Understanding the Zero-State Response

Understanding the Zero-State ResponseSwapnil Sunil Jain15 January, 2007

Understanding the Zero-State ResponseBy definition we know that the $\text{ZSR}(t)$ is the response of the system due to the input $x(t)$ when all the initial conditions are set to zero.

Now, let a LTI system $𝕊$ with input $x(t)$ and output $y(t)$ be described by the following differential equation

$a(D)y(t)=b(D)x(t)$(1)

where

$a(D)=D^n+a_1{D}^{n-1}+\mathrm{\dots }+a_n$

$b(D)=b_0{D}^m+b_1{D}^{m-1}+\mathrm{\dots }+b_n$

and the initial conditions: $y^{(n-1)}(0^{-}),y^{(n-2)}(0^{-}),\mathrm{\dots },y(0^{-})$.

To find the zero-state response, we first set all the initial conditions to zero i.e.

$y^{(n-1)}(0^{-})=0,$

$y^{(n-2)}(0^{-})=0,$

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$y(0^{-})=0$

Then we note that x(t) can be written in the following way (due to the shifting property of $\delta (t)$):

$x(t)={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\delta (t-u)x(u)𝑑u$(2)

Now, we define a new function h(t) (known as the impulse response of the LTI system $𝕊$) the following way

$\delta (t)\stackrel{𝕊}{⟶}h(t)$(3)

Then, due to the time-invariance property of the LTI system $𝕊$, it is true that, for some constant $t_0,$

$\delta (t-t_0)\stackrel{𝕊}{⟶}h(t-t_0)$(4)

and due to the linearity property of the LTI system $𝕊$, it is also follows that, for some constant $C_1$,

$C_1\delta (t)\stackrel{𝕊}{⟶}{C}_{1}h(t)$(5)

Then, it follows from (2), (4) and (5) that

$x(t)={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\delta (t-u)x(u)𝑑u\stackrel{𝕊}{⟶}y(t)={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}h(t-u)x(u)𝑑u$(6)

Now, we define the output $y(t)$ in (6) as the zero-state response of x(t) i.e.

$\text{ZSR}(t)={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}h(t-u)x(u)𝑑u$

and we also define a new binary operation ’*’, called the convolution, as

$h(t)*x(t)\equiv {\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}h(t-u)x(u)𝑑u$

Using this notation,

$\text{ZSR}(t)=h(t)*x(t)$(7)

Thus, in order to find ${\text{ZSR}}_{(t)}$ for a given input $x(t)$, we need to find $h(t)$. In order to do this, we first define a function $z(t)$ that satisfies the following equality:

$a(D)z(t)=\delta (t)$(8)

Then the following also holds:

$b_m[a(D)D^0(z(t))]=b_m[D^0(\delta (t))]$

$b_{m-1}[a(D)D^1(z(t))]=b_{m-1}[D^1(\delta (t))]$

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$b_{m-k}[a(D)D^k(z(t))]=b_{m-k}[D^k(\delta (t))]$

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$b_1[a(D)D^{m-1}(z(t))]=b_0[D^{m-1}(\delta (t))]$

$b_0[a(D)D^m(z(t))]=b_0[D^m(\delta (t))]$

Adding all the above equations together we get

$a(D)[b_0{D}^m(z(t))+b_1{D}^{m-1}(z(t))+\mathrm{\dots }+b_m{D}^0(z(t))]$

$=b_0{D}^m(\delta (t))+b_1{D}^{m-1}(\delta (t))+\mathrm{\dots }+b_m{D}^0(\delta (t))$

or equivalently,

$a(D)[b(D)z(t)]=b(D)\delta (t)$

Since the above equation has the same form as (1) with $x(t)=\delta (t)$, it follows that $h(t)=b(D)z(t)$ (since h(t), by definition, is the output when the input is the impulse function). Hence,

$h(t)=b(D)z(t)$(9)

where $z(t)$ is given by

$a(D)z(t)=\delta (t)$(10)

Thus, in order to find $h(t)$, we need to find $z(t)$ first. We will do this with the help of an example. Given,

$a(D)=D^2+5D+6$

$\Rightarrow (D^2+5D+6)z(t)=\delta (t)$

$\Rightarrow z^{\prime \prime }(t)+5z^{\prime }(t)+6z(t)=\delta (t)$

with initial conditions $z^{\prime }(0^{-})=0,z(0^{-})=0$. In order to get $\delta (t)$ on the R.H.S, $z^{\prime \prime }(t)$ must be $\delta (t)$ plus some ordinary function $m(t)$ i.e.

$z^{\prime \prime }(t)=\delta (t)+m(t)$

$\Rightarrow z^{\prime }(t)=u(t)+{\displaystyle {\int }_{0^{-}}^t}m(a)𝑑a$

$\Rightarrow z(t)=r(t)+{\displaystyle {\int }_{0^{-}}^t}𝑑b{\displaystyle {\int }_{0^{-}}^b}m(a)𝑑a$

For $t\ge 0^{+}$, we have

$z^{\prime \prime }(t)+5z^{\prime }(t)+6z(t)=0$

since $\delta (t)=0$ for $t\ge 0^{+}$. Furthermore, we now get new initial conditions at $t=0^{+}$. Thus,

${z^{\prime }(t)|}_{t=0^{+}}=1+0=1$

${z(t)|}_{t=0^{+}}=0+0=0$

In general, for $t\ge 0^{+}$, the initial condition involving the n-1 derivative would be equal to 1 and all the other initial conditions would be 0. Thus, if

$a(D)z(t)=\delta (t)\mathit{\hspace{1em}}\text{with initial conditions }{z}^{(n-1)}(0^{-})=0,z^{(n-2)}(0^{-})=0,\mathrm{\dots },z(0^{-})=0$

then for $t\ge 0^{+}$,

$a(D)z(t)=0\mathit{\hspace{1em}}\text{with initial conditions }{z}^{(n-1)}(0^{+})=1,z^{(n-2)}(0^{+})=0,\mathrm{\dots },z(0^{+})=0$

In summary, we found out that the zero-state response is given by

$\text{ZSR}(t)=h(t)*x(t)$(11)

where $h(t)$ is the impulse response of the system $𝕊$ and is given by

$h(t)=b(D)z(t)$(12)

where $z(t)$ is the solution of the differential equation

$a(D)z(t)=0$(13)

with initial conditions $z^{(n-1)}(0^{+})=1,z^{(n-2)}(0^{+})=0,\mathrm{\dots },z^{\prime }(0^{+})=0,z(0^{+})=0$.