2
Signals and Systems
2.1
SIGNALS
We will be considering two basic types of signals: (i) Continuous-time signals and (ii) Discretetime signals. In case of continuous-time signals the independent variable is continuous, and
thus these signals are defined for continuum of values of the independent variable. On the
other hand, discrete-time signals are defined only at discrete instants of times, and
consequently, for these signals, the independent variable takes on only a discrete set of values.
To distinguish between continuous-time and discrete-time signals, we will use the symbol t to
denote continuous-time independent variable and n to denote discrete-time independent
variable. Similarly (.) is used to denote continuous while [.] is used to denote discrete valued
quantities.
A discrete-time signal is defined only at discrete instant of time, otherwise zero. Thus
independent variable has discrete values and let t = nT, n is integer (0, ± 1, ± 2, ...) and T is the
sampling time. Thus a discrete time signal is defined as f [nT], and for the sake of convenience,
it is denoted by f [n]. A continuous time signal f (t) and discrete time signal f [n] are shown in
Fig. 2.1(a) and (b) respectively.
Illustrations of continuous-time signals f (t) and discrete-time signals f [n] are being
made throughout the chapter for better concept and understanding.
49
50
NETWORKS AND SYSTEMS
x(t)
t
0
( a)
x[n]
x[0]
x[–1]
–9 –8 –7
x[1]
x[2]
–5 –4 –3 –2
–6
10
–1 0
1
2
3
4
5
6
7
8
n
9
( b)
Fig. 2.1. Graphical representations of (a) Continuous-time and (b) Discrete-time signals
Properties of Discrete Signals
Conjugate Symmetric
For a complex valued signal f (t), it is said to be conjugate symmetric, if
f (–t) = f * (t)
where ‘ * ’ denotes complex conjugate. That is, if
f(t) = a + jb
then
f* (t) = a – jb
where, a and b are the real and imaginary part of f (t).
It may be noted that for the function f (t) to be conjugate symmetric its real part has to
be even and imaginary part has to be an odd function.
The Continuous-Time Unit Step and Unit Impulse Functions
Continuous-time unit step function u(t) shown in Fig. 2.2 is defined as
u(t) =
RS0,
T1,
t<0
t>0
(2.1)
u(t)
1
0
t
Fig. 2.2. Continuous-time unit step function
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SIGNALS AND SYSTEMS
The continuous-time unit step is the running integral of the unit impulse
z
t
−∞
δ( τ ) dτ
(2.2)
The continuous-time unit impulse is the first derivative of the continuous-time unit
step as
du(t)
δ(t) =
(2.3)
dt
In practical sense, the unit step u∆(t) shown in Fig. 2.3 rises from the value 0 to the
value 1 in a short-time interval of length ∆. The unit step can be thought of as an idealisation
of u∆(t) for Lt ∆. Formally, u(t) is the limit of u∆(t) as ∆ → 0 and the derivative becomes the
∆→0
impulse in practical sense as shown in Fig. 2.4.
(t)
u(t)
1
1
0 0 t
Fig. 2.3
t
Fig. 2.4
The Unit Impulse and Unit Step Functions
We will see how we can use unit impulse signal as basic building blocks for the construction
and representation of other signals.
2.1.1 The Discrete-Time Unit Impulse and Unit Step Sequences
The unit impulse (or unit sample), which is defined by
δ[n] =
is as shown in Fig. 2.5
RS0,
T1,
n≠0
n=0
(2.4)
[n]
1
n
0
Fig. 2.5. Discrete-time unit impulse (sample)
The discrete-time unit step, is defined as
RS0,
T1,
n<0
n≥0
The unit step sequence is shown in Fig. 2.6
u[n] =
(2.5)
u[n]
1
0
Fig. 2.6. Discrete-time unit step sequence
n
Signals and Systems
u(t) =
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NETWORKS AND SYSTEMS
Further, the discrete-time unit impulse is the first difference of the discrete-time step as
δ[n] = u[n] – u[n – 1]
(2.6)
The discrete-time unit step is the running sum of the unit sample as
n
u[n] =
∑ δ[m]
(2.7)
m=−∞
Equation (2.7) is illustrated graphically in Fig. 2.6. Since the only non-zero value of the
unit sample is at the point at which its argument is zero, we see from the figure that the
running sum in Eqn. (2.7) is 0 for n < 0 and 1 for n ≥ 0. Furthermore, by changing the variable
of summation from m to k = n – m in Eqn. (2.7), i.e., m = n – k, now for m = – ∞ ⇒ n – k =
– ∞ ⇒ k = ∞ + n = ∞ and for m = n ⇒ n – k = n ⇒ k = 0; now we find that the discretetime unit step can also be written in terms of the unit sample as
0
u[n] =
∑ δ[n − k]
(2.8)
k=∞
∞
or equivalently,
u[n] =
∑ δ[n − k ]
(2.9)
k=0
Even and Odd Signals
A continuous signal f (t) is referred to as an even signal if it is identical to its time-reversed
counterpart, i.e., with its reflection about the origin. A continuous signal f (t) is said to be even if
f (t) = f (–t); for all t
(2.10)
and signal f (t) is said to be odd if
f (–t) = – f (t); for all t
(2.11)
This may be noted that an odd continuous time signal will be zero at origin, i.e., f (0) = 0
at t = 0. Examples of even and odd continuous-time
x(t)
signals are shown in Fig. 2.7.
Decomposition of continuous signal f(t) can be
done as:
f (t) = fe (t) + fo(t)
(2.12)
where, fe(t) is the even and fo(t) is the odd component of
continuous signal f(t). Obviously, the even function has
the property
fe(–t) = fe(t)
(2.13)
0
t
(a)
x(t)
and the odd function has the property
fo(–t) = – fo(t)
(2.14)
Replacing t by –t in the expression of f(t), we get
f (–t) = fe (–t) + fo(–t) = fe(t) – fo(t)
t
(2.15)
Solving from the expression of f (t) and f (– t), we
get
1
fe(t) = [f (t) + f (–t)]
2
0
(2.16)
(b)
Fig. 2.7. (a) An even continuous-time signal;
(b) An odd continuous-time signal
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SIGNALS AND SYSTEMS
and
fo(t) =
1
[f (t) – f (– t)]
2
(2.17)
The discrete signal f [n] is said to be even if
f [n] = f [– n];
for all n
(2.18)
f [–n] = – f [n]; for all n
(2.19)
that is, an odd discrete-time signal will be zero at origin.
