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Discretize_Continuous_State_Space_Model_of_LTI_System.pdf
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Continuous State Space Model of Linear Time invariant System

x˙=Ax(t)+Bu(t)y(t)=Cx(t)+Du(t),t≥0{\dot x} = Ax(t)+Bu(t) \\ y(t) = Cx(t)+ Du(t), t\geq 0, (1)

 

The matrix exponential function as

ddteAt=AeAt=eAtA{d\over dt}e^{At} =Ae^{At}=e^{At}A , (2)

 

Multiplying both sides of (1) by

e−Ate^{-At}, we obtain(3)

e−Atx˙(t)=e−AtAx(t)+e−AtBu(t)e^{-At}{\dot x(t)} = e^{-At}Ax(t)+e^{-At}Bu(t), (3)

 

Substituting (2) into (3)

e−Atddtx(t)=−ddte−Atx(t)+e−AtBu(t)e^{-At}{d\over dt}x(t) = -{d\over dt}e^{-At} x(t) + e^{-At}Bu(t)

→ddt[e−Atx(t)]=e−AtBu(t)\rightarrow {d\over dt} [e^{-At}x(t) ] = e^{-At}Bu(t), (4)

 

Integrating the result of multiplying

dtdt both side of (4)

e−Atx(t)∣0t=e−Atx(t)−x(0)=∫0te−AτBu(τ)dτe^{-At}x(t)|^{t}_{0} =e^{-At}x(t)-x(0) = \int^{t}_{0}e^{-A\tau}Bu(\tau)d\tau, (5)

 

Rewriting the equation (5), we get the solution of the state-space eq of LTI system.

x(t)=eAtx(0)+∫0teA(t−τ)Bu(τ)dτx(t) = e^{At}x(0) +\int^{t}_{0}e^{A(t-\tau)}Bu(\tau)d\tau, (6)

 

Discretize the continuous state space model

In an analog control system, the control input is updated continuously, while in a digital control system, it is held constant within each sampling period.

To accurately represent this behavior in a system model, discretization must be applied. we use the Zero-Order-Hold(ZOH) method in this contents.

x(t)=eA(t−t0)x(t0)+∫t0teA(t−τ)Bu(τ)dτx(t) = e^{A(t-t_0)}x(t_0) +\int^{t}_{t_0}e^{A(t-\tau)}Bu(\tau)d\tau, (1)

 

To transition to a discrete-time representation, we replace the continuous-time variable

t, t0t,\ t_0 with the discrete-time variable

T(k+1), TkT(k+1), \ Tk

t=T(k+1),t0=Tk,T:sampling timet=T(k+1), \quad t_0=Tk ,\quad T:sampling\ time

 

Rewriting equation (1) under the assumption that the input

uu remained during sampling time. so input

uu is constant.

x(T(k+1))=eATx(t0)+∫TkT(k+1)eA(T(k+1)−τ)Bu(k)dτx(T(k+1)) = e^{AT}x(t_0)+\int^{T(k+1)}_{Tk}e^{A(T(k+1)-\tau)}Bu(k)d\tau (2)

 

For simplicity, introduce a change of variable

σ=T(k+1)−τ\sigma = T(k+1)-\tau

 

Thus, by substituting

dτ=−dσd\tau=-d\sigma, equation (2) transforms into (3)

x(T(k+1))=eATx(t0)+∫T0eAσBu(τ)(−dσ)x(T(k+1)) = e^{AT}x(t_0)+\int^{0}_{T}e^{A\sigma}Bu(\tau)(-d\sigma) (3)

 

Rearranging (3), we get the final form in continuous-time

x(T(k+1))=eATx(t0)+∫0TeAσdσBu(k)x(T(k+1)) = e^{AT}x(t_0)+\int^{T}_{0}e^{A\sigma}d\sigma Bu(k) (4)

 

Ad=eAT=I+AT+(AT)22!...Bd=∫0TeAσdσB=A−1[eAT−I]B=T+AT22!+A2T33!...A_d =e^{AT} = I+AT+{(AT)^2\over 2! } ... \\ B_d =\int^{T}_{0}e^{A\sigma}d\sigma B=A^-1[e^{AT}-I]B=T+{AT^2\over 2!}+{A^2T^3 \over 3!} ...

 

We can represent the eq (4) in discrete-time.

x(k+1)=Adx(k)+Bdu(k)x(k+1)=A_dx(k)+B_du(k)

 

Given these following system, we compare how well the continuous and discrete models describe GT. GT is continuous model use very small dt to describe GT.

Assume GT is continuous model that dt is 0.000001.

 

The test where dt is 0.1sec was simulated for 5 sec. and we plot simple results of three cases (GT, continuous model, discrete model)

 

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