Signals and Systems
Linear and Time-Invariant (LTI)
Discrete-time Systems
Ertem Tuncel
Professor & Chair of Electrical and Computer Engineering
University of CA, Riverside
Why LTI systems?
• Linear and time-invariant systems are
especially easy to analyze and design.
• In a lot of cases, they are good enough to do
the "signal processing" job.
• Amenable to frequency analysis in the Fourier
domain.
The impulse response
• An LTI system's response to an impulse input
is called its impulse response.
LTI SYSTEM
• Because the system is LTI,
LTI SYSTEM
The impulse response
• Not only that, but also
LTI SYSTEM
• Now, extending this all the way,
LTI SYSTEM
The impulse response
• If only all signals came in the form
• Then we would figure out the output for any
input in terms of the impulse response.
• But they DO come in that form!!! Recall that
The impulse response
which implies
• This sum is known as the convolution sum.
• Summary: If you know the impulse response
of an LTI system, you know everything there
is to know!
The convolution sum
• This operation is also denoted as
• As before, two ways to interpret this formula:
• An infinite summation of shifted impulse
responses h[n-k] each scaled with x[k].
• For every n, an infinite sum of the samples of
the product signal x[k] h[n-k].
The convolution sum
• Example: Find if
x[n] h[n]
2
1 1 1 1
-1 1 2 3 4
n -1 1 2 3 4
n
• Method 1: Accumulate x[k] h[n-k]'s.
x[n] x[1] h[n-1]
1 1 1
1
2
-1 1 2 3 4
n
-1 1 2 3 4
n
x[2] h[n-2] 2 2 2
h[n] -1 1 2 3 4 5
n
1 1 1
n y[n] 3 3
-1 1 2 3 4 2
1
-1 1 2 3 4 5
n
• Method 2: Calculate sum for each n.
x[n]
x[k] • Change the running variable to k
2 • To plot h[-k], flip h[k] around the
1
y-axis.
-1 1 2 3 4
nk
• To plot h[n-k], shift h[-k] to the
right by n units.
h[-k]
h[n]
h[k]
• For each n, sum up all samples
1 1 1 1 1 of the product signal x[k]h[n-k].
-3 -2 -1 1 2 3 4
nk
x[k] • To plot h[n-k], shift h[-k] to the
2 right by n units.
1
-1 1 2 3 4
k • For each n, sum up all samples
of the product signal x[k]h[n-k].
h[-k] • For n = 0, the sum yields
1 1 1
k
-3 -2 -1 1 2 3 4
y[n]
-1 1 2 3 4
n
x[k] • To plot h[n-k], shift h[-k] to the
2 right by n units.
1
-1 1 2 3 4
k • For each n, sum up all samples
of the product signal x[k]h[n-k].
h[1-k] • For n = 1, the sum yields
1 1 1
k
-3 -2 -1 1 2 3 4
y[n]
-1 1 2 3 4
n
x[k] • To plot h[n-k], shift h[-k] to the
2 right by n units.
1
-1 1 2 3 4
k • For each n, sum up all samples
of the product signal x[k]h[n-k].
h[2-k] • For n = 2, the sum yields
1 1 1
k
-3 -2 -1 1 2 3 4
y[n] 3
-1 1 2 3 4
n
x[k] • To plot h[n-k], shift h[-k] to the
2 right by n units.
1
-1 1 2 3 4
k • For each n, sum up all samples
of the product signal x[k]h[n-k].
h[3-k] • For n = 3, the sum yields
1 1 1
k
-3 -2 -1 1 2 3 4
y[n] 3 3
-1 1 2 3 4
n
x[k] • To plot h[n-k], shift h[-k] to the
2 right by n units.
1
-1 1 2 3 4
k • For each n, sum up all samples
of the product signal x[k]h[n-k].
h[4-k] • For n = 4, the sum yields
1 1 1
k
-3 -2 -1 1 2 3 4
y[n] 3 3
2
1
-1 1 2 3 4
n
x[k] • To plot h[n-k], shift h[-k] to the
2 right by n units.