Decomposition of discrete signal f [n] can be done as in the case of continuous time case
as follows:
f [n] = fe [n] + fo[n]
where, fe[n] is the even and fo[n] is the odd component of discrete signal f [n].
Obviously, the even function as usual has the property
fe [–n] = fe [n]
and the odd function as usual has the property
fo[–n] = – fo[n]
Replacing n by – n in the expression of f [n], we get
f [–n] = fe [–n] + fo[–n] = fe[n] – fo[n]
Solving from the expression of f [n] and f [– n], we get
and
fe[n] =
1
[ f [ n] + f [ − n]
2
(2.20)
fo[n] =
1
[ f [ n] − f [− n]
2
(2.21)
Examples of even and odd discrete-time signals are shown in Fig. 2.8.
1
,n<0
2
1, n = 0
1
,n>0
2
ev{x[n]} =
x[n] = 1, n > 0
0, n < 0
1
1
–3 –2 –1 0 1 2 3
n
–3 –2 –1 0 1 2 3
–
od{x[n]} =
–3 –2 –1
– 12
1
2
1
,n<0
2
0, n = 0
1
,n>0
2
1
2
0 1 2 3
n
Fig. 2.8. Example of the even-odd decomposition of a discrete-time signal
n
Signals and Systems
and signal f [n] is said to be odd if
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NETWORKS AND SYSTEMS
Periodic Signal
A periodic continuous-time signal x(t) has the property that for a positive value of T, for which
x(t) = x(t + T)
(2.22)
for all value of t. Then x(t) is periodic with time period T.
A periodic continuous-time signal x(t), periodic with time period T is shown in Fig. 2.9.
Then one can write
x(t) = x(t + mT)
(2.23)
for any integer m. Thus x(t) is also periodic with period 2T, 3T, ... . The fundamental period T0
is the smallest positive value of T for which Eqn. (2.22) holds.
Similarly, a periodic discrete-time signal x[n] has the property that for a positive integer
N, for which
x[n] = x[n + N]
(2.24)
for all values of n, then the discrete time signal x[n] is periodic with period N if it is unchanged
by a time shift of N. Further, x[n] is periodic with period 2N, 3N, ..., that is; for any integer
value m,
x[n] = x[n + mN]
(2.25)
The fundamental period N0 is the smallest positive value of N for which Eqn. (2.24)
holds. A periodic discrete-time signal x [n] with fundamental period N0 = 3 is shown in
Fig. 2.10.
x(t)
–2T
–T
0
T
2T
Fig. 2.9. A continuous-time periodic signal
x[n]
n
Fig. 2.10. A discrete-time periodic signal with fundamental period N0 = 3
A signal closely related to the periodic complex exponential is the sinusoidal signal
x(t) = A cos (ω0t + φ),
(2.26)
as illustrated in Fig. 2.11. With seconds as the units of t, the units of φ and ω0 are radian and
radians/second, respectively. ω0 = 2πf0 =
2π
, where f0 is in cycles/second or hertz(Hz). The
T0
sinusoidal signal is periodic with fundamental period T0 = 1/f0.
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SIGNALS AND SYSTEMS
x(t) = A cos (0t + )
T0 =
A
2
0
A cos Fig. 2.11. Continuous-time sinusoidal signal
Time Scaling: If f(t) is a continuous-time signal, the y(t) is obtained by scaling the
independent variable time by factor a is referred as the time scaling and is defined by
y(t) = f (at)
(2.27)
Now if a > 1, the output signal y(t) is compressed version of f (t) on time axis and if a < 1,
the output signal y(t) is rarefied version of f (t) on time axis.
Illustrations of continuous-time signals x(t), x(2t) and x(t/2) are through examples in
Figs. 2.12 and 2.13 that are related by linear scale changes in the independent variable. Suppose
tape recorder is having normal speed x(t) for which sound is clear. Now if the speed is doubled
that is, x(2t), as in Fig. 2.12, the tape recorder is moving very fast, that is, sound is not
distinguishable because, in the same space of time double the words are played, hence
compressed. On the contrary, if the speed is half, that is, the signal x(t/2) as in Fig. 2.13, the
recorder is moving very slowly, that is in the same space of time half the words are played,
that is, rarefaction occurs.
x(t)
x(2t) = x()
x(2t)
x(t) vs t
x() vs , i.e., x(2t) vs 2t
x(2t) vs t
–2
–2
–1
0
0
0
t
2t = 2
2
1
t = /2
(a)
x(2t)
x(2t) vs t in
conventional time
format
–1 0
1
t
(b)
Fig. 2.12. ( a) Continuous-time signal (b) Compressed signal after time scaling where a > 1
Another example to illustrate the linear scale change of independent time t. Let x2(t) =
x(αt), α > 1 and x3(t) = x(βt), β < 1 as in Fig. 2.14. Suppose, x1(t) = cos ωt, then x2(t) = x(αt) = cos
ω(αt) = cos (αω)t. The signals cos ωt and cos (αω)t is to be compared in the same time scale.
The signal cos (αω)t is having the higher frequency as that of cos ωt. Hence time period of x(αt)
Signals and Systems
t
0
56
NETWORKS AND SYSTEMS
is lesser as that of x1(t), that is, x(αt) is compressed as compared to x1(t). In the same line of
argument, the signal x3(t) = x(βt) is rarefied compared to signal x1(t).
x(t)
(a )
x(t) vs t
–2
–2
–4
x() vs , i.e., x(t/2) vs t/2
x(t/2) vs t
0
0
0
x(t/2) = x()
x(t/2)
2
2
4
t
t/2 = 2 = t
x(t/2)
(b )
x(t/2) vs t
in conventional
time format
–4
0
4 t
Fig. 2.13. ( a) Continuous-time signals (b) Rarefied signal related by time scaling with a < 1
x1(t) = cos 1t
0
t
T1
(a)
x2(t) = x(t) = cos t, > 1 = cos 2t
0
t
T2
x3(t) = x(t) = cos t = cos t
(b)
x3(t) = x(t) = cos t, < 1 = cos 3t
0
T3
t
(c)
Fig. 2.14. Relationship between the fundamental frequency and period for continuous-time sinusoidal signals, here ω1 <
ω2 and ω1 > ω3 which implies that T1 > T2 and T1 < T3
57
SIGNALS AND SYSTEMS
For better understanding, another figurative illustration has been made in Fig. 2.15.
