sys2-jupyter-notebooks/exam_examples/V03.1.ipynb

+%% Cell type:markdown id:04105e65-6705-4c92-a009-48c54bd38ad4 tags:
+
+# <span style='color:OrangeRed'>V3 STABILTÄT DISKRETER SYSTEME - TEIL 1</span>
+
+%% Cell type:markdown id:ffe7d73f-7986-44b6-a9fd-a678b9aaa182 tags:
+
+<div style="font-family: 'times'; font-size: 13pt; text-align: justify">
+
+Wir analysieren hier die Stabilität eines diskreten Systems mit dem Jury-Verfahren. Am Ende validieren wir das Ergebnis mit einer Sprungantwort. Wir betrachten der Einfachheit wegen nur ein System zweiter Ordnung.
+
+%% Cell type:code id:3a336fb9-2f73-44ce-ab67-91b96ccfc1c1 tags:
+
+``` octave
+% Necessary to use control toolbox
+pkg load control
+clear all
+```
+
+%% Cell type:markdown id:53866cc1-a0ea-4d77-8d9c-c8345704d874 tags:
+
+<div style="font-family: 'times'; font-size: 13pt; text-align: justify">
+Es ist das folgende System in Form einer Übertragungsfunktion gegeben:
+
+%% Cell type:code id:a0b9d775-f176-483e-aaca-43da284aa2e4 tags:
+
+``` octave
+a1 = 1
+a2 = 2
+b1 = 0.1
+b2 = 0.2
+num = [a1 a2]
+den = [1 b1 b2]
+G = tf(num,den, 1)
+```
+
+%% Output
+
+    a1 =  1
+    a2 =  2
+    b1 =  0.10000
+    b2 =  0.20000
+    num =
+    
+       1   2
+    
+    den =
+    
+       1.00000   0.10000   0.20000
+    
+    
+    Transfer function 'G' from input 'u1' to output ...
+    
+                z + 2
+     y1:  -----------------
+          z^2 + 0.1 z + 0.2
+    
+    Sampling time: 1 s
+    Discrete-time model.
+
+%% Cell type:markdown id:1cb98bc6-138b-45bc-9178-74f8493d4171 tags:
+
+<div style="font-family: 'times'; font-size: 13pt; text-align: justify">
+
+Berechnet man die Nullstellen des Nenners, erhält man die Pole des Systems:
+
+%% Cell type:code id:03459a21-1072-416b-a04b-5de7b9a30813 tags:
+
+``` octave
+p = roots(den)
+q = abs(p)
+```
+
+%% Output
+
+    p =
+    
+      -0.05000 + 0.44441i
+      -0.05000 - 0.44441i
+    
+    q =
+    
+       0.44721
+       0.44721
+    
+
+%% Cell type:markdown id:071800da-0e16-4189-bb65-5746c0415165 tags:
+
+<div style="font-family: 'times'; font-size: 13pt; text-align: justify">
+
+Für Stabilität, müssen sich die Pole im Inneren des Einheitskreises befinden. Das bedeutet, dass die Werte von q betragsmäßig kleiner als 1 sein müssen, was hier offensichtlich erfüllt ist.
+
+%% Cell type:markdown id:c9e8bd3b-e932-4710-9a6a-220f67d882e9 tags:
+
+<div style="font-family: 'times'; font-size: 13pt; text-align: justify">
+Wir wollen nun die Stabilität mit dem Jury-Kriterium nachweisen. Dafür werten wir zunächst die Übertragungsfunktion in 1 und -1 aus.
+
+%% Cell type:code id:5adb101b-4286-428a-a4bd-eeb57e860e9c tags:
+
+``` octave
+F1 = 1+b1+b2
+Fm1 = 1-b1+b2
+```
+
+%% Output
+
+    F1 =  1.3000
+    Fm1 =  1.1000
+
+%% Cell type:markdown id:eae75561-2795-400d-802d-f649281ff316 tags:
+
+<div style="font-family: 'times'; font-size: 13pt; text-align: justify">
+F1 und Fm1 müssen beide positiv sein. Dies ist erfüllt.
+
