SS-Bab-01 polinema

8/11/2019 SS-Bab-01 polinema

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Signals and Systems with MATLAB Applications, Second Edition 1-1Orchard Publications

Chapter 1

Elementary Signals

his chapter begins with a discussion of elementary signals that may be applied to electric net-

works. The unit step, unit ramp, and delta functions are introduced. The sampling and sifting

properties of the delta function are defined and derived. Several examples for expressing a vari-

ety of waveforms in terms of these elementary signals are provided.

1.1 Signals Described in Math Form

Consider the network of Figure 1.1 where the switch is closed at time .

Figure 1.1. A switched network with open terminals.

We wish to describe in a math form for the time interval . To do this, it is conve-

nient to divide the time interval into two parts, , and .

For the time interval ,the switch is open and therefore, the output voltage is zero. In

other words,

(1.1)

For the time interval ,the switch is closed. Then, the input voltage appears at the output,

i.e.,

(1.2)

Combining (1.1) and (1.2) into a single relationship, we get

(1.3)

We can express (1.3) by the waveform shown in Figure 1.2.

t 0=

+

+

vou t

vSt 0=

R

open terminals

vou t t +<


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Chapter 1 Elementary Signals

1-2 Signals and Systems with MATLAB Applications, Second EditionOrchard Publications

Figure 1.2. Waveform for as defined in relation (1.3)

The waveform of Figure 1.2 is an example of a discontinuous function. A function is said to be dis-

continuousif it exhibits points of discontinuity, that is, the function jumps from one value to anotherwithout taking on any intermediate values.

1.2 The Unit Step Function

A well-known discontinuous function is the unit step function *that is defined as

(1.4)

It is also represented by the waveform of Figure 1.3.

Figure 1.3. Waveform for

In the waveform of Figure 1.3, the unit step function changes abruptly from to at .

But if it changes at instead, it is denoted as . Its waveform and definition are as

shown in Figure 1.4 and relation (1.5).


* In some books, the unit step function is denoted as , that is, without the subscript 0. In this text, however, we

will reserve the designation for any input when we discuss state variables in a later chapter.

0

vou tvS

t

vou t

u0 t( )

u0 t( )

u t( )

u t( )

u0 t( )0 t 0

=

u0 t( )

0

1

t

u0 t( )

u0 t( ) 0 1 t 0=

t t0= u0 t t0( )

1

t00

u0 t t0( )t

u0 t t0( )


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The Unit Step Function

(1.5)

If the unit step function changes abruptly from to at , it is denoted as . Its

waveform and definition are as shown in Figure 1.5 and relation (1.6).


(1.6)

Example 1.1

Consider the network of Figure 1.6, where the switch is closed at time .

Figure 1.6. Network for Example 1.1

Express the output voltage as a function of the unit step function, and sketch the appropriate

waveform.

Solution:

For this example, the output voltage for ,and for . Therefore,

(1.7)

and the waveform is shown in Figure 1.7.

u0 t t0( )0 t t0

=

0 1 t t 0= u0 t t0+( )

tt0 0

1u0 t t0+( )

u0 t t0+( )

u0 t t0+

( )

0 t t0

=

t T=

+

+

vou t

vSt T=

R

openterminals

vou t

vou t 0= t T< vou t vS= t T>

vou t vSu0 t T( )=


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Figure 1.7. Waveform for Example 1.1

Other forms of the unit step function are shown in Figure 1.8.

Figure 1.8. Other forms of the unit step function

Unit step functions can be used to represent other time-varying functions such as the rectangular

pulse shown in Figure 1.9.

Figure 1.9. A rectangular pulse expressed as the sum of two unit step functions

Thus, the pulse of Figure 1.9(a) is the sum of the unit step functions of Figures 1.9(b) and 1.9(c) is

represented as .

Tt

0

vSu0 t T( )vou t

0t

t

t t

0

00

0

0

0

t

tt

0 0t t

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

A A A

A A

A

A A AAu0 t( )

A u0 t( ) A u0 t T( )A u0 t T+( )

Au0 t T+( ) Au0 t T( )

A u0 t( )A u0 t T+( ) A u0 t T( )

0 0 0t t t1

1

1u0 t( )

u0 t 1( )a( ) b( )

c( )

u0 t( ) u0 t 1( )


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The unit step function offers a convenient method of describing the sudden application of a voltage

or current source. For example, a constant voltage source of applied at ,can be denoted

as . Likewise, a sinusoidal voltage source that is applied to a circuit at

, can be described as . Also, if the excitation in a circuit is a rect-

angular, or triangular, or sawtooth, or any other recurring pulse, it can be represented as a sum (dif-ference) of unit step functions.

Example 1.2

Express the square waveform of Figure 1.10 as a sum of unit step functions. The vertical dotted lines

indicate the discontinuities at and so on.

