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Exponential
{s | s − a ∈ RoC(x)}
scaling
(Exercise 11)
t
y(t) = R x(τ)dτ
ˆ
Y (s) = ˆ
X (s)/s
RoC(y) ⊃
Integration
−∞
RoC(x) ∩ {s | Re{s} > 0} (Section 13.3.1)
ˆ
Y (s) = s ˆ
X (s)
RoC(y) ⊃ RoC(x)
Differentiation
(page 13.5.2)
d
y(t) =
x(t)
dt
Table 13.4: Properties of the Laplace transform. In this table, a, b are complex
constants, c and τ are real constants.
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13. LAPLACE AND Z TRANSFORMS
13.3.1
Integration
Let y be defined by
t
Z
∀ t ∈ R, y(t) =
x(τ)dτ.
−∞
The Laplace tranform is
∀ s ∈ RoC(y), ˆY(s) = ˆX(s)/s,
where
RoC(y) ⊃ RoC(x) ∩ {s | Re{s} > 0}.
This follows from the convolution property in table 13.4. We recognize that
y(t) = (x ∗ u)(t),
where u is the unit step. Hence, from the convolution property,
ˆ
Y (s) = ˆ
X (s) Û (s)
and
RoC(y) ⊃ RoC(x) ∩ RoC(u).
Û and RoC(u) are given in example 12.15, from which the property follows.
13.3.2
Sinusoidal signals
Sinusoidal signals have Laplace transforms with poles on the imaginary axis, as illustrated
in the following example.
Example 13.10: Let the causal sinusoidal signal y be given by
∀ t ∈ R, y(t) = sin(ω0t)u(t),
where ω0 is a real number and u is the unit step. Euler’s relation implies that
1
y(t) =
[eiω0t u(t) − e−iω0tu(t)].
2i
Lee & Varaiya, Signals and Systems
553
13.3. PROPERTIES OF THE LAPLACE TRANSFORM
Using (12.18) and linearity,
1
1
ˆ
Y (s)
=
−
1
2i
s + iω0
s − iω0
ω0
=
.
s2 + 2
ω0
This has no finite zeros and two poles, one at s = iω0 and the other at s = −iω0.
Both of these poles lie on the imaginary axis, as shown in Figure 13.3. The region
of convergence is the right half of the complex plane. Note that if this were the
impulse response of an LTI system, that system would not be stable. The region of
convergence does not include the imaginary axis.
13.3.3
Differential equations
We can use the differentiation property in table 13.4 to solve differential equations with
constant coefficients.
Example 13.11: In the tuning fork example of example 2.16, the displacement y
of a tine is related to the acceleration of the tine by
¨
y(t) = − 2
ω0y(t),
where ω0 is a real constant. Let us assume that the tuning fork is initially at rest,
and an external input x (representing say, a hammer strike) adds to the acceleration
as follows,
¨
y(t) = − 2
ω0y(t) + x(t).
We can use Laplace transforms to find the impulse response of this LTI system.
Taking Laplace transforms on both sides, using linearity and the differentiation
property,
∀ s ∈ RoC(y) ∩ RoC(x), s2 ˆY(s) = − 2
ω ˆ
0Y (s) + ˆ
X (s).
From this, we can find the transfer function of the system,
ˆ
Y (s)
1
ˆ
H(s) =
=
.
ˆ
X (s)
s2 + 2
ω0
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13. LAPLACE AND Z TRANSFORMS
Comparing this with example 13.10, we see that this differs only by a scaling by ω0
from the Laplace transform in that example. Thus, the pole-zero plot of the tuning
fork is shown in Figure 13.3, and the impulse response is given by
∀ t ∈ R, h(t) = sin(ω0t)u(t)/ω0.
Interestingly, this implies that the tuning fork is not stable. This impulse response
is not absolutely integrable. However, this model of the tuning fork is idealized. It
fails to account for loss of energy due to friction. A more accurate model would be
stable.
The above example can be easily generalized to find the transfer function of any LTI
system described by a differential equation. In fact, Laplace transforms offer a powerful
and effective way to solve differential equations.
In the previous example, we inverted the Laplace transform by recognizing that it matched
the example before that. In the next chapter, we will give a more general method for
inverting a Laplace transform.
13.4
Frequency response and pole-zero plots
Just as with Z transforms, the pole-zero plot of a Laplace transform can be used to get a
quick estimate of key properties of an LTI system. Consider a stable continuous-time LTI
system with impulse response h, frequency response H, and rational transfer function ˆ
H.
We know that the frequency response and transfer function are related by
∀ω ∈ R, H(ω) = ˆH(iω).