The signal x(t) vs t is shown in Fig. 2.15 (a). The compressed version x(2t) vs t is shown in
Fig. 2.15 (b) and the rarefied version x(t/2) vs t is shown in Fig. 2.15(c).
x(t)
x(2t)
–2
2
0
–1
t
0
1
t
(b)
(a)
x(t/2)
2
–4
4
0
t
(c)
Fig. 2.15
Reflection: For a continuous time signal f(t), its reflection about vertical axis becomes
y(t) = f(–t)
(2.28)
A continuous-time signal f (t) is shown in Fig. 2.16 (a) and its reflection f (– t) about
t = 0 is shown in Fig. 2.15 (b). The procedure is like this: by just changing the abscissa with
proper manipulation as is evident in Fig. 2.16 (a); we obtain the desired curve. But for the
conventional look the curve is as shown in Fig. 2.16 (b).
f(t)
f(–t) = f()
f(–t)
f(t) vs t
f(–t) = f() vs f(–t) vs t
–3
–3
3
–2
–2
2
–1
–1
1
0
0
0
Fig. 2.16 (a)
1
+1
–1
2
+2
–2
3
+3
–3
t
–t = t = –
Signals and Systems
2
2
58
NETWORKS AND SYSTEMS
f(–t)
f(–t) vs t
–4
–3
–2
–1
0
1
2
3
t
Fig. 2.16 (b)
Similarly, for a discrete-time signal f [n] its reflection about vertical axis becomes
y [n] = f [–n]
(2.29)
This may be noted for even signal, its reflection is same as that of the original signal.
A discrete-time signal f [n] is shown in Fig. 2.16 (c) and its reflection f [– n] about n = 0
is shown in Fig. 2.16 (d). As in the case of continuous case, the abscissa is only changed at each
sequence and finally for the conventional look, the final curve of f [– n] vs n is redrawn as
shown in Fig. 2.16 (d).
f[n]
f[–n] = f[k]
f[–n]
f[n] vs n
f[–n] = f[k] vs k
f[–n] vs k
0
0
0
–8
–8
8
8
8
–8
n
(–n) = k
n = –k
Fig. 2.16 ( c)
f[–n]
f[–n] vs n
–8
0
Fig. 2.16 (d )
8
n
Time Shifting
If for a continuous time signal f(t), its time shifted version y(t) can be written as
y(t) = f(t – t0)
(2.30)
where, t0 is the time shift.
If t0 > 0, then shifting is in the right in time scale, that is, the waveform is shifted to the
right.
If t0 < 0, then shifting is in the left in time scale, that is, the waveform is shifted to the
left.
For discrete time signal f [n], its time shifted version y[n] can be written as
y [n] = f [n – n0]
(2.31)
where, n0 is the integer and the shift in right in time axis if n0 > 0 and the shift in left in time
axis if n0 < 0.
Continuous-time signals related by a time shift is shown in the Fig. 2.17. In this figure
t0 < 0 (say, – 5), so that f(t – t0 ) is an advanced version of f(t), that means each point in f(t)
occurs at an earlier time in f (t – t0 ). This advanced version is obtained through manipulation
in Fig. 2.17(a) in abscissa only and ultimately the conventional version is shown in Fig. 2.17 (b).
f(t)
f(t – t0) = x()
f(t – t0)
t0 < 0 = –5
f(t) vs t
f(t – t0) = x() vs f(t – t0) vs t
0
0
t0 = –5
t
t – t0 = t = + t0
(a )
f(t – t0)
t0 < 0 = –5
f(t – t0) vs t
t0 = –5
0
t
(b )
Fig. 2.17
A time-shift in discrete time signal is illustrated in Fig. 2.18 in which we have two
signals f [n] and f [n – n0] that are identical in shape, but are shifted relative to each other.
Here f [n – n0] is a delayed version of f [n] for n0 > 0. For given f [n] vs n; we have to draw
f [n – n0] vs n for n0 is having positive integer. The procedure is explained as below. Draw
f [n – n0] = f [k] vs [n – n0] = k. Then in the same figure draw f [n – n0] vs n = k + n0 by only
algebraic manipulation in the time scale in abscissa as shown in Fig. 2.18 (a) and redrawn
f [n – n0] vs n in Fig. 2.18 (b) for the conventional look which is nothing but the delayed version
of f [n] by n0 > 0.
Signals and Systems
59
SIGNALS AND SYSTEMS
60
NETWORKS AND SYSTEMS
f[n]
f[(n – n0)] = x[k]
f[(n – n0)]
f[n] vs n
f[n – n0] = x [k] vs k
f[n – n0] vs n
n
n0
n0
0
0
n0
[n – n0] = k
n = [k + n0]
( a)
f[n – n0]
0
f[n – n0] vs n
n
n0
(b )
Fig. 2.18. Discrete-time signals related by time-shift. For n0 > 0, [n – n 0] is a delayed version of f [n];
(that is, at n = n0 instant f [n – n0 ] is having the same value what f [n = 0]th instant was)
EXAMPLE 1
Decompose the given step function f (t) into its odd and even parts systematically through
a sequence of diagrammatic approach.
f(–t)
f(t)
1
0
(a)
1
t
0
( b)
t
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SIGNALS AND SYSTEMS
fe(t) =
[f(t) + f(–t)]
2
f0(t) =
1/2
[f(t) – f(–t)]
2
1/2
t
0
t
0
(c)
(d )
Fig. 2.19 Decomposition of signal f(t) into even and odd components
(a) Unit step function f(t)
(b) Folded step function f(– t)
(c) Even part of unit step function f(t)
(d ) Odd part of unit step function f(t).
