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4 ANALOG SYSTEM

By Betty Dunn,2014-05-27 15:04
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4 ANALOG SYSTEM

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4.1 Frequency Response

Having equipped ourselves with tools for representing and analyzing

signals in the time and frequency domain, we are now ready to see how these can be used to find and

study the outputs of common analog systems.

From determining the spectra of the input and output, the problems

below illustrate how the "frequency response" of an analog system is defined and how it can be obtained.

53

54

Analog Sysfem

System - Frequency Response

Relate

r(/)

to

x(7)for

the system below. Find the frequency response.

4,r

y(t)= *Q

ffil-',r

-t)

vf)=

-:)]: "'* rk( )l= n "' x(f) Fsiqffi'ffi: bl l--4-:1q&:1\$q:qc i'' x(r) l&ffi-&S--+ v(r) , t rl ='V),: Frequency response = e l3ffiY"q*:tfsl

xlf ) lgi-ry

r[y(r)]- e['(,

System - Delay Line Filter

Find the frequency response by relating

y(l)to x(r)and also by Fourier transforming

the block diagram.

yQ): t *Q)+t*Q -z)

r (7) = t rl*(t)l+ s E[:r(r - :)]

=7x(f)+te i='x("f) =Q +e"-'*\r(1)

Y(f)

r(fl

H(

=t

x(fl +s"-i" x(f)

=Q

+u t")x(1)

,tn=Y(l),=j+Be,3'

x\f)

n= Yg) =7 -8e 'Jx\f )

Analog System

55

System - Gapacitor

Find the frequency response of the circuit, taking the current as

input.

lF

3F

ifr)

+-

"(,

)

4f:t+

U : zrlffl

=

:r,rt"(z)l

=

ltatt(r)

H(f)=ffi=#=

lmpedance of capacitor

jgJ+

+_

'vlf)

System'RG Circuit

Find the frequency response of the circuit, taking

x(l)as input

A+ +l +

5f)

1

H(f\:Y(f\ = x(,fl

j'3

s+j l

1+

jal5

ro3

Analog System

4.2 Output from Frequency Response

The frequency response is a very useful characteristic of a system.

Being the ratio of the output to input spectra, it gives the amplitude

gain and phase difference of the output with respect to the input at

a particular frequency. Using this, the following two problems show

how the frequency response can be used to find the system output.

System - System Output

Find

y(r) given that x(t) = 1 a gn'n' .

5f) I

rAl+l+

"(r)

3F

+

y(r)

Ht!'!=.

Y(fl iu3 x(f) -L

ja3

I

x(t)=7 ag"is'

4

x(.r)=ru1i.za(r

t

])

t-

vU)=

n(r)x( t" | = H f tll 6v)-

sdf

/

+ n( Llrr{ r] )l - ",onu t r t

*

r_

I

s(t)a(r y\t) = a(o)t*

il= n(r,)a(f - /,)

8e'o' -I

I

o[]''lr"'"'

\.lE )

t-

Sets' iQno\rs r* { zo e \rs '\. 2r ) 7

3"i(s'-u" 'r::)

-

itss

Jr

- 6s

,

Analog Sysfem

57

System - Partial Fraction and Laplace Transform

Find the output of a system w iIhH(f | =

i

rgiven

that the input is

x(r) le -lo,

t>0

" r

y(,)

x(f\:1 ja '" 1+

I

r (7) = a (y)x

f

rl :

6;7ft;fA

I

(r

7rz)(: +

ja)

\)- jd+ p- j,t-

a

b

3o+b-iao+iab t

0ray: lr;

-

Ll

I .. a=-dnob=

t

' '0)=p lt- j,

tr rr j

w

rr r=-{' lle-,,r>0 lie,,,r>0 '' -'t3- jtt) 2|.0. r<0 2|.0. r<0

r

Ft I

lln''. t>01 l-_ 'Llo t

convenient to substitue j at with s in some analysis, resulting in

the

It is more

transform x(s)=

rt'(t)l= l*Q),

" at

rather than

x(f)= rk(r)l=

j"'at

J,Q)e

ln the last problem, the technique of partial fraction is used. This is a common method used to break up terms multiplied together into sum of terms for which the

inverse Fourier transform can be easily found from table-look-up. The problem also shows why we sometimes like to use Laplace transform instead of Fourier transform for this type of manipulation. Specifically, although both will lead to the same results, it is easier to write s instead of jat

.

4.3 Frequency Response ldentification

Since the frequency response is an important system characteristic, it is often necessary to "identify" or find it for, say, an unknown system that we are investigating.

' The following problem shows how this can be done by measuring the response or

output of a system subject to a known input.

58

Analog Sysfem

System - ldentification

The signal x(t)is applied to a system, giving the output below . Find its frequency response.

{,r:{"'1:: ].0. r<0 H

d x(ft=ii

I

-[[""', r>o.l l-<

Ilo. ,.0]

d'j(r)

wtt\t_- I .- I r|| 1+jdt 4+ja

,\ 'e'.t>0'e-a',t20 A L/ -i -y{i}-i [0. r<0 [0. r<0

11 H(f l=Y(!1,=l-i:Ltia ' x\f) I I+ ja

=t-1+ 4+ja- 4+jdt

ja

3

4.4 lmpulse Response

Another way to find the frequency response of an unknown system is to apply an impulse input and measure the response or output. As an example, to measure the acoustical reverberation characteristics of a room or concert hall, we can apply an impulse signal to a loudspeaker and measure the resulting sound at various strategic

locations. Since the spectrum of an impulse is 1, the frequency response is given simply by the spectrum of the "impulse response" for an impulse input. The following problem

depicts this relationship.

