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 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

Reply by Taras_96●January 15, 20072007-01-15

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