EXAMPLE 2
Decompose the given function f (t) into its odd and even parts through a systematic
sequence of diagrammatic approach.
f(–t)
f(t)
1
1
1/2
1/2
1
–1
0
t
–1
1
0
–1/2
t
–1/2
(a)
(b)
f0(t) =
[f(t) + f(–t)]
fe(t) =
2
[f(t) – f(–t)]
2
1/2
1/2
1
–1
–1
0
1
t
(c)
0
t
–1/2
(d )
Fig. 2.20 Decomposition of signal f(t) of Example 2 into even and odd components
(a) Original function f (t)
(b) Folded function f (–t )
(c) Even part of the given function f (t )
(d ) Odd part of the given function f (t )
Signals and Systems
–1/2
62
NETWORKS AND SYSTEMS
EXAMPLE 3
Decompose the given square pulse f (t) into even and odd components.
f(–t)
f(t)
1
1
0
1
–1
t
( a)
0
t
( b)
fe(t)
f0(t)
1/2
1/2
–1
–1
0
1
0
t
1
t
–1/2
( c)
(d )
Fig. 2.21 Decomposition of signal f(t) of Example 3 into even and odd components
(a) Original function f (t)
(b) Folded function f (– t )
(c) Even part of the given function f (t )
(d) Odd part of the given function f (t )
Types of Sequences
The discrete-time sequences may be represented in a number of ways. Some of the alternative
representations that are often more convenient to use. These are shown in the following
illustrations
1. Functional representation
f [n] =
RS1
T0
for n = 0, 1, 2, 3, ...
otherwise
as shown in Fig. 2.22 (a)
2. Representation based on length the discrete-time signal may be a finite length or finite
length sequences. The finite length also called finite duration sequence is only defined
only for a finite time interval.
A finite duration sequence can be represented
f [n] = {... 1, 1, 1, 2, 1, 1, 1, ...}
as shown in Fig. 2.22 (b)
↑
where the sign origin is indicated by the symbol ↑
An infinite duration sequence can be represented as
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SIGNALS AND SYSTEMS
f [n] = {... – 1, – 1, – 1, 2, 1, 1, 1, ...} as shown in Fig. 2.22 (c)
↑
A sequence which is zero for n < 0 can be represented as
f [n] = {0, 1, 2, 0, 1, 2, 2, 0, 0} as shown in Fig. 2.22 (d)
↑
{ 1,0, nn < 00
Signals and Systems
f[n] =
2
1
1
–3
–2
0
–1
1
2
n
3
–3
–2
–1
0
(a)
1
n
3
2
(b)
2
1
–3
–2
1
–1
1
0
2
3
n
0
1
2
3
4
5
6
7
8 n
–1
(c)
(d )
Fig. 2.22
EXAMPLE 4
Consider the discrete-time signal f [n] depicted in Fig. 2.23 (a). Further, we have depicted
five time-shifted, scaled unit impulse sequences in Figs. 2.23 (b) to ( f ) respectively which are
expressed mathematically as
RS f [− 2] , n = − 2
T0 , n ≠ − 2
R f [− 1] , n = − 1
f [– 1] δ[n + 1] = S
T0 , n ≠ − 1
R f [0] , n = 0
f [0] δ[n] = S
T0 , n ≠ 0
Rf [1] , n = 1
f [1] δ[n – 1] = S
T0 , n ≠ 1
Rf [2] , n = 2
f [2] δ[n – 2] = S
T0 , n ≠ 2
f [– 2] δ[n + 2] =
as
Therefore, the five sequences in the figure equals x[n] for – 2 ≤ n ≤ 2 and can be written
x[n] = f [– 2] δ[n + 2] + f [– 1] δ [n + 1] + f [0] δ [n] + f [1] δ [n – 1] + f [2] δ [n – 2]
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NETWORKS AND SYSTEMS
2
=
∑
k= −2
f [k] δ[n – k]
(2.32)
Please note the given impulse function δ[n] is given in Fig. 2.5 and f [n] is shown in
Fig. 2.23 (a). Hence we get all components of x[n] as shown through Fig. 2.23 (b) to ( f )
f[n]
–4
2 3
–1
–3 –2
f[–2] [n + 2]
0 1
4
n
–4 –3 –2 –1 0 1 2 3 4
( a)
n
(b )
f[0] [n]
f[–1] [n + 1]
–1
–4 –3 –2
0 1 2 3 4
n
–4 –3 –2 –1 0 1 2 3 4
(c )
n
(d )
f[2] [n – 2]
f[1] [n – 1]
2
–4 –3 –2 –1 0 1 2 3 4
n
–4 –3 –2 –1 0 1
( e)
3 4
n
(f )
Fig. 2.23. Decomposition of a discrete-time signal into a weighted sum of shifted impulses
EXAMPLE 5
For a discrete-time LTI system, the given impulse response h [n] and input f [n] are as
shown in Fig. 2.24 (a). We are going to show how output y[n] is generated.
For this case, f [0] and f [1] are only non-zero. Then the output y [n] is
y[n] = f [0] h[n – 0] + f [1] h[n – 1]
= 0.5 h[n] + 2 h[n – 1]
The two components 0.5 h[n] and 2 h[n – 1] are shown in Fig. 2.24 (b) separately. By
summing for each value of n, we obtain output y[n] which is shown in Fig. 2.24 (c).
65
SIGNALS AND SYSTEMS
2
f[n]
h[n]
1
0.5
0
1
n
2
0
n
1
2h(n – 1)
2
0.5h(n)
0.5
0
1
n
2
0
1
( b)
2
3
n
(c)
2.5
2
y(n)
0.5
0
1
2
3
n
(d )
Fig. 2.24
Impulse Response
The impulse response is defined as the system output due to an impulse sequence δ[n] at the
system input, where
RS
T
1, n = 0
δ[n] = 0 , n ≠ 0
Define the response to δ[n] as h[n], the impulse response sequence as
δ[n] → h[n]
If we multiply the impulse sequence by a constant c and apply to the system, then by
linearity, the output is also multiplied by c, that is,
c δ[n] → c h[n]
If we shift the position of this sequence, then by shift invariance of the system, we also
shift the output sequence by the same amount, that is,
c δ[n ± k] → c h[n ± k]
Let us see how we can represent any arbitrary input sequence in terms of impulse
response sequence as follows:
f [n] = ... + f [–2] δ [n + 2] + f [–1] δ [n + 1] + f [0] δ [n + 0] + f [1] δ [n –1] + ...
