {}

2/ Interest and accumulation

Let us consider a simple model of the accumulation of wealth through savings. At the end of year n you deposit x[n] dollars in the bank which pays an annual interest of r. Your accumulation at the end of year n is y[n] dollars. Therefore,

y [ n + 1 ] = y [ n ] + ry [ n ] + x [ n ] y [ n + 1 ] = y [ n ] + ry [ n ] + x [ n ] size 12{y \[ n+1 \] =y \[ n \] + ital "ry" \[ n \] +x \[ n \] } {}

We can rewrite this equation as

y[n+1] − (1+r)y[n] = x[n]

This difference equation can be realized in a block diagram as shown below.

Figure 26
Figure 26 (graphics28.png)

D is a unit delay unit.

3/ Discretized CT system

An important application of DT systems is a numerical simulation of a CT system. For example, consider the CT lowpass filter shown below.

Figure 27
Figure 27 (graphics29.png)

The differential equation is

dv o ( t ) dt = 1 RC v o ( t ) + 1 RC v i ( t ) dv o ( t ) dt = 1 RC v o ( t ) + 1 RC v i ( t ) size 12{ { { ital "dv" rSub { size 8{o} } \( t \) } over { ital "dt"} } = - { {1} over { ital "RC"} } v rSub { size 8{o} } \( t \) + { {1} over { ital "RC"} } v rSub { size 8{i} } \( t \) } {}

To solve this equation numerically in a computer, the CT signals are discretized and the derivative is approximated.

To discretize the signals, we can define DT signals as samples of CT signals, i.e.,

vi[n]=vi(t)t=nT=vi(nT)vi[n]=vi(t)t=nT=vi(nT) size 12{v rSub { size 8{i} } \[ n \] =v rSub { size 8{i} } \( t \) \lline rSub { size 8{t= ital "nT"} } =v rSub { size 8{i} } \( ital "nT" \) } {} and vo[n]=vo(t)t=nT=vo(nT)vo[n]=vo(t)t=nT=vo(nT) size 12{v rSub { size 8{o} } \[ n \] =v rSub { size 8{o} } \( t \) \lline rSub { size 8{t= ital "nT"} } =v rSub { size 8{o} } \( ital "nT" \) } {}

Figure 28
Figure 28 (graphics30.png)

The derivative can be approximated as follows

dv o ( t ) dt t = nT v o ( ( n + 1 ) T ) = v o ( nT ) T dv o ( t ) dt t = nT v o ( ( n + 1 ) T ) = v o ( nT ) T size 12{ { { ital "dv" rSub { size 8{o} } \( t \) } over { ital "dt"} } \lline rSub { size 8{t= ital "nT"} } approx { {v rSub { size 8{o} } \( \( n+1 \) T \) =v rSub { size 8{o} } \( ital "nT" \) } over {T} } } {}

Figure 29
Figure 29 (graphics31.png)

Therefore, we can approximate the differential equation as

v o ( ( n + 1 ) T ) = v o ( nT ) T = 1 RC v o ( nT ) + 1 RC v i ( nT ) v o ( ( n + 1 ) T ) = v o ( nT ) T = 1 RC v o ( nT ) + 1 RC v i ( nT ) size 12{ { {v rSub { size 8{o} } \( \( n+1 \) T \) =v rSub { size 8{o} } \( ital "nT" \) } over {T} } = - { {1} over { ital "RC"} } v rSub { size 8{o} } \( ital "nT" \) + { {1} over { ital "RC"} } v rSub { size 8{i} } \( ital "nT" \) } {}

which can be written as

v o [ n + 1 ] = ( 1 T RC ) v o [ n ] + ( T RC ) v i [ n ] v o [ n + 1 ] = ( 1 T RC ) v o [ n ] + ( T RC ) v i [ n ] size 12{v rSub { size 8{o} } \[ n+1 \] = \( 1 - { {T} over { ital "RC"} } \) v rSub { size 8{o} } \[ n \] + \( { {T} over { ital "RC"} } \) v rSub { size 8{i} } \[ n \] } {}

