1
SIGNALS
CHAPTER
1
Signals
1.1
INTRODUCTION
A signal is a piece of information. In olden times, the signals were conveyed by burning fire (fire
flame attached to a pole is seen from a long distance) or by drum beats (sound of drum can also
be heard at a large distance) or by somaphore (signal sent by different coloured flags), and by
various other means. As the technology developed, we needed long or complex variety of various
signals and those which can be easily sent to long distances. The ideal choice is the electrical
signals. Once the signal is complicated and intricate we require a system to process such signals
to extract the information contained in the signal. The development of different systems to cater
to the processing needs of different signals is obvious. In this chapter, we shall introduce the
mathematic representation of various signals and systems, together with their classifications.
Basic signals have also been defined, which are needed to formulate complex signals.
1.2
CLASSIFICATION OF SIGNALS
A signal represents some physical quantity or a parameter. In general, a signal gives the behaviour
of a phenomenon which in turn depends on time. Mathematically, signals are denoted by symbols
such as: x, y, z, f, g, h, ......, etc. and are functions of time t or may be constant.
A signal x(t) is a continuous signal if the independent variable t (t is usually time) is continuous.
If the time is not continuous, but has discrete values: 0, T, 2T, ......, nT etc., we write the signal as
x(nT), n = 0, 1, 2, ...... . It is customary to write x (nT), which we call discrete signal, by the symbol,
x[n], which implies x[0], x[1], ......, x[n]. Also, the other symbols used for discrete signals x[n] are:
xn = x [n] = x [nT]
...(1.1)
or
x0, x1, x2, ......, xn
At times, we ignore T (i.e., assume T as unity) and x[n] would automatically mean x[nT];
given x[5], for example, we know it means x [5T]. We shall see that assuming T = 1 will simplify
most of the calculations. We call x[n] as samples since it is the value of parameter x at discrete
times nT and nT are the sampling instants. In this book we shall, at times, also use x(n) to indicate
a discrete signal, as it is still very popularly used by some authors.
1
2
SIGNALS AND SYSTEMS
Fig. 1.1. The signals (a) a continuous-time signal x(t)
(b) a discrete-time signals x[n] or x(n)
Figure 1.1 shows typically a continuous-time signal x(t), and a discrete-time signal x[n]. Note
that x(t) is a continuous curve. If there are finite number of breaks in the curve x(t), we still call
it continuous signal with zero values at the breaks. Time t can take any value from
– ∞ to + ∞. The discrete signal x[n] may be written as:
x[n] = {......, – 0.6, 1.3, 0.7, 1, 1.5, – 1, 0.5, 1, ......,}
↑
The arrow below the value 1 in above sequence shows that n = 0 here. Note that x[n] can be
easily defined though by a lengthy sequence. But x(t) cannot be written in this fashion, except
when x(t) has a curve which can be expressed in a mathematical expression, such as x(t) = sin 2t
or x(t) = 1 – t2 ......, etc. It is this property of x[n] which makes it computer friendly since we can
readily store all the samples x[n] in the computer memory. A continuous signal x(t) cannot be
exactly stored in a computer hard disc except by taking infinitely many values of x at close
points t.
Now we classify the signals in a broad sense.
1.2.1 Analog and Digital Signals
A typical signal x(t), where the argument t is continuous time and t can assume values in the
range – ∞ ≤ t ≤ ∞, is an analog signal x(t) shown in Fig. 1.1(a) is an analog signal.
The signal x[n], where the argument n assumes discrete values, is an example of a digital or
discrete signal. In general n can take only finite discrete value. x[n] shown in Fig. 1.1 (b) is a digital
(or discrete-time) signal.
1.2.2 Real and Complex Signals
Let
where j =
if
x(t) = x1(t) + j x2(t)
...(1.2)
− 1 and x1(t) and x2(t) are real valued, then x(t) is a complex analog signal. Similarly,
...(1.3)
x[n] = x1[n] + j x2[n]
where x1[n] and x2[n] are real valued, then x[n] given by (1.3) is an example of a complex digital
signal. If a signal does not have imaginary part i.e., if x2(t) and x2[n] are zero, then the signal is
termed as a real signal.
3
SIGNALS
1.2.3 Deterministic and Random/Stochastic Signals
A signal which is completely determined, and there is no uncertainty about its value or behaviour
is called deterministic. For example
x(t) = A e–5t
...(1.4)
is completely known if A and the value of t are known. If we are not sure what A is, may be it
is 0 or 1, or – 0.2,......, we cannot use x(t). In general, deterministic signals are used in most of the
Computer Applications.