Signals and Systems
(a)
66
NETWORKS AND SYSTEMS
In other words, we take each point of the sequence f [ j] and multiply it by a shifted
version of the impulse sequence δ [n – j]. Because shifted unit impulse δ [n – j] has the value
unity for n = j and zero otherwise. Thus
∞
f [n] =
∑
f [ j] δ [n – j]
j=−∞
Some of the problems are solved in a different way which is simple and elegant.
We have explained the procedure for computing the response [n], both mathematically
and graphically, given the impulse input δ[n] which convolves with another function f [n] at
different discrete instant of time n for – 2 < n < 2 and can be summed up to get the total
response x[n].
On the same line of thought, the convolution integral can be applied to find the output
response y[n] of any system having the impulse response h[n] subjected to any input signal
x[n] can be written as
∞
y[n1] =
∑
x[ k] h[ n1 − k]
k= − ∞
The convolution process can be summarised in four steps:
Folding: Fold h [k] about origin and obtain h [– k]
Time shifting: Shift h[k] by n0 unit to the right to obtain h [– (k – n0)] = h [n0 – k]
Multiplication: Multiply x[k] by h[n1 – k] to obtain f [k] = x[k] h [n1 – k]
Summation: Sum all the values of the product f [k] to obtain value of output at n = n0
EXAMPLE 6
Determine the output y[n] of an LTI system with impulse response
h[n] = {6, 5, 4, 3, 2, 1} as shown in Fig. 2.25 (b)
↑
Subjected to the input is
x[n] = {1, 1, 1, 1, 1} as shown in Fig. 2.25 (a)
↑
Solution: Given h[n] and x[n]. Get under Folding h[– n] as obtained and shown in
Fig. 2.25 (c). Now we have to find
y[n] =
∞
∑
x[ k] h[ n − k]
k= − ∞
for –∞ < k < ∞.
∞
The output at n = 0 as y[0] =
∑
k =− ∞
x[ k] h[ − k] = 6; see Fig. 2.25 (d)
67
SIGNALS AND SYSTEMS
∞
The output at n = 1 as y[1] =
∑
x[ k] h[1 − k] = 5 + 6 = 11; see Fig. 2.25 (e)
k= − ∞
[n]
0
1
2
3
4
5
6
7
8
9
x[k] h [n – k]y[n]
6
11
15
18
14
10
6
3
1
0
Fig. 2.25 (.)
(d)
(e)
(f )
(g)
(h)
(i)
j
k
l
m
Finally we see that, the product sequence contains all zeros for n < – 1 and n > 8 and
written as
y[n] = [...0, 6, 11, 15, 18, 14, 10, 6, 3, 1, 0...]
6
5
h[k]
h[–k]
6
5
4
4
3
3
2
1
2
1
x[k]
1
0 1 2 3 4 5
k
(a )
0 1 2 3 4 5 6
k
–6 –5 –4 –3 –2 –1 0
( b)
( c)
h[–k]
6
5
x[k]
y0[k] = x[k] h[–k]
∞
6
Linear
system
4
1
k
3
x[k]
y(0) =k Σ
= x[k] h[–k] = 6
=–∞
y[k]
2
1
0 1 2 3 4 5
k
–6 –5 –4–3 –2 –1 0
k
–2 –1 0 1 2 3 4 5
k
(d )
h[–(k–1)] = h[1–k]
x[k]
y1[k] = x[k] h[1– k]
6
5
Linear
system
x[k]
∞
y(1) =k Σ
= x[k] h[k] = 11
=–∞
y[k]
1
0 1 2 3 4 5
k
–4 –3 –2 –1 0 1
( e)
k
–2 –1 0 1 2 3 4 5
k
Signals and Systems
In a similar line of action we can get y[2] = 4 + 5 + 6 = 15; see Fig. (f ) and so on. The
output at different instants is tabled below and shown in Fig. (d) through (m).
68
NETWORKS AND SYSTEMS
h[2–k]
x[k]
y2[k] = x[k] h[2–k]
Linear
system
1
x[k]
0 1 2 3 4 5
k
–3 –2 –1 0 1 2 3 4 5
y[2] = Σ x[k] h[2–k]
6
5
4
y[k]
–2 –1 0 1 2 3 4 5
k
k
(f )
y3[k] = x[k] h[3–k]
h[3–k]
x[k]
6
6
5
5
Linear
system
4
3
2
1
1
0 1 2 3 4 5
k
x[k]
–2 –1 0 1 2 3 4 5
4
3
y[k]
k
y[3] = Σ x[k]
h[3–k] = 18
–2 –1 0 1 2 3 4 5
k
( g)
y4[k] = x[k] h[4–k]
h[4–k]
x[k]
5
Linear
system
1
x[k]
0 1 2 3 4 5
k
–1 0 1 2 3 4 5
y[k]
y[4] = Σ x[k]
h[4–k] = 14
4
3
2
–2 0 1 2 3 4 5 6
k
k
( h)
y5[k] = x[k] h[5–k]
h[5–k]
x[k]
6
5
4
3
2
1
1
0 1 2 3 4 5
k
y[5] = Σ x[k] h[5–k] = 10
Linear
system
x[k]
0 1 2 3 4 5
k
(i )
y[k]
4
3
2
1
0 1 2 3 4 5 6
k
69
SIGNALS AND SYSTEMS
y6[k] = x[k] h[6–k]
h[6–k]
6
y[6] = Σ x[k] h[6–k] = 6
5
Linear
system
4
3
2
1
1
0 1 2 3 4 5
k
x[k]
0 1 2 3 4 5 6
y[k]
3
2
1
k
0 1 2 3 4 5 6
k
(j)
y7[k] = x[k] h[7–k]
h[7–k]
x[k]
y[7] = x[k] h[7–k] = 3
Linear
system
1
x[k]
y[k]
2
1
0 1 2 3 4 5
k
0 1 2 3 4 5 6 7
–1 0 1 2 3 4 5
k
k
(k )
y8[k] = x[k] h[8–k]
h[8–k]
x[k]
y[8] = x[k] h[8–k] = 1
Linear
system
1
x[k]
y[k]
1
0 1 2 3 4 5
0 1 2 3 4 5 6 7 8
k
k
–1 0 1 2 3 4 5
k
(l )
h[9–k]
x[k]
y9[k] = x[k] h[9–k]
= y[9] = 0
Linear
system
1
x[k]
0 1 2 3 4 5
k
0 1 2 3 4 5 6 7 8 9
(m)
Fig. 2.25
y[k]
k
–1 0 1 2 3 4 5
k
Signals and Systems
x[k]
70
NETWORKS AND SYSTEMS
EXAMPLE 7
Let us consider an example where we wish to convolve h and x, where
R|F 1I ,
h[n] = SGH 2 JK
|T 0 ,
k
and
k≥0
k<0
x[n] = {3, 2, 1}
R|FG 1IJ ,
h[n] = SH 2 K
|T 0 ,
k
Solution: We get
k≥0
k<0
FG
H
= ..., 0, 1,
IJ
K
1 1 1
, , , ...
