Understanding the Zero-State Response

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

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

+

…

+

a

n

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.

y

(

n

-

1

)

⁢

(

0

-

)

=

0

,

y

(

n

-

2

)

⁢

(

0

-

)

=

0

,

.

.

.

y

⁢

(

0

-

)

=

0

Then we note that x(t) can be written in the following way (due to the shifting

property

of

δ

⁢

(

t

)

):

x

⁢

(

t

)

=

∫

-

∞

∞

δ

⁢

(

t

-

u

)

⁢

x

⁢

(

u

)

⁢

𝑑

u

(2)

Now, we define a new function h(t) (known as the

impulse response

of the LTI system

𝕊

) the following way

1

1

Simply put, the impulse response of a system

𝕊

is the output we get when we drive the input by the impulse function

δ

⁢

(

t

)

δ

⁢

(

t

)

⁢

⟶

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

,

δ

⁢

(

t

-

t

0

)

⁢

⟶

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

⁢

δ

⁢

(

t

)

⁢

⟶

𝕊

⁢

C

1

⁢

h

⁢

(

t

)

(5)

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

2

2

This 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.

x

⁢

(

t

)

=

∫

-

∞

∞

δ

⁢

(

t

-

u

)

⁢

x

⁢

(

u

)

⁢

𝑑

u

⁢

⟶

𝕊

⁢

y

⁢

(

t

)

=

∫

-

∞

∞

h

⁢

(

t

-

u

)

⁢

x

⁢

(

u

)

⁢

𝑑

u

(6)

Now, we define the output

y

⁢

(

t

)

in (6) as the zero-state response of x(t) i.e.

ZSR

⁢

(

t

)

=

∫

-

∞

∞

h

⁢

(

t

-

u

)

⁢

x

⁢

(

u

)

⁢

𝑑

u

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

convolution

, as

h

⁢

(

t

)

*

x

⁢

(

t

)

≡

∫

-

∞

∞

h

⁢

(

t

-

u

)

⁢

x

⁢

(

u

)

⁢

𝑑

u

Using this notation,

ZSR

⁢

(

t

)

=

h

⁢

(

t

)

*

x

⁢

(

t

)

(7)

Thus, in order to find

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

)

=

δ

⁢

(

t

)

(8)

Then the following also holds:

b

m

⁢

[

a

⁢

(

D

)

⁢

D

0

⁢

(

z

⁢

(

t

)

)

]

=

b

m

⁢

[

D

0

⁢

(

δ

⁢

(

t

)

)

]

b

m

-

1

⁢

[

a

⁢

(

D

)

⁢

D

1

⁢

(

z

⁢

(

t

)

)

]

=

b

m

-

1

⁢

[

D

1

⁢

(

δ

⁢

(

t

)

)

]

.

.

.

b

m

-

k

⁢

[

a

⁢

(

D

)

⁢

D

k

⁢

(

z

⁢

(

t

)

)

]

=

b

m

-

k

⁢

[

D

k

⁢

(

δ

⁢

(

t

)

)

]

.

.

.

b

1

⁢

[

a

⁢

(

D

)

⁢

D

m

-

1

⁢

(

z

⁢

(

t

)

)

]

=

b

0

⁢

[

D

m

-

1

⁢

(

δ

⁢

(

t

)

)

]

b

0

⁢

[

a

⁢

(

D

)

⁢

D

m

⁢

(

z

⁢

(

t

)

)

]

=

b

0

⁢

[

D

m

⁢

(

δ

⁢

(

t

)

)

]

Adding all the above equations together we get

a

⁢

(

D

)

⁢

[

b

0

⁢

D

m

⁢

(

z

⁢

(

t

)

)

+

b

1

⁢

D

m

-

1

⁢

(

z

⁢

(

t

)

)

+

…

+

b

m

⁢

D

0

⁢

(

z

⁢

(

t

)

)

]

=

b

0

⁢

D

m

⁢

(

δ

⁢

(

t

)

)

+

b

1

⁢

D

m

-

1

⁢

(

δ

⁢

(

t

)

)

+

…

+

b

m

⁢

D

0

⁢

(

δ

⁢

(

t

)

)

or equivalently,

3

3

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

a

⁢

(

D

)

⁢

[

b

⁢

(

D

)

⁢

z

⁢

(

t

)

]

=

b

⁢

(

D

)

⁢

δ

⁢

(

t

)

Since the above equation has the same form as (1) with

x

⁢

(

t

)

=

δ

⁢

(

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

)

=

δ

⁢

(

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

+

5

⁢

D

+

6

⇒

(

D

2

+

5

⁢

D

+

6

)

⁢

z

⁢

(

t

)

=

δ

⁢

(

t

)

⇒

z

′′

⁢

(

t

)

+

5

⁢

z

′

⁢

(

t

)

+

6

⁢

z

⁢

(

t

)

=

δ

⁢

(

t

)

with initial conditions

z

′

⁢

(

0

-

)

=

0

,

z

⁢

(

0

-

)

=

0

. In order to get

δ

⁢

(

t

)

on the R.H.S,

z

′′

⁢

(

t

)

must be

δ

⁢

(

t

)

plus some ordinary function

m

⁢

(

t

)

i.e.

z

′′

⁢

(

t

)

=

δ

⁢

(

t

)

+

m

⁢

(

t

)

⇒

z

′

⁢

(

t

)

=

u

⁢

(

t

)

+

∫

0

-

t

m

⁢

(

a

)

⁢

𝑑

a

⇒

z

⁢

(

t

)

=

r

⁢

(

t

)

+

∫

0

-

t

𝑑

b

⁢

∫

0

-

b

m

⁢

(

a

)

⁢

𝑑

a

For

t

≥

0

+

, we have

z

′′

⁢

(

t

)

+

5

⁢

z

′

⁢

(

t

)

+

6

⁢

z

⁢

(

t

)

=

0

since

δ

⁢

(

t

)

=

0

for

t

≥

0

+

. Furthermore, we now get new initial conditions at

t

=

0

+

. Thus,

z

′

⁢

(

t

)

|

t

=

0

+

=

1

+

0

=

1

z

⁢

(

t

)

|

t

=

0

+

=

0

+

0

=

0

In general, for

t

≥

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

)

=

δ

⁢

(

t

)

with initial conditions

⁢

z

(

n

-

1

)

⁢

(

0

-

)

=

0

,

z

(

n

-

2

)

⁢

(

0

-

)

=

0

,

…

,

z

⁢

(

0

-

)

=

0

then for

t

≥

0

+

,

a

⁢

(

D

)

⁢

z

⁢

(

t

)

=

0

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

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

,

…

,

z

′

⁢

(

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

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