Download The Inverting Integrator

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The Inverting Integrator
The circuit shown below is the inverting integrator.
C
i2 (s)
vin (s)
R
v-
ideal
i1 (s)
v+
oc
vout
(s )
+
Since the circuit uses the inverting configuration, we can
conclude that the circuit transfer function is:
oc
vout
(s )
G (s ) 
vin (s )
Z (s )
 2
Z 1 (s )


1 s C 
R
1
s RC
In other words, the output signal is related to the input as:
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oc
vout
(s ) 
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1 vin (s )
RC
s
From our knowledge of Laplace Transforms, we know this means
that the output signal is proportional to the integral of the
input signal!
Taking the inverse Laplace Transform, we find:
1
v (t ) 
RC
oc
out
t
v
in
(t ) dt 
0
For example, if the input is:
vin (t )  sinωt
then the output is:
v
oc
out
1
(t ) 
RC
t
 sinωt dt 
0
1 1
cosωt
RC ω
1

cosωt
ω RC

We likewise could have determined this result using Fourier
Analysis (i.e., frequency domain):
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G (ω ) 
oc
vout
(ω )
vin (ω )



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Z 2(ω )
Z 1(ω )
1 j ω C 
j
ω RC
R
Thus, the magnitude of the transfer function is:
G (ω ) 
j
ω RC

1
ω RC
And since:
 
 
j 
j  e  2  cos  2  j sin  2
the phase of the transfer function is:
G (ω )  π
2
 90
Given that:
radians
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oc
vout
(ω )  G (ω ) vin (ω )
and:
oc
vout
(ω )  G (ω )  vin (ω )
we find for the input:
vin (t )  sinωt
where:
vin (ω )  1
and
vin (ω )  0
that the output of the inverting integrator is:
oc
vout
(ω )  G (ω ) vin (ω )

1
ω RC
and:
oc
vout
(ω )  G (ω )  vin (ω )
 90  0
 90
Therefore:
1
sin ωt  90
ω RC
1

cosωt
ω RC
oc
vout
(t ) 
Exactly the same result as before!


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If you are still unconvinced that this circuit is an integrator,
consider this time-domain analysis.
i2 (t)
C
+ vc -
vin(t)
R
v-
-
i  0
i1 (t)
ideal
v+
oc
vout
(t )
+
From our elementary circuits knowledge, we know that the
voltage across a capacitor is:
1
vc (t ) 
C
t
 i (t ) dt 
2
0
and from the circuit we see that:
oc
oc
vc (t )  v (t )  vout
(t )  vout
(t )
therefore the output voltage is:
1
v (t )  
C
oc
out
t
 i (t ) dt 
2
0
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From KCL, we likewise know that:
i1(t )  i2(t )
and from Ohm’s Law:
i1(t ) 
Therefore:
vin (t )  v (t ) vin (t )

R1
R1
i2 (t ) 
vin (t )
R1
and thus:
t
1
v (t ) 
i2(t ) dt 

C 0
oc
out
1

RC
The same result as before!
t
v
in
0
(t ) dt 