2 4 8
We get h[k – n], and x[n] = {3, 2, 1} ; for different values of k, convolve graphically as in
previous example 6, h[k – n] and x[n] = {3, 2, 1} from the following expression:
∞
y[k] =
∑
x[ n] h[ k − n]
k = −∞
The result obtained as
RS
T
y[k] = 3,
UV
W
7 11 11 11
11
, ,
,
, ..., k , ...
2 4 8 16
2
There is another alogrithm that we can use to evaluate discrete convolutions for the
same example we finished just now. Suppose that we wish to convolve h and x, where
R|F 1I ,
h[n] = SGH 2 JK
|T 0 ,
k
and
k≥0
k<0
x[n] = {3, 2, 1}
Construct a matrix with h bordering the top of
the matrix and x the left side of the matrix, as shown
in Fig. 2.26. The entries in the matrix are the products
of the corresponding row and column headers. To find
the convolution of the two sequences, we need only ‘‘fold
and add’’ according to the dotted diagonal lines. The
3
7
=
2
2
which is the sum of the terms contained between the
first and second diagonal lines. We can continue in this
way and obtain the output sequence as
first term, y[0] = 3. The second term y[1] = 2 +
h
x
1
1/2
1/4
1/8
3
3
3/2
3/4
3/8
2
2
1
1/2
1/4
1
1
1/2
1/4
1/8
Fig. 2.26. Matrix representation of
convolution summation
71
SIGNALS AND SYSTEMS
RS
T
y[k] = 3,
UV
W
7 11 11 11
11
,
,
,
, ..., k , ...
2 4 8 16
2
New Approach for Solving Problems
A continuous time signal f (t) is shown in Fig. 2.27 (a). Example of a transformation of the
independent variable of continuous signal f (t) is shown as f (t – t0). The signals f(t) and
f(t – t0) are identical in shape but that are displaced or shifted relative to each other. The
signal f(t – t0) means the signal f(t) is shifted right by t0, or moved forward by t0 and obviously
the signal f(t + t0) means the signal f (t) is shifted left by t0 or moved backward by t0. Few
examples are handled with for better understanding.
EXAMPLE 8
Given signal f (t) vs t, draw
(i) f (t + 1) vs t
(iii) f
(ii) f (– t + 1) vs t
FG 3 tIJ vs t
H2 K
(iv) f
FG 3 t + 1IJ vs t
H2 K
Solution: (i) Given the signal f (t) vs t in Fig. 2.27 (a).
f(t)
1
0
1
2
t
Fig. 2.27 (a)
From the given signal f (t) vs t we get f (T ) vs T and then f (t + 1) vs t is obtained as
depicted in Fig. 2.27 (b) in the following:
f(t + 1)
f(T)
f(t)
1
0
0
–1
1
1
0
2
2
1
Fig. 2.27(b)
t
T
(t + 1 = T); T – 1 = t
Signals and Systems
This algorithm is not always a satisfactory method, because the result is not easily
placed in closed form. Only in simple cases, one can determine the closed form solution of y[k].
72
NETWORKS AND SYSTEMS
Now f (t + 1) vs t with conventional time scale with formal look, looks like what is shown
in Fig. 2.27 (c).
f(t + 1)
–1
1
0
2 t
Fig. 2.27(c)
Figure 2.27 (c) is basically the curve f (t + 1) vs t is the time shifted curve f (t + 1) which
corresponds to an advance (shift to the left) of the given signal f (t) by one unit along t-axis. It
means that, at t = – 1, f (t + 1) should have the same value of the given curve f (t) at t = 0.
(ii) Given the curve f (t) vs t in Fig. 2.27 (a). Then obtain the curve f (– t + 1) vs t through
the following sequence of operations as shown in Fig. 2.27 (d).
f(t), f(T), f(–T), f(–t + 1)
0
0
0
1
1
1
–1
0
2
t
2
T
–2
–T
–1 (– t + 1 = T); – T + 1 = t
Fig. 2.27(d )
Now f (– t + 1) vs t with conventional time scale with formal look, looks like as given in
Fig. 2.27 (e).
f(–t + 1)
–1
0
1
t
Fig. 2.27(e)
Figure 2.27 (e) is the curve f (– t + 1) is a time shift and time reversal of given curve f (t).
