Motor for velocity control
Jω + cω = KT i
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State-space representation
▶ motion equation
Jω = KT i − cω ⇒ ω = −c
Jω +
KT
Ji
▶ equilibrium state
0 = ω∗ = −c
Jω∗ +
KT
Ji∗ ⇒ ω∗ =
KT
ci∗
▶ state equation
ω − ω∗︸ ︷︷ ︸x
= −c
J︸︷︷︸a
(ω − ω∗)︸ ︷︷ ︸x
+KT
J︸︷︷︸b
(i − i∗)︸ ︷︷ ︸u
▶ output equationω − ω∗︸ ︷︷ ︸
y
= 1︸︷︷︸c
(ω − ω∗)︸ ︷︷ ︸x
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Response of 1st-order system
Given the input u,
what response y is expected?
x=0 means an equilibrium state.
3 / 23
Interrelation among responses
4 / 23
Stability of 1st-order system
Disturbed equilibrium state x(0) = 0
can return to x = 0?
5 / 23
1st-order unforced system (c=1)
▶ 1st-order unforced systemx(t) = ax(t)
▶ responsex(t) = eatx(0)
▶ asymptotically stablea < 0 ⇒ x(t) → 0 (t → ∞)
▶ stable but not asymptotically stablea = 0 ⇒ x(t) → x(0) (t → ∞)
▶ unstablea > 0 ⇒ x(t) → ∞ (t → ∞)
6 / 23
Specify a for #1,#2,#3
7 / 23
Using SCILAB (resp3a.sce)
function dx=f(t,x),
dx=a*x,
endfunction
a=-1;
x0=1; t0=0; t=0:0.1:5;
x=ode(x0,t0,t,f);
plot(t,x), mtlb_grid
8 / 23
Response of 1st-order system (c=1)
▶ 1st-order system: x(t) = ax(t) + bu(t)
x(t) = eatx(0) +
∫ t
0
ea(t−τ)bu(τ) dτ
▶ solving x(t) = ax(t) + bu(t)e−at(x(t)− ax(t)) = e−atbu(t)
⇒ e−at x(t)− ae−atx(t) = e−atbu(t)⇒ d
dt(e−atx(t)) = e−atbu(t)
⇒[e−atx(t)
]t′0=
∫ t′
0
e−atbu(t) dt
⇒ e−at′x(t ′)− x(0) =
∫ t′
0
e−atbu(t) dt
⇒ x(t ′) = eat′x(0) + eat
′∫ t′
0
e−atbu(t) dt
9 / 23
Response of 1st-order system
y(t) = ceatx(0)︸ ︷︷ ︸zero−input resp.
+
∫ t
0
g(t − τ )u(τ ) dτ︸ ︷︷ ︸zero−state resp.
impulse response : g(t) = ceatb
10 / 23
Which is the zero-input response?
11 / 23
Using SCILAB (resp3a.sce)
function dx=f(t,x),
dx=a*x+b*u(t),
endfunction
function ut=u(t),
ut=sin(10*t),
endfunction
a=-1; b=1;
x0=1; t0=0; t=0:0.01:5;
x=ode(x0,t0,t,f);
plot(t,x),mtlb_grid
12 / 23
Step response, Impulse response
▶ stable 1st-order system:
x(t) = − 1
Tx(t) +
K
Tu(t)
▶ impluse response: g(t) =K
Te−
1T t
▶ step response:
x(t) =
∫ t
0
K
Te−
1T (t−τ) · 1 dτ = K (1− e−
1T t)
▶ gain constant/steady state gain: K = x(∞)
▶ time constant (t s.t. x(t) = (1− 1e )K ): T
▶ relationship:d
dtx(t) = g(t)
13 / 23
Step responses
14 / 23
Using SCILAB (resp4a.sce)
function dx=f(t,x),
dx=a*x+b*u(t),
endfunction
function ut=u(t),
ut=1,
endfunction
a=-1; b=1;
x0=1; t0=0; t=0:0.01:5;
x=ode(x0,t0,t,f);
plot(t,x),mtlb_grid
15 / 23
Impulse responses
16 / 23
Using SCILAB (resp2a.sce)
function dx=f(t,x),
dx=a*x,
endfunction
a=-1; b=1;
x0=b; t0=0; t=0:0.1:5;
x=ode(x0,t0,t,f);
plot(t,x), mtlb_grid
17 / 23
Laplace Transformation
▶ Laplace Transformation
x(s) :=
∫ ∞
0
x(t)e−st dt
x(t) x(s) x(t) x(s)
δ(t) 1 1 1s
t 1s2 tn n!
sn+1
eat 1s−a eatx(t) x(s − a)
sinωt ω2
s2+ω2 cosωt ss2+ω2
x(t) sx(s)− x(0)∫ t
0 x(τ) dτ1s x(s)
x(0) lims→∞ sx(s) x(∞) lims→0 sx(s)▶ time lag y(t) = x(t − L)
y(s) =
∫ ∞
0
x(t − L)e−st dt = e−sLx(s)18 / 23
Frequency response
▶ stable 1st-order system: x(t) = − 1
Tx(t) +
K
Tu(t)
▶ zero-state resp. in the case of u(t) = sinωt:x(t) =
K√1+(ωT )2
(sin(ωt − tan−1 ωT ) + e−1Tt sin(tan−1 ωT ))
⇓x(t) ≃ K√
1 + (ωT )2︸ ︷︷ ︸|g(jω)|
sin(ωt − tan−1 ωT )︸ ︷︷ ︸∠g(jω)
▶ transfer function:
x(s)− x(0) = − 1
Tx(s) +
K
Tu(s) ⇒ g(s) =
x(s)
u(s)=
K
Ts + 1▶ frequency response: gain |g(jω)|, phase ∠g(jω)▶ bandwidth (ωb s.t.|g(jωb)| = |g(j0)|√
2): ωb =
1T
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Derivation
Using
∫eax sin bx dx =
eax
a2 + b2(a sin bx − b cos bx),
x(t) =
∫ t
0ea(t−τ)b sinωτ dτ = eatb
∫ t
0e−aτ sinωτ dτ
= eatb[e−aτ
a2 + ω2(−a sinωτ − ω cosωτ)]t0
=eatb
a2 + ω2(e−at(−a sinωt − ω cosωt) + ω)
=b√
a2 + ω2(sinωt
−a√a2 + ω2
− cosωtω√
a2 + ω2+ e+at ω√
a2 + ω2)
=b√
a2 + ω2(sinωt cos θ − cosωt sin θ + e+at sin θ)
=b√
a2 + ω2(sin(ωt − θ) + e+at sin θ) (θ = tan−1 ω
−a)
=b√
a2 + ω2(sin(ωt + ϕ)− e+at sinϕ) (ϕ = −θ)
20 / 23
Bode Diagram
21 / 23
Using SCILAB (resp5a.sce)
function g=f(a,b,w),
g=b./(%i*w-a),
endfunction
a=-1; b=1;
w=logspace(-1,1);
g=f(a,b,w);
ga=20*log10(abs(g));
ph=180/%pi*atan(imag(g),real(g));
subplot(211),plot2d(w,ga,logflag=’ln’),mtlb_grid
subplot(212),plot2d(w,ph,logflag=’ln’),mtlb_grid
22 / 23
Problem LEC01
For x(x) = ax(t) + bu(t), derive the time responsex(t) when the the following input is specified.
u(t) =
{1Tst (0 ≤ t ≤ Ts)
1 (t ≥ Ts)
23 / 23
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