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2.1 SIGNALS

We will be considering two basic types of signals: (i) Continuous-time signals and (ii) Discrete-time signals. In case of continuous-time signals the independent variable is continuous, andthus these signals are defined for continuum of values of the independent variable. On theother hand, discrete-time signals are defined only at discrete instants of times, andconsequently, 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 todenote continuous-time independent variable and n to denote discrete-time independentvariable. Similarly (.) is used to denote continuous while [.] is used to denote discrete valuedquantities.

A discrete-time signal is defined only at discrete instant of time, otherwise zero. Thusindependent variable has discrete values and let t = nT, n is integer (0, ± 1, ± 2, ...) and T is thesampling 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 inFig. 2.1(a) and (b) respectively.

Illustrations of continuous-time signals f (t) and discrete-time signals f [n] are beingmade throughout the chapter for better concept and understanding.

49

��������������� �

�

50 NETWORKS AND SYSTEMS

x(t)

0 t

(a)

n109876543210–1–6

–9 –8 –7 –5 –4 –3 –2

x[1]x[2]

x[–1]

x[0]

x[n]

(b)

Fig. 2.1. Graphical representations of (a) Continuous-time and (b) Discrete-time signals

Properties of Discrete Signals

Conjugate SymmetricFor 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 tobe even and imaginary part has to be an odd function.

The Continuous-Time Unit Step and Unit Impulse FunctionsContinuous-time unit step function u(t) shown in Fig. 2.2 is defined as

u(t) = 0 01 0,,

tt

<>

���

(2.1)

1

u(t)

0 t

Fig. 2.2. Continuous-time unit step function

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ms

The continuous-time unit step is the running integral of the unit impulse

u(t) = − ∞�t

dδ τ τ( ) (2.2)

The continuous-time unit impulse is the first derivative of the continuous-time unitstep as

δ(t) = du t

dt( )

(2.3)

In practical sense, the unit step u∆(t) shown in Fig. 2.3 rises from the value 0 to thevalue 1 in a short-time interval of length ∆. The unit step can be thought of as an idealisation

of u∆(t) for Lt∆ → 0

∆. Formally, u(t) is the limit of u∆(t) as ∆ → 0 and the derivative becomes the

impulse in practical sense as shown in Fig. 2.4.

1

u (t)�

0 t�

1�

� (t)�

0 t�

Fig. 2.3 Fig. 2.4

The Unit Impulse and Unit Step FunctionsWe will see how we can use unit impulse signal as basic building blocks for the constructionand 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] = 0 01 0,,

nn

≠=

���

(2.4)

is as shown in Fig. 2.5

0

1

n

�[n]

Fig. 2.5. Discrete-time unit impulse (sample)

The discrete-time unit step, is defined as

u[n] = 0 01 0,,

nn

<≥

���

(2.5)

The unit step sequence is shown in Fig. 2.6

0

1

n

u[n]

Fig. 2.6. Discrete-time unit step sequence

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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

u[n] = m

n

m= − ∞∑ δ[ ] (2.7)

Equation (2.7) is illustrated graphically in Fig. 2.6. Since the only non-zero value of theunit sample is at the point at which its argument is zero, we see from the figure that therunning sum in Eqn. (2.7) is 0 for n < 0 and 1 for n ≥ 0. Furthermore, by changing the variableof 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 discrete-time unit step can also be written in terms of the unit sample as

u[n] = k

n k= ∞∑ −

0

δ[ ] (2.8)

or equivalently, u[n] = k

n k=

∞

∑ −0

δ[ ] (2.9)

Even and Odd SignalsA continuous signal f (t) is referred to as an even signal if it is identical to its time-reversedcounterpart, 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) = 0at t = 0. Examples of even and odd continuous-timesignals are shown in Fig. 2.7.