(iii) Given the signal f (t) vs t in Fig. 2.27 (a). As in earlier case, get f (T ) vs T ; then the
curve f [(3/2)t] vs t is obtained through the following sequence of operations as depicted in
Fig. 2.28 (a):
73
SIGNALS AND SYSTEMS
0
0
0
1
1
2/3
2
t
T
2
4/3 [T=(3/2)t]; (2/3)T=t
Fig. 2.28 (a)
Now f [(3/2)t] vs t with conventional time scale looks like as redrawn in Fig. 2.28 (b)
f[(3/2)t]
0
2/3
4/3
t
Fig. 2.28 (b)
(iv) Given the signal f(t) vs t in Fig. 2.27 (a). Now, get f (T) vs T ; then the curve f [(3/2) + 1]
vs t is obtained through the following sequence of operations as depicted in Fig. 2.29 (a) :
f(t), f(T), f[(3/2)t + 1]
1
1
2
0
1
2
–2/3
0
2/3;
0
t
T
[T=(3/2)t+1]; [(2T–2)/3] = t
Fig. 2.29 (a)
Now f [(3/2)t + 1] vs t with conventional time scale looks like as redrawn in Fig. 2.29 (b).
f[(3/2)t + 1]
1
–2/3
0
2/3
Fig. 2.29 (b)
t
Signals and Systems
f(t), f(T), f[(3/2)t]
74
NETWORKS AND SYSTEMS
It may be noted that Fig. 2.29 (b) is the signal f [(3/2)t + 1] obtained by time shifting
and scaling of signal f(t).
EXAMPLE 9
Given f (t) vs t in Fig. 2.30 (a). Our task is to draw f (2 – t/3) vs t. First draw f (T) vs T
which is same as f (t) vs t. Next draw f (2 – t/3) vs t where (2 – t/3) = T from which we get
t = (6 – 3T ) as shown in Fig. 2.30 (a).
f(t), f(T)
t
3
f 2–
2
1
–1
9
0
6
1
3
2
0
3
–3
t, T
6–t
= T; t = 6 – 3 T
3
Fig. 2.30 (a)
The more conventional form of the graph f (2 – t/3) vs t is redrawn in conventional time
format from the above Fig. 2.30 (a) and as shown in Fig. 2.30 (b).
f(2 – t/3)
2
1
–3
0
3
6
9
t
–1
Fig. 2.30 (b)
EXAMPLE 10
Given f (t) vs t as in Fig. 2.30 (a), our task is to draw f (t – 2) vs t. Proceed as follows.
Draw f (T ) vs T. Then put T = t – 2, hence t = T + 2. We are changing the horizontal scale only to
get the curve of f (t – 2) vs t.
75
SIGNALS AND SYSTEMS
f(t), f(T)
f(t – 2)
2
–2
0
–1
1
0
2
1
3
2
4
3
5
t, T
t=T+2
Fig. 2.31(a)
The common form of the curve f (t – 2) vs t is drawn in the conventional way as
advancement of f (t) by 2 units in right-hand side, otherwise, the proposed method is alright to
get the same result by manipulating the time axis. The conventional form of f (t – 2) is redrawn
in Fig. 2.31 (b).
f(t – 2)
2
1
0
1
2
3
4
5
t
Fig. 2.31(b)
In the same line of reasoning, one can draw f(t + 2) vs t as follows:
f(t), f(T)
f(t + 2)
2
1
–1
–3
0
–2
1
–1
2
0
Fig. 2.32 (a)
3
1
t, T
t=T–2
Signals and Systems
1
76
NETWORKS AND SYSTEMS
The conventional form of f (t + 2) vs t is redrawn from Fig. 2.32 (a) in Fig. 2.32 (b).
f(t + 2)
2
1
–3
–2
–1
0
1
t
Fig. 2.32 (b)
EXAMPLE 11
Given f (t) vs t as in Fig. 2.31 (a), we have to draw f(1 – t) vs t.
Now proceed as follows. Draw f (T) vs T. Now to get f(1 – t) = f (T) vs t, put T = 1 – t,
hence t = – T + 1. We are changing the time scale to get the desired graph f (1 – t) vs t as in
Fig. 2.33 (a).
f(t), f(T)
f(1 – t)
2
1
–1
2
0
1
1
0
2
–1
3
–2
t, T
t = –T +1
Fig. 2.33 (a)
From the above Fig. 2.33 (a) which is absolutely correct but for the conventional form of
the desired graph f (1 – t) vs t is redrawn as shown in Fig. 2.33 (b) from the above figure.
f(1 – t)
2
1
–2
–1
0
1
Fig. 2.33 (b)
2
t
77
SIGNALS AND SYSTEMS
EXAMPLE 12
Given f (t) vs t as shown in Fig. 2.31(a) Draw the curve f (2t + 2) vs t.
Solution: Our task is to draw f (2t + 2) vs t. First draw f (t) vs t which is same as f (T) vs
T. Next draw f (2t + 2) = f (T ) vs t where t = (T – 2)/2 as in Fig. 2.34 (a).
2
1
–1
–3/2
0
–1
1
–1/2
2
0
3
1/2
t, T
t = [T – 2]/2
Fig. 2.34 (a)
The conventional form of the graph f (2t + 2) vs t is redrawn in Fig. 2.34 (b) from the
Fig. 2.34 (a).
f(2t + 2)
2
1
–3/2
–1
–1/2
0
1/2
t
–1
Fig. 2.34 (b)
EXAMPLE 13
A discrete signal x[n] is shown in Fig. 2.35 (a). Sketch and label the following signals:
(i) x[n – 2]
(iv) x[ 2n + 1]
(ii) x [4 – n]
(iii) x[2n]
(v) x[n] u [2 – n].
1
–2
x[n]
0
1
Fig. 2.35 (a)
3
n
Signals and Systems
f(t), f(T)
f(2t + 2)
78
NETWORKS AND SYSTEMS
Solution: (i) Given x[n] vs n in Fig. 2.35. We have to draw x [n – 2] vs n. First draw x[n]
vs n which is the same as x[k] vs k. Next draw x[n – 2] vs n where n – 2 = k or n = k + 2 as drawn
in Fig. 2.35 (b). The conventional form of the graph x[n – 2] vs n is redrawn in Fig. 2.35 (c).
x[n], x[k]
x[n – 2]
1
–2 –1 0
0 1 2
1
3
2
4
3
5
n, k
n=k+2
Fig. 2.35 (b)
Hence the desired waveform x[n – 2] vs n is shown in Fig. 2.35 (c).
x[n – 2]
1
0
1
2
3
4
5
n
Fig. 2.35 (c)
For other problems, try to solve by your own in the same line.