If a signal is not fully determined at any instant of time t or n, the signal is called random or
stochastic. For example, in a digital signal
x(n) = n2 + 5n
...(1.5)
if n = number of heads, when a coin is tossed, say, 100 times, then the value of n is not exactly
known. It could be 30, 41, 71, or any value. In such a case, the value of x(n) in Eq. (1.5), is not
exactly calculable. This is the case of a random signal. Such signals play important role in
communication and control. We shall not deal with random signals in this book. Note that the
random signals are usually defined by a few important properties such as mean, variance, auto
and cross correlations etc.
1.2.4 Periodic and Aperiodic Signals
If
x(t) = x(t + kT), k = 0, ± 1, ± 2, ± 3, ......
...(1.6)
then x(t) is a periodic signal with period T.
x(t)
It implies that if in the argument t of x(t)
we add or subtract integral multiples of
T, the signal value remains the same. x(t)
...
...
shown in Fig. 1.2 is periodic with period
T. After and before time T the signal
0
t
–2T
–T
T
2T
repeats. As we cannot draw infinite
segments of basic x(t), we show three dots
(...) on both sides of one or two segments
of x(t), as shown in Fig. 1.2.
Similarly, we can also have periodic
Fig. 1.2. A periodic analog signal x(t) with period T
digital signals.
Here,
x[n] = x[n + kN], k = 0, ± 1, ±2, ......,
...(1.7)
x[n] is periodic with period N samples. A periodic digital signal is shown in Fig. 1.3 with period
N = 3. Note that x[0] = x[– 3] = x[3] = 3.5. Similarly, x[– 2] = x[1] = x[4] = 2. We also add three
dots (...) to avoid drawing a long periodic sequence x[n].
We can generate a periodic signal, given one “lobe” (or one basic expression) of the signal as
follows :
∞
Continuous-time periodic signal xp(t) =
∑
x (t − iTa )
...(1.8)
i=−∞
where x(t) is one lobe of the complete periodic signal xp (t) and T0 is its period.
Similarly, discrete-time periodic signal xp[n] is given by
∞
xp[n] =
∑
i = −∞
x [n − iN]
...(1.9)
4
SIGNALS AND SYSTEMS
x(n)
3.5
3.5
3.5
2
2
2
...
...
–4
–3
–2
–1
–1
–1
0
1
2
3
4
5
–1
n
–1
Fig. 1.3. A digital periodic signal x[n] with period N = 3
where x[n] is the signal for one period (N samples) and xp[n] is the periodic version of x[n] with
period N.
For example
2π ⎞ ⎞
x(t) = A sin (2πωt) is periodic in t with period 2π/ω sec since A sin ⎛⎜ 2πω ⎛⎜ t +
⎟ = A sin
ω ⎠ ⎟⎠
⎝
⎝
(2πωt + 2π) = A sin (2πωt).
Similarly,
2πn ⎤
⎡ 2π (n + N ) ⎤
x[n] = B sin ⎡⎢
is periodic in n with period N samples, since B sin ⎢
⎥⎦
N
⎣ N ⎦⎥
⎣
2π n ⎤
reduces to B sin ⎡⎢
⎣ N ⎥⎦
1.2.5 Even and Odd Signals
An analog signal xe(t) is an even signals if
xe(t) = xe (– t)
...(1.10a)
An analog signal x0(t) is an odd signal if
x0(t) = – x0(– t)
...(1.10b)
Thus for even signals making the argument t negative does not affect the signal whereas in
odd signal, the signal value x0(t) becomes – x0(t) when t is changed to (– t).
Similar definitions can be extended to digital sequence i.e.,
xe[n] = xe[– n]
for even signal xe[n]
...(1.10 c)
x0[n] = – x0[– n]
for odd signal x0 [n]
...(1.10 d)
Note that we have used subscript e or o to x(t) and x[n] to indicate that xe(t) and x0[n] are even
signals and odd signals, respectively.
xe(t)
–t0
0
(a)
xo(t)
t0
t
–t1
0
(b)
t1
t
5
SIGNALS
Fig. 1.4. Examples of even [xe(t), xe[n]] and odd [xo(t), xo[n]] signals
Note Important
We may also note some additional properties of even and odd signals, as follows:
1. odd function + odd function = odd function
2. even function + even function = even function
3. (odd function) × (odd function) = even function
4. (even function) × (even function) = even function
5. (even function) × (odd function) = odd function
6. For an even function xe(t)
∫
t0
− t0
xe (t) dt
= 2
∫
t0
0
xe (t) dt
...(1.11 a)
7. For an odd function x0(t)
∫
t0
− t0
x0(t) dt
= 0
...(1.11 b)
8. For discrete even signal xe[n]
n0
n0
∑
xe [n] = 2
∑
xe [n] − xe [0]
(Note the relation)
...(1.11 c)
n=0
n = − n0
9. For discrete odd signal x0[n]
n0
∑
x0 [n] = 0
...(1.11 d)
n = − n0
1.2.6 Derivation of Even and Odd Parts of a Generic Signal
Let a generic signal be x(t), for which we are required to find its even part xe(t) and odd part x0(t).
