Chapter 2: Sinusoidal Signals

Sinusoids play a fundamental role in signal and sys-

tem theory.

x(t) = A cos(ωot + φ)

three independent parameters:

A: (real) amplitude
ωo: radian frequency
φ: phase

Review Tables 2.1 & 2.2 on the properties of sinu-

soids.
It is conventional to use the cosine form rather than
sine.

1
 Many physical systems generate signals that can be
modeled as sinusoids. Two examples:

• tuning fork
• clay whistle

TUNING FORK A-440 Waveform

T≈8.15−5.85
=2.3 ms

Time (sec)

f =1/T
=1000 /2. 3
¿ 435 Hz

2
 Let us look at the sinusoidal signal more closely:

x(t) = A cos(ωot + φ) = A cos(2πfot + φ)

three independent parameters:

A: (real) amplitude
This real-valued parameter determines the range of
oscillations.
cos θ oscillates between −1 and 1
A cos θ oscillates between −A and A

ωo: radian frequency or fo: cyclic frequency

These two parameters are dependent on each other.
ωo = 2πfo or fo = ωo/2π
Both are a measure of how rapidly the sinusoid os-
cillates.
The higher the frequency, the more rapidly the sig-
nal waveform changes.
units of ωo: radians/unit time
units of fo: 1/(unit time)
If time units are in seconds, units of fo will be Hertz
(Hz).
The argument of the sinusoid is in radians unless
otherwise stated.

3
 periodic vs. aperiodic (non-periodic) signals
Periodic signals are characterized by a repeating
pattern that repeats itself infinitely many times.
expressing this mathematically:
x(t + nTo) = x(t) ∀t ∈ R ∀n ∈ Z and some To
To: period of the signal
The duration of one cycle of the sinusoid.
The shortest time interval over which the pattern
repeats itself is called the fundamental period. Its
reciprocal is the fundamental frequency fo.
The frequency of the sinusoid determines its period.
To = 1/fo or fo = 1/To
fo indicates the number of cycles per unit time.
To prove that sinusoids are periodic:
x(t) = A cos(ωot + φ)
x(t + nTo) = A cos[ωo(t + nTo) + φ]
= A cos(ωot + ωonTo + φ)
= A cos(ωot + n2π |{z}
foTo +φ)
1
= A cos(ωot + 2πn + φ)
= A cos(ωot + φ)
= x(t) ∀t ∈ R ∀n ∈ Z
4
 There is an inverse relationship between time and
frequency.
What happens if fo = 0 or if ωo = 0?
fo = 0 is a perfectly acceptable value.

x(t) = A cos(ωot + φ)
= A cos(2πfot + φ)
= A cos(2π0t + φ)
= A cos(0 + φ)
= A cos(φ) (a constant)

The sinusoid of zero frequency is no longer time de-

pendent. It becomes a constant. The value of the
constant is determined by the amplitude and phase.
Such constant signals are called d.c. signals. We can
include constants in the same class as sinusoids. Con-
stants can be considered as degenerate sinusoids.

d.c.: direct current

a.c.: alternating current

Both terms are used in a more general sense than

their literal meaning.

5
 Time Shift/Phase Shift
The phase shift parameter, together with fo, de-
termines the time locations of the maxima and the
minima of the sinusoid.
time-shifting a signal:
x(t): original signal
x(t − t1): time-shifted signal, shifted by t1
if t1 > 0, shift is to the right −→ delay
if t1 < 0, shift is to the left −→ advance
if t1 = 0, no shift
for the cosine waveform:

xo(t) = A cos(ωot)
(a cosine waveform with zero phase shift)
xo(t − t1) = A cos[ωo(t − t1)]
= A cos(ωot −ω
| {zot}1) φ : a constant
φ
= A cos(ωot + φ)

Based on this, we get:

 
t1
φ = −ωot1 = −2πfot1 = −2π
To
 
t1
φ = −2π
To
6
 This is useful for converting a given time shift to a
phase shift.
To determine the phase shift for a given sinusoidal
plot w.r.t. time, choose the peak nearest to the origin
(t = 0). The nearest peak can be to the right or to
the left of the origin and will give you the phase shift
φ in the principal interval:

−π < φ ≤ π or (−π, π]

To convert a phase shift to a time shift, we use the

same formula the other way around to get:
φ φ
t1 = − =−
ωo 2πfo

By the end of this part, two capabilities are ex-

pected:
• Given the closed-form expression for a sinusoid,
make an accurate plot.
• Given an accurate plot for a sinusoid, write the
closed-form expression.

