SME
Signals & Systems
Laplace and z-transforms
Use this for “Find x(t) when X(s) = (s + 3) ÷ ((s + 1)(s + 2)).”
Diagram
Poles (×) at s = −1, −2; zeros (○) at s = −3.
Each pole gives one term: x(t) = 2e−t − e−2t.
This is an example. Type your own values to start a new problem.
X(s)
=
s
+
(s
+
)(s
+
)
=
A
s
+
+
B
s
+
t
Time
example
x(t)
Signal at t
calculated
Formulas
A
=
c
−
a
b
−
a
2
=
3 − 1
2 − 1
B
=
c
−
b
a
−
b
−1
=
3 − 2
1 − 2
x(
t
)
=
A
×
e
−
a
×
t
+
B
×
e
−
b
×
t
0.6004
=
2
×
e
−1 × 1
+
(−1)
×
e
−2 × 1
Assumptions
•
X(s) = (s + c) ÷ ((s + a)(s + b)) with two different poles, −a and −b, left of the jω axis.
•
Cover-up rule: a residue is the rest of X(s) at its pole, with that pole’s factor covered.
•
From the transform table, 1 ÷ (s + a) ↔ e−atu(t), so each pole gives one decaying term.
•
With s + c on top the residues add to 1: a quick check.
Step-by-step
WE KNOW
c = 3 s⁻¹, a = 1 s⁻¹, b = 2 s⁻¹, t = 1 s
FIND
residue at −a (A), residue at −b (B), signal at t (x(t))
Step 1 · Find residue at −a (A)
Use A = (c − a) ÷ (b − a)
Cover
up
(s
+
a
)
and
put
s
=
−a
into
what
is
left:
−
a
+
c
−
a
+
b
.
A
=
3 − 1
2 − 1
A
=
2
Step 2 · Find residue at −b (B)
Use B = (c − b) ÷ (a − b)
Cover
up
(s
+
b
)
and
put
s
=
−
b
into
what
is
left:
−
b
+
c
−
b
+
a
.
B
=
3 − 2
1 − 2
B
=
1
−1
B
=
−1
Step 3 · Find signal at t (x(t))
Use x(t) = A × e−a × t + B × e−b × t
Each
fraction
turns
into
its
exponential;
add
the
two
terms
at
t
.
x(
t
)
=
2
×
e
−1 × 1
+
(−1)
×
e
−2 × 1
x(
t
)
=
2
×
0.3679
+
(−1)
×
0.1353
x(
t
)
=
0.7358
+
(−0.1353)
x(
t
)
=
0.6004
Check
Put the numbers back in. Both sides should match.
2
=
3 − 1
2 − 1
✓
−1
=
3 − 2
1 − 2
✓
0.6004
=
2
×
e
−1 × 1
+
(−1)
×
e
−2 × 1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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