Decomposition of continuous signal f(t) can bedone as:

f (t) = fe(t) + fo(t) (2.12)

where, fe(t) is the even and fo(t) is the odd component ofcontinuous signal f(t). Obviously, the even function hasthe property

fe(–t) = fe(t) (2.13)

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) (2.15)

Solving from the expression of f (t) and f (– t), weget

fe(t) = 12

[f (t) + f (–t)] (2.16)

Fig. 2.7. (a) An even continuous-time signal;(b) An odd continuous-time signal

0

x(t)

t

( )a

x(t)

t

( )b

0

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and fo(t) = 12

[f (t) – f (– t)] (2.17)

The discrete signal f [n] is said to be even if

f [n] = f [– n]; for all n (2.18)

and signal f [n] is said to be odd if

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 caseas 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

fe[n] = 12

[ [ ] [ ]f n f n+ − (2.20)

and fo[n] = 12

[ [ ] [ ]f n f n− − (2.21)

Examples of even and odd discrete-time signals are shown in Fig. 2.8.

1, n > 00, n < 0

0

1

n321–1–2–3

x[n] =

0

1

n321–1–2–3

ev{x[n]} =

12

12 , n < 0

1, n = 012 , n > 0

0 n321

–1–2–3

od{x[n]} =

12

12 , n < 0

0, n = 012 , n > 0

12–

–

Fig. 2.8. Example of the even-odd decomposition of a discrete-time signal

54 NETWORKS AND SYSTEMS

Periodic SignalA 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 T0is 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 integerN, 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 unchangedby a time shift of N. Further, x[n] is periodic with period 2N, 3N, ..., that is; for any integervalue 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 inFig. 2.10.

–2T –T 0 T 2T

x(t)

Fig. 2.9. A continuous-time periodic signal

n

x[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 = 2

0

πT

, where f0 is in cycles/second or hertz(Hz). The

sinusoidal signal is periodic with fundamental period T0 = 1/f0.

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t

x(t) = A cos ( t + )� �0

A

A cos �

T =0T =02��0

2��0

0

Fig. 2.11. Continuous-time sinusoidal signal

Time Scaling: If f(t) is a continuous-time signal, the y(t) is obtained by scaling theindependent 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. Supposetape recorder is having normal speed x(t) for which sound is clear. Now if the speed is doubledthat is, x(2t), as in Fig. 2.12, the tape recorder is moving very fast, that is, sound is notdistinguishable because, in the same space of time double the words are played, hencecompressed. On the contrary, if the speed is half, that is, the signal x(t/2) as in Fig. 2.13, therecorder 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)

�

–2–2–1

000

221

tx(t) vs t

x( ) vs , ., x(2t) vs 2t� � i.e

x(2t) vs t

2t = �t = �/2

( )a

x(2t)

–1 0 1 tx(2t) vs t in

conventional timeformat

( )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) = cosω(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)

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 ofargument, the signal x3(t) = x(βt) is rarefied compared to signal x1(t).

(a)

–2–2–4

000

224

tx(t) vs t

x( ) vs , ., x(t/2) vs t/2� � i.e

x(t/2) vs t

t/2 = �2 = t�

x(t)x(t/2) = x( )�x(t/2)

(b)

–4 0 4 tx(t/2) vs t

in conventionaltime format

x(t/2)

Fig. 2.13. (a) Continuous-time signals (b) Rarefied signal related by time scaling with a < 1

x (t) = cos t1 1�

T1t

( )a

0

T2

x (t) = x( t) = cos t, > 1 = cos t2 2� �� � �

x (t) = x( t) = cos t = cos t3 � �� �

t

( )b

0

x (t) = x( t) = cos t, < 1 = cos t3 � �� � �3

T3t

( )c

0

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

SIGNALS AND SYSTEMS 57

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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 inFig. 2.15 (b) and the rarefied version x(t/2) vs t is shown in Fig. 2.15(c).

x(t)

2

2–2 t0

( )a

x(2t)

2

1–1 t

( )b

0

x(t/2)

2

4–4 t0

( )c

Fig. 2.15

Reflection: For a continuous time signal f(t), its reflection about vertical axis becomesy(t) = f(–t) (2.28)

