Classification of signals and systems Lecture notes
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Class Note for Signals and Systems
Stanley Chan
University of California, San DiegoChapter 1
Fundamentals of Signals
1.1 What is a Signal?
A signal is a quantitative description of a physical phenomenon, event or process.
Some common examples include:
1. Electrical current or voltage in a circuit.
2. Daily closing value of a share of stock last week.
3. Audio signal: continuous-time in its original form, or discrete-time when stored
on a CD.
More precisely, a signal is a function, usually of one variable in time. However, in
general, signals can be functions of more than one variable, e.g., image signals.
In this class we are interested in two types of signals:
1. Continuous-time signal x(t), where t is a real-valued variable denoting time,
i.e., t2R. We use parenthesis () to denote a continuous-time signal.
2. Discrete-time signal xn, where n is an integer-valued variable denoting the
discrete samples of time, i.e., n2 Z. We use square brackets to denote a
discrete-time signal. Under the de nition of a discrete-time signal, x1:5 is not
de ned, for example.
56 CHAPTER 1. FUNDAMENTALS OF SIGNALS
1.2 Review on Complex Numbers
We are interested in the general complex signals:
x(t)2C and xn2C;
where the set of complex numbers is de ned as
p
C =fzjz =x +jy; x;y2R; j = 1:g
A complex number z can be represented in Cartesian form as
z =x +jy;
or in polar form as
j
z =re :
Theorem 1. Euler's Formula
j
e = cos +j sin: (1.1)
Using Euler's formula, the relation between x, y, r, and is given by
( (
p
2 2
x =r cos r = x +y ;
and
y
1
y =r sin = tan :
x
Figure 1.1: A complex numberz can be expressed in its Cartesian formz =x +jy, or in its polar
j
form z =re .1.3. BASIC OPERATIONS OF SIGNALS 7
A complex number can be drawn on the complex plane as shown in Fig. 1.1. The
y-axis of the complex plane is known as the imaginary axis, and the x-axis of the
complex plane is known as the real axis. A complex number is uniquely de ned by
j
z =x +jy in the Cartesian form, or z =re in the polar form.
Example. Convert the following complex numbers from Cartesian form to polar
form: (a) 1 + 2j; (b) 1 j.
For (a), we apply Euler's formula and nd that
p p
2
1
2 2
r = 1 + 2 = 5; and = tan 63:64 :
1
Therefore,
p
j63:64
1 + 2j = 5e :
For (b), we apply Euler's formula again and nd that
p p
1
1
2 2
r = 1 + ( 1) = 2; and = tan = 45 :
1
Therefore,
p
j=4
1 j = 2e :
Recall that: in radian = 180 in degree.
j
Example. Calculate the value of j .
j j=2 j =2
j = (e ) =e 0:2078.
1.3 Basic Operations of Signals
1.3.1 Time Shift
For any t 2R and n 2Z, time shift is an operation de ned as
0 0
x(t) x(t t )
0
(1.2)
xn xn n :
0
If t 0, the time shift is known as \delay". If t 0, the time shift is known as
0 0
\advance".
Example. In Fig. 1.2, the left image shows a continuous-time signal x(t). A time-
shifted version x(t 2) is shown in the right image.8 CHAPTER 1. FUNDAMENTALS OF SIGNALS
Figure 1.2: An example of time shift.
1.3.2 Time Reversal
Time reversal is de ned as
x(t) x( t)
(1.3)
xn x n;
which can be interpreted as the \
ip over the y-axis".
Example.
Figure 1.3: An example of time reversal.
1.3.3 Time Scaling
Time scaling is the operation where the time variablet is multiplied by a constanta:
x(t) x(at); a 0 (1.4)
Ifa 1, the time scale of the resultant signal is \decimated" (speed up). If 0a 1,
the time scale of the resultant signal is \expanded" (slowed down).
1.3.4 Combination of Operations
In general, linear operation (in time) on a signalx(t) can be expressed asy(t) =x(at
b); a;b2R: There are two methods to describe the output signal y(t) =x(at b).1.3. BASIC OPERATIONS OF SIGNALS 9
Figure 1.4: An example of time scaling.
Method A: \Shift, then Scale" (Recommended)
1. De ne v(t) =x(t b),
2. De ne y(t) =v(at) =x(at b).
Method B: \Scale, then Shift"
1. De ne v(t) =x(at),
b
2. De ne y(t) =v(t ) =x(at b).
a
Example.
For the signal x(t) shown in Fig. 1.5, sketch x(3t 5).
Figure 1.5: Example 1. x(3t 5).
Example.