We give one example of each in what follows.

7
PROBLEM 2.1:

McClellan, Schafer and Yoder, Signal Processing First, ISBN 0-13-065562-7. This page should not be copied or electronically transmitted unless prior written
Pearson Prentice Hall, Inc. Upper Saddle River, NJ 07458.
c 2003 permission has been obtained from the authors. November 28, 2007

8
 ex: The sinusoid x(t) = A cos(ωot + φ) is plot-
ted below. Find A, ωo, φ using the plot and write a
closed-form expression for the sinusoid.

By inspection, we observe that A = 2.

Easiest is to measure the period peak-to-peak:
To = 6 sec → fo = 16 Hz → ωo = 2πfo = 2π 6 =
π
3 rad/s
Identifying the nearest peak to the origin (t = 0):
t1 = 1 sec (time shift)
Is this a delay or an advance?
Using theformula
 that we have derived:
t1
φ = −2π To = −2π 6 = − π3 rad.
1

Therefore, we can express the given sinusoid in two

equivalent forms:
hπ πi hπ i
x(t) = 2 cos t − or x(t) = 2 cos (t − 1)
3 3 3

9
 Review of Complex Numbers and Operations
with Complex Numbers:

• study Appendix A

A complex number z is an ordered pair of real num-

bers.
Cartesian form:
z = (x, y) or z = x + jy
where
x = Re{z} is the real part of z
y = Im{z} is the imaginary part of z
√
and j = −1.
In EE, we use j to avoid confusion with i, often
used for current in circuit theory where there is a lot
of analysis with complex numbers. Mathematicians,
physicists, and other branches of science usually use
√
i = −1.
polar form:
z = r∠θ
r: the vector length or the magnitude of z, |z|
θ: angle made with the real axis of the complex plane
(argument of z), arg z
10
 The complex plane is a 2-D plane where the hori-
zontal axis is for the real part x of the complex num-
ber and the vertical axis is for its imaginary part y.

Im{z} z=∞
y z = x + jy = r∠θ = rejθ
unit
circle
1
r r sin θ

θ
r cos θ x Re{z}

conversion from polar to Cartesian:

x = r cos θ
y = r sin θ

conversion from Cartesian to polar:

p
r = x2 + y 2
θ = atan2(y, x)

(atan2 is the inverse tangent or arctangent function

with two arguments instead of one which removes the
ambiguity about the quadrant of the angle)

11
 Euler’s Formula:

ejθ = cos θ + j sin θ

ejθ is called a complex exponential and

p
jθ
|e | = cos2 θ + sin2 θ = 1

(complex numbers on the unit circle)

More generally,

z = rejθ = r cos θ + jr sin θ

where r can be greater than or less than 1.

magnitude of z:
p √
jθ 2
|z| = |re | = r cos θ + r sin θ = r2 = r
2 2 2

In other words, the complex number z = rejθ can

be located anywhere on the complex plane.
(inside, outside, or on the unit circle)

12
Proof of Euler’s Formula:

PROBLEM 2.4:

McClellan, Schafer and Yoder, Signal Processing First, ISBN 0-13-065562-7. This page should not be copied or electronically transmitted unless prior written
Pearson Prentice Hall, Inc. Upper Saddle River, NJ 07458.
c 2003 permission has been obtained from the authors. November 28, 2007

13
 Complex Operations:
Cartesian form is more convenient for performing
addition and subtraction.

addition:
Suppose we have N arbitrary complex numbers zi in
Cartesian form where i = 1, . . . , N .

