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

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

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 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 } 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 λ i λ k 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 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

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 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 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 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 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"} } } } {}

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 R 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 = 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} } } } } } } {}

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 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 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 e st size 12{e rSup { size 8{ ital "st"} } } {} we can solve for Y which has the form

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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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