# 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 UnderstandingTheZeroStateResponse 2013-03-11 19:27:00 2013-03-11 19:27:00 swapnizzle (13346) (0) 1 swapnizzle (0) Definition