z1 = x1 + jy1
z2 = x2 + jy2
..
zN = xN + jyN

N
X
zsum = z1 + z2 + . . . + zN = zi
i=1
= (x1 + jy1) + (x2 + jy2) + . . . + (xN + jyN )
= (x1 + x2 + . . . + xN ) + j(y1 + y2 + . . . + yN )
XN N
X
= xi + j yi
i=1 i=1

• real part of the sum is the sum of the real parts

• imaginary part of the sum is the sum of the imag-
inary parts

14
 subtraction:
First, demonstrate subtraction with two complex
numbers z1 and z2 in Cartesian form where
z1 = x1 + jy1
z2 = x2 + jy2
z1 − z2 = (x1 − x2) + j(y1 − y2)

More generally, we can perform signed addition for

N complex numbers expressed in Cartesian form:

z1 = x1 + jy1
z2 = x2 + jy2
..
zN = xN + jyN

zsum = z1 + z2 − z3 − z4 + . . . + zN
= (x1 + jy1) + (x2 + jy2) − (x3 + jy3) − (x4 + jy4)
+ . . . + (xN + jyN )
= (x1 + x2 − x3 − x4 + . . . + xN )
+j(y1 + y2 − y3 − y4 + . . . + yN )

• Follow the same order of signs in the real and

imaginary parts of the signed addition.

15
 The polar form is more convenient for performing
multiplication and division.
Recall law of exponents for ordinary exponentials
from calculus: exey = e(x+y). This law is also appli-
cable to complex exponentials:
multiplication:
z1 = r1∠θ1 = r1ejθ1
z2 = r2∠θ2 = r2ejθ2
z3 = r3∠θ3 = r3ejθ3
..
zN = rN ∠θN = rN ejθN

zproduct = z1.z2.z3 . . . zN
= r1ejθ1 .r2ejθ2 .r3ejθ3 . . . rN ejθN
= r1.r2.r3 . . . rN ej(θ1+θ2+θ3+...+θN )

Demonstrate division with two complex numbers z1

and z2 where

z1 = r1∠θ1 = r1ejθ1
z2 = r2∠θ2 = r2ejθ2

z1 r1ejθ1 jθ1 1 −jθ2 r1 j(θ1−θ2)

zquotient = = = r1e . e = e
z2 r2ejθ2 r2 r2

16
 complex conjugation:
If a complex number z is expressed in Cartesian
form: z = x + jy, its complex conjugate is z ∗ =
x − jy.
z + z ∗ = 2Re{z} = 2x z − z ∗ = 2jIm{z} = 2jy
If a complex number z is expressed in polar form:
z = rejθ , its complex conjugate is z ∗ = re−jθ .
z.z ∗ = |z|2 = (x + jy)(x − jy) = x2 − (j 2) y 2
|{z}
−1
2 2 2
= x + y = r (magnitude squared)
e−jθ} = r2
Or, z.z ∗ = |z|2 = rejθ .re−jθ = r.r. e|jθ{z
1
A complex number and its complex conjugate are
symmetric about the real axis of the complex plane.
complex conjugate of a sum:
zsum = z1 + z2 + . . . + zN
∗
zsum = (z1 + z2 + . . . + zN )∗ = z1∗ + z2∗ + . . . + zN∗
Complex conjugate of a sum is the sum of the com-
plex conjugates.
complex conjugate of a product:
zpro = z1.z2 . . . zN
∗
zpro = (z1.z2 . . . zN )∗ = z1∗.z2∗ . . . .zN∗
Complex conjugate of a product is the product of
the complex conjugates.

