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0.15 Lecture 16:the z transform - method of solution  (Page 2/3)

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The equilibrium equation is

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.

2/ Forward Euler algorithm

a/ Discretization

Define

v i [ n ] = v i ( t ) ∣ t = nT = v i ( nT ) and v o [ n ] = v o ( t ) ∣ t = nT = v o ( 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} } \( ital "nT" \) matrix {{} # {} } ital "and" matrix {{} # {} } v rSub { size 8{o} } \[ n \]=v rSub { size 8{o} } \( t \) \lline rSub { size 8{t= ital "nT"} } =v rSub { size 8{o} } \( ital "nT" \) } {}

and approximate the derivative by the forward Euler algorithm

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"} } \rline rSub { size 8{t= ital "nT"} } approx { {v rSub { size 8{o} } \( \( n+1 \) T \) - v rSub { size 8{o} } \( ital "nT" \) } over {T} } } {}

b/ Difference equation

Substituting the approximation for the derivative into the differential equation, we obtain

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} } \( ital "nT" \) } {}

This equation can be written as a difference equation

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

This equation was solved iteratively for v i [ n ] = u [ n ] size 12{v rSub { size 8{i} } \[ n \] =u \[ n \]} {} and v o [ 0 ] = 0 size 12{v rSub { size 8{o} } \[ 0 \] =0} {} to yield the solution

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 \[ n \]} {}

where α = T/RC. We now solve this difference equation using the Z-transform method.

c/ System function

The Z-transform of the difference equation

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} } \[ n \] } {}

is

z V ~ o ( z ) = ( 1 − α ) V ~ o ( z ) + α V ~ i ( z ) size 12{z {V} cSup { size 8{ "~" } } rSub { size 8{o} } \( z \) = \( 1 - α \) {V} cSup { size 8{ "~" } } rSub { size 8{o} } \( z \) +α {V} cSup { size 8{ "~" } } rSub { size 8{i} } \( z \) } {}

so that the system function is

H ~ ( z ) = V o ~ ( z ) V i ~ ( z ) = α z − ( 1 − α ) = αz − 1 1 − ( 1 − α ) z − 1 size 12{ {H} cSup { size 8{ "~" } } \( z \) = { { {V rSub { size 8{o} } } cSup { size 8{ "~" } } \( z \) } over { {V rSub { size 8{i} } } cSup { size 8{ "~" } } \( z \) } } = { {α} over {z - \( 1 - α \) } } = { {αz rSup { size 8{ - 1} } } over {1 - \( 1 - α \) z rSup { size 8{ - 1} } } } } {}

The region of convergence of H ~ ( z ) size 12{ {H} cSup { size 8{ "~" } } \( z \) } {} needs to be specified from the system description. Because we are simulating a CT system consisting of an RC circuit, we are dealing with a causal system, one that does not respond before it is stimulated. Then we know that the ROC is |z|>|1 − α|.

d/ Input and output Z-transforms

The input is v i [ n ] = u [ n ] size 12{v rSub { size 8{i} } \[ n \] =" u" \[ n \]} {} . Therefore,

V ~ i ( z ) = 1 1 − z − 1 for ∣ z ∣ > 1 size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{i} } \( z \) = { {1} over {1 - z rSup { size 8{ - 1} } } } matrix { {} # {}} ital "for" matrix { {} # {}} \lline z \lline>1} {}

Therefore,

V ~ o ( z ) = 1 1 − z − 1 αz − 1 1 − ( 1 − α ) z − 1 for ∣ z ∣ > 1 ∧ ∣ z ∣ > ∣ 1 − α ∣ size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{o} } \( z \) = left [ { {1} over {1 - z rSup { size 8{ - 1} } } } right ] left [ { {αz rSup { size 8{ - 1} } } over {1 - \( 1 - α \) z rSup { size 8{ - 1} } } } right ]matrix { {} # {}} ital "for" matrix { {} # {}} \lline z \lline>1&\lline z \lline>\lline 1 - α \lline } {}

