We are now looking for a more consise way to characterize discrete-time systems.
Let’s recall the unit sample (also called unit impulse) signal (Figure 1)
δ(n)=1,n=0=0,n≠0δ(n)=1,n=0=0,n≠0alignl { stack {
size 12{δ \( n \) =1, matrix {
{} # {} # {}
} n=0} {} #
size 12{ matrix {
{} # {}
} =0, matrix {
{} # {} # {}
} n <> 0} {}
} } {}
(1)
When the sample is shifted to time index (or sample) k in the future
(k>0)(k>0) size 12{ \( k>0 \) } {} the signal is
δ
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=
1,
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alignl { stack {
size 12{δ \( n - k \) =1, matrix {
{} # {}
} n=k} {} #
size 12{ matrix {
{} # {} # {} # {}
} 0, matrix {
{} # {}
} n <> k} {}
} } {}
When the sample is shifted to the past at index –k
(k>0)(k>0) size 12{ \( k>0 \) } {}, the signal is
δ(n+k)δ(n+k) size 12{δ \( n+k \) } {}
δ
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alignl { stack {
size 12{δ \( n+k \) =1, matrix {
{} # {}
} n= - k} {} #
size 12{ matrix {
{} # {} # {} # 0, matrix {
{} # {}
} {}
} n <> - k} {}
} } {}
Now,let’s express a signal in terms of unit samples. In Figure 2 the value of
x(n)x(n) size 12{x \( n \) } {} at
n=1n=1 size 12{n=1} {} is 3, so we can write ( Equation )
x
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n
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1
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=
x
(
1
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δ
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=
3
x
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x
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1
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δ
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x
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size 12{x \( n=1 \) =x \( 1 \) matrix {
{} # {}
} δ \( n - 1 \) =3 matrix {
{} # {}
} x \( 1 \) =3} {}
Similarly at
n=2n=2 size 12{n=2} {}
x
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δ
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x
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x
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=
2
size 12{x \( n=2 \) =x \( n \) matrix {
{} # {}
} δ \( n - 2 \) =2 matrix {
{} # {}
} x \( 1 \) =2} {}
Thus a signal
x(n)x(n) size 12{x \( n \) } {} can be expressed as
x(n)=∑k=−∞∞x(k)δ(n−k)x(n)=∑k=−∞∞x(k)δ(n−k) size 12{x \( n \) = Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {x \( k \) δ \( n - k \) } } {}
(2)
Impulse response (or impulsive response) of a discrete-time (or digital) system is defined as the output (response), denoted by
h(n)h(n) size 12{h \( n \) } {}, when the input is an unit sample
δ(n)δ(n) size 12{δ \( n \) } {}. The impulse response may be real or complex, but usually assumed real. Figure 3 is an example.
When excited by an unit sample
δ(n)δ(n) size 12{δ \( n \) } {}, the impulse response
h(n)h(n) size 12{h \( n \) } {}of a system may last a finite duration, or forever even before the input is applied. In the former case the system is Finite Duration Impulse Response (FIR), and in the latter case the system is Infinite Duration Impulse Response (IIR). Many authors do not include the word duration in the names. It’s just a matter of choice.
Alternatively, the systems can be classfied as Recursive or Nonrecursive instead of FIR or IIR. We’ll see the difference later.
Causality (see section )of a system is reffected on its impulse resopnse: For causal systems
h(n)=0h(n)=0 size 12{h \( n \) =0} {} at
n<0n<0 size 12{n<0} {} (or
n≤−1n≤−1 size 12{n <= - 1} {}), otherwise they are noncausal. Both systems of Figure are noncausal.
From the definition of impulse response we can apply an unit sample to the system concerned and obtain the output experimentally, which is the impulse response. Otherwise we derive it from the difference equation as presented here. There are still other ways to obtain the impulse response.
Find the impulse response of system whose input-output signal difference equation is given by
y
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8y
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8y
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size 12{y \( n \) =0 "." 8y \( n - 1 \) +x \( n \) } {}
Solution
Replacing
x(n)x(n) size 12{x \( n \) } {} by
δ(n)δ(n) size 12{δ \( n \) } {} then
y(n)y(n) size 12{y \( n \) } {} is just
h(n)h(n) size 12{h \( n \) } {}:
h
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=
0
.