1
-1 1 2 3 4
k • For each n, sum up all samples
of the product signal x[k]h[n-k].
h[5-k] • For n = 5, the sum yields
1 1 1
k
-3 -2 -1 1 2 3 4
y[n] 3 3
2
1
-1 1 2 3 4 5
n
The convolution sum
• Example: Find if
• Now, since x[n] has infinitely many non-zero
samples, Method 1 won't work.
• For Method 2, the infinite sum becomes
• If n < 0, the above sum is zero since it contains
u[n], u[n-1], u[n-2], ..., all of which is zero.
• Otherwise, u[n-k] = 0 only when k > n. Thus,
Digression: Power sums
• How do we compute ?
• Here is the trick:
• In other words,
• What about the infinite sum ?
• Converges only if , and to
Back to the example
• Example: Find if
• We had found
A more visual solution
x[k]
1 1 1 1 1
-2 -1 1 2 3 4
k
h[k]
1
0.5
-2 -1 1 2 3 4
k
A more visual solution
x[k]
1 1 1 1 1
-2 -1 1 2 3 4
k
h[-k]
1
0.5
-4 -3 -2 -1 1 2 3 4
k
A more visual solution
x[k]
1 1 1 1 1
-2 -1 1 2 3 4
k
h[1-k]
1
0.5
-4 -3 -2 -1 1 2 3 4
k
A more visual solution
x[k]
1 1 1 1 1
-2 -1 1 2 3 4
k
h[2-k]
1
0.5
-4 -3 -2 -1 1 2 3 4
k
A more visual solution
x[k]
1 1 1 1 1
-2 -1 1 2 3 4
k
h[3-k]
1
0.5
-4 -3 -2 -1 1 2 3 4
k
Properties of convolution
• Commutativity:
• Proof:
Properties of convolution
• Associativity:
• Proof:
Properties of convolution
• Linearity:
implies
• Proof: Follows from the fact that
convolution of the input with the impulse
response yields the output for linear and
time-invariant systems.
Properties of convolution
• The same logic leads to Time-invariance:
implies
• Thanks to commutativity, we also have
Properties of convolution
• Time-reversal:
implies
• Proof:
Properties of convolution
• Identity element:
• Proof:
• All the terms in the above sum is zero,
except at k = 0, where it is equal to x[n].
System properties revisited
• For an LTI system, we can tell whether the
system is memoryless, causal, stable, or
invertible just by analyzing the impulse
response.
• It may be more convenient to write the
convolution sum as
• In general, this indicates that y[n] depends
on all samples of x[n].
System properties revisited
• Write this more openly as
PAST
PRESENT
FUTURE
System properties revisited
• For the system to be memoryless, the
present value of y[n] must depend only on
the present value of x[n].
• That is the same as
• In other words, the impulse response must
be of the form for some c.
System properties revisited
• For the system to be causal, the present
value of y[n] must depend only on the
present and past values of x[n].
• That is the same as
• In other words, the impulse response must
be of the form for some
g[n].
System properties revisited
• For stability, let us analyze :
• Now, if is bounded by B for all n,
System properties revisited
• Therefore, a sufficient condition for stability is
• It is also necessary because otherwise, we
could just select to obtain
System properties revisited
• The system has an LTI inverse if and only if
there exists a signal g[n] such that
for all x[n].
• This is equivalent to
• If such g[n] exists, it is the impulse response of
the inverse system.
Examples
• Example: Determine if the system is
memoryless, causal, stable, or invertible if its
impulse response is given by
HAS
• Memory: h[n] is not of the form MEMORY
CAUSAL
• Causality: h[n] is of the form
• Stability:
STABLE
Examples
• Example: Determine if the system is
memoryless, causal, stable, or invertible if its
impulse response is given by
• Invertibility: Observe that
Examples
• Invertibility: Observe that
• Now, can we rewrite this as
for some g[n]?
• Yes. Take :
INVERTIBLE
Examples
• Example: Determine if the system is
memoryless, causal, stable, or invertible if its
impulse response is given by
HAS
• Memory: h[n] is not of the form MEMORY
NON-
• Causality: h[n] is not of the form CAUSAL
• Stability:
UNSTABLE
Examples
• Invertibility: If g[n] exists such that
what would be the result of ?
• Due to time invariance, it must be
• Due to the fact that , it must be
• Contradiction!!!!
• No such g[n] can exist.
NOT INVERTIBLE