Let αα size 12{α} {} = T/(RC). Then the difference equation is

v o [ n + 1 ] = ( 1 α ) v o [ n ] + αv i [ n ] v o [ n + 1 ] = ( 1 α ) v o [ n ] + αv i [ n ] size 12{v rSub { size 8{o} } \[ n+1 \] = \( 1 - α \) v rSub { size 8{o} } \[ n \] +αv rSub { size 8{i} } \[ n \] } {}

This equation can be solved iteratively for a given input and initial condition. Assume that vi[n]=u[n]vi[n]=u[n] size 12{v rSub { size 8{i} } \[ n \] =u \[ n \] } {}and that vo[0]=0vo[0]=0 size 12{v rSub { size 8{o} } \[ 0 \] =0} {}then

vo[1]=αvo[1]=α size 12{v rSub { size 8{o} } \[ 1 \] =α} {},

vo[2]=(1α)α+αvo[2]=(1α)α+α size 12{v rSub { size 8{o} } \[ 2 \] = \( 1 - α \) α+α} {},

vo[3]=(1α)2α+(1α)α+αvo[3]=(1α)2α+(1α)α+α size 12{v rSub { size 8{o} } \[ 3 \] = \( 1 - α \) rSup { size 8{2} } α+ \( 1 - α \) α+α} {},

v o [ k ] = α k = 0 n 1 ( 1 α ) k = 1 ( 1 α ) n v o [ k ] = α k = 0 n 1 ( 1 α ) k = 1 ( 1 α ) n size 12{v rSub { size 8{o} } \[ k \] =α Sum cSub { size 8{k=0} } cSup { size 8{n - 1} } { \( 1 - α \) rSup { size 8{k} } } =1 - \( 1 - α \) rSup { size 8{n} } } {}

[We have made use of an important formula for the summation of a geometric series which we will prove shortly.] Hence, the solution is

v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] size 12{v rSub { size 8{o} } \[ n \] = \( 1 - \( 1 - α \) rSup { size 8{n} } \) u \[ n \] } {}

The solution

v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] size 12{v rSub { size 8{o} } \[ n \] = \( 1 - \( 1 - α \) rSup { size 8{n} } \) u \[ n \] } {}

is shown plotted below for αα size 12{α} {} = 0.25.

Figure 30
Figure 30 (graphics32.png)

4/ Important side issue — the summation of finite geometric series

Suppose

S = n = l k a n = a l + a l + 1 + . . . + a k S = n = l k a n = a l + a l + 1 + . . . + a k size 12{S= Sum cSub { size 8{n=l} } cSup { size 8{k} } {a rSup { size 8{n} } } =a rSup { size 8{l} } +a rSup { size 8{l+1} } + "." "." "." +a rSup { size 8{k} } } {}

Then

αS = α n = l k a n = a l + 1 + a l + 2 + . . . + a k + 1 αS = α n = l k a n = a l + 1 + a l + 2 + . . . + a k + 1 size 12{αS=α Sum cSub { size 8{n=l} } cSup { size 8{k} } {a rSup { size 8{n} } } =a rSup { size 8{l+1} } +a rSup { size 8{l+2} } + "." "." "." +a rSup { size 8{k+1} } } {}

Hence,

( 1 α ) S = ( 1 α ) n = l k a n = a l a k + 1 ( 1 α ) S = ( 1 α ) n = l k a n = a l a k + 1 size 12{ \( 1 - α \) S= \( 1 - α \) Sum cSub { size 8{n=l} } cSup { size 8{k} } {a rSup { size 8{n} } } =a rSup { size 8{l} } - a rSup { size 8{k+1} } } {}

and

S = n = l k a n = a l a k + 1 ( 1 α ) S = n = l k a n = a l a k + 1 ( 1 α ) size 12{S= Sum cSub { size 8{n=l} } cSup { size 8{k} } {a rSup { size 8{n} } } = { {a rSup { size 8{l} } - a rSup { size 8{k+1} } } over { \( 1 - α \) } } } {}

Conclusion

  • Difference equations arise in a variety of interesting contexts.
  • Simple difference equations can be solved iteratively.
  • We will develop more systematic and efficient methods to solve arbitrary difference equations.