+%% Cell type:markdown id:fb81055e-1e14-4401-afb2-f90884c549be tags:
+
+<div style="font-family: 'times'; font-size: 13pt; text-align: justify">
+Die zweite Bedingung ist, dass der Betrag von a2 kleiner als 1 sein muss.
+
+%% Cell type:code id:acf57137-23eb-4be3-87ef-eb9d08076f07 tags:
+
+``` octave
+pkg load symbolic
+
+syms z
+syms w
+F1 = (a1*z+a2)/(z^2 + b1*z +b2)
+F2 = subs(F1,z,(1+w)/(1-w))
+simplify(F2)
+```
+
+%% Output
+
+    Symbolic pkg v2.9.0: Python communication link active, SymPy v1.5.1.
+    warning: passing floating-point values to sym is dangerous, see "help sym"
+    warning: called from
+        double_to_sym_heuristic at line 50 column 7
+        sym at line 379 column 13
+        mtimes at line 63 column 5
+    warning: passing floating-point values to sym is dangerous, see "help sym"
+    warning: called from
+        double_to_sym_heuristic at line 50 column 7
+        sym at line 379 column 13
+        plus at line 61 column 5
+    F1 = (sym)
+    
+         z + 2
+      ───────────
+       2   z    1
+      z  + ── + ─
+           10   5
+    
+    F2 = (sym)
+    
+                  w + 1
+              2 + ─────
+                  1 - w
+      ─────────────────────────
+                              2
+      1     w + 1      (w + 1)
+      ─ + ────────── + ────────
+      5   10⋅(1 - w)          2
+                       (1 - w)
+    
+    ans = (sym)
+    
+         ⎛ 2          ⎞
+      10⋅⎝w  - 4⋅w + 3⎠
+      ─────────────────
+          2
+      11⋅w  + 16⋅w + 13
+    
+
+%% Cell type:markdown id:56641183-2cb4-48c2-b61d-7f612a56d0a7 tags:
+
+<div style="font-family: 'times'; font-size: 13pt; text-align: justify">
+Alle Zeichen im Nenner müssen positiv sein.
+
+%% Cell type:markdown id:ab36a55a-6239-4214-9d36-85f4d111c61e tags:
+
+<div style="font-family: 'times'; font-size: 13pt; text-align: justify">
+Zum Schluss überprüfen wir die Ergebnisse mit der Simulation einer Sprungantwort.
+
+%% Cell type:code id:578daff8-f4fd-4001-bb09-aea89d809eec tags:
+
+``` octave
+addpath("../Octsim");
+tini = 0 # Start time
+tfinal = 20 # End time
+dt = 0.1 # Time Step
+nflows = 2 #Number of data flows in the schematic, Zahlenwert entspricht nicht dem in Matlab (hier um 1 größer)
+Ts = 1 # Sampling time for discrete time
+
+c1{1} = StepSource(1,0,1,0.1); #StepSource(self,out,startv,endv,ts)
+c1{2} = DTTransferFunction(1,2,num,den,Ts); #DTTransferFunction(self,inp,out,num,den,Ts)
+
+% Instance of the simulation schematic
+sc1 = Schema(tini,tfinal,dt,nflows);
+
+sc1.AddListComponents(c1);
+
+% Run the schematic and plot
+out1 = sc1.Run([1 2]);
+plot(out1(1,:),out1(2,:),out1(1,:),out1(3,:));
+ylim([-2, 4]);
+```
+
+%% Output
+
+    tini = 0
+    tfinal =  20
+    dt =  0.10000
+    nflows =  2
+    Ts =  1
+
+
+
+%% Cell type:code id:9e1a668f-07ac-442a-bb62-950dacd49832 tags:
+
+``` octave
+```

sys2-jupyter-notebooks/exam_examples/figures/DCClosedLoop.svg

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