This is a proper rational function and we expand it in a partial fraction expansion as follows

V ~ o ( z ) = 1 1 − z − 1 − 1 1 − ( 1 − α ) z − 1 for ∣ z ∣ > 1 ∧ ∣ z ∣ > ∣ 1 − α ∣ size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{o} } \( z \) = { {1} over {1 - z rSup { size 8{ - 1} } } } - { {1} over {1 - \( 1 - α \) z rSup { size 8{ - 1} } } } matrix { {} # {}} ital "for" matrix { {} # {}} \lline z \lline>1&\lline z \lline>\lline 1 - α \lline } {}

Because the ROC is outside a circle enclosing all the poles, the solution is

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 \[ n \]} {}

e/ Step response

CT and DT system step responses for α = T/RC, RC = 1,

v o ( t ) = ( 1 − e − t ) u ( t ) and v o [ n ] = ( 1 − ( 1 − α ) n ) u [ n ] size 12{v rSub { size 8{o} } \( t \) = \( 1 - e rSup { size 8{ - t} } \) u \( t \) matrix { {} # {}} ital "and" matrix { {} # {}} v rSub { size 8{o} } \[ n \] = \( 1 - \( 1 - α \) rSup { size 8{n} } \) u \[ n \]} {}

The DT solution is increasingly accurate as T is made smaller. Note that when

T>2, the DT solution diverges. Why?

f/ Natural frequencies of the CT and DT systems

The natural frequency of the CT system is s = −1/(RC) and the natural frequency of the DT system is z = 1− α = 1− T/(RC) .

g/ Relation of natural frequencies in s- and z-planes

The CT system has a natural frequency of s = −1/RC and the DT system for the forward Euler algorithm has the natural frequency z = 1− (T/RC) = 1+sT . Thus, the forward Euler algorithm maps the s-plane into the z-plane by the mapping z = 1+sT . The mapping is shown below.

s → z 0 → 1 − 2 T → − 1 jω → 1 + jωT alignl { stack { size 12{s matrix {{} # {} } rightarrow matrix {{} # {} } z} {} #size 12{0 matrix { {} # {}} rightarrow matrix { {} # {}} 1} {} # size 12{ - { {2} over {T} } matrix {{} # {} } rightarrow matrix {{} # {} } - 1} {} #size 12{jω matrix { {} # {}} rightarrow matrix { {} # {}} 1+jωT} {} } } {}

The conclusion is that for s < − 2 T size 12{"s "<{ { - 2} over {T} } } {} , z<−1 and the DT approximation to the CT system is unstable. Therefore, to extend the range of stability, decrease T.

3/ Backward Euler algorithm

The forward Euler algorithm is one of many possible approximations to the derivative. Another is called the backward Euler algorithm.

a/ Discretization

Define

v i [ n ] = v i ( t ) ∣ t = nT = v i ( nT ) and v o [ n ] = v o ( t ) ∣ t = nT = v o ( 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} } \( ital "nT" \) matrix {{} # {} } ital "and" matrix {{} # {} } v rSub { size 8{o} } \[ n \]=v rSub { size 8{o} } \( t \) \lline rSub { size 8{t= ital "nT"} } =v rSub { size 8{o} } \( ital "nT" \) } {}

and approximate the derivative by the backward Euler algorithm

dv o ( t ) dt ∣ t = nT ≈ v o ( nT ) − v o ( ( n − 1 ) T ) T size 12{ { { ital "dv" rSub { size 8{o} } \( t \) } over { ital "dt"} } \rline rSub { size 8{t= ital "nT"} } approx { {v rSub { size 8{o} } \( ital "nT" \) - v rSub { size 8{o} } \( \( n - 1 \) T \) } over {T} } } {}

b/ Difference equation and system function

Substituting the approximation for the derivative into the differential equation as before, we now obtain
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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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