A continuous-time signal f (t) is shown in Fig. 2.16 (a) and its reflection f (– t) aboutt = 0 is shown in Fig. 2.15 (b). The procedure is like this: by just changing the abscissa withproper manipulation as is evident in Fig. 2.16 (a); we obtain the desired curve. But for theconventional look the curve is as shown in Fig. 2.16 (b).

f(t)f(–t) = f( )f(–t)

�

–3–3

3

–2–2

2

–1–1

1

000

1+1–1

2+2–2

3+3–3

t

–t =t = –

��

f(t) vs tf(–t) = f( ) vsf(–t) vs t

� �

Fig. 2.16 (a)

58 NETWORKS AND SYSTEMS

f(–t)

–4 –3 –2 –1 0 1 2 3 tf(–t) vs t

Fig. 2.16 (b)

Similarly, for a discrete-time signal f [n] its reflection about vertical axis becomesy [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 = 0is shown in Fig. 2.16 (d). As in the case of continuous case, the abscissa is only changed at eachsequence and finally for the conventional look, the final curve of f [– n] vs n is redrawn asshown in Fig. 2.16 (d).

f[n]f[–n] = f[k]f[–n]

–8–88

000

88

–8

n

(–n) = kn = –k

f[n] vs nf[–n] = f[k] vs kf[–n] vs k

Fig. 2.16 (c)

f[–n]

–8 0 8 nf[–n] vs n

Fig. 2.16 (d )

SIGNALS AND SYSTEMS 59

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Time ShiftingIf 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 theright.

If t0 < 0, then shifting is in the left in time scale, that is, the waveform is shifted to theleft.

For discrete time signal f [n], its time shifted version y[n] can be written asy [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 timeaxis if n0 < 0.

Continuous-time signals related by a time shift is shown in the Fig. 2.17. In this figuret0 < 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 manipulationin Fig. 2.17(a) in abscissa only and ultimately the conventional version is shown in Fig. 2.17(b).

f(t)f(t – t ) = x( )f(t – t )

0

0

�

0 tt – t =t = + t

�0� 0

t < 0 = –50

f(t) vs tf(t – t ) = x( ) vsf(t – t ) vs t

0

0

�� 0t = –50

(a)

f(t – t )0

t = –50 t

t < 0 = –50

0f(t – t ) vs t0

(b )

Fig. 2.17

A time-shift in discrete time signal is illustrated in Fig. 2.18 in which we have twosignals 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 drawf [n – n0] vs n for n0 is having positive integer. The procedure is explained as below. Drawf [n – n0] = f [k] vs [n – n0] = k. Then in the same figure draw f [n – n0] vs n = k + n0 by onlyalgebraic manipulation in the time scale in abscissa as shown in Fig. 2.18 (a) and redrawnf [n – n0] vs n in Fig. 2.18 (b) for the conventional look which is nothing but the delayed versionof f [n] by n0 > 0.

60 NETWORKS AND SYSTEMS

f[n]f[(n – n )] = x[k]0

f[(n – n )]0

00n0

n0n0

n[n – n ] = kn = [k + n ]

0

0

f[n] vs nf[n – n ] = x [k] vs kf[n – n ] vs n

0

0

(a)

f[n – n ]0

0

n0

nf[n – n ] vs n0

(b)

Fig. 2.18. Discrete-time signals related by time-shift. For n0 > 0, [n – n0] 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 througha sequence of diagrammatic approach.

f(t)

1

0 t

f(–t)

1

0 t

(a) (b)

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ms

0 t

1/2

[f(t) + f(–t)]2

f (t) =e[f(t) – f(–t)]

2

0 t

1/2

–1/2

f (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 systematicsequence of diagrammatic approach.

f(t)

1

1/2

–11

t0

–1/2

f(–t)

1

1/2

–1 1 t0

–1/2

(a) (b)

[f(t) + f(–t)]

1/2

–1 1 t0

f (t) =e 2

f (t) =0

–1

1t0

1/2

–1/2

[f(t) – f(–t)]

2

(c) (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 )

