natural and forced responses versus zero-state and zero-input responses

Started by Taras_96 January 15, 2007
Hi everyone,

Pg 79 of "Analogue and Digital Signal Processing" by Ambardar states
that "Note that the natural and roced component y_n(t) and y_f(t) do
not, in general, crrespond to the zero-input and zero-state response,
respectively, even though each pair adds up to the total response".

Seeing that the natural response is obtained by setting all input's to
zero that are in the differential equation that describes the system, I
do not see how the natural response can differ to the zero-input
response. How are the zero-input and natural responses different?

Thanks

Taras

Taras_96 wrote:
> Hi everyone, > > Pg 79 of "Analogue and Digital Signal Processing" by Ambardar states > that "Note that the natural and roced component y_n(t) and y_f(t) do > not, in general, crrespond to the zero-input and zero-state response, > respectively, even though each pair adds up to the total response". > > Seeing that the natural response is obtained by setting all input's to > zero that are in the differential equation that describes the system, I > do not see how the natural response can differ to the zero-input > response. How are the zero-input and natural responses different? > > Thanks > > Taras >
Take for example the following differential equation: dy/dt + 5y(t) = 10, y(0)=7. If we solve for the natural response we get: dy/dt + 5y(t) = 0 s+5=0 s = -5 y_nat(t) = Aexp(-5t) If we solve for the forced response we get: y_force(t) = B d/dt{B} + 5{B} = 10 0+5B = 10 B=2 y_force(t) = 2 y(t) = y_nat(t) + y_force(t), y(0)=7 y(t) = Aexp(-5t) + 2 y(0) = 7 Aexp(0) + 2 = 7 A=5 y(t) = 5exp(-5t) + 2 ----------------------------- Now solve it using the zero-state and zero-input method. First let's solve the zero-input: dy/dt + 5y(t) = 0, y(0)=7 s+5= 0 s = -5 y(t) = Aexp(-5t) y(0) = 7 Aexp(0) = 7 A=7 y_zi(t) = 7exp(-5t) ********* Now solve for the zero-state: dy/dt + 5y(t) = 10, y(0)=0 First get the natural solution: dy/dt + 5y(t) = 0 s+5=0 s = -5 y_nat(t) = Aexp(-5t) Now get the forced y(t) = B d/dt{B} + 5B = 10 0+5B=10 B=2 y_zs(t) = 2+Aexp(-5t), y_zs(0)=0 y_zs(0)=0 2+Aexp(0)=0 2+A=0 A = -2 y_zs(t) = 2-2exp(-5t) y(t) = y_zi(t) + y_zs(t) y(t) = 7exp(-5t) + 2 - 2exp(-5t) y(t) = 5exp(-5t) + 2 ------------------ Hopefully that demonstrates to you better than an explanation. So as you can see when you solve by doing the zero-state, zero-input method you more or less end up solving the differential eqn twice! So why would anyone ever use the zero-state, zero-input method? 1) It provides better insight into the system. You can see what component of the output is due to initial conditions of the system and what component comes from what you're putting into the sytem. 2) When you use other methods of analysis such as Fourier transforms you are not taking into account the initial conditions. So for example if you're given h(t) and f(t) and you do a convolution (or equivalently multiplication in frequency) then then the output solution y(t) that you have solved is the zero-state solution. Therefore if you wanted the complete solution you'd need to compute the zero-input solution using an alternate method (such as solving the diff eqn). I hope that clarifies! Brad
On 15 Jan 2007 01:19:42 -0800, "Taras_96" <taras.di@gmail.com> wrote:

>Hi everyone, > >Pg 79 of "Analogue and Digital Signal Processing" by Ambardar states >that "Note that the natural and roced component y_n(t) and y_f(t) do >not, in general, crrespond to the zero-input and zero-state response, >respectively, even though each pair adds up to the total response". > >Seeing that the natural response is obtained by setting all input's to >zero that are in the differential equation that describes the system, I >do not see how the natural response can differ to the zero-input >response. How are the zero-input and natural responses different? > >Thanks > >Taras
Hi, you can always E-mail Prof.Ambardar. http://www.ece.mtu.edu/pages/faculty/Ambardar.html [-Rick-]