1.1.1 Create continuous time transfer function given the poles, zeros and gain
Problem: Find the transfer function \(H(s)\) given zeros \(s=-1,s=-2\), poles \(s=0,s=-4,s=-6\) and gain 5.
1.1.1.1 Mathematica
Clear["Global`*"];
num = (s+1)(s+2);
den = (s)(s+4)(s+6);
gain = 5;
sys = TransferFunctionModel[
gain (num/den),s]
|
Out[30]= TransferFunctionModel[{
{{5*(1 + s)*(2 + s)}},
s*(4 + s)*(6 + s)}, s] |
1.1.1.2 Matlab
clear all;
m_zeros = [-1 -2];
m_poles = [0 -4 -6];
gain = 5;
sys = zpk(m_zeros,m_poles,gain);
[num,den] = tfdata(sys,'v');
printsys(num,den,'s')
|
num/den =
5 s^2 + 15 s + 10
-------------------
s^3 + 10 s^2 + 24 s
|
1.1.1.3 Maple
restart;
alias(DS=DynamicSystems):
zeros :=[-1,-2]:
poles :=[0,-4,-6]:
gain :=5:
sys :=DS:-TransferFunction(zeros,poles,gain):
#print overview of the system object
DS:-PrintSystem(sys);
exports(sys);
#To see fields in the sys object, do the following
# tf, inputcount, outputcount, statecount, sampletime,
# discrete, systemname, inputvariable, outputvariable,
# statevariable, parameters, systemtype, ModulePrint
tf:=sys[tf]; #reads the transfer function
Matrix(1,1,{(1,1)=(5*s^2+15*s+10)/(s^3+10*s^2+24*s)})
numer(tf[1,1]); #get the numerator of the tf
\[ \text {tf} = \left [ {\begin {array}{c} {\frac {5\,{s}^{2}+15\,s+10}{{s}^{3}+10\,{s}^{2}+24\,s}}\end {array}} \right ] \]
denom(tf[1,1]); #get the denominator of the tf
\[ s*(s^2+10*s+24) \]
1.1.2 Convert transfer function to state space representation
1.1.2.1 problem 1
Problem: Find the state space representation for the continuous time system defined by the
transfer function \[ H(s)=\frac {5s}{s^{2}+4s+25}\]
1.1.2.2 Mathematica
Clear["Global`*"];
sys = TransferFunctionModel[(5 s)/(s^2+4 s+25),s];
ss = MinimalStateSpaceModel[StateSpaceModel[sys]]
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(a=Normal[ss][[1]])//MatrixForm
(b=Normal[ss][[2]])//MatrixForm
(c=Normal[ss][[3]])//MatrixForm
(d=Normal[ss][[4]])//MatrixForm
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1.1.2.3 Matlab
clear all;
s = tf('s');
sys_tf = (5*s) / (s^2 + 4*s + 25);
[num,den] = tfdata(sys_tf,'v');
[A,B,C,D] = tf2ss(num,den)
|
A =
-4 -25
1 0
B =
1
0
C =
5 0
D =
0
|
1.1.2.4 Maple
restart;
alias(DS=DynamicSystems):
sys := DS:-TransferFunction(5*s/(s^2+4*s+25)):
sys := DS:-StateSpace(sys);
exports(sys); #to see what fields there are
a := sys:-a;
|
\[ \left [{\begin {array}{cc} 0 & 1 \\ -25 & -4 \end {array}} \right ] \] |
b:=sys:-b;
|
\[ \left [ {\begin {array}{c} 0\\ 1\end {array}} \right ] \] |
c:=sys:-c;
| \[ \left [ {\begin {array}{cc} 0&5\end {array}} \right ] \] |
d:=sys:-d;
|
\[ \left [ {\begin {array}{c} 0\end {array}} \right ] \] |
1.1.2.5 problem 2
Problem: Given the continuous time S transfer function defined by \[ H(s)=\frac {1+s}{s^{2}+s+1}\] convert to state space
and back to transfer function.