62 NETWORKS AND SYSTEMS

EXAMPLE 3

Decompose the given square pulse f (t) into even and odd components.

f(t)

1

0 1 t

f(–t)

1

0 t–1

(a) (b)

f (t)e

1/2

0 1 t–1

f (t)0

1/2

0 1 t–1

–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 alternativerepresentations that are often more convenient to use. These are shown in the followingillustrations

1. Functional representation

f [n] = 1 0 1 2 30

forotherwise

n =���

, , , , ... as shown in Fig. 2.22 (a)

2. Representation based on length the discrete-time signal may be a finite length or finitelength sequences. The finite length also called finite duration sequence is only definedonly 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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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)

↑

0 1 2

1

n

1, n 00, n < 0

�

3–3 –2 –1

{f[n] =

( )a

0 1 2

2

n3–3 –2 –1

( )b

1

0 1 2 n3–3 –2 –1

( )c

– 1

1

4 5 6 87

1

2

3

( )d

1 2 n0

Fig. 2.22

EXAMPLE 4

Consider the discrete-time signal f [n] depicted in Fig. 2.23 (a). Further, we have depictedfive time-shifted, scaled unit impulse sequences in Figs. 2.23 (b) to ( f ) respectively which areexpressed mathematically as

f [–2] δ[n + 2] = f n

n[ ] ,,− = −

≠ −���

2 20 2

f [–1] δ[n + 1] = f n

n[ ] ,,− = −

≠ −���

1 10 1

f [0] δ[n] = f n

n[0] ,,

=≠

���

00 0

f [1] δ[n – 1] = f n

n[ ] ,,1 1

0 1=≠

���

f [2] δ[n – 2] = f n

n[ ] ,,2 2

0 2=≠

���

Therefore, the five sequences in the figure equals x[n] for – 2 ≤ n ≤ 2 and can be writtenas

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]

64 NETWORKS AND SYSTEMS

= k = −∑

2

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 inFig. 2.23 (a). Hence we get all components of x[n] as shown through Fig. 2.23 (b) to ( f )

–4–3 –2

–10 1

2 34 n

f[n]

–4 –3 –2 –1 0 1 2 3 4 n

f[–2] [n + 2]�

(a) (b)

–4 –3 –2

–1

0 1 2 3 4 n

f[–1] [n + 1]�

–4 –3 –2 –1 0 1 2 3 4 n

f[0] [n]�

(c) (d )

–4 –3 –2 –1 0 1 2 3 4 n

f[1] [n – 1]�

–4 –3 –2 –1 0 1

2

3 4 n

f[2] [n – 2]�

(e) (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 asshown 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. Bysumming for each value of n, we obtain output y[n] which is shown in Fig. 2.24 (c).

SIGNALS AND SYSTEMS 65

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0 1 2

1

n

h[n]

0 1

0.5

n

f[n]2

( )a

0 1 2

0.5

n

0.5h(n)

0 1 2

2

n

2h(n – 1)

3

(b) (c)

0 1 2

2.5

n

y(n)

3

2

0.5

(d )

Fig. 2.24

Impulse ResponseThe impulse response is defined as the system output due to an impulse sequence δ[n] at thesystem input, where

δ[n] = 1 00 0

,,nn

=≠

���

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 bylinearity, 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 alsoshift 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 impulseresponse 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] + ...

66 NETWORKS AND SYSTEMS

In other words, we take each point of the sequence f [ j] and multiply it by a shiftedversion of the impulse sequence δ [n – j]. Because shifted unit impulse δ [n – j] has the valueunity for n = j and zero otherwise. Thus

f [n] = j = − ∞

∞

∑ f [ j] δ [n – 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 mathematicallyand graphically, given the impulse input δ[n] which convolves with another function f [n] atdifferent discrete instant of time n for – 2 < n < 2 and can be summed up to get the totalresponse x[n].