X. GENERAL LINEAR DIFFERENCE EQUATION

For a DT system, such as the ones shown previously, we can write the relation between an input variable x[n] and an output variable y[n] by a general difference equation of the form

k = 0 K a k y [ n + k ] = l = 0 L b l x [ n + 1 ] k = 0 K a k y [ n + k ] = l = 0 L b l x [ n + 1 ] size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } y \[ n+k \] } = Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{l} } x \[ n+1 \] } } {}

We seek a solution of this system for an input that is of the form

x [ n ] = Xz n u [ n ] x [ n ] = Xz n u [ n ] size 12{x \[ n \] = ital "Xz" rSup { size 8{n} } u \[ n \] } {}

where X and z are in general complex quantities and u[n] is the unit step function. We will also assume that the system is at rest for n < 0, so that y[n] = 0 for n < 0.

We will seek the solution for n > 0 by finding the homogeneous (unforced solution) and then a particular solution (forced solution).

1/ Homogeneous solution

a/ Geometric (exponential) solution

Let the homogeneous solution be yh[n]yh[n] size 12{y rSub { size 8{h} } \[ n \] } {}, then the homogeneous equation is

k = 0 K a k y [ n + k ] = 0 k = 0 K a k y [ n + k ] = 0 size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } y \[ n+k} \] =0} {}

To solve this equation we assume a solution of the form

y h [ n ] = n y h [ n ] = n size 12{y rSub { size 8{h} } \[ n \] =Aλ rSup { size 8{n} } } {}

Since

y h [ n + k ] = n + k y h [ n + k ] = n + k size 12{y rSub { size 8{h} } \[ n+k \] =Aλ rSup { size 8{n+k} } } {}

we have

k = 0 K a k n + k = 0 k = 0 K a k n + k = 0 size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } Aλ rSup { size 8{n+k} } } =0} {}

b/ Characteristic polynomial

The equation can be factored to yield

( k = 0 K a k λ k ) n = 0 ( k = 0 K a k λ k ) n = 0 size 12{ \( Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } λ rSup { size 8{k} } \) } Aλ rSup { size 8{n} } =0} {}

We are not interested in the trivial solution

y h [ n ] = n = 0 y h [ n ] = n = 0 size 12{y rSub { size 8{h} } \[ n \] =Aλ rSup { size 8{n} } =0} {}

Therefore, we can divide by nn size 12{Aλ rSup { size 8{n} } } {} to obtain the characteristic polynomial

k = 0 K a k λ k = 0 k = 0 K a k λ k = 0 size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } λ rSup { size 8{k} } } =0} {}

This polynomial of order K has K roots which can be exposed by writing the polynomial in factored form

k = 1 K ( λ λ k ) = 0 k = 1 K ( λ λ k ) = 0 size 12{ Prod cSub { size 8{k=1} } cSup { size 8{K} } { \( λ - λ rSub { size 8{k} } \) } =0} {}

c/ Natural frequencies

The roots of the characteristic polynomial {λ1,λ2,...λK}{λ1,λ2,...λK} size 12{ lbrace λ rSub { size 8{1} } ,λ rSub { size 8{2} } , "." "." "." λ rSub { size 8{K} } rbrace } {} are called natural frequencies. These are frequencies for which there is an output in the absence of an input.

If the natural frequencies are distinct, i.e., if λiλkλiλk size 12{λ rSub { size 8{i} } <> λ rSub { size 8{k} } } {}for ikik size 12{i <> k} {} the most general homogeneous solution has the form

y h [ n ] = k = 1 K A k λ k n y h [ n ] = k = 1 K A k λ k n size 12{y rSub { size 8{h} } \[ n \] = Sum cSub { size 8{k=1} } cSup { size 8{K} } {A rSub { size 8{k} } λ rSub { size 8{k} rSup { size 8{n} } } } } {}

Two-minute miniquiz problem

Problem 10-1 — Natural frequencies of the ladder network

The electric ladder network shown below has the following difference equation

v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = rgv i [ n ] v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = rgv i [ n ] size 12{ - v rSub { size 8{o} } \[ n+1 \] + \( 2+ ital "rg" \) v rSub { size 8{o} } \[ n \] - v rSub { size 8{o} } \[ n - 1 \] = ital "rgv" rSub { size 8{i} } \[ n \] } {}

Figure 31
Figure 31 (graphics33.png)

Find the natural frequencies.

