# lecture21

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```					EECS 20 N—March 9, 2001
Lecture 21:
Time Responses
Laurent El Ghaoui
1
announcements: recall
• reading assignment: Chapter 6 of Lee and Varaiya
• no lab next week (3/12–3/16)
• still, go to lab sessions for problem set discussions
• this week’s lab due in two weeks only
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outline
• time-invariance
• linearity
• impulse response
• response to sinusoids
look at discrete-time systems only
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delaying signals
given a discrete-time signal, we can associate the same signal, delayed
by a time T :
a discrete−time sequence
5
4
response

3
2
1
0
0   1           2           3          4         5         6    7   8   9
time (seconds)
a discrete−time sequence, delayed
5
original sequence
4           delayed sequence
response
3
2
1
0
0   1           2           3          4         5         6    7   8   9
time (seconds)
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delay operator
formally, if x is a function of time, then we deﬁne a new signal DN (x)
by
∀ n ∈ Ints, DN (x)(n) = x(n − N )
here, N ∈ Ints is the delay
DN is called the delay operator
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time-invariance
if y is the zero-state response to input x of the system
s(n + 1)   = As(n) + bx(n), n = 0, 1, 2, . . .
y(n)   = cT s(n) + dx(n)
then, for any N ∈ Ints, the response to DN x is DN y
i.e., if we denote the zero-state response to input x by y = S(x), then
S(DN (x)) = DN (S(x))
this property is called time-invariance
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time-invariance and output response
for the system (denoted S)
s(n + 1)   = As(n) + bx(n), n = 0, 1, 2, . . .
y(n)   = cT s(n) + dx(n)
recall the zero-state output response formula:
n−1
(S(x))(n) = y(n) =         cT An−1−m bx(m) + dx(n)
m=0
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if we delay x by N , the output is delayed likewise:
n−1−N
DN (Sx)(n) = y(n − N )      =            cT An−1−m bx(m) + dx(n − N )
m=0
n−1
=          cT An−1−m bx(m − N ) + dx(n − N )
m=0
= S(DN (x))(n)
we’ll see a more transparent proof next
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example of a non-time-invariant system

 1 if n = 0,


y(n) = d(n)x(n), where d(n) =           2 if n = 1


0 if n > 1

is not time-invariant, example:
n    d x     D1 (x) S(x) D1 (S(x)) S(D1 (x))
0   1   1     0        1       0             0
1   2   2     1        4       1             2
2   0   3     2        0       4             0
3   0   4     3        0       0             0
4   0   5     4        0       0             0
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a general result
the system with time-varying state-space matrices
s(n + 1) =      A(n)s(n) + b(n)x(n), n = 0, 1, 2, . . .
y(n)   =    c(n)T s(n) + d(n)x(n)
is in general not time-invariant
when A(n), b(n), c(n) and d(n) are independent of n, then the system
is time-invariant
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impulse response
impulse response is the zero-state response with impulse input:

 0 if n < 0


δ(n) =       1 if n = 0,


0 if n > 0

we get

 0               if n < 0,

n−1                               
y(n) =         cT An−1−m bδ(m) + dδ(n) =         d           if n = 0,

m=0                               
cT An−1 b   if n > 0

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convolution sum
the zero-state output response
n−1
y(n) =         cAn−1−m bx(m) + dx(n)
m=0
can be written as a convolution sum
n−1                      m=+∞
y(n) =         h(n − m)x(m) =            h(n − m)x(n)
m=0                      m=−∞
where h is the sequence