On the same line of thought, the convolution integral can be applied to find the outputresponse y[n] of any system having the impulse response h[n] subjected to any input signalx[n] can be written as

y[n1] = k

x k h n k=− ∞

∞

∑ −[ ] [ ]1

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] = k

x k h n k= − ∞

∞

∑ −[ ] [ ]

for –∞ < k < ∞.

The output at n = 0 as y[0] = k

x k h k=− ∞

∞

∑ −[ ] [ ] = 6; see Fig. 2.25 (d)

SIGNALS AND SYSTEMS 67

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nal

s an

d S

yste

ms

The output at n = 1 as y[1] = k

x k h k=− ∞

∞

∑ −[ ] [ ]1 = 5 + 6 = 11; see Fig. 2.25 (e)

In a similar line of action we can get y[2] = 4 + 5 + 6 = 15; see Fig. (f ) and so on. Theoutput at different instants is tabled below and shown in Fig. (d) through (m).

[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 andwritten as

y[n] = [...0, 6, 11, 15, 18, 14, 10, 6, 3, 1, 0...]

k0 1 2 3 4 5 k

1

x[k]

12

3

4

5

6

–5–4–3 –2 –1 0 k–6

12

3

4

5

6h[k]

0 1 2 3 4 5 6

h[–k]

(a) (b) (c)

y(0) = = x[k] h[–k] = 6Σ∞

k = – ∞

–5 –4–3 –2 –1 0 k0 1 2 3 4 5 k –2 –1 0 1 2 3 4 5 k

1

6Linearsystem

x[k] y[k]

x[k]

–6

12

3

4

56

y [k] = x[k] h[–k]0h[–k]

(d )

y(1) = = x[k] h[k] = 11Σ∞

k = – ∞

–4 –3 –2 –1 0 1 k0 1 2 3 4 5 k –2 –1 0 1 2 3 4 5 k

1

h[–(k–1)] = h[1–k]

5

6

y [k] = x[k] h[1– k]1

Linearsystem

x[k] y[k]

x[k]

(e)

68 NETWORKS AND SYSTEMS

0 1 2 3 4 5 k –3 –2 –1 0 1 2 3 4 5 k –2 –1 0 1 2 3 4 5 k

1

h[2–k]

45

6 y[2] = x[k] h[2–k]Σy [k] = x[k] h[2–k]2

Linearsystem

x[k] y[k]

x[k]

(f )

–2 –1 0 1 2 3 4 5 k0 1 2 3 4 5 k

1

h[3–k]

y[3] = x[k]h[3–k] = 18

Σ

y [k] = x[k] h[3–k]3

Linearsystem

x[k] y[k]

x[k]

12

4

5

6

3

–2 –1 0 1 2 3 4 5 k

4

5

6

3

(g)

–1 0 1 2 3 4 5 k0 1 2 3 4 5 k

1

h[4–k]y[4] = x[k]

h[4–k] = 14Σ

y [k] = x[k] h[4–k]4

Linearsystem

x[k] y[k]

x[k]

–2 0 1 2 3 4 5 6 k

4

5

3

2

(h)

0 1 2 3 4 5 k0 1 2 3 4 5 k

1

h[5–k]

y[5] = x[k] h[5–k] = 10Σ

y [k] = x[k] h[5–k]5

Linearsystem

x[k] y[k]

x[k]

0 1 2 3 4 5 6 k

432

12

4

5

6

3

1

(i )

SIGNALS AND SYSTEMS 69

Sig

nal

s an

d S

yste

ms

0 1 2 3 4 5 k0 1 2 3 4 5 k

1

h[6–k]

y[6] = x[k] h[6–k] = 6Σ

y [k] = x[k] h[6–k]6

Linearsystem

x[k] y[k]

x[k]

0 1 2 3 4 5 6 k

32

12

4

5

6

3

1

6

( j )

0 1 2 3 4 5 k0 1 2 3 4 5 k

1

h[7–k]

y[7] = x[k] h[7–k] = 3�

y [k] = x[k] h[7–k]7

Linearsystem

x[k] y[k]

x[k]

–1 0 1 2 3 4 5 k

21

6 7

(k )