| Mathematica
Clear["Global`*"];
sys = TransferFunctionModel[(s + 1)/(s^2 + 1 s + 1), s];
ss = MinimalStateSpaceModel[StateSpaceModel[sys]]
|
\[ \left ( {\begin {array}{cc|c|c} 0 & 1 & 0 \\ -1 & -1 & 1 \\ \hline 1 & 1 & 0 \\ \end {array}} \right )_{} \] |
TransferFunctionModel[ss]
|
\[ \frac {s+1}{s^2+s+1} \] |
| Matlab
clear all;
s = tf('s');
sys = ( s+1 )/(s^2 + s + 1);
[num,den]=tfdata(sys,'v');
%convert to state space
[A,B,C,D] = tf2ss(num,den)
|
A =
-1 -1
1 0
B =
1
0
C =
1 1
D =
0
|
%convert from ss to tf
[num,den] = ss2tf(A,B,C,D);
sys=tf(num,den)
|
sys =
s + 1
-----------
s^2 + s + 1
|
1.1.3 Create a state space representation from A,B,C,D and find the step response
Problem: Find the state space representation and the step response of the continuous time
system defined by the Matrices A,B,C,D as shown
A =
0 1 0 0
0 0 1 0
0 0 0 1
-100 -80 -32 -8
B =
0
0
5
60
C =
1 0 0 0
D =
0
| Mathematica
Clear["Global`*"];
a = {{0,1,0,0},
{0,0,1,0},
{0,0,0,1},
{-100,-80,-32,-8}
};
b = {{0} ,{0}, {5},{60}};
c = {{1,0,0,0}};
d = {{0}};
sys = StateSpaceModel[{a,b,c,d}];
y = OutputResponse[sys,UnitStep[t],{t,0,10}];
p = Plot[Evaluate@y, {t, 0, 10},
PlotRange -> All,
GridLines -> Automatic,
GridLinesStyle -> LightGray]
|
|
| Matlab
clear all;
A = [0 1 0 0;
0 0 1 0;
0 0 0 1;
-100 -80 -32 -8];
B = [0;0;5;60];
C = [1 0 0 0];
D = [0];
sys = ss(A,B,C,D);
step(sys);
grid;
title('unit step response');
|
|
| Maple
restart:
alias(DS=DynamicSystems):
alias(size=LinearAlgebra:-Dimensions);
a:=Matrix([[0,1,0,0],[0,0,1,0],
[0,0,0,1],[-100,-90,-32,-8]]):
b:=Matrix([[0],[0],[5],[6]]):
c:=Matrix([[1,0,0,0]]):
d:=Matrix([[1]]):
sys:=DS:-StateSpace(a,b,c,d);
DS:-ResponsePlot(sys, DS:-Step(),
duration=10,color=red,legend="step");
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1.1.4 Convert continuous time to discrete time transfer function, gain and phase
margins
Problem: Compute the gain and phase margins of the open-loop discrete linear time system
sampled from the continuous time S transfer function defined by \[ H(s)=\frac {1+s}{s^{2}+s+1}\] Use sampling period of 0.1
seconds.
| Mathematica
Clear["Global`*"];
tf = TransferFunctionModel[(s+1)/(s^2+1 s+1),s];
|
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ds = ToDiscreteTimeModel[tf, 0.1, z,
Method -> "ZeroOrderHold"]
|
|
{gm, pm} = GainPhaseMargins[ds];
|
gm
Out[28]={{31.4159,19.9833}}
pm
Out[29]={{1.41451,1.83932},{0.,Pi}}
|
max = 100;
yn = OutputResponse[ds,
Table[UnitStep[n],{n, 0, max}]];
ListPlot[yn,
Joined -> True,
InterpolationOrder -> 0,
PlotRange -> {{0, max}, {0, 1.4}},
Frame -> True,
FrameLabel ->{{"y[n]", None},
{"n", "unit step response"}},
ImageSize -> 400,
AspectRatio -> 1,BaseStyle -> 14]
|
|
| Matlab
clear all; close all;
Ts = 0.1; %sec, sample period
s = tf('s');
sys = ( s + 1 ) / ( s^2 + s + 1);
%convert s to z domain
sysd = c2d(sys,Ts,'zoh');
tf(sysd)
|
0.09984 z - 0.09033
----------------------
z^2 - 1.895 z + 0.9048
Sample time: 0.1 seconds
Discrete-time transfer function.