Analog Sysfem

59

System - lmpulse Response

The impulse response ft(l)of a system hQ)and

H(/)is

its output to an impulse input. How are

a(f)retated?

"f

'(')=a(')

H(.f):

'',''''=''''

)= {y(r)l= n[r(,)]

@l-'(')=111-*r\ --

r-) x(r)=r

H:rf

4.5 Gonvolution

The output of a system can be found by Fourier transforming the input, multiplying with the frequency response, and inverse transforming the result. lt is sometimes more

convenient if we can obtain the output from the input in the time domain. New insight into how the input and output are related can often be gained in the process. However, how can the output be found from the input in the time domain? The next problem shows how this is achieved through a process called "convolution". We will

see that the output is given by convolving the input with the impulse response, and multiplication in the frequency domain corresponds to convolution In the time domain.

The problem that follows shows another realization of the symmetrical properties of

the fonrrard and inverse Fourier transform equations. Specifically, convolution in the

frequency domain also corresponds to multiplication in the time domain.

60

Analog Sysfem

System - Time Gonvolution and Frequency Multiplication

The output y(r)=

r-' [r(11ot

a system can be found from

Y(f): x(f

)H(7).

tt can atso

be found from x(r)and nQ)

=r

'[A(1)]Oirectty. Derive this convolution operation.

/,)=F'[aU)x(t)] - f_n(1)x(;p'^a7

=

' x(f)

I

I

= F['(r)]

I

-

H

(.1''l

t-

= f-x(r)e

j"'

ctr

I

x(r)e "'d r le'' ctl

- J[ *{'fLJ,I

= =

u11

t" " e "'d/ ,tr

_

-,

Y

1

I

Exchange order of integration

f ^ I ^ J.'(.] ) H(rv' " "4r d,

h(t')

: F'ln (t\= f",a(flei'

aJ

f.*l'lnk-'la'

System - Frequency Convolution and Time Multiplication

Muliiplication in the frequency domain, v( f) = U(f)X(7) corresponds

to convolution in the time domain, y(t) = n(t). *(t). Show that the

reciprocal relationship is also valid.

y(,)= t (');..(,)

r(r) = rly(')l

= F[a(,)"(')l

= f_nQ)xQ)e i^ at

| - dr --:a =J hdLJ.rtY'le''dI'), t 'i ! = l. x( f 'Lf htrte -'e ,

"(r)=

F

'lx(f)l- f_x(.il",*

af '

Exchange order of integration

ar

)af

= =

[ xU'U"h(rte '- -'a,)ar'

f_x(r')s(t

-

H(f\

=e'[a(r)]

- f-n(t)" i'' at

ntf'

Analog Sysfern

61

4.6 Graphical Convolution

Convolution is often carried out graphically so that the

relationship between the input and output can be better seen.

The following problems give examples on how this is done through

a flip, right-shift, multiply and area finding process. Note that

getting

flipping:r(r) withrespecttotheverticalaxis,whilegetting

corresponds to shifting r) to the right by 0.5. "(System - Graphical

Convolution

Find the output y(r)

OV

"(-r)

trom

"(r)

corresponds to

"(0.:-r)

trom

x(-r)

* convolving ft(r)and x(r), or y(r) = f-n(r)*(t r)ar.

05 hhl

Area = y(o.s)=

t

ft

(')*(o.s -

'),tr

=

2,0.s

-t

Area = y(2) =

-

ln(r)'(z - r)ar = 2x2 :

)

4

hlr)x\2- r)

hG)

62

Analog Sysfem

System - Gonvolution of Square Pulses

Find the output y(r) by convolving /r(r)and :r(r) or y(,)

= f-n(')*(t - ,)a'.

02

02

"tr@

lltfu

nrea = y(o.s)=

f

a(')"(o.s - r)a' =1x 0.5 = 0.5

Area= y(z)=

f.n(r),(z-)ar =tx2

=2

,

ir(r) -*.'e"

-q2 \>

n(r)x(z- r)

System - Gonvolution with lmpulse

Find the output y(r) by convolving /r(r)and ;r(r) or y(r)

xhl= 5(r 'I

2\

f

l,

Area

I

/,(r)

f

= f-n(r)*(t .

')a'.

-/l

!

1

27 jr(-r) T tI

o4t

'L 1

r(:-r)=d(r-l) ? l.t(5

: ln(')d(' * t)ar - nO = o.zs

01

4

r)=a(r-:)

t , 3r

I

Area = y(s) = [-n(t)x(s

- r)a'

- ln(,)a('

-

z)a

r

: n(z) = o.t s

v(rl- hlt 2l

2

--*-j

61

Analog Sysfem

63

System - Gonvolution with lmpulse Train

Find the output y(r) by convolvi ng /z(r) and ;r (t), or y(t)

: f-n(r)r(t - r)a t

36t-l

, . .A',,^_ 1

hQ)

,

t

,(t)=ia(t-t,)

ll tt

@

v

I

-3

y(t)= x(t)* n(t): n(t).

0

3

6

7

ialr

-t,y= inpy. a\t -rnl= Lf_rA -":u(r

-3n)dr = L'

,t,-tA

d(r-3n)is

nonzero only at r

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