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Faltning Differensekvationer Differentialekvationer
Modellering i Tidsplanet
En introduktion till signaler och system

Faltning Differensekvationer Differentialekvationer
Innehåll
Faltning
Differensekvationer
Differentialekvationer

Faltning Differensekvationer Differentialekvationer
Innehåll
Faltning
Differensekvationer
Differentialekvationer

Faltning Differensekvationer Differentialekvationer
Filtrering med viktade medelvärden
N-punkts MA-filter
y[n] = 1 N−1
N (x[n] + x[n − 1] + . . . + x[n − N + 1]) = ∑ 1
x[n − i]
N
i=0
EWMA-filter (Exponentially Weighted Moving Average)
y[n] = ∑ N−1
i=0 a · bi x[n − i]
{ 0 < b < 1
a = 1−b
1−b N
y[n] = ∑ N−1
i=0 w ix[n − i] w i = ab i
successivt minskande vikt till äldre indata

Faltning Differensekvationer Differentialekvationer
B MA=ones(1,11)/11;
y MA=filter(B MA,1,d);
y MA(1:10)=[];
n=11:length(d);
Jämförelse MA-EWMA
b=0.7;
a=(1-b)/(1-bˆ11);
B EWMA=a*b.ˆ(0:10);
y EWMA = filter(B EWMA,1,d);
y EWMA(1:10) = [];
plot(d);
plot(n, y MA, ’b’);
plot(n, y EWMA, ’r’);
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0 10 20 30 40 50

Faltning Differensekvationer Differentialekvationer
Allmänt
Allmänt system då x[n] = 0, n < 0
y[n] =
n∑
w i x[n − i], n ≥ 0
i=0
Puls och pulssvar
{ 1 n = 0
n∑
δ[n] =
0 n ≠ 0 → h[n] = w i δ[n − i] = w n
Faltningsrepresentation
i=0
y[n] = h[n] ∗ x[n] =
n∑
h[i]x[n − i]
i=0

Faltning Differensekvationer Differentialekvationer
Faltning med conv i Sysquake eller Matlab
n = 0:40;
x = sin(0.2*n);
h = sin(0.5*n);
y = conv(x, h);
subplot 311
stem(n, h(1:length(n)))
title(’h[n]’)
subplot 312
stem(n, x(1:length(n)))
title(’x[n]’)
subplot 313
stem(n, y(1:length(n)))
title(’y[n]’)
h[n]
1
0
-1
0 10 20 30 40
x[n]
1
0
-1
0 10 20 30 40
y[n]
0
0 10 20 30 40

Faltning Differensekvationer Differentialekvationer
Innehåll
Faltning
Differensekvationer
Differentialekvationer

Faltning Differensekvationer Differentialekvationer
Exempel: avbetalning på banklån
y[0] = banklån från början
y[n] = kvar av lånet månad n
r = årsräntan, r/12 månatsräntan
x[n] = avbetalning månad n
Lånets storlek månad n
y[n] = (1 + r/12)y[n − 1] − x[n]
Rekursion
y[1] = (1 + r/12)y[0] − x[1]
y[2] = (1 + r/12)y[1] − x[2]
y[3] = (1 + r/12)y[2] − x[3]
.

Faltning Differensekvationer Differentialekvationer
Första ordningens differensekvation
Inför parametrar a = −(1 + r/12), b = −1
y[n] + ay[n − 1] = bx[n], n = 1, 2, . . .
Lösning
y[1]
y[2]
y[3]
y[n]
= −ay[0] + bx[1]
= −ay[1] + bx[2] = −a(−ay[0] + bx[1]) + bx[2]
= a 2 y[0] − abx[1] + bx[2]
= −ay[2] + bx[3] = −a(a 2 y[0] − abx[1] + bx[2]) + bx[3]
= −a 3 y[0] + a 2 bx[1] − abx[2] + bx[3]
.
= (−a) n y[0] + ∑ n
i=1 (−a)n−i bx[i]

Faltning Differensekvationer Differentialekvationer
Differensekvationer
N:te ordningens differensekvation
N∑
y[n] + a i y[n − i] =
i=1
M∑
b i x[n − i]
i=0
Rekursiv form
⎡
⎤
y[n − N]
y[n − N + 1]
y[n] = −[ a N a N−1 . . . a 1 ] ⎢
⎥
⎣ .
⎦
⎡
y[n − 1]
⎤
x[n − M]
x[n − M + 1]
+[ b M b M−1 . . . b 0 ] ⎢
⎥
⎣ .
⎦
x[n]

Faltning Differensekvationer Differentialekvationer
Exempel: andra-ordningens system
Bestäm svar då
y[n] − 1.5y[n − 1] + y[n − 2] = 2x[n − 2],
⎧
⎨
⎩
y[−2] = 2
y[−1] = 1
x[n] = u[n]
Rekursion
y[0] = 1.5y[−1] − y[−2] + 2x[−2] = 1.5(1) − (2) + 2(0) = −0.5
y[1] = 1.5y[0] − y[−1] + 2x[−1] = 1.5(−0.5) − (1) + 2(0) = −1.75
y[2] = 1.5y[1] − y[0] + 2x[0] = 1.5(−1.75) − (−0.5) + 2(1) = −0.125
y[3] = 1.5y[2] − y[1] + 2x[1] = 1.5(−0.125) − (−1.75) + 2(1) = 3.562