We can write
x(t) = xe(t) + x0(t)
...(1.12)
change t to – t or both sides of Eq. (1.12). Therefore
x (– t) = xe(– t) + x0(– t)
...(1.13)
But xe(– t) = + xe(t) and x0(– t) = – x0(t) due to Eq. (1.10), (i.e. by definition).
Using these values in Eq. (1.13), we get
x (– t) = xe(t) – x0(t)
...(1.14)
6
SIGNALS AND SYSTEMS
adding (1.12) and (1.14), and subtracting (1.14) from (1.12), we readily get
xe(t) =
1
[x(t) + x(– t)]
2
...(1.14 a)
1
[x(t) – x(– t)]
...(1.14 b)
2
By similar procedure, a generic digital signal x[n] can be split in its even part xe[n] and odd
part x0[n] by
x0(t) =
xe[n] =
1
(x[n] + x[– n])
2
...(1.15 a)
x0[n] =
1
(x[n] – x[– n])
2
...(1.15 b)
Note that the odd functions x0(t) and x0[n] must have zero value at argument t = 0 or n = 0 i.e.,
x0(t)|t =0 = x0(0) = 0
...(1.16 a)
x0(n)|n=0 = x0(0) = 0
...(1.16 b)
sin (− x) sin x
sin x
is an even function since
we shall see in later chapters that
=
−x
x
x
calculations of various transforms becomes easy if a signal is purely even or purely odd.
sin c (x) =
1.2.7
Energy Signals and Power Signals
Let v(t) and i(t) be the voltage and current, respectively, across a resistance, then the power p(t)
per ohm is defined as
v(t) ⋅ i(t) 2
= i (t)
R
Total energy, E, and average power P on a per-ohm is
p(t) =
Energy E =
∫
∞
−∞
Power P = lim
T →∞
i 2(t) dt joules
1
T
∫
T/2
− T/2
...(1.17)
...(1.18)
i 2(t) dt watts
...(1.19)
For an arbitrary continuous x(t), we extend this concept and define.
Normalised energy content of x(t) = E Δ
∫
∞
−∞
|x(t)|2 dt
...(1.20)
and
Normalised average power P of x(t) is defined as
P Δ
lim
T →∞
1
T
∫
T/2
− T/2
| x (t)|2 dt
...(1.21)
Extending similar concept for digital signal x[n], we have
Normalised energy content E of x [n] defined as
∞
E Δ
∑
n=−∞
|x[n]|2
...(1.22)
7
SIGNALS
and normalised power P of x[n] is defined as
P Δ lim
N→∞
1.2.8
1
2N + 1
N
∑ |x[n]|
2
...(1.23)
n=−N
Tests for Energy Signals and Power Signals
We categorise the signals as energy/power signal by using the following tests:
1. If x(t) or x[n] have 0 < E < ∞ and P = 0, then x(t) or x[n] are the Energy signals.
2. If x(t) or x[n] have 0 < P < ∞, and therefore E = ∞, then x(t) or x[n] are Power signals.
3. If the signals x(t) or x[n] do not satisfy above properties then these signals are neither
energy signals nor power signals.
4. If a signal is periodic with finite energy content per period, then it is a power signals. Its
average power P is the average power of one period only.
EXAMPLE 1.1: If g(t) = 7e–2t–3 find out the value of
(a) g(3), (b) g(2 – t), (c) g((t/10) + 4), (d) g(jt), (e)
–3)/2).
g(jt) + g( − jt)
(f) g((jt – 3)/2) + g((– jt
2
SOLUTION:
(a) g(3) = 7e– 2(3)– 3 = 7e–9 Ans.
(b) g(2 – t) = 7e–2(2 – t) – 3 = 7e–7 + 2t Ans.
(c) g(t/10 + 4) = 7e
t
−2⎛⎜ + 4 ⎞⎟ − 3
⎝ 10
⎠
= 7e–t/5 – 8 – 3 = 7e–(t/5) – 11 Ans.
(d) g(jt) = 7e–2(jt) – 3 = 7e–j 2t – 3 Ans.