Solution

The natural frequencies are determined by the homogeneous solution

v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = 0 v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = 0 size 12{ - v rSub { size 8{o} } \[ n+1 \] + \( 2+ ital "rg" \) v rSub { size 8{o} } \[ n \] - v rSub { size 8{o} } \[ n - 1 \] =0} {}

Substituting a solution of the form v0[n]=nv0[n]=n size 12{v rSub { size 8{0} } \[ n \] =Aλ rSup { size 8{n} } } {} results in

{} n + 1 + ( 2 + rg ) n n 1 1 = 0 n + 1 + ( 2 + rg ) n n 1 1 = 0 size 12{ - Aλ rSup { size 8{n+1} } + \( 2+ ital "rg" \) Aλ rSup { size 8{n} } - Aλ rSup { size 8{n - 1} } - 1=0} {}

which yields the characteristic polynomial

λ 2 ( 2 + rg ) λ + 1 = 0 λ 2 ( 2 + rg ) λ + 1 = 0 size 12{λ rSup { size 8{2} } - \( 2+ ital "rg" \) λ+1=0} {}

whose roots are the characteristic frequencies

λ 1,2 = ( 1 + rg 2 ) ± ( 1 + rg 2 ) 2 1 λ 1,2 = ( 1 + rg 2 ) ± ( 1 + rg 2 ) 2 1 size 12{λ rSub { size 8{1,2} } = \( 1+ { { ital "rg"} over {2} } \) +- sqrt { \( 1+ { { ital "rg"} over {2} } \) rSup { size 8{2} } - 1} } {}

d/ Natural frequencies of the ladder network

The electric ladder network shown below has the following difference equation

v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = rgv i [ n ] v o [ n + 1 ] + ( 2 + rg ) v o [ n ] v o [ n 1 ] = rgv i [ n ] size 12{ - v rSub { size 8{o} } \[ n+1 \] + \( 2+ ital "rg" \) v rSub { size 8{o} } \[ n \] - v rSub { size 8{o} } \[ n - 1 \] = ital "rgv" rSub { size 8{i} } \[ n \] } {}

Figure 32
Figure 32 (graphics34.png)

The natural frequencies are:

λ 1,2 = ( 1 + rg 2 ) ± ( 1 + rg 2 ) 2 1 λ 1,2 = ( 1 + rg 2 ) ± ( 1 + rg 2 ) 2 1 size 12{λ rSub { size 8{1,2} } = \( 1+ { { ital "rg"} over {2} } \) +- sqrt { \( 1+ { { ital "rg"} over {2} } \) rSup { size 8{2} } - 1} } {}

The natural frequencies depend on rg,

λ 1,2 = ( 1 + rg 2 ) ± ( 1 + rg 2 ) 2 1 λ 1,2 = ( 1 + rg 2 ) ± ( 1 + rg 2 ) 2 1 size 12{λ rSub { size 8{1,2} } = \( 1+ { { ital "rg"} over {2} } \) +- sqrt { \( 1+ { { ital "rg"} over {2} } \) rSup { size 8{2} } - 1} } {}

The following plot shows the locus of natural frequencies as rg increases.

Figure 33
Figure 33 (graphics35.png)

Both natural frequencies lie along the positive, real λ-axis. When rg = 0, λ1,2λ1,2 size 12{λ rSub { size 8{1,2} } } {}= 1. As rg increases, one natural frequency decreases toward λ = 0 and the other natural frequency increases. What is the physical significance of this pattern?

Note that the product of the two natural frequencies is 1,

λ 1 λ 2 = ( 1 + rg 2 + ( 1 + rg 2 ) 2 1 ) × ( 1 + rg 2 ( 1 + rg 2 ) 2 1 ) = 1 λ 1 λ 2 = ( 1 + rg 2 + ( 1 + rg 2 ) 2 1 ) × ( 1 + rg 2 ( 1 + rg 2 ) 2 1 ) = 1 size 12{λ rSub { size 8{1} } λ rSub { size 8{2} } = \( 1+ { { ital "rg"} over {2} } + sqrt { \( 1+ { { ital "rg"} over {2} } \) rSup { size 8{2} } - 1} \) times \( 1+ { { ital "rg"} over {2} } - sqrt { \( 1+ { { ital "rg"} over {2} } \) rSup { size 8{2} } - 1} \) =1} {}

Therefore, λ1=1λ2λ1=1λ2 size 12{λ rSub { size 8{1} } = { {1} over {λ rSub { size 8{2} } } } } {}and the form of the homogeneous solution is

v o [ n ] = A 1 λ 1 n + A 2 λ 1 n v o [ n ] = A 1 λ 1 n + A 2 λ 1 n size 12{v rSub { size 8{o} } \[ n \] =A rSub { size 8{1} } λ rSub { size 8{1} rSup { size 8{n} } } +A rSub { size 8{2} } λ rSub { size 8{1} rSup { size 8{ - n} } } } {}

The two terms of the homogeneous solution are shown below.