Gm =
19.9833
Pm =
105.3851
Wcg =
31.4159
Wcp =
1.4145
|
step(sysd,10);
[Gm,Pm,Wcg,Wcp] = margin(sysd)
|
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1.1.5 Convert differential equation to transfer functions and to state space
Problem: Obtain the transfer and state space representation for the differential equation
\[ 3\frac {d^{2}y}{dt^{2}}+2\frac {dy}{dt}+y\left ( t\right ) = u(t) \]
| Mathematica
Clear[y, u, t];
eq = 3 y''[t] + 2 y'[t] + y[t] == u[t];
ss = StateSpaceModel[ {eq}, {{y'[t], 0},
{y[t], 0}}, {{u[t], 0}}, {y[t]}, t];
ss = MinimalStateSpaceModel[ss]
|
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tf = Simplify[TransferFunctionModel[ss]]
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| Matlab
clear all;
syms y(t) Y
eq = 3*diff(y,2)+2*diff(y)+y;
lap = laplace(eq);
lap = subs(lap,'y(0)',0);
lap = subs(lap,'D(y)(0)',0);
lap = subs(lap,'laplace(y(t),t,s)',Y);
H = solve(lap-1,Y);
pretty(H)
|
1
--------------
2
3 s + 2 s + 1
|
[num,den] = numden(H);
num = sym2poly(num);
den = sym2poly(den);
[A,B,C,D] = tf2ss(num,den)
|
A =
-0.6667 -0.3333
1.0000 0
B =
1
0
C =
0 0.3333
D =
0
|
| Maple
restart;
alias(DS=DynamicSystems):
ode:=3*diff(y(t),t$2)+2*diff(y(t),t)+y(t)=
Heaviside(t):
sys:=DS:-DiffEquation(ode,
'outputvariable'=[y(t)],
'inputvariable'=[Heaviside(t)]):
sys:=DS:-TransferFunction(sys):
sys:-tf(1,1);
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\[ \frac {1}{3{s}^{2}+2\,s+1} \] |
sys:=DS:-StateSpace(sys):
#show the state space matrices
{sys:-a,sys:-b,sys:-c,sys:-d};
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\[ \left \{ \left [ {\begin {array}{cc} 0&1\\ \noalign {\medskip }-1/3&-2/3 \end {array}} \right ] , \left [ {\begin {array}{c} 0\\ \noalign {\medskip } 1\end {array}} \right ] , \left [ {\begin {array}{cc} 1/3&0\end {array}} \right ] , \left [ {\begin {array}{c} 0\end {array}} \right ] \right \} \] |
1.1.6 Convert from continuous transfer function to discrete time transfer function
Problem: Convert \[ H\left (s\right ) = \frac {1}{s^2+10 s+10} \]
To \(Z\) domain transfer function using sampling period of \(0.01\) seconds and using the zero order hold
method.
| Mathematica
Clear["Global`*"]
sys = TransferFunctionModel[
1/(s^2 +10 s+20),s];
sysd= ToDiscreteTimeModel[sys,0.01,z,
Method->"ZeroOrderHold"]
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Matlab
s = tf('s');
sys = 1/(s^2 + 10*s + 20);
T = 0.01; %seconds
sysz = c2d(sys,T,'zoh')
|
sysz =
4.837e-05 z + 4.678e-05
-----------------------
z^2 - 1.903 z + 0.9048
Sample time: 0.01 seconds
Discrete-time transfer function.
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1.1.7 Convert a Laplace transfer function to an ordinary differential equation
Problem: Give a continuous time transfer function, show how to convert it to an
ordinary differential equation. This method works for non-delay systems. The transfer
function must be ratio of polynomials. For additional methods see this question at
stackexchange
| Mathematica
tfToDiff[tf_, s_, t_, y_, u_]:=
Module[{rhs,lhs,n,m},
rhs = Numerator[tf];
lhs = Denominator[tf];
rhs = rhs /. m_. s^n_. -> m D[u[t], {t, n}];
lhs = lhs /. m_. s^n_. -> m D[y[t], {t, n}];
lhs == rhs
]
tf = (5s)/(s^2+4s+25);
tf = C0 s/(R0 C0 s + 1);
eq=tfToDiff[tf, s, t, y, u]
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25 + 4 y'[t]+y''[t]==5 u'[t]
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