(e) (g(jt) + g(– jt))/2 = (7e–j2t – 3 + 7e+j2t – 3)/2
= 7e–3 [(e–j2t + e+j2t)/2] = 7e–3 cos 2t Ans.
(f) g((jt – 3)/2) + g((– jt – 3)/2)
As
g((jt – 3)/2) = 7e–2[(jt – 3)/2] – 3 = 7e–jt + 3 – 3 = 7e–jt
...(A)
g((– jt – 3)/2) = 7e+jt
(change j to – j in (A))
∴
g((jt – 3)/2) + g((– jt – 3)/2) = 7(e–jt + e+jt) = 14 cos t Ans.
EXAMPLE 1.2: If g(x) = x2 – 4x + 4, write out the values of
(a) g(z) , (b) g(u + v), (c) g(ejt), (d) g(g(t)), (e) g(2).
SOLUTION: (a) g(z) = z2 – 4z + 4 (Replace x by z only in g(x)) Ans.
(b)
g(u + v) = (u + v)2 – 4(u + v) + 4
= u2 + v2 + 2uv – 4u – 4v + 4 Ans.
jt
(c)
g(e ) = (ejt)2 – 4(ejt) + 4
= (ejt – 2)2 Ans.
(d)
g(g(t)) = (t2 – 4t + 4)2 – 4(t2 – 4t + 4) + 4
(replacing x by (t2 – 4t + 4) in g(x))
4
3
2
2
= (t – 8t – 32t + 8t + 16t + 16) – 4(t2 – 4t + 4) + 4
= t4 – 8t3 + 20t2 – 16t + 4 Ans.
(e)
g(2) = (2)2 – 4(2) + 4
= 4 – 8 + 4 = 0 Ans.
8
SIGNALS AND SYSTEMS
EXAMPLE 1.3: Which of the following signals are periodic, and if one is, what is its fundamental
period.
SOLUTION: (a) g(t) = 7 sin(400 πt) (given signal)
If the argument of sine function is increased by 2π/decreased by 2π, or their integral multiples,
the value remains the same
Here, if t is increased by
1
, then argument becomes
200
1 ⎞
400π ⎛⎜ t +
⎟ = 400 t + 2π
200 ⎠
⎝
1
Ans.
200
(b) g(t) = 3 + t2 (given signal)
As g(t) incereases monotonically, g(t) is not periodic. Ans.
(c) g(t) = cos 60 πt – j sin 60 πt
(given signal)
= e–i 60πt (by Euler’s theorem)
∴ g(t) is periodic with T0 =
T0 =
1
30
1 ⎞
± j2π
ä 60π ⎛⎜ t +
=1
⎟ = 60πt + 2π and e
30 ⎠
⎝
∴ g(t) is periodic with period T0 = 1
Ans.
30
(d) g(t) = 10 sin (12πt) + 4 cos (18πt)
(given signal)
T01 = 1 , T02 =
6
1 1
3
LCM of ,
i.e., of
and
6 9
18
1
9
(T01 and T02 are determined as in part (a))
2
6 1
which is
=
18
18 3
1
g(t) is periodic with period T0 =
Ans.
3
1⎞
⎛
(Check by changing t to ⎜ t + 3 ⎟ in the given signal we get back g(t))
⎝
⎠
(e) g(t) = 10 sin (12 πt) + 4 cos (8t)
(given signal)
T01 =
∴
1
1
, T02 =
6
4π
2π
T01/T02 = 1 1 =
, an irrational value ∴ g(t) is not periodic.
3
6 4π
jω t
EXAMPLE 1.4: Let x(t) = e 0 and x[n] = x(nTS) = e jω0nTs . Find the condition on value of TS so that
x[n] is periodic.
SOLUTION: Let x[n] be periodic with fundamental period N0
i.e.,
x[n + N0] = x[n]
e jω0 (n + N0 )TS = e jω0nTS
i.e.,
e jω0nTS ⋅ e jω0N0TS = e jω0nTS
∴
e jω0N0TS = 1
9
SIGNALS
or
∴
∴
∴
ω0 N0 TS = m · 2π (a multiple of 2π)
(2πf0) N0 TS = m · 2π (cancel 2π both sides)
1
N0 TS = m
T0
⎛
1⎞
⎜∵ f0 = T ⎟
0⎠
⎝
TS
m
T0 = N0 = rational number
(because m, and N0 are both integers)
Ts
Sampling interval
or
is a rational number. Note
T0
Fundamental period
that the above condition is also true for sinusoidal signal x(t) = cos (ω0 t + θ).