EXAMPLE 14
Consider the signals h[n – 1] and u[n + 3] as shown in Fig. 2.36 (a) and (b) respectively.
Sketch and label carefully each of the signals for ultimately achieving the signal h[n – 1]
{u[n + 3]– u [– n]}.
h[n – 1]
u[n + 3]
1
–3 –2 –1
0
1
2
3
4
5
–5 –4 –3 –2 –1
0
1
Fig. 2.36 (a)
2
3
4
n
n
Fig. 2.36 (b)
79
SIGNALS AND SYSTEMS
u[n + 3] – u[–n]
u[–n]
1
–5 –4 –3 –2 –1 0
–4 –3 –2 –1
0
1
2
3
1
4
Fig. 2.36 (c)
2
3
4
5
n
Fig. 2.36 (d )
1
2
3
4
Signals and Systems
h[n – 1]{u[n + 3] –u[– n]}
5
Fig. 2.36 (e)
Solution: From the given graph of u [n + 3] vs n, get the graph u [n] vs n and then
u[– n] vs n as in Fig. 2.36 (c). Get the graph {u[n + 3] – u[– n]} vs n as shown in Fig. 2.36 (d).
Now multiply h[n – 1] vs n and {u[n + 3] – u[– n]} vs n to get the desired graph h[n – 1]
{u [n + 3] – u[– n]} as shown in Fig. 2.36 (e).
EXAMPLE 15
Consider the signals h[n] and x[–n] as in Fig. 2.37 (a) and (b) respectively. Sketch and
label h[n]x [– n].
Solution: The solution is illustrated in Fig. 2.37 (c) which is self-explanatory.
2
h[n]
x[–n]
1
1/2
–4
0
4
–4
n
0
Fig. 2.37 (a)
1
Fig. 2.37 (b)
1/2
h[n] x[–n]
–4 –3 –2 –1
0
1
–1/2
–1
–1
–3/2
Fig. 2.37 (c)
n
n
80
NETWORKS AND SYSTEMS
EXAMPLE 16
A two-dimensional signal d(x, y) can often be usually visualised as a picture where the
brightness of the picture at any point is used to represent the value of d(x, y) at that point.
Figure 2.38 (a) has depicted a picture representing the signal d(x, y) which takes on the value
1 in the shaded portion of the (x, y)-plane and zero elsewhere.
(i) Sketch d(x + 1, y – 2) for given
RS
T
1,
d(x, y) = 0,
(ii) Sketch d(x/2, 2y) for given
for − 1 ≤ x ≤ 1 and − 1 ≤ y ≤ 1
otherwise
RS
T
1, for − 1 ≤ x ≤ 1 and − 1 ≤ y ≤ 1
d(x, y) = 0, otherwise
y
1
–1
x
1
–1
d(x, y) =
{ 1,0,
for –1 ≤ x ≤ 1 and – 1 ≤ y ≤ 1
otherwise
Fig. 2.38 (a)
Solution: (i) Given the signal d(x, y) as in Fig. 2.38 (a). We have to sketch the following
d(x + 1, y – 2) for given:
d(x, y) =
RS1,
T0,
for − 1 ≤ x ≤ 1 and − 1 ≤ y ≤ 1
otherwise
For new d(x + 1, y – 2) = d( , ∆) we have to find the corresponding shaded portion for
which d( , ∆) is 1 and zero otherwise. We should proceed as follows:
–1≤
≤1
which implies
–1≤x+1≤1
–1–1≤x+1–1≤1–1
–2≤x≤0
Similarly for
1≤∆≤1
which implies
81
SIGNALS AND SYSTEMS
–1≤y–2≤1
–1+2≤y–2+2≤1+2
1≤y≤3
Hence
d(x + 1, y – 2) =
RS1 ;
T0 ,
− 2 ≤ x ≤ 0 and 1 ≤ y ≤ 3
otherwise
Signals and Systems
The sketch is shown in Fig. 2.38 (b).
y
3
d(x + 1, y + 2)
1
–2
x
0
Fig. 2.38 ( b)
(ii)
Given
d(x/2, 2y) ≡ d(
d(x, y) =
RS1,
T0,
, ∆)
− 1 ≤ x ≤ 1 and − 1 ≤ y ≤ 1
otherwise
Now given
–1 ≤ x ≤ 1
Consider
–1 ≤
≤1
That is,
–1 ≤ x/2 ≤ 1
–2 ≤ x ≤ 2
and given
–1 ≤ y ≤ 1
Consider
1≤∆≤1
That is,
–1 ≤ 2y ≤ 1
–1/2 ≤ y ≤ 1/2
Hence
d(x/2, 2y) =
RS− 2 ≤ x ≤ 2 and
T0, otherwise
− 1/ 2 ≤ y ≤ 1/ 2
82
NETWORKS AND SYSTEMS
The sketch is shown in Fig. 2.39.
y
1/2
d(x/2, y/2)
–2
2
x
–1/2
Fig. 2.39
(iii) Sketch the following d(x – y, x + y) for given d(x, y)
=
RS1,
T0,
for − 1 ≤ x ≤ 1 and − 1 ≤ y ≤ 1
otherwise
Now given
–1 ≤ x ≤ 1
That is,
–1 ≤
That is,
–1 ≤ x – y ≤ 1
≤1
or,
–1 ≤ x – y ⇒ y = x + 1
and further, given
–1 ≤ y ≤ 1
that is,
–1 ≤ ∆ ≤ 1
–1 ≤ x + y ≤ 1
–1 ≤ x + y ⇒ y = – x – 1
and
x+y≤1 ⇒ y=–x+1
The four linear lines are drawn and the shaded portion that is, 1 for d(x – y, x + y) which
is shown in Fig. 2.40.
y
y = –x + 1
y=x+1
1
y=x–1
–1
1
x
d(x – y, x + y)
–1
y = –x – 1
Fig. 2.40