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4 ANALOG SYSTEM

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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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