That is, the frequency response is equal to the Laplace transform evaluated on the imag-
inary axis. The imaginary axis is in the region of convergence because the system is
stable.
Assume that ˆ
H is a rational polynomial, in which case we can express it in terms of the
first-order factors of the numerator and denominator polynomials,
(s − q
ˆ
1) · · · (s − qM )
H(s) =
,
(s − p1) · · · (s − pN)
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555
13.4. FREQUENCY RESPONSE AND POLE-ZERO PLOTS
with zeros at q1, · · · , qM and poles at p1, · · · , pN. The zeros and poles may be repeated
(i.e., they may have multiplicity greater than one). The frequency response is therefore
∀
(iω − q1) · · · (iω − qM)
ω ∈ R,
H(ω) =
.
(iω − p1) · · · (iω − pN)
The magnitude response is
|
∀
iω − q1| · · · |iω − qM|
ω ∈ R,
|H(ω)| =
.
|iω − p1|···|iω − pN|
Each of these factors has the form
|iω − b|
where b is the location of either a pole or a zero. The factor |iω − b| is just the distance
from iω to b in the complex plane.
Of course, iω is a point on the imaginary axis. If that point is close to a zero at location
q, then the factor |iω − q| is small, so the magnitude response will be small. If that point
is close to a pole at p, then the factor |iω − p| is small, but since this factor is in the
denominator, the magnitude response will be large. Thus,
The magnitude response of a stable LTI system may be estimated from the pole-zero
plot of its transfer function. Starting at iω = 0, trace upwards and downwards along the
imaginary axis to increase or decrease ω. If you pass near a zero, then the magnitude
response should dip. If you pass near a pole, then the magnitude response should rise.
Example 13.12: Consider an LTI system with transfer function given by
∀
s
s ∈ {s | Re{s} > Re{a}},
H(s) =
.
(s − a)(s − a∗)
Suppose that a = c + iω0. Figure 13.4 shows three pole-zero plots for ω0 = 1
and three values of c, namely c = −1, c = −0.5, and c = −0.1. The magnitude
frequency responses can be calculated and plotted using the following Matlab code:
1
w = [-10:0.05:10];
2
a1 = -1.0 + i;
3
H1 = i*w./((i*w - a1).*(i*w-conj(a1)));
4
a2 = -0.5 + i;
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13. LAPLACE AND Z TRANSFORMS
5
H2 = i*w./((i*w - a2).*(i*w-conj(a2)));
6
a3 = -0.1 + i;
7
H3 = i*w./((i*w - a3).*(i*w-conj(a3)));
8
plot(w, abs(H1), w, abs(H2), w, abs(H3));
The plots are shown together at the bottom of Figure 13.4. The plot with the higher
peaks corresponds to the pole-zero plot with the poles closer to the imaginary axis.
13.5
The inverse transforms
There are two inverse transforms. The inverse Z transform recovers the discrete-time sig-
nal x from its Z transform ˆ
X . The inverse Laplace transform recovers the continuous-time
signal x from its Laplace transform ˆ
X . We study the inverse Z transform in detail. The
inverse Laplace transform is almost identical. The general approach is to break down
a complicated rational polynomial into a sum of simple rational polynomials whose in-
verse transforms we recognize. We consider only the case where ˆ
X can be expressed as a
rational polynomial.
13.5.1
Inverse Z transform
The procedure is to construct the partial fraction expansion of ˆ
X , which breaks it down
into a sum of simpler rational polynomials.
Example 13.13: Consider a Z transform given by
−
∀
1
1
1
z ∈ RoC(x),
ˆ
X (z) =
=
+
.
(13.6)
(z − 1)(z − 2)
z − 1
z − 2
This sum is called the partial fraction expansion of ˆ
X , and we will see below how
to find it systematically. We can write this as
∀ z ∈ RoC(x),
ˆ
X (z) = ˆ
X1(z) + ˆ
X2(z),
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557
13.5. THE INVERSE TRANSFORMS
where ˆ
X1(z) = −1/(z − 1) and ˆ
X2(z) = 1/(z − 2) are the two terms.
To determine the inverse Z transforms of the two terms, we need to know their
regions of convergence. Recall from the linearity property that RoC(x) includes the
intersection of the regions of convergence of the two terms,
RoC(x) ⊃ RoC(x1) ∩ RoC(x2).
(13.7)
Once we know these two regions of convergence, we can use table 13.1 to obtain
the inverse Z transform of each term. By the linearity property the sum of these
inverses is the inverse Z transform of ˆ
X .