0 1 2 3 4 5 k

1

y[8] = x[k] h[8–k] = 1�

y [k] = x[k] h[8–k]8

Linearsystem

x[k] y[k]

x[k]

–1 0 1 2 3 4 5 k1 2 3 4 5 6 7 8 k

h[8–k]

0

1

(l )

0 1 2 3 4 5 k

1

y [k] = x[k] h[9–k]= y[9] = 0

9

Linearsystem

x[k] y[k]

x[k]

–1 0 1 2 3 4 5 k1 2 3 4 5 6 7 8 9 k

h[9–k]

0

(m)

Fig. 2.25

70 NETWORKS AND SYSTEMS

EXAMPLE 7

Let us consider an example where we wish to convolve h and x, where

h[n] = 12

0

0 0

����

≥

<

���

��

k

k

k

,

,

and x[n] = {3, 2, 1}

Solution: We get h[n] = 12

0

0 0

����

≥

<

���

��

k

k

k

,

, = ..., , , , , , ...0 1

12

14

18

���

�

We get h[k – n], and x[n] = {3, 2, 1} ; for different values of k, convolve graphically as inprevious example 6, h[k – n] and x[n] = {3, 2, 1} from the following expression:

y[k] = k

x n h k n= −∞

∞

∑ −[ ] [ ]

The result obtained as

y[k] = 372

114

118

1116

112

, , , , , ..., , ...k���

� �

There is another alogrithm that we can use to evaluate discrete convolutions for thesame example we finished just now. Suppose that we wish to convolve h and x, where

h[n] = 12

0

0 0

����

≥

<

���

��

k

k

k

,

,

and x[n] = {3, 2, 1}

Construct a matrix with h bordering the top ofthe matrix and x the left side of the matrix, as shownin Fig. 2.26. The entries in the matrix are the productsof the corresponding row and column headers. To findthe convolution of the two sequences, we need only ‘‘foldand add’’ according to the dotted diagonal lines. The

first term, y[0] = 3. The second term y[1] = 2 + 32

= 72

which is the sum of the terms contained between thefirst and second diagonal lines. We can continue in thisway and obtain the output sequence as

1 1/2 1/4 1/8

3 3/2 3/4 3/83

2 1 1/2 1/42

1 1/2 1/4 1/81

x

h

Fig. 2.26. Matrix representation ofconvolution summation

SIGNALS AND SYSTEMS 71

Sig

nal

s an

d S

yste

ms

y[k] = 372

114

118

1116

112

, , , , , ..., , ...k���

� �

This algorithm is not always a satisfactory method, because the result is not easilyplaced in closed form. Only in simple cases, one can determine the closed form solution of y[k].

New Approach for Solving ProblemsA continuous time signal f (t) is shown in Fig. 2.27 (a). Example of a transformation of theindependent variable of continuous signal f (t) is shown as f (t – t0). The signals f(t) andf(t – t0) are identical in shape but that are displaced or shifted relative to each other. Thesignal f(t – t0) means the signal f(t) is shifted right by t0, or moved forward by t0 and obviouslythe signal f(t + t0) means the signal f (t) is shifted left by t0 or moved backward by t0. Fewexamples are handled with for better understanding.

EXAMPLE 8

Given signal f (t) vs t, draw

(i) f (t + 1) vs t (ii) f (– t + 1) vs t

(iii) f32

t���� vs t (iv) f

32

1t +���

� vs t

Solution: (i) Given the signal f (t) vs t in Fig. 2.27 (a).

f(t)

1

0 21 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 asdepicted in Fig. 2.27 (b) in the following:

f(t + 1)f(T)f(t)

1

00

–1

221

110

tT

(t + 1 = T); T – 1 = t

Fig. 2.27(b)

72 NETWORKS AND SYSTEMS

Now f (t + 1) vs t with conventional time scale with formal look, looks like what is shownin Fig. 2.27 (c).

f(t + 1)

–1 0 t1 2

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) whichcorresponds to an advance (shift to the left) of the given signal f (t) by one unit along t-axis. Itmeans 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 throughthe following sequence of operations as shown in Fig. 2.27 (d).