8h
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8h
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size 12{h \( n \) =0 "." 8h \( n - 1 \) +δ \( n \) } {}
Remember that
δ(n)=1δ(n)=1 size 12{δ \( n \) =1} {} at
n=0n=0 size 12{n=0} {}, otherwise zero, and assume a causal system, i.e.
h(n)=0h(n)=0 size 12{h \( n \) =0} {} for
n<0n<0 size 12{n<0} {}, we have
h
(
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=
0
.
8h
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1
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+
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=
1
h
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8h
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0
.
8
h
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8h
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8
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8h
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8
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.
h
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8
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8h
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8h
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8
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8h
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8
2
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8h
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8
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alignl { stack {
size 12{h \( 0 \) =0 "." 8h \( - 1 \) +δ \( 0 \) =1} {} #
size 12{h \( 1 \) =0 "." 8h \( 0 \) +δ \( 1 \) =0 "." 8} {} #
size 12{h \( 2 \) =0 "." 8h \( 1 \) +δ \( 2 \) =0 "." 8 rSup { size 8{2} } } {} #
h \( 3 \) =0 "." 8h \( 2 \) +δ \( 3 \) =0 "." 8 rSup { size 8{3} } {} #
"." "." "." "." {} #
h \( n \) =0 "." 8 rSup { size 8{n} } u \( n \) {}
} } {}
The system is IIR and stable (since
h(n)h(n) size 12{h \( n \) } {} converges). Usually we don’t get the result in closed form as above. See also Example
Resersely when the impulse response of a system is known we can derive its difference equation. Following is an illustrative example.
The impulse response of a system is periodic with the period of 3 indexes
h
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=
[
1,
2,
3
;
1,
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;
1,
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3
;
1,
2
.
.
.
]
h
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=
[
1,
2,
3
;
1,
2,
3
;
1,
2,
3
;
1,
2
.
.
.
]
size 12{h \( n \) = \[ 1, matrix {
} 2, matrix {
} 3; matrix {
} 1, matrix {
} 2, matrix {
} 3; matrix {
} 1, matrix {
} 2, matrix {
} 3; matrix {
} 1, matrix {
} 2 "." "." "." \] } {}
Find its input-output signal difference equation.
Solution
Delay the given impulse response 3 samples
h
(
n
−
3
)
=
[
0,
0,
0
;
1,
2,
3
;
1,
2,
3
;
1,
2,
3
;
1,
2
.
.
.
]
h
(
n
−
3
)
=
[
0,
0,
0
;
1,
2,
3
;
1,
2,
3
;
1,
2,
3
;
1,
2
.
.
.
]
size 12{h \( n - 3 \) = \[ 0, matrix {
} 0, matrix {
} 0; matrix {
} 1, matrix {
} 2, matrix {
} 3; matrix {
} 1, matrix {
} 2, matrix {
} 3; matrix {
} 1, matrix {
} 2, matrix {
} 3; matrix {
} 1, matrix {
} 2 "." "." "." \] } {}
Take the difference
h
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n
)
−
h
(
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−
3
)
=
[
1,2,3
;
0,0,0,0,0
.
.
.
]
=
δ
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+
2δ
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+
3δ
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h
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[
1,2,3
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0,0,0,0,0
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]
=
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2δ
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alignl { stack {
size 12{h \( n \) - h \( n - 3 \) = \[ 1,2,3;0,0,0,0,0 "." "." "." \] } {} #
size 12{ matrix {
matrix {
matrix {
matrix {
matrix {
{} # {}
} {} # {}
} {} # {} # {}
} {} # {} # {}
} {} # {} # {}
} =δ \( n \) +2δ \( n - 1 \) +3δ \( n - 2 \) } {}
} } {}
Then
h
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=
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size 12{h \( n \) =h \( n - 3 \) +δ \( n \) +2δ \( n - 1 \) +3δ \( n - 2 \) } {}
By definition on of impulse response, the difference equation is
y
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=
y
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2x
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y
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size 12{y \( n \) =y \( n - 3 \) +x \( n \) +2x \( n - 1 \) +3x \( n - 2 \) } {}