ˆ
X given by (13.6) has one pole at z = 1 and one pole at z = 2. From section 12.2.3
we know that RoC(x) is bordered by these poles, so it has one of three forms:
1. RoC(x) = {z ∈ C | |z| < 1},
2. RoC(x) = {z ∈ C | 1 < |z| < 2}, or
3. RoC(x) = {z ∈ C | |z| > 2}.
Suppose we have case (1), which implies that x is anti-causal. From (13.7), the
region of convergence of the term −1/(z − 1) must be {z ∈ C | |z| < 1}. The
only other possibility is {z ∈ C | |z| > 1}, which would violate (13.7) unless the
intersection is empty (which would not be an interesting case). Thus, from table
13.1, the inverse Z transform of the first term must be the anti-causal signal x1(n) =
u(−n), for all n ∈ Z.
For the second term, 1/(z − 2), its region of convergence could be either {z ∈
C | |z| < 2} or {z ∈ C | |z| > 2}. Again, the second possibility would violate
(13.7), so we must have the first possibility. This results in x2(n) = −2n−1u(−n),
from the last entry in table 13.1. Hence, the inverse Z transform is
∀ n ∈ Z, x(n) = u(−n) − 2n−1u(−n).
If RoC(x) is given by case (2), we rewrite (13.6) slightly as
z
1
ˆ
X (z) = −z−1
+
.
z − 1
z − 2
The inverse Z transform of the first term is obtained from table 13.1, together with
the delay property in table 13.2. The inverse Z transform of the second term is the
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13. LAPLACE AND Z TRANSFORMS
same as in case (1). We conclude that in case (2) the inverse Z transform is the
two-sided signal
∀n, x(n) = −u(n − 1) − 2n−1u(−n).
In case (3), we write (13.6) as
z
z
ˆ
X (z) = −z−1
+ z−1
,
z − 1
z − 2
and conclude that the inverse Z transform is the causal signal
∀n, x(n) = −u(n − 1) + 2n−1u(n − 1).
We can generalize this example. Consider any strictly proper rational polynomial
A(z)
a
ˆ
M zM + · · · + a1z + a0
X (z) =
=
.
B(z)
zN + bN−1zN−1 + · · · + b1z + b0
The numerator is of order M, the denominator is of order N. ‘Strictly proper’ means that
M < N. We can factor the denominator,
a
ˆ
M zM + · · · + a1z + a0
X (z) =
.
(13.8)
(z − p1)m1(z − p2)m2 · · · (z − pk)mk
Thus ˆ
X has k distinct poles at pi, each with multiplicy mi. Since the order of the denomi-
nator is N, it must be true that
k
N = ∑ mi .
(13.9)
i=1
The partial fraction expansion of (13.8) is
k
R
R
R
î1
i2
im
X (z) =
i
∑
+
+ · · · +
.
(13.10)
(z − p
(z − p
(z − p
i=1
i)
i)2
i)mi
A pole with multiplicity mi contributes mi terms to the partial fraction expansion, so the
total number of terms is N, the order of the denominator, from (13.9). The coefficients Ri j
are complex numbers called the residues of the pole pi.
We assume that the poles p1, · · · , pN are indexed so that |p1| ≤ · · · |pN|. The RoC(x) must
have one of the following three forms:
Lee & Varaiya, Signals and Systems
559
13.5. THE INVERSE TRANSFORMS
1. RoC = {z ∈ C | |z| < |p1|},
2. RoC = {z ∈ C | |p j−1| < |z| < |p j|}, for j ∈ {2, · · · , k}, or
3. RoC = {z ∈ C | |z| > |pk|} .
As in example 13.13, each term in the partial fraction expansion has two possible regions
of convergence, only one of which overlaps with RoC(x). Thus, if we know RoC(x), we
can determine the region of convergence of each term of the partial fraction expansion,
and then use table 13.1 to find its inverse.
The following example illustrates how to find the residues.
Example 13.14: We will find the inverse Z transform of
2z + 3
R
R
ˆ
1
2
X (z) =
=
+
.
(z − 1)(z + 2)
z − 1
z + 2
The residues R1, R2 can be found by matching coefficients on both sides. Rewrite
the right-hand side as
(R1 + R2)z + (2R1 − R2) .
(z − 1)(z + 2)
Matching the coefficients of the numerator polynomials on both sides we conclude
that R1 + R2 = 2 and 2R1 − R2 = 3. We can solve these simultaneous equations to
determine that R1 = 5/3 and R2 = 1/3.