Figure 34
Figure 34 (graphics36.png)

Thus, the homogeneous solution consists of a linear combination of decaying and growing geometric functions whose rates of decrease and increase are the same. Increasing the quantity rg increases the magnitude of the rate of change of voltage.

2/ Particular solution

Next we find a particular solution to the difference equation

k = 0 K a k y p [ n + k ] = l = 0 L b l x [ n + 1 ] k = 0 K a k y p [ n + k ] = l = 0 L b l x [ n + 1 ] size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } y rSub { size 8{p} } \[ n+k \] } = Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{l} } x \[ n+1 \] } } {}

for n > 0 and

x [ n ] = Xz n x [ n ] = Xz n size 12{x \[ n \] = ital "Xz" rSup { size 8{n} } } {}

where z does not equal one of the natural frequencies. We assume that

y p [ n ] = Yz n y p [ n ] = Yz n size 12{y rSub { size 8{p} } \[ n \] = ital "Yz" rSup { size 8{n} } } {}

and we solve for Y . Substitution for both x[n] and y[n] in the difference equation yields, after factoring,

( k = 0 K a k z k ) Yz n = ( l = 0 L b l x l ) Xz n ( k = 0 K a k z k ) Yz n = ( l = 0 L b l x l ) Xz n size 12{ \( Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } z rSup { size 8{k} } \) ital "Yz" rSup { size 8{n} } = \( Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{l} } x rSup { size 8{l} } } \) ital "Xz" rSup { size 8{n} } } } {}

After dividing both sides of the equation by znzn size 12{z rSup { size 8{n} } } {} we can solve for Y which has the form

Y = H ˜ ( z ) X Y = H ˜ ( z ) X size 12{Y= { tilde {H}} \( z \) X} {}

where

H ˜ ( z ) = l = 0 L b l z l k = 0 K a k z k H ˜ ( z ) = l = 0 L b l z l k = 0 K a k z k size 12{ { tilde {H}} \( z \) = { { Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{l} } z rSup { size 8{l} } } } over { Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } z rSup { size 8{k} } } } } } {}

3/ System function

H ˜ ( z ) = l = 0 L b l z l k = 0 K a k z k zeros poles H ˜ ( z ) = l = 0 L b l z l k = 0 K a k z k zeros poles size 12{ { tilde {H}} \( z \) = { { Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{l} } z rSup { size 8{l} } } } over { Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } z rSup { size 8{k} } } } } `` { { dlarrow `` ital "zeros"} over { dlarrow `` ital "poles"} } } {}

  • is called the system function
  • is a rational function in z that has poles and zeros
  • has poles that are the natural frequencies of the system
  • is a skeleton of the difference equation
  • characterizes the relation between x[n] and y[n]

a/ Example—reconstruction of difference equation from H˜(z)H˜(z) size 12{ { tilde {H}} \( z \) } {}

Suppose

H ˜ ( z ) = Y X = z z + 1 H ˜ ( z ) = Y X = z z + 1 size 12{ { tilde {H}} \( z \) = { {Y} over {X} } = { {z} over {z+1} } } {}

what is the difference equation that relates y[n] to x[n]? Crossmultiply the equation and multiply both sides by znzn size 12{z rSup { size 8{n} } } {}to obtain

( z + 1 ) Y z n = z X z n ( z + 1 ) Y z n = z X z n size 12{ \( z+1 \) Y`z rSup { size 8{n} } =z`X`z rSup { size 8{n} } } {}

which yields

( z n + 1 Y + z n Y = z n + 1 X ( z n + 1 Y + z n Y = z n + 1 X size 12{ \( z rSup { size 8{n+1} } Y`+z rSup { size 8{n} } Y=z rSup { size 8{n+1} } `X} {}

from which we can obtain the difference equation

y[n+1] + y[n] = x[n+1]

b/ Pole-zero diagram

H~H~ size 12{ {H} cSup { size 8{ "~" } } } {}characterizes the difference equation and H~H~ size 12{ {H} cSup { size 8{ "~" } } } {} is characterized by K + L + 1 numbers: K poles, L zeros, and one gain constant. Except for the gain constant, H~H~ size 12{ {H} cSup { size 8{ "~" } } } {}is characterized by a pole-zero diagram which is a plot of the locations of poles and zeros in the complex-z plane.