For the signal x(t) shown in Fig. 1.6, sketch x(1 t).10 CHAPTER 1. FUNDAMENTALS OF SIGNALS
Figure 1.6: Example 2. x( t + 1).
Figure 1.7: Decimation and expansion.
1.3.5 Decimation and Expansion
Decimation and expansion are standard discrete-time signal processing operations.
Decimation.
Decimation is de ned as
y n =xMn; (1.5)
D
for some integers M. M is called the decimation factor.
Expansion.
Expansion is de ned as
(
n
x ; n = integer multiple of L
L
y n = (1.6)
E
0; otherwise:
L is called the expansion factor.1.4. PERIODICITY 11
Figure 1.8: Examples of decimation and expansion for M = 2 and L = 2.
1.4 Periodicity
1.4.1 De nitions
De nition 1. A continuous time signal x(t) is periodic if there is a constant T 0
such that
x(t) =x(t +T ); (1.7)
for all t2R.
De nition 2. A discrete time signal xn is periodic if there is an integer constant
N 0 such that
xn =xn +N; (1.8)
for all n2Z.
Signals do not satisfy the periodicity conditions are called aperiodic signals.
Example. Consider the signal x(t) = sin( t), 0. It can be shown that
0 0
2
+
x(t) =x(t +T ), where T =k for any k2Z :
0
2
x(t +T ) = sin t +k
0
0
= sin ( t + 2k)
0
= sin( t) =x(t):
0
Therefore, x(t) is a periodic signal.12 CHAPTER 1. FUNDAMENTALS OF SIGNALS
De nition 3. T is called the fundamental period of x(t) if it is the smallest value
0
2
of T 0 satisfying the periodicity condition. The number = is called the
0
T
0
fundamental frequency of x(t).
De nition 4. N is called the fundamental period of xn if it is the smallest value
0
2
of N 0 where N2Z satisfying the periodicity condition. The number
= is
0
N
0
called the fundamental frequency of xn.
j3t=5
Example. Determine the fundamental period of the following signals: (a) e ;
j3n=5
(b) e .
j3t=5
For (a), we letx(t) =e . Ifx(t) is a periodic signal, then there existsT 0 such
that x(t) =x(t +T ). Therefore,
x(t) =x(t +T )
3 3
j t j (t+T )
5 5
) e =e
3
j T
5
) 1 =e
3
j2k j T +
5
) e =e ; for somek2Z :
10
) T = : (k = 1)
3
j3n=5
For (b), we let xn =e . Ifxn is a periodic signal, then there exists an integer
N 0 such that xn =xn +N. So,
xn =xn +N
3 3
j n j (n+N)
5 5
) e =e
3
j2k j N +
5
) e =e ; for somek2Z
10k
) N =
3
) N = 10: (k = 3):
1.4.2 A More Dicult Example
Consider the following two signals
2
t
x(t) = cos ;
8
2
n
xn = cos :
8
We will show thatx(t) is aperiodic whereasxn is periodic with fundamental period
N = 8.
01.4. PERIODICITY 13
2 2
t n
(a) cos (b) cos
8 8
2 2
t n
Figure 1.9: Di erence betweenx(t) = cos andxn = cos . Note thatx(t) is aperiodic,
8 8
whereas xn is periodic.
Fig. 1.9 plots the signals
2
t
x(t) = cos
8
for 8t 8 and
2
n
xn = cos
8
for n = 8; 7;:::; 8. It is clear that x(t) is aperiodic, since the values of t 0 for
whichx(t) = 0 form a sequence which the di erence between consecutive elements is
monotonically decreasing.
On the other hand, xn is periodic, with fundamental period N = 8. To see this,
0
consider the periodicity condition xn =xn +N, which becomes:
2 2
cos (n +N) =8 = cos n =8 ;
for all n2Z. This means
2 2
(n +N) n
= + 2k;
8 8
for somek2Z, wherek may depend on a particular time instantn. We can simplify14 CHAPTER 1. FUNDAMENTALS OF SIGNALS
this condition by dividing both sides of the equation by =8 to yield
8
2 2
(n +N) =n + (2k);
or
2 2 2
n + 2nN +N =n + 16k;
implying
2
2nN +N = 16k;
2
for some k2Z. Next, we want to nd an N such that 2nN +N is divisible by 16
for all n2Z. Now we claim: N = 8 satis es this condition, and no smaller N 0
does.
Setting N = 8, we get
2
2nN +N = 16n + 64;
which, for anyn2Z, is clearly divisible by 16. SoN = 8 is a period ofxn. You can
2
check directly that, for any 1N 8, there is a valuen2Z such that 2nN +N is
not divisible by 16. For example if we consider N = 4, we get
2
2nN +N = 8n + 16;
which, for n = 1, is not divisible by 16. So N = 4 is not a period of xn.