Alternatively, we can find residue R1 by multiplying both sides by (z − 1) and eval-
uating at z = 1. That is,
2z + 3
5
R1 =
=
.
z + 2
3
z=1
Similarly, we can find R2 by we multiplying both sides by z + 2 and evaluating at
z = −2, to get
2z + 3 |
= R2,
z − 1 z=−2
so R2 = 1/3. Thus the partial fraction expansion is
5/3
1/3
ˆ
X (z) =
+
.
z − 1
z + 2
RoC(x) is either
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13. LAPLACE AND Z TRANSFORMS
1. {z ∈ C | |z| < 1},
2. {z ∈ C | 1 < |z| < 2}, or
3. {z ∈ C | |z| > 2}.
Knowing which case holds, we can find the inverse Z transform of each term from
table 13.1. In the first case, x is the anti-causal signal
∀n, x(n) = −5u(−n) − 1(−2)n−1u(−n).
3
3
In the second case it is the two-sided signal
∀
5
n,
x(n) =
u(n − 1) − 1 (−2)n−1u(−n).
3
3
In the third case it is the causal signal
∀
5
1
n,
x(n) =
u(n − 1) + (−2)n−1u(n − 1).
3
3
If some pole of ˆ
X has multiplicity greater than one, it is slightly more difficult to carry out
the partial fraction expansion. The following example illustrates the method.
Example 13.15: Consider the expansion
2z + 3
R
R
R
ˆ
1
21
22
X (z) =
=
+
+
.
(z − 1)(z + 2)2
z − 1
z + 2
(z + 2)2
Again we can match coefficients and determine the residues. Alternatively, to ob-
tain R1 we multiply both sides by (z − 1) and evaluate the result at z = 1, to get
R1 = 5/9. To obtain R22 we multiply both sides by (z + 2)2 and evaluate the result
at z = −2, to get R22 = 1/3.
To obtain R21 we multiply both sides by (z + 2)2,
2z + 3
(z + 2)2R1
=
+ R21(z + 2) + R22,
z − 1
z − 1
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13.5. THE INVERSE TRANSFORMS
and then differentiate both sides with respect to z. We evaluate the result at z = −2,
to get
d 2z + 3
= R21.
dz z − 1 z=−2
Hence R21 = −5/9. So the partial fraction expansion is
2z + 3
5/9
1/3
=
− 5/9 +
.
(z − 1)(z + 2)2
z − 1
z + 2
(z + 2)2
Knowing the RoC, we can now obtain the inverse Z transform of ˆ
X . For instance,
in the case where RoC = {z ∈ C | |z| > 2}, the inverse Z transform is the causal
signal
∀
5
1
n,
x(n) =
u(n − 1) − 5 (−2)n−1u(n − 1) + (n − 1)(−2)n−2u(n − 2).
9
9
3
In example 13.15, we used the next to the last entry in table 13.1 to find the inverse transform of the term (1/3)/(z + 2)2. That entry in the table is based on a generalization
of the geometric series identity, given by (12.9). The first generalization is
∞
∞
1
∑ (n + 1)an = ( ∑ an)2 =
.
(13.11)
(1 − a)2
n=0
n=0
The series above converges for any complex number a with |a| < 1 (see Exercise 3 of
Chapter 12). The broader generalization, for any integer k ≥ 1, is
1 ∞
1
∑ (n + k)(n + k − 1) · · · (n + 1)an =
,
(13.12)
k!
(1 − a)k+1
n=0
for any complex number a with |a| < 1.
Consider then a Z transform ˆ
X that has a pole at p of multiplicity m and no zeros. Since
the pole p cannot belong to RoC, the RoC is either
{z ∈ C | |z| > |p|} or {z ∈ C | |z| < |p|}.
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In the first case we expand ˆ
X in a series involving only the terms z−n, n ≥ 0,
1
ˆ
X (z)
=
(z − p)m
z−m
=
(1 − pz−1)m
1
∞
=
z−m
(m + n − 1) · · · (n + 1)(pz−1)n, using (13.12)
(
∑
m − 1)! n=0
1
∞
=
(k − 1) · · · (k − m + 1)pk−mz−k, defining k = n + m,
(
∑
m − 1)! k=m
and the series converges for any z with |z| > |p|. We can match the coefficients of the
powers of z in the Z transform definition,
∞
ˆ
X (z) = ∑ x(k)z−k,
k=−∞
from which we can recognize that
∀
0,
k < m
k ∈ Z,
x(k)
=
1
(k − 1) · · · (k − m + 1)pk−m, k ≥ m
(m−1)!
1
=
(k − 1) · · · (k − m + 1)pk−mu(k − m).
(m − 1)!
(13.13)
In the second case, RoC