Understanding the Zero-State Response

Understanding the Zero-State Response Swapnil Sunil Jain 15 January, 2007

Understanding the Zero-State Response By definition we know that the $\mbox{ZSR}(t)$ 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 $\mathbb{S}$ with input $x(t)$ and output $y(t)$ be described by the following differential equation

$\displaystyle a(D)y(t)=b(D)x(t)$

(1)

where

	$\displaystyle a(D)=D^{n}+a_{1}D^{n-1}+...+a_{n}$
	$\displaystyle b(D)=b_{0}D^{m}+b_{1}D^{m-1}+...+b_{n}$

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

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

	$\displaystyle y^{(n-1)}(0^{-})=0,$
	$\displaystyle y^{(n-2)}(0^{-})=0,$
	$\displaystyle.$
	$\displaystyle.$
	$\displaystyle.$
	$\displaystyle y(0^{-})=0$

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

$\displaystyle x(t)=\int_{-\infty}^{\infty}\delta(t-u)x(u)du$

(2)

Now, we define a new function h(t) (known as the impulse response of the LTI system $\mathbb{S}$ ) the following way¹¹Simply put, the impulse response of a system $\mathbb{S}$ is the output we get when we drive the input by the impulse function $\delta(t)$

$\displaystyle\delta(t)\overset{\mathbb{S}}{\longrightarrow}h(t)$

(3)

Then, due to the time-invariance property of the LTI system $\mathbb{S}$ , it is true that, for some constant $t_{0},$

$\displaystyle\delta(t-t_{0})\overset{\mathbb{S}}{\longrightarrow}h(t-t_{0})$

(4)

and due to the linearity property of the LTI system $\mathbb{S}$ , it is also follows that, for some constant $C_{1}$ ,

$\displaystyle C_{1}\delta(t)\overset{\mathbb{S}}{\longrightarrow}C_{1}h(t)$

(5)

Then, it follows from (2), (4) and (5) that²²This result makes sense if we think of the integral as an infinite sum and treat x(u) as some constant with respect to time t.

$\displaystyle x(t)=\int_{-\infty}^{\infty}\delta(t-u)x(u)du\overset{\mathbb{S}% }{\longrightarrow}y(t)=\int_{-\infty}^{\infty}h(t-u)x(u)du$

(6)

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

$\displaystyle\mbox{ZSR}(t)=\int_{-\infty}^{\infty}h(t-u)x(u)du$

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

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

Using this notation,

$\displaystyle\mbox{ZSR}(t)=h(t)*x(t)$

(7)

Thus, in order to find $\mbox{ZSR}_{(t)}$ 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:

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

(8)

Then the following also holds:

	$\displaystyle b_{m}[a(D)D^{0}(z(t))]=b_{m}[D^{0}(\delta(t))]$
	$\displaystyle b_{m-1}[a(D)D^{1}(z(t))]=b_{m-1}[D^{1}(\delta(t))]$
	$\displaystyle.$
	$\displaystyle.$
	$\displaystyle.$
	$\displaystyle b_{m-k}[a(D)D^{k}(z(t))]=b_{m-k}[D^{k}(\delta(t))]$
	$\displaystyle.$
	$\displaystyle.$
	$\displaystyle.$
	$\displaystyle b_{1}[a(D)D^{m-1}(z(t))]=b_{0}[D^{m-1}(\delta(t))]$
	$\displaystyle b_{0}[a(D)D^{m}(z(t))]=b_{0}[D^{m}(\delta(t))]$

Adding all the above equations together we get

	$\displaystyle a(D)[b_{0}D^{m}(z(t))+b_{1}D^{m-1}(z(t))+...+b_{m}D^{0}(z(t))]$
	$\displaystyle=b_{0}D^{m}(\delta(t))+b_{1}D^{m-1}(\delta(t))+...+b_{m}D^{0}(% \delta(t))$

or equivalently,³³One could have also easily come up with the following result by simply operating b(D) on both sides of (8)

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

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,

$\displaystyle h(t)=b(D)z(t)$

(9)

where $z(t)$ is given by

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

(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,

	$\displaystyle a(D)=D^{2}+5D+6$
	$\displaystyle\Rightarrow(D^{2}+5D+6)z(t)=\delta(t)$
	$\displaystyle\Rightarrow z^{\prime\prime}(t)+5z^{\prime}(t)+6z(t)=\delta(t)$

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.

	$\displaystyle z^{\prime\prime}(t)=\delta(t)+m(t)$
	$\displaystyle\Rightarrow z^{\prime}(t)=u(t)+\int_{0^{-}}^{t}m(a)da$
	$\displaystyle\Rightarrow z(t)=r(t)+\int_{0^{-}}^{t}db\int_{0^{-}}^{b}m(a)da$

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

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

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

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

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

$\displaystyle a(D)z(t)=\delta(t)\quad\mbox{with initial conditions }z^{(n-1)}(% 0^{-})=0,z^{(n-2)}(0^{-})=0,...,z(0^{-})=0$

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

$\displaystyle a(D)z(t)=0\quad\mbox{with initial conditions }z^{(n-1)}(0^{+})=1% ,z^{(n-2)}(0^{+})=0,...,z(0^{+})=0$

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

$\displaystyle\mbox{ZSR}(t)=h(t)*x(t)$

(11)

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

$\displaystyle h(t)=b(D)z(t)$

(12)

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

$\displaystyle a(D)z(t)=0$

(13)

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

Title	Understanding the Zero-State Response
Canonical name	UnderstandingTheZeroStateResponse
Date of creation	2013-03-11 19:27:00
Last modified on	2013-03-11 19:27:00
Owner	swapnizzle (13346)
Last modified by	(0)
Numerical id	1
Author	swapnizzle (0)
Entry type	Definition