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Y=H(s)X

where

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 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 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 e st size 12{e rSup { size 8{ ital "st"} } } {} to obtain

( 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 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 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 ) 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 = b M a M size 12{K= { {b rSub { size 8{M} } } over {a rSub { size 8{M} } } } } {}

  • { z 1 , z 2 , . . . , z M } 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.
  • { p 1 , p 2 , . . . , p M } 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 ) = 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 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 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 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 } = size 12{ ital "jI" lbrace s rbrace =jϖ} {} and the abscissa is R { s } = σ size 12{R lbrace s rbrace =σ} {} where

s = σ + size 12{s=σ+jϖ} {}

Example — system function of a network

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 ) 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 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 ) 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 ) 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 e st size 12{e rSup { size 8{ ital "st"} } } {} to obtain

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

Questions & Answers

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Source:  OpenStax, Signals and systems. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10803/1.1
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