EXAMPLE 1.5: If x1(t) and x2(t) are periodic signals with fundamental periods T1 and T2, respectively,
under what conditions x(t) = x1(t) + x2(t) is periodic. What is the fundamental period of x(t)?
SOLUTION: Given x1(t + T1) = x1(t), periodic with fundamental period T1
and
x2(t + T2) = x2(t), periodic with fundamental period T2
Also,
x1(t) = x1(t + mT1), m an integer
and
x2(t) = x2(t + kT2) k an integer
∴
x(t) = x1(t) + x2(t)
i.e.,
x(t) = x1(t + mT1) + x2(t + kT2)
...(A)
Let x(t) have a fundamental period of T
Then
x(t + T) = x1(t + T) + x2 (t + T)
...(B)
Comparing (A) and (B), we must have
T = mT1 = kT2
Thus, x[n] is periodic with period N0 if
T1
T2
=
k
= rational number
m
∴ k and m are integers.
The fundamental period T of sum of two periodic signals x1(t) and x2(t) is LCM of T1 and T2.
T
If 1 is an irrational number then x(t) = x1(t) + x2(t) cannot be periodic.
T2
EXAMPLE 1.6: Consider the sinusoidal signal x(t) = cos 15t
(a) Find the value of sampling interval TS such that x[n] = x(nTS) is periodic sequence.
(b) Find the fundamental period of x[n] = x(nTS) if Ts = 0.1 p second.
SOLUTION: Since x(t) = cos 15t, replace t by nTS, and get x[n]
(a) As
x[n] = x(nTS)
∴
x[n] = cos (15 nTS)
Let N0 be the period of x[n]. Then we have
x[n + N0] = x[n]
∴
cos (15(n + N0)TS) = cos (15nTS)
or
cos[15nTS + 15N0TS] = cos (15nTS)
∴
15 N0 TS = 2π m, m an integer
or
TS =
n 2π
N 0 15
Ans.
′
⎛ 2π ⎞
i.e., N0 = ⎜
⎟ multiple
⎝ 15 TS ⎠
10
SIGNALS AND SYSTEMS
(b) If T0 = 0.1π, then from above result
m 2π
, m and N0 integers
N0 15
0.1π =
N0
2
1
4
=
×
=
m
15 0.1 3
4
N0 =
m, m an integer
3
as N0 must be an integer, hence the smallest value of m = 3.
Then N0 = 4 Ans.
i.e., fundamental period of cos (1.5p n) is N0 = 4 Ans.
EXAMPLE 1.7: Determine whether or not each of the following signals are periodic. If periodic,
determine its fundamental period, of each signal given as (a), (b), (c) and (d). (IP Univ. 2007, 2008)
∴
SOLUTION:
π
(a) x(t) = cos ⎛⎜ t + ⎞⎟
4⎠
⎝
(given signal)
2π
Periodic with T0 = 2π Ans. ⎛⎜ T0 =
= 2π ⎞⎟
1
⎝
⎠
(b) x(t) = sin 2π t.
3
(given signal)
⎛
⎞
Periodic with T0 = 3 Ans. ⎜ T0 = 2π = 3 ⎟
π
2
/3
⎝
⎠
(c) x(t) = cos π t + sin π t
3
4
⎛
T01 6
⎞
⎜ and T is 8 rational ⎟
02
⎝
⎠
LCM of 6 and 8 is 24. ∴ x(t) is periodic with T0 = 24
Periodic with T01 = 6 Periodic with T02 = 8
(d) x(t) = cos t + sin
As
As
As
(given signal)
2t
cos t = cos ω1t ∴ period =
2 t = sin ω2t, period =
sin
Ans. (T0 = 24)
2π 2π
=
= 2π ≡ T01
ω1 1
2π 2π
=
= 2π ≡ T02
ω2
2
T01
2π
=
= 2 is irrational number ∴ x(t) is not periodic, since we cannot find
T02
2π
LCM of (2π) and ( 2π) .
2
(e) x(t) = sin t
Ans. x(t) not periodic
(given signal)
1
(1 – cos 2t). As cos 2t has period T0 = π
2
∴ x(t) is periodic with T0 = π Ans.
=
(f) x(t) = e
j
LM π t − 1OP
N2 Q
As x(t) = e–j . e
π
j t
2
⎛ from e j(ω0t − 1) = e − j ⋅ e j(ω0t) , here ω = π ⎞
⎜
⎟
0
2⎠
⎝
2π 2π
π
=
, Period T0 =
=4
e jω0t , where ω0 =
ω0 π/2
2
(given signal)
≡ e–j
i.e. x(t) is periodic with T0 = 4 Ans.