Figure 1.10. Square waveform for Example 1.2

Solution:

Line segment has height , starts at , and terminates at . Then, as in Example 1.1, this

segment is expressed as

(1.8)

Line segmenthas height ,starts at and terminates at . This segment is expressed

as

(1.9)

Line segmenthas height , starts at and terminates at . This segment is expressed as

(1.10)

Line segmenthas height ,starts at , and terminates at . It is expressed as

(1.11)

Thus, the square waveform of Figure 1.10 can be expressed as the summation of (1.8) through (1.11),

that is,

24V t 0=

24u0 t( )V v t( ) Vm tVcos=

t t0= v t( ) Vm tcos( )u0 t t0( )V=

T 2T 3T ,,

t

v t( )

3T

A

0

A

T 2T

A t 0= t T=

v1 t( ) A u0 t( ) u0 t T( )[ ]=

A t T= t 2T=

v2 t( ) A u0 t T( ) u0 t 2T( )[ ]=

A t 2T= t 3T=

v3 t( ) A u0 t 2T( ) u0 t 3T( )[ ]=

A t 3T= t 4T=

v4 t( ) A u0 t 3T( ) u0 t 4T( )[ ]=


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(1.12)

Combining like terms, we get(1.13)

Example 1.3

Express the symmetric rectangular pulse of Figure 1.11 as a sum of unit step functions.

Figure 1.11. Symmetric rectangular pulse for Example 1.3

Solution:

This pulse has height , starts at , and terminates at . Therefore, with reference to

Figures 1.5 and 1.8 (b), we get

(1.14)

Example 1.4

Express the symmetric triangular waveform of Figure 1.12 as a sum of unit step functions.

Figure 1.12. Symmetric triangular waveform for Example 1.4

Solution:

We first derive the equations for the linear segmentsandshown in Figure 1.13.

v t( ) v1 t( ) v2 t( ) v3 t( ) v4 t( )+ + +=

A u0 t( ) u0 t T( )[ ] A u0 t T( ) u0 t 2T( )[ ]=

+A u0 t 2T( ) u0 t 3T( )[ ] A u0 t 3T( ) u0 t 4T( )[ ]

v t( ) A u0 t( ) 2u0 t T( ) 2u0 t 2T( ) 2u0 t 3T( ) + +[ ]=

t

A

T 2 T 20

i t( )

A t T 2= t T 2=

i t( ) Au0 tT

2

---+ Au0 t

T

2

---

A u0 tT

2

---+ u0 t

T

2

---

= =

t

1

0T 2

v t( )

T 2


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Figure 1.13. Equations for the linear segments of Figure 1.12

For line segment,

(1.15)

and for line segment,

(1.16)

Combining (1.15) and (1.16), we get

(1.17)

Example 1.5

Express the waveform of Figure 1.14 as a sum of unit step functions.

Figure 1.14. Waveform for Example 1.5.

Solution:

As in the previous example, we first find the equations of the linear segmentsandshown in Fig-

ure 1.15.

t

1

0T 2

v t( )

T 2

2

T--- t 1+

2

T--- t 1+

v1 t( )2

T--- t 1+

u0 tT

2---+

u0 t( )=

v2 t( )2

T--- t 1+

u0 t( ) u0 tT

2---

=

v t( ) v1 t( ) v2 t( )+=

2

T--- t 1+

u0 tT

2---+

u0 t( )2

T--- t 1+

u0 t( ) u0 tT

2---

+=

1

2

3

1 2 30

t

v t( )


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Figure 1.15. Equations for the linear segments of Figure 1.14

Following the same procedure as in the previous examples, we get

Multiplying the values in parentheses by the values in the brackets, we get

or

and combining terms inside the brackets, we get

(1.18)

Two other functions of interest are the unit ramp function, and the unit impulse or delta function. We

will introduce them with the examples that follow.

Example 1.6

In the network of Figure 1.16 is a constant current source and the switch is closed at time .

Figure 1.16. Network for Example 1.6

1

2

3

1 2 30

2t 1+

v t( )

t

t 3+

v t( ) 2t 1+( ) u0 t( ) u0 t 1( )[ ] 3 u0 t 1( ) u0 t 2( )[ ]+=

+ t 3+( ) u0 t 2( ) u0 t 3( )[ ]

v t( ) 2t 1+( )u0 t( ) 2t 1+( )u0 t 1( ) 3u0 t 1( )+=

3u0 t 2( ) t 3+( )u0 t 2( ) t 3+( )u0 t 3( )+

v t( ) 2t 1+( )u0 t( ) 2t 1+( ) 3+[ ]u0 t 1( )+=

+ 3 t 3+( )+[ ]u0 t 2( ) t 3+( )u0 t 3( )

v t( ) 2t 1+( )u0 t( ) 2 t 1( )u0 t 1( ) t u0 t 2( ) t 3( )u0 t 3( )+=

iS t 0=

vC t( )

t 0=iS

R

C

+


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Express the capacitor voltage as a function of the unit step.

Solution:

The current through the capacitor is ,and the capacitor voltage is

* (1.19)

where is a dummy variable.

Since the switch closes at , we can express the current as

(1.20)

and assuming that for , we can write (1.19) as

(1.21)

or

(1.22)

Therefore, we see that when a capacitor is charged with a constant current, the voltage across it is a

linear function and forms a rampwith slope as shown in Figure 1.17.

Figure 1.17. Voltage across a capacitor when charged with a constant current source.

* Since the initial condition for the capacitor voltage was not specified, we express this integral with at the

lower limit of integration so that any non-zero value prior to would be included in the integration.

vC t( )

iC t( ) iS cons t tan= = vC t( )

vC t( )1

C---- iC ( ) d

t

=

t 0

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