The Inverting Integrator

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The Inverting Integrator The circuit shown below is the inverting integrator. &  i 2 "s # '  v in "s #

v * 

-

ideal 

i 1 "s # v + 

oc  vout  "s #

+

Since the circuit uses the inverting conf config igur urat atio ion n we can can conclude that the circuit transfer function is! oc  vout  "s # $ "s # = vin "s # % 2 "s # = %1 "s # ( 1 s & ) = '  =

1

-

s '& 

(n other words the out)ut signal is related to the in)ut as!

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1 vin "s # '&  s -

oc  out "

s # =

,ro- our nowledge of Laplace Transforms  we now this -eans that the out)ut signal is )ro)ortional to the integral of the in)ut signal Taing the inverse a)lace Transfor- we find! oc  vout "t #

1

=

t 

vin "t  � # dt � '&  -

 

0

,or ea-)le if the input is! # = sin vin "tt

then the output is! v

1

-

t 

tt # = dt   � sin '&  0

oc  " out 

=

=

1

-

1

-

'&  1

 '& 

cost 

cost 

e liewise could have deter-ined this result using Fourier Analysis "i.e. freuenc5 do-ain#!

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oc  v  " # out  $  " #= v  " # in  %  " # 2 = %  " # 1 = -

=

( 1  6 &  ) ' 

 6   '& 

Thus the magnitude of the transfer function is! " # $ 

=

=

 6   '&  1

 '& 

nd since!  6

 6  e ( 2)  

=

=

cos

( 2) 



6  sin (   2 )

the phase of the transfer function is! �$  " # = 9  radians 2 o = 80

$iven that!

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oc  " #$=  " v # in  "   # v out

and! oc  �v " # =$� " # v �in "    # out

we find for the input! # = sin vin "tt  where! " # v in

=

1

v� in" #   = 0

and

that the output of the inverting integrator is! oc  " #$=  " v# in "    # v out =

1

 '& 

and! oc  �v " # =$� " # v �in "    # out =

80o  0

=

80o

Therefore! 1

oc  "tt # =  sin ( vout   '&  =

1

 '& 

cost 

:actl5 the same result as before



80o )

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(f 5ou are still unconvinced that this circuit is an integrator consider this time-domain anal5sis. i 2 "t #

& 

+ v c  * ' 

v * 

v in "t # i 1 "t #

i 

-

-

0

=

v + 

ideal 

oc 

vout "t #

+

,ro- our ele-entar5 circuits knowledge  we now that the voltage across a ca)acitor is! vc "t # =

t 

1

# dt i "t � � &  2

 

0

and fro- the circuit we see that! oc oc   vc "t # = v "t # - vout "t # = -vout "t  #  -

therefore the output voltage is! oc  vout  "t #

1

=

-

t 

i "t � # dt � &  2

0

 

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,ro- <& we liewise now that! i1"t # = i2"t #  and fro- =h->s aw! i1"t #

vin "t # v "t # '1 -

=

-

=

vin "t  #  ' 1

Therefore! vin "t # i2 "t # = ' 1 and thus! oc  vout  "t # =

=

1

-

&  1

-

t 

i "t �   # dt � � 2

0

t 

#d t   vin "t � � '&  0

The same result as before