0001

22

–2–1

11

–10

t

f(t), f(T), f(–T), f(–t + 1)

T–T

(– 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 inFig. 2.27 (e).

–1 10 t

f(–t + 1)

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 thecurve f [(3/2)t] vs t is obtained through the following sequence of operations as depicted inFig. 2.28 (a):

SIGNALS AND SYSTEMS 73

Sig

nal

s an

d S

yste

ms

000

22

4/3

11

2/3

tT

f(t), f(T), f[(3/2)t]

[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)

0 4/32/3 t

f[(3/2)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) :

1

f(t), f(T), f[(3/2)t + 1]

0 21 t

0

–2/3

2

2/3; [T=(3/2)t+1]; [(2T–2)/3] = t

1

0

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]

–2/3 2/30 t

1

Fig. 2.29 (b)

74 NETWORKS AND SYSTEMS

It may be noted that Fig. 2.29 (b) is the signal f [(3/2)t + 1] obtained by time shiftingand 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 Twhich is same as f (t) vs t. Next draw f (2 – t/3) vs t where (2 – t/3) = T from which we gett = (6 – 3T ) as shown in Fig. 2.30 (a).

f(t), f(T)

2

1

–1 0 1 2 39 6 3 0 –3

t, T= T; t = 6 – 3 T6 – t

3

f 2 – t3

Fig. 2.30 (a)

The more conventional form of the graph f (2 – t/3) vs t is redrawn in conventional timeformat 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

–1

t

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 toget the curve of f (t – 2) vs t.

SIGNALS AND SYSTEMS 75

Sig

nal

s an

d S

yste

ms

f(t), f(T)f(t – 2)

2

1

–1 0 1 2 31 2 3 4 5

t, Tt = T + 2

–20

Fig. 2.31(a)

The common form of the curve f (t – 2) vs t is drawn in the conventional way asadvancement of f (t) by 2 units in right-hand side, otherwise, the proposed method is alright toget the same result by manipulating the time axis. The conventional form of f (t – 2) is redrawnin Fig. 2.31(b).

f(t – 2)

2

1

t

0

1 2 3 4 5

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 0 1 2 3–3 –2 –1 0 1

t, Tt = T – 2

Fig. 2.32 (a)

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).

2

1

–3 –2 –1 0 1 t

f(t + 2)

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 inFig. 2.33 (a).

f(t), f(T)f(1 – t)

2

1

–1 0 1 2 32 1 0 –1 –2

t, Tt = –T +1

Fig. 2.33 (a)

From the above Fig. 2.33 (a) which is absolutely correct but for the conventional form ofthe desired graph f (1 – t) vs t is redrawn as shown in Fig. 2.33 (b) from the above figure.

2

1

–2 0 1 2 t–1

f(1 – t)

Fig. 2.33 (b)

SIGNALS AND SYSTEMS 77

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nal

s an

d S

yste

ms

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) vsT. Next draw f (2t + 2) = f (T ) vs t where t = (T – 2)/2 as in Fig. 2.34 (a).

f(t), f(T)f(2t + 2)

2

1

–1 0 1 2 3–3/2 –1 –1/2 0 1/2

t, Tt = [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 theFig. 2.34 (a).

2

1

–3/2 –1 –1/2 0 1/2 t

f(2t + 2)

–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] (ii) x [4 – n] (iii) x [2n]

(iv) x[ 2n + 1] (v) x[n] u [2 – n].

1x[n]

–2 0 1 3 n

Fig. 2.35 (a)

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 drawnin Fig. 2.35 (b). The conventional form of the graph x[n – 2] vs n is redrawn in Fig. 2.35 (c).

1

x[n], x[k]x[n – 2]

–2 0–1 3 n, k1 20 21 53 4 n = k + 2

Fig. 2.35 (b)

Hence the desired waveform x[n – 2] vs n is shown in Fig. 2.35 (c).

1

x[n – 2]

n0 21 53 4

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]}.