1.4.3 Periodicity and Scaling
1. Suppose x(t) is periodic, and let a 0. Is y(t) =x(at) periodic?
Yes, and if T is the fundamental period of x(t), then T =a is the fundamental
0 0
period of y(t).
+
2. Suppose xn is periodic, and let m2Z . Is yn =xmn periodic?
Yes, and if N is the fundamental period of xn, then the fundamental period
0
N of yn is the smallest positive integer such that mN is divisible by N , i.e.
0
mN 0 ( mod N ):
0
Example 1: N = 8, m = 2, then N = 4.
0
Example 2: N = 6, m = 4, then N = 3.
01.5. EVEN AND ODD SIGNALS 15
1.5 Even and Odd Signals
1.5.1 De nitions
De nition 5. A continuous-time signal x(t) is even if
x( t) =x(t) (1.9)
and it is odd if
x( t) = x(t): (1.10)
De nition 6. A discrete-time signal xn is even if
x n =xn (1.11)
and odd if
x n = xn: (1.12)
Remark: The all-zero signal is both even and odd. Any other signal cannot be both
even and odd, but may be neither. The following simple example illustrate these
properties.
2
Example 1: x(t) =t 40 is even.
3
Example 2: x(t) = 0:1t is odd.
0:4t
Example 3: x(t) =e is neither even nor odd.
2 3 0:4t
(a) x(t) =t 40 (b) x(t) = 0:1t (c) x(t) =e
Figure 1.10: Illustrations of odd and even functions. (a) Even; (b) Odd; (c) Neither.16 CHAPTER 1. FUNDAMENTALS OF SIGNALS
1.5.2 Decomposition Theorem
Theorem 2. Every continuous-time signal x(t) can be expressed as:
x(t) =y(t) +z(t);
where y(t) is even, and z(t) is odd.
Proof. De ne
x(t) +x( t)
y(t) =
2
and
x(t) x( t)
z(t) = :
2
Clearlyy( t) =y(t) andz( t) = z(t). We can also check thatx(t) =y(t)+z(t).
Terminology: The signal y(t) is called the even part of x(t), denoted byEvfx(t)g.
The signal z(t) is called the odd part of x(t), denoted byOddfx(t)g.
t
Example: Let us consider the signal x(t) =e .
t t
e +e
Evfx(t)g = = cosh(t):
2
t t
e e
Oddfx(t)g = = sinh(t):
2
Similarly, we can de ne even and odd parts of a discrete-time signal xn:
xn +x n
Evfxng =
2
xn x n
Oddfxng = :
2
It is easy to check that
xn =Evfxng +Oddfxng
Theorem 3. The decomposition is unique, i.e., if
xn =yn +zn;
thenyn is even andzn is odd if and only ifyn =Evfxng andzn =Oddfxng.1.6. IMPULSE AND STEP FUNCTIONS 17
Proof. If yn is even and zn is odd, then
x n =y n +z n =yn zn:
Therefore,
xn +x n = (yn +zn) + (yn zn) = 2yn;
xn+x n xn x n
implying yn = =Evfxng. Similarly zn = =Oddfxng.
2 2
The converse is trivial by de nition, asEvfxng must be even andOddfxng must
be odd.
1.6 Impulse and Step Functions
1.6.1 Discrete-time Impulse and Step Functions
De nition 7. The discrete-time unit impulse signal n is de ned as
(
1; n = 0;
n = (1.13)
0; n =6 0:
De nition 8. The discrete-time unit step signal n is de ned as
(
1; n 0;
un = (1.14)
0; n 0:
It can be shown that
n =un un 1
1
X
un = n k
k=0
1
X
un = ukn k:
k= 118 CHAPTER 1. FUNDAMENTALS OF SIGNALS
(a) n (b) un
Figure 1.11: De nitions of impulse function and a step function.
1.6.2 Property of n
Sampling Property
By the de nition of n, n n = 1 if n =n , and 0 otherwise. Therefore,
0 0
(
xn; n =n
0
xnn n =
0
0; n =6 n
0
=xn n n : (1.15)
0 0
As a special case whenn = 0, we havexnn =x0n. Pictorially, when a signal
0
xn is multiplied with n, the output is a unit impulse with amplitude x0.