17
 complex conjugate of a ratio:
zquotient = zz12
 ∗
∗ z1 z1∗
zquotient = z2 = z ∗
2

Complex conjugate of a ratio is the ratio of the com-

plex conjugates.

taking the magnitude of a complex ratio:

first way:
z1 x1 +jy1
z= =
z2
 x2 +jy2
 √2 2
|x1 +jy1 | x +y
|z| = r =  xx12+jy 1
+jy2  = = √ 12 12

|x2 +jy2 | x2 +y2

Magnitude of a ratio is the ratio of magnitudes.

second way:
x1 +jy1 x1 −jy1
z= x2 +jy2 z∗ = x2 −jy2
1 x1 −jy1 x21 −j 2 y12 x21 +y12
∗
z.z = |z| = 2
r2 = xx12+jy
+jy2 . x2 −jy2 = x22 −j 2 y22
= x22 +y22

This gives magnitude squared.

To get thermagnitude of z, take the square root:
√2 2
x21 +y12 x1 +y1
|z| = r = 2 2 = √ 2 2
x2 +y2 x2 +y2

The second way usually takes longer.

18
 Make sure you can also do the following polar-to-
Cartesian and Cartesian-to-polar conversions very fast
because we need them often:
ej0 = ej2π = e−j2π = 1 ej2πk = 1 (k integer)
ejπ = e−jπ = −1
π 3π
ej 2 = e−j 2 = j
π 3π
e−j 2 = ej 2 = −j
π
ej 4 = 1+j
√
2
−j π4
e = 1−j
√
2
etc.
Simple form of Euler’s formula can be used to prove
all of the above.

√
Since j = −1, we also have:
1
j 2 = −1 j 3 = −j j4 = 1 j = −j − 1j = j

Im{z}
z=∞
unit
circle
1

−1 1
Re{z}

19
 Complex Exponential Signals:
The complex number z = rejθ is a constant.
(no time dependence)
Now, we introduce time dependence to z in its angle
part: z = rejθ(t) where θ(t) = ωot + φ. We still keep
its magnitude r constant.
This way, we obtain a complex exponential signal:

z(t) = rej(ωot+φ)

Renaming r as A and using Euler’s formula:

z(t) = Aej(ωot+φ) = A cos(ωot + φ) + jA sin(ωot + φ)

• real part of z(t) is an ordinary cosine function

• imaginary part of z(t) is an ordinary sine function
They differ only by a phase shift of π2 .

|z(t)| = A
arg z(t) = θ(t) = ωot + φ

Thus, we have

x(t) = Re{z(t)} = Re{Aej(ωot+φ)} = A cos(ωot + φ)

y(t) = Im{z(t)} = Im{Aej(ωot+φ)} = A sin(ωot + φ)

20
 Real & Imaginary Part Plots

PHASE DIFFERENCE = π /2

It may seem that we have complicated matters by

introducing the imaginary part to obtain the complex
exponential signal and then by taking its real part to
get a cosine. However, many calculations are simpli-
fied by using the properties of (complex) exponentials
such as the law of exponents.
Now we introduce a new, complex quantity, called
phasor or complex amplitude X.

z(t) = Aej(ωot+φ)
= Aejωotejφ (using the law of exponents in reverse)
jφ jωo t
= Ae
|{z} e (changing the order of the exponentials)
∆
=X
jωo t
= Xe

21
 The phasor X is a complex constant which contains
the real amplitude A and the phase angle φ of the
complex exponential signal z(t). It is independent of
time. We use it to represent the complex exponential
signal z(t).
In electrical engineering, complex exponentials are
used to greatly simplify the design and analysis of
electrical circuits.
at t = 0: 
jωo 0
z(t) = Xejωot = X e|{z} =X
 
t=0 t=0 1
The phase shift φ defines where the phasor is point-
ing at t = 0. Thus, X could be considered as the
starting point of the complex exponential signal z(t)
on the complex plane at t = 0.
• interpret z(t) as a rotating vector
on the complex plane
• if ωo > 0, as t ↑, θ(t) = ωot + φ ↑.
rotation is counter-clockwise (CCW).
• if ωo < 0, as t ↑, θ(t) = ωot + φ ↓.
rotation is clockwise (CW).

22
 demo: rotating phasors (single phasor)
Note that the phasor itself is constant and does not
rotate on the complex plane. It is the complex ex-
ponential signal z(t) that rotates because it changes
with time.