 0           if n < 0,


h(n) =        d        if n = 0,


cAn−1 b if n > 0

h is the impulse response!
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result
for a linear, time-invariant system, the zero-state response to an
arbitrary input is the convolution of the input with the impulse
response
n−1                    m=+∞
y(n) =         h(n − m)x(m) =          h(n − m)x(n)
m=0                    m=−∞
we denote this with the convolution operator:
y=h x
what does this mean?
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LTI systems and convolution
assume we are given a discrete-time system y = S(x) that satisﬁes the
following properties:
• it is linear
• it is time-invariant
we say the system is LTI
given a signal x, what is the form of the output?
claim: the output signal can be written as
y = h x,
where h is the impulse response
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to get the idea
start with x(0), what is y(0)? by linearity, there exist a number h(0)
(independent of x(0)!) such that
y(0) = h(0)x(0)
now, what is the response at time n = 1? by linearity, there exist
numbers h(1) and h(0, 1) (independent of x(0), x(1)!) such that
y(1) = h(0, 1)x(1) + h(1)x(0)
by time-invariance, we must have h(0, 1) = h(0), so that
y(1) = h(0)x(1) + h(1)x(0)
similarly, at time n = 2 y(2) takes the form
y(2) = h(0)x(2) + h(1)x(1) + h(2)x(0)
and so on . . .
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signals as sums of impulses
let’s take an arbitrary signal x, can decompose it in a series of
impulses:
+∞
x(n) =          x(m)δ(n − m)
m=−∞
this is a weighted sum of delayed impulses
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response to a delayed impulse
take the delayed impulse as input:

 0 if n < m,


(Dm δ)(n) = δ(n − m) =           1 if n = m,


0 if n > m

what is the output response?
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delayed impulse response
by time-invariance, the answer is: simply delay the impulse
response!
thus, the response to the delayed impulse (with delay m) is
(Dm h)(n) = h(n − m)
by linearity, response to arbitrary input x is weighted sum of delayed
impulse responses
+∞
y(n)   =     m=−∞     x(m)(Dm h)(n)
+∞
=     m=−∞     x(m)h(n − m)
the ﬁnal result: y = h x, uses both time-invariance and linearity
18
complex numbers
a complex number is one of the form
z = a + i · b,
√
where a, b are reals, and i =     −1 (so that i2 = −1)
a is called the real part of z, b the imaginary part
the magnitude of z is deﬁned by
|z| =      a2 + b 2
it is the Euclidean distance of the 2-D vector
    
a
        
b
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polar representation of complex numbers
√
let |z| =       a2 + b 2 ,
a       b
z = a + i · b = |z| ·                   +i·         ,
|z|     |z|
since
2               2
a               b             a2 + b 2
+               =            = 1,
|z|             |z|              |z|2
any such number can be represented as
z = |z| · (cos θ + i · sin θ),
where θ is such that
a             b
cos θ =       , sin θ =
|z|           |z|
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complex exponentials
we can represent a complex number as
z = ρeiθ
where
• ρ = |z| is the magnitude
• the complex exponential is deﬁned by
eiθ = cos θ + i · sin θ
where
+∞
un
eu =
n=0
n!
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sinusoidal inputs
a sinusoidal input is one of the form
x(n) = sin(nω + φ),
where φ ∈ [0 2π[ is the phase, and ω ∈ Reals is the frequency.
can also use a cos in the above (change φ)
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complex exponentials
it is more convenient to work with complex exponentials, for example
x(n) = einω
or, with a non-zero phase:
x(n) = ei(nω+φ) = cos(nω + φ) + i · sin(nω + φ)
these signals model a lot of real-life signals (more later)
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response to a complex exponential
consider the output response:
n−1                    m=+∞
y(n) =         h(n − m)x(m) =          h(n − m)x(n)
m=0                    m=−∞
what is the response to a complex exponential
x(n) = einω ?
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response to complex exponential
m=+∞           imω
y(n)   =       m=−∞ h(n − m)e
m=+∞           i(m−n)ω
=        m=−∞ h(n − m)e            einω
= H(w)einω ,
where H(w) is the complex number
m=+∞
H(w) =         m=−∞     h(n − m)ei(m−n)ω
m=+∞
=      m=−∞     h(m)eimω
H is called the frequency response
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