–3 –2 –1 0

1 2 3 4 5

h[n – 1]

n–5 –4 –3 –1 0 1 2 3 4

u[n + 3]

–2

1

n

Fig. 2.36 (a) Fig. 2.36 (b)

SIGNALS AND SYSTEMS 79

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nal

s an

d S

yste

ms

–4 –3 –1 0 1 2 3 4

u[–n]

–2

–4 –3 –2 0

1 2 3 4 5

u[n + 3] – u[–n]

–1

1

n–5

Fig. 2.36 (c) Fig. 2.36 (d )

1 2 3 4 5

h[n – 1]{u[n + 3] –u[– n]}

Fig. 2.36 (e)

Solution: From the given graph of u [n + 3] vs n, get the graph u [n] vs n and thenu[– 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 andlabel h[n]x [– n].

Solution: The solution is illustrated in Fig. 2.37 (c) which is self-explanatory.

–4

0 4

h[n]2

n–4 0 1

x[–n]

1/21

n

Fig. 2.37 (a) Fig. 2.37 (b)

0 1

h[n] x[–n]

–1

–3/2

–1–1/2

–4 –3 –2 –1

1/2

n

Fig. 2.37 (c)

80 NETWORKS AND SYSTEMS

EXAMPLE 16

A two-dimensional signal d(x, y) can often be usually visualised as a picture where thebrightness 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 value1 in the shaded portion of the (x, y)-plane and zero elsewhere.

(i) Sketch d(x + 1, y – 2) for given

d(x, y) = 1 1 1 1 10,,

for andotherwise

− ≤ ≤ − ≤ ≤���

x y

(ii) Sketch d(x/2, 2y) for given

d(x, y) = 1 1 1 1 10,,

for andotherwise

− ≤ ≤ − ≤ ≤���

x y

1

1

–1

–1x

y

d(x, y) = { 1, for –1 x 1 and – 1 y 10, otherwise

≤ ≤ ≤ ≤

Fig. 2.38 (a)

Solution: (i) Given the signal d(x, y) as in Fig. 2.38 (a). We have to sketch the followingd(x + 1, y – 2) for given:

d(x, y) = 1 1 1 1 10

,,

for andotherwise

− ≤ ≤ − ≤ ≤���

x y

For new d(x + 1, y – 2) = d( , ∆) we have to find the corresponding shaded portion forwhich 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 ≤ 0Similarly for

1 ≤ ∆ ≤ 1

which implies

SIGNALS AND SYSTEMS 81

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nal

s an

d S

yste

ms

– 1 ≤ y – 2 ≤ 1

– 1 + 2 ≤ y – 2 + 2 ≤ 1 + 2

1 ≤ y ≤ 3

Hence d(x + 1, y – 2) = 1 2 0 1 30

;,

− ≤ ≤ ≤ ≤���

x yandotherwise

The sketch is shown in Fig. 2.38 (b).

x

y

3

1

–2 0

d(x + 1, y + 2)

Fig. 2.38 (b)

(ii) d(x/2, 2y) ≡ d( , ∆)

Given d(x, y) = 1 1 1 1 10

,,

− ≤ ≤ − ≤ ≤���

x yandotherwise

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) = − ≤ ≤ − ≤ ≤���

2 2 1 2 1 20

x yand /otherwise

/,

82 NETWORKS AND SYSTEMS

The sketch is shown in Fig. 2.39.

x–2

1/2

y

–1/2

d(x/2, y/2)

2

Fig. 2.39

(iii) Sketch the following d(x – y, x + y) for given d(x, y)

= 1 1 1 1 10,,

for andotherwise

− ≤ ≤ − ≤ ≤���

x y

Now given –1 ≤ x ≤ 1

That is, –1 ≤ ≤ 1

That is, –1 ≤ x – y ≤ 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) whichis shown in Fig. 2.40.

y

y = –x + 1y = x + 1

1

1

–1

–1

y = x – 1

x

y = –x – 1

d(x – y, x + y)

Fig. 2.40