Figure 35
Figure 35 (graphics37.png)

4/ Total solution

The general solution is

y [ n ] = k = 1 K A k λ k n + X H ˜ ( z ) z n for n > 0 y [ n ] = k = 1 K A k λ k n + X H ˜ ( z ) z n for n > 0 size 12{y \[ n \] = Sum cSub { size 8{k=1} } cSup { size 8{K} } {A rSub { size 8{k} } λ rSub { size 8{k} rSup { size 8{n} } } } +X { tilde {H}} \( z \) z rSup { size 8{n} } `` ital "for"``n>0} {}

and

y[n] = 0 for n < 0.

The general solution can be written compactly as follows

y [ n ] = ( k = 1 K A k λ k n + X H ˜ ( z ) z n ) u [ n ] y [ n ] = ( k = 1 K A k λ k n + X H ˜ ( z ) z n ) u [ n ] size 12{y \[ n \] = \( Sum cSub { size 8{k=1} } cSup { size 8{K} } {A rSub { size 8{k} } λ rSub { size 8{k} rSup { size 8{n} } } } +X { tilde {H}} \( z \) z rSup { size 8{n} } \) u \[ n \] } {}

5/ Initial conditions

To completely determine the total solution we need to determine the K coefficients { A1,A2,...AKA1,A2,...AK size 12{A rSub { size 8{1} } ,A rSub { size 8{2} } , "." "." "." A rSub { size 8{K} } } {} }. These are determined from K initial conditions which must be specified. These conditions result in a set of K algebraic equations that need to be solved to obtain the initial conditions so that the total solution can be specified. We shall find another, and simpler, method to determine the total solution later.

Example — discretized CT system

We have previously considered the discretized approximation to a lowpass filter.

Figure 36
Figure 36 (graphics38.png)

The equilibrium equation is

dv o ( t ) dt = 1 RC v o ( t ) + 1 RC v i ( t ) dv o ( t ) dt = 1 RC v o ( t ) + 1 RC v i ( t ) size 12{ { { ital "dv" rSub { size 8{o} } \( t \) } over { ital "dt"} } = - { {1} over { ital "RC"} } v rSub { size 8{o} } \( t \) + { {1} over { ital "RC"} } v rSub { size 8{i} } \( t \) } {}

We know that the unit step response of this network starting from initial rest is

v o ( t ) = ( 1 e t / RC ) u ( t ) v o ( t ) = ( 1 e t / RC ) u ( t ) size 12{v rSub { size 8{o} } \( t \) = \( 1 - e rSup { size 8{ - t/ ital "RC"} } \) u \( t \) } {}

We showed that a discretized approximation to this system yields the difference equation

v o [ n + 1 ] = ( 1 β ) v o + αv i [ n ] v o [ n + 1 ] = ( 1 β ) v o + αv i [ n ] size 12{v rSub { size 8{o} } \[ n+1 \] = \( 1 - β \) v rSub { size 8{o} } +αv rSub { size 8{i} } \[ n \] } {}

where α = T/(RC). We will determine the solution by finding the homogeneous and particular solution. But all the information we need is contained in the system function which we can obtain by substituting vi[n]=Viznvi[n]=Vizn size 12{v rSub { size 8{i} } \[ n \] =V rSub { size 8{i} } z rSup { size 8{n} } } {}and vo[n]=Voznvo[n]=Vozn size 12{v rSub { size 8{o} } \[ n \] =V rSub { size 8{o} } z rSup { size 8{n} } } {}into the difference equation to obtain

so that

H ˜ ( z ) = V ˜ o ( z ) V ˜ i ( z ) = α z ( 1 α ) H ˜ ( z ) = V ˜ o ( z ) V ˜ i ( z ) = α z ( 1 α ) size 12{ { tilde {H}} \( z \) = { { { tilde {V}} rSub { size 8{o} } \( z \) } over { { tilde {V}} rSub { size 8{i} } \( z \) } } = { {α} over {z - \( 1 - α \) } } } {}