Figure 1.12: Illustration of xnn =x0n.1.6. IMPULSE AND STEP FUNCTIONS 19
Shifting Property
1
P
Since xnn =x0n and n = 1, we have
n= 1
1 1 1
X X X
xnn = x0n =x0 n =x0;
n= 1 n= 1 n= 1
and similarly
1 1
X X
xnn n = xn n n =xn : (1.16)
0 0 0 0
n= 1 n= 1
In general, the following result holds:
(
b
X
xn ; if n 2 a; b
0 0
xnn n = (1.17)
0
0; if n 62 a; b
0
n=a
Representation Property
Using the sampling property, it holds that
xkn k =xnn k:
Summing the both sides over the index k yields
1 1 1
X X X
xkn k = xnn k =xn n k =xn:
k= 1 k= 1 k= 1
This result shows that every discrete-time signal xn can be represented as a linear
combination of shifted unit impulses
1
X
xn = xkn k: (1.18)
k= 1
For example, the unit step function can be expressed as
1
X
un = ukn k:
k= 1
Why do we use these complicated representation ofxn? Because, when we consider
linear time-invariant systems (Chapter 2), it will allow us to determine the system
response to any signal xn from the impulse response.20 CHAPTER 1. FUNDAMENTALS OF SIGNALS
Figure 1.13: Representing of a signal xn using a train of impulses n k.
1.6.3 Continuous-time Impulse and Step Functions
De nition 9. The dirac delta function is de ned as
(
0; if t6= 0
(t) = ;
1; if t = 0
where
Z
1
(t)dt = 1:
1
De nition 10. The unit step function is de ned as
(
0; t 0
u(t) =
1; t 0:
1.6.4 Property of (t)
The properties of (t) are analogous to the discrete-time case.1.6. IMPULSE AND STEP FUNCTIONS 21
Sampling Property
x(t)(t) =x(0)(t): (1.19)
To see this, note that x(t)(t) = x(0) when t = 0 and x(t)(t) = 0 when t6= 0.
Similarly, we have
x(t)(t t ) =x(t )(t t ); (1.20)
0 0 0
for any t 2R.
0
Shifting Property
The shifting property follows from the sampling property. Integrating x(t)(t) yields
Z Z Z
1 1 1
x(t)(t)dt = x(0)(t)dt =x(0) (t)dt =x(0): (1.21)
1 1 1
Similarly, one can show that
Z
1
x(t)(t t )dt =x(t ) (1.22)
0 0
1
Representation Property
The representation property is also analogous to the discrete-time case:
Z
1
x(t) = x( )(t )d ; (1.23)
1
where the special case of u(t) is given by
Z
1
u(t) = u( )(t )d :
1
d
As an example of the properties, let us consider u(t).
dt
Z
1
u(t) = u( )(t )d ; (representation property)
1
Z
1
= (t )d ; because u( ) = 0 if 0
0
Z
t
= ()d; let =t :
1
Then by fundamental theorem of calculus, we have
Z
t
d d
u(t) = ()d =(t): (1.24)
dt dt
122 CHAPTER 1. FUNDAMENTALS OF SIGNALS
1.7 Continuous-time Complex Exponential Func-
tions
De nition 11. A complex exponential function is de ned as
at
x(t) =Ce ; where C;a2C (1.25)
1.7.1 Real-valued Exponential
We rst consider the case of real-valued exponential functions, i.e.,C2R anda2R.
1 1
1 1
t t
2 2
(a) x(t) = e (b) x(t) = e
2 2
Figure 1.14: Real exponential functions.
When a = 0, then x(t) =C, which is a constant function.
1.7.2 Periodic Complex Exponential
Let us consider the case where a is purely imaginary, i.e., a =j , 2R. Since C
0 0
is a complex number, we have
j
C =Ae ;
where A;2R. Consequently,
j t j j t
0 0
x(t) =Ce =Ae e
j( t+)
0
=Ae =A cos( t +) +jA sin( t +):
0 01.7. CONTINUOUS-TIME COMPLEX EXPONENTIAL FUNCTIONS 23
The real and imaginary parts of x(t) are given by
Refx(t)g =A cos( t +)
0
Imfx(t)g =A sin( t +):
0
We can think of x(t) as a pair of sinusoidal signals of the same amplitude A, and
0
phase shift with one a cosine and the other a sine.
j t j t
0 0
(a)RefCe g (b)ImfCe g
j t
0
Figure 1.15: Periodic complex exponential function x(t) =Ce (C = 1, = 2).
0
j t
0
Claims. x(t) =Ce is periodic with
2
1. fundamental period: T = .
0
jj
0
2. fundamental frequency:jj.
0
The second claim is the immediate result from the rst claim. To show the rst claim,
we need to showx(t+T ) =x(t) and no smallerT can satisfy the periodicity criteria.
0 0
2
j (t+ )
0
j t j2
j j 0
0
x(t +T ) =Ce =Ce e
0
j t
0
=Ce =x(t):
It is easy to show that T is the smallest period
0
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