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Time Domain State and Output Equations

Module by: Thanos Antoulas, JP Slavinsky

Summary: Time Domain State and Output Equation stuff

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

Note:

xx and uu are functions of time, and the notation ddtx t1 x implies ddtxt t1 x t .

Given the state equations and output equations

ddtxt=Ax+Bu t1 x t A x B u (1)
y=Cx+Du y C x D u (2)

we would like to be able to find solutions for xx and yy in terms of the initial state of the system and the system's input. To find such solutions, we begin with an examination of a scalar (n=1n1, m=1m1) state equation:

ddtxt=ax+bu t1 x t a x b u (3)

If we looked at a special case of this equation, one where the input uu was 00, we'd have ddtxt=ax t1 x t a x . We've seen this many times before; to solve this, we need a function whose derivative is directly proportional to itself. This function is the exponential function. Therefore, in solving the more general case presented by the state equation, we expect that the exponential function will also come into play.

Starting with the state equation, we can collect like terms, multiply through by ⅇ-at a t , and rewrite the left-hand side of the derivative equation in terms of the derivative. (We take this last step after noticing that the left-hand side of the derivation equation looks like the derivative product rule has already been applied to it.)

ⅇ-atddtxt−aⅇ-atxt=bⅇ-atut a t t1 x t a a t x t b a t u t (4)
ddtⅇ-atxt=ⅇ-atbut t1 a t x t a t b u t (5)

Since we are searching for xtxt instead of its derivative, we will integrate both sides from t0t0 to tt.

∫t0tddtⅇ-atxtdt=∫t0tⅇ-atbutdt t t0 t t1 a t x t t t0 t a t b u t (6)

In the left-hand side of this equation, the integral and the derivative counteract each other and we are left with the difference of the function ⅇ-atxt a t x t evaluated at the upper and lower integration limits. To avoid confusion, the variable of integration will be changed from tt (which is now a constant limit in the integral) to ττ.

ⅇ-atxt−ⅇ-at0xt0=∫t0tⅇ-aτbuτdτ a t x t a t0 x t0 τ t0 t a τ b u τ (7)

We now move the xt0xt0 term to the other side and divide through by ⅇ-atat. This leaves us with a solution for the state variable xtxt in the scalar case:

xt=ⅇat−t0xt0+∫t0tⅇat−τbuτdτ x t a t t0 x t0 τ t0 t a t τ b u τ (8)

What happens if we let t0t0 go to -∞∞? The first term on the right-hand side will go to zero since x-∞=0 x 0 . Then, if we say that ht=ⅇatb ht a t b , the second term can be rewritten as

∫-∞tht−τuτdτ τ t h t τ u τ (9)

This is the convolution equation h*uhu.

For the scalar case, the solution to the output equation ytyt has the same basic form as the solution to the state equation:

yt=cⅇat−t0xt0+∫t0tcⅇat−τbuτdτ+ⅆut y t c a t t0 x t0 τ t0 t c a t τ b u τ ⅆ u t (10)

Again, we can see the convolution in the second term.

The general matrix forms of the solutions for the state and output equations follow the same pattern. The only differences are that we use the matrix exponential instead of the scalar exponential, and that we use the matrices AA, BB, CC, and DD instead of the scalars aa, bb, cc, and dd.

xt=ⅇAtx0+∫0tⅇAt−τBuτdτ x t A t x 0 τ 0 t A t τ B u τ (11)
yt=CⅇAtx0+∫0tCⅇAt−τBuτdτ+Dut y t C A t x 0 τ 0 t C A t τ B u τ D u t (12)

The convolution term is easy to see in the solution for xtxt. However, it takes a little regrouping to find it in the solution for ytyt. If we pull the DD matrix into the integral (by multiplying it by the impulse function, δδ, we once again have the integral of a function of t−τtτ being multiplied by the input (convolution).

yt=CⅇAtx0+∫0tCⅇAt−τB+Dδt−τuτdτ y t C A t x 0 τ 0 t C A t τ B D δ t τ u τ (13)

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