The natural frequency is λ=1αλ=1α size 12{λ=1 - α} {} so that the solution has the form

v o [ n ] = ( A ( 1 α ) n + H ˜ ( 1 ) ( 1 ) n ) u [ n ] v o [ n ] = ( A ( 1 α ) n + H ˜ ( 1 ) ( 1 ) n ) u [ n ] size 12{v rSub { size 8{o} } \[ n \] = \( A \( 1 - α \) rSup { size 8{n} } + { tilde {H}} \( 1 \) \( 1 \) rSup { size 8{n} } \) u \[ n \] } {}

where we have made use of the fact that u[n]= 1nu[n]u[n]= 1nu[n] size 12{u \[ n \] =" 1" rSup { size 8{n} } u \[ n \] } {}. Since H~H~ size 12{ {H} cSup { size 8{ "~" } } } {}(1) = 1,

v o [ n ] = ( A ( 1 α ) n + 1 ) u [ n ] v o [ n ] = ( A ( 1 α ) n + 1 ) u [ n ] size 12{v rSub { size 8{o} } \[ n \] = \( A \( 1 - α \) rSup { size 8{n} } +1 \) u \[ n \] } {}

Finally, the initial condition, vo[0]= 0vo[0]= 0 size 12{v rSub { size 8{o} } \[ 0 \] =" 0"} {}, implies that A = −1 so that the solution is

v o [ n ] = ( 1 1 α ) n ) u [ n ] v o [ n ] = ( 1 1 α ) n ) u [ n ] size 12{v rSub { size 8{o} } \[ n \] = \( 1 - 1 - α \) rSup { size 8{n} } \) u \[ n \] } {}

which is the same result we obtained earlier by solving the difference equation iteratively.

We compare the step response of the CT system with the DT approximation

vo(t)=(1et/RC)u(t)vo(t)=(1et/RC)u(t) size 12{v rSub { size 8{o} } \( t \) = \( 1 - e rSup { size 8{ - t/ ital "RC"} } \) u \( t \) } {}and vo[n]=(1(1TRC)n)u(t)vo[n]=(1(1TRC)n)u(t) size 12{v rSub { size 8{o} } \[ n \] = \( 1 - \( 1 - { {T} over { ital "RC"} } \) rSup { size 8{n} } \) u \( t \) } {}

The solutions are shown below for RC = 1 and T/(RC) = 0.1.

Figure 37
Figure 37 (graphics39.png)

XI. CONCLUSIONS

  • Systems are typically described by an arrangement of subsystems each of which is defined by a functional relation. Systems are classified according to such properties as: memory, causality, stability, linearity, and time-invariance. Linear, time-invariant systems (LTI) are special systems for which powerful mathematical methods of description are available.

Logic for an analysis method for LTI systems

  • H(s) characterizes system size 12{ drarrow } {} compute H(s) efficiently.
  • In steady state, response to XestXest size 12{"Xe" rSup { size 8{"st"} } } {}is H(s)XestH(s)Xest size 12{H \( s \) "Xe" rSup { size 8{"st"} } } {}
  • Represent arbitrary x(t) as superpositions of XestXest size 12{"Xe" rSup { size 8{"st"} } } {}on s.
  • Compute response y(t) by superposing H(s)Xest H(s)Xest size 12{H \( s \) "Xe" rSup { size 8{"st "} } } {}on s.

1/ Structural versus functional descriptions

Just as with CT systems, DT systems can be described either structurally, with a block diagram or a network diagram, or functionally by a system function.

H~(z)H~(z) size 12{ {H} cSup { size 8{ "~" } } \( z \) } {}characterizes system

2/ Steady-state response to zn is particularly simple

Since the steady-state response to a complex geometric (exponential) is so simple, it is desirable to represent arbitrary DT functions as sums (integrals) of building-block complex geometric (exponential) functions chosen so that steady-state dominates. The steady-state response to each complex geometric (exponential) is readily computed. For a DT LTI system, the response to an arbitrary input can be computed by superposition.

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