23
 Inverse Euler Formulas:
derivation:
Using Euler’s formula, we have:
ejθ = cos θ + j sin θ (1)
e−jθ = cos(−θ) + j sin(−θ) = cos θ − j sin θ (2)
Adding the two expressions together, the imaginary
parts cancel out and we get:
ejθ + e−jθ = 2 cos θ
Leaving cos θ on the left-hand side, we have:
ejθ + e−jθ
cos θ =
2
Similarly, we can subtract (2) from (1). This time,
the real parts cancel out and we get:
ejθ − e−jθ = 2j sin θ
Leaving sin θ on the left-hand side, we have:
ejθ − e−jθ
sin θ =
2j
Now, let us use the first one of inverse Euler formu-
las by choosing θ as θ(t) = ωot + φ:
ej(ωot+φ) + e−j(ωot+φ)
cos(ωot + φ) =
2
24
Multiplying both sides by the (real) amplitude A:
+ e−j(ωot+φ)
 j(ωot+φ) 
e
A cos(ωot + φ) = A
2
Expanding the right-hand side using the law of expo-
nents in reverse:
A jφ jωot A −jφ −jωot
A cos(ωot + φ) = e e + e e
2 2
Rewriting using the definition of the phasor:
∆
X = Aejφ
X jωot X ∗ −jωot
A cos(ωot + φ) = e + e
2 2
∗
z(t) z (t)
= +
2 2
= Re{z(t)}

• The real cosine signal with frequency ωo is com-

posed of two complex exponentials:
−one with positive frequency (ωo)
with complex amplitude X2 = A2 ejφ.
−one with negative frequency (−ωo)
∗
with complex amplitude X2 = A2 e−jφ.
• The real cosine signal can be represented as the
sum of two complex conjugate terms rotating in opposite
directions on the complex plane as time evolves.

25
 demo: rotating phasors (two phasors)

Following a very similar analysis for the sine func-

tion, it is easy to show that:
X −j π jωot X ∗ j π −jωot
A sin(ωot + φ) = e 2e + e 2e
2 2
= Im{z(t)}
26
 Phasor Addition:
• phasor representation is useful when we need to
add two or more sinusoidal signals.
• assume that the sinusoids to be added have the
same frequency but different amplitudes and phases.
N
X
Ak cos(ωot+φk ) = A cos(ωot+φ) where N ∈ Z
k=1

ADD SINUSOIDS

 Sum Sinusoid has SAME Frequency

Any sinusoid can be expressed in the form:

A cos(ωot+φ) = Re{z(t)} = Re{Aej(ωot+φ)} = Re{Aejφejωot}

27
 applications in:
• telecommunications
• electromagnetics (e.g., antennas)
• circuit theory
• music
essence of the phasor addition rule:
N
X
Ak ejφk = Aejφ
k=1

proof of the phasor addition rule:

N
X N
X
Ak cos(ωot + φk ) = Re{Ak ej(ωot+φk )}
k=1 k=1
( N )
X
= Re Ak ejφk ejωot
k=1
(sum of the real parts is = to the real part of the sum)
( N ! )
X
= Re Ak ejφk ejωot
k=1
factored out ejωot since all N sinusoids have the same freq.
jφ

jω t
= Re Ae e o
n o
j(ωo t+φ)
= Re Ae (law of exp.)
= A cos(ωot + φ)

28
PROBLEM 2.15:

McClellan, Schafer and Yoder, Signal Processing First, ISBN 0-13-065562-7. This page should not be copied or electronically transmitted unless prior written
Pearson Prentice Hall, Inc. Upper Saddle River, NJ 07458.
c 2003 permission has been obtained from the authors. November 28, 2007
29
 exercise to do at home:
Find the resultant signal (in cosine form) obtained
by summing x(t) and y(t) where x(t) = 6 cos(20t+5)
and y(t) = 3 sin(20t). Note that the 5 in the argu-
ment of the sinusoid is in radians here. Verify your
numerical result by making a phasor diagram on
the complex plane. Identify also the period and the
cyclic frequency of the resultant signal.

30