Faltning Differensekvationer Differentialekvationer
Implementering (Sysquake, Matlab)
function y = recur(a, b, n, x, x0, y0)
% y = recur(a, b, n, x, x0, y0)
N = length(a);
M = length(b)-1;
y = [y0 zeros(1,length(n))];
x = [x0 x];
a1 = a(length(a):-1:1);
b1 = b(length(b):-1:1);
for i = N+1:N+length(n),
y(i) = -a1*y(i-N:i-1)’ + b1*x(i-N:i-N+M)’;
end
y = y(N+1:N+length(n));

Faltning Differensekvationer Differentialekvationer
Exempel i Sysquake
a = [-1.5 1];
b = [0 0 2];
y0 = [2 1];
x0 = [0 0];
n = 0:20;
x = ones(size(n));
y = recur(a, b, n, x, x0, y0);
stem(n, y)
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6
4
2
0
0 5 10 15 20

Faltning Differensekvationer Differentialekvationer
Innehåll
Faltning
Differensekvationer
Differentialekvationer

Faltning Differensekvationer Differentialekvationer
Elektriskt system
Kirchoff’s lag (summa spänning i
lopen är noll)
Utnyttja i = C dy
dt
y + Ri − x = 0
Konstitutiva samband
v(t) = Ri(t)
i(t) = C dv(t)
dt
v(t) = L di(t)
dt
y + RC dy
dt − x = 0
Första ordningens
differentialekvation
RC dy(t)
dt
+ y(t) = x(t)

Faltning Differensekvationer Differentialekvationer
Mekaniskt system
Newton (riktning uppåt positiv):
Mÿ = −k d ẏ − k s y + x
Andra ordningens
differentialekvation
• Newton’s lag: Mÿ = ∑ i F i
• Fjäderkraft: F s = k s y
• Dämpkraft: F d = k d ẏ
• Extern kraft: x(t)
Mÿ(t) + k d ẏ(t) + k s y(t) = x(t)

Faltning Differensekvationer Differentialekvationer
Euler-approximation
dy(t)
dt
= −ay(t) + bx(t)
Euler-approximation
dy(t)
dt
∣ ≈
t=nT
y(nT + T ) − y(nT )
T
Differentialekvation → differensekvation
y(nT +T )−y(nT )
T
= −ay(nT ) + bx(nT )
y[n+1]−y[n]
T
= −ay[n] + bx[n]
y[n + 1] = (1 − aT )y[n] + bTx[n]
y[n] = (1 − aT )y[n − 1] + bTx[n − 1]

Faltning Differensekvationer Differentialekvationer
Jämför med exakt lösning
Differensekvation med lösning
y[n] = (1 − aT )y[n − 1]
y[n] = (1 − aT ) n y[0]
Differentialekvation med lösning
ẏ(t) = −ay(t)
y(t) = e −at y(0)
y[n] = e −aTn y[0] = (e −aT ) n y[0]
Jämför
e −aT = 1 − aT + a2 T 2
2
Approximation bra om aT ≪ 1
− a3 T 3
6
+ . . .

Faltning Differensekvationer Differentialekvationer
Approximation av andra-derivator
Euler-approximation of första-derivata
dy(t)
y(nT + T ) − y(nT )
dt ∣ ≈
t=nT
T
Euler-approximation av andra-derivata
=
y[n + 1] − y[n]
T
d 2 y(t)
dt 2
=
∣
∣∣t=nT
≈ dy
dt | t=nT +T − dy
dt | t=nT
T
(y[n+2]−y[n+1])/T −(y[n+1]−y[n])/T
T
= y[n+2]−2y[n+1]+y[n]
T 2

Faltning Differensekvationer Differentialekvationer
Differentialekvation
dy(t)
dt
Euler-approximation
+ 1
RC y(t) = 1
RC x(t)
Exempel: RC-krets
Jämför exakt lösning
t = 0:0.04:8;
y2 = 1 -exp(-t);
plot(t,y2,’r’)
y[n] = (1− T
RC )y[n−1]+ T
RC x[n−1]
R=1; C=1; T=0.2;
a= -(1-T/(R*C)); b =[0 T/(R*C)];
y0 = 0; x0 = 1;
n=1:40;
x=ones(size(n));
y1 = recur(a, b, n , x, x0, y0);
n = [0, n];
y1 = [y0, y1];
plot(n*T, y1, ’bo’);
1
0.8
0.6
0.4
0.2
0
0 1 2 3 4 5 6 7 8

Faltning Differensekvationer Differentialekvationer
Alternativ
Simulering av linjärt system ( d n
dt n y(t) = y (n) (t))
a 0 y (n) + a 1 y (n−1) + . . . + a n y = b 0 x (m) + . . . + b m x
A = [ a 0 a 1 . . . a n ], B = [ b 0 b 1 . . . b m ]
y = lsim(B, A, x, t, ’r’);
RC-krets-exemplet:
x = ones(size(t));
lsim(1, [1 1], x, t);
Oskiljaktig från analytisk lösning
Stegsvars-simulering
step(1,[1 1])

Faltning Differensekvationer Differentialekvationer
Simulering av olinjära differentialekvationer
Avancerad numerisk
differentialekvationslösare ode45
(t,y) = ode45(fun,[t0,tend],y0)
Lös Van der Pol’s ekvation
ẍ = µ(1 − x 2 )ẋ − x
För x(0) = 2, ẋ(0) = 0, µ = 1.
Inför y 1 = x, y 2 = ẋ
( ) ( )
ẏ1 y2
ẏ = =
ẏ2 µ(1 − y1 2)y 2 − y 1
function yp = VanDerPol(t,y,mu)
yp =[y(2); mu*(1-y(1)ˆ2)*y(2) - y(1)];
(t,y)=ode45(@VanDerPol,[0,10], [2;0],[],1);
plot(t’,y’,’rb’)
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