2/23/2011
The Inverting Integrator lecture
1/8
The Inverting Integrator
The circuit shown below is the inverting integrator.
C
i2 (s)
vin (s)
R
i1 (s)
Jim Stiles
v-
-
ideal
v+
oc
vout
(s )
+
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
2/8
It’s the inverting configuration!
Since the circuit uses the inverting configuration, we can conclude that the
circuit transfer function is:
oc
(1 s C ) = −1
vout
(s )
Z (s )
=− 2
=−
G (s ) =
vin (s )
Z 1 (s )
R
s RC
In other words, the output signal is related to the input as:
oc
vout
(s ) =
−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!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
3/8
The circuit integrates the input
Taking the inverse Laplace Transform, we find:
v (t ) =
oc
out
−1
RC
t
∫vin (t ′) dt ′
0
For example, if the input is:
vin (t ) = sinωt
then the output is:
v
Jim Stiles
oc
out
(t ) =
−1
RC
t
∫ sinωt dt ′ =
0
−1 −1
RC ω
cosωt =
The Univ. of Kansas
1
ω RC
cosωt
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
4/8
Or, in the Fourier domain
We likewise could have determined this result using Fourier Analysis (i.e.,
frequency domain):
oc
1 jω C )
(
vout
(ω )
Z 2(ω )
j
G (ω ) =
=−
=−
=
vin (ω )
Z 1(ω )
R
ω RC
Thus, the magnitude of the transfer function is:
G (ω ) =
And since:
j
1
=
ω RC
ω RC
( )
( )
j π
j = e ( 2) = cos π 2 + j sin π 2
the phase of the transfer function is:
∠G (ω ) = π
Jim Stiles
2
radians = 90D
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
5/8
Magnitude and phase
Given that:
oc
vout
(ω ) = G (ω ) vin (ω )
and:
oc
∠vout
(ω ) = ∠G (ω ) + ∠vin (ω )
we find for the input:
where:
vin (t ) = sinωt
vin (ω ) = 1
and
∠vin (ω ) = 0
that the output of the inverting integrator is:
oc
(ω ) = G (ω ) vin (ω ) =
vout
and:
Jim Stiles
1
ω RC
oc
∠vout
(ω ) = ∠G (ω ) + ∠vin (ω ) = 90D + 0 = 90D
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
6/8
See, it’s an integrator
Therefore:
oc
vout
(t ) =
=
1
ω RC
1
ω RC
(
sin ωt + 90D
)
cosωt
Exactly the same result as before!
If you are still unconvinced that this circuit is an integrator, consider this timedomain analysis.
i ( t)
2
C
+ vc -
vin(t)
R
i 1 ( t)
Jim Stiles
v-
-
i− = 0
v+
ideal
oc
vout
(t )
+
The Univ. of Kansas
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
7/8
The time-domain solution
From our elementary circuits
knowledge, we know that the voltage
across a capacitor is:
vc (t ) =
1
t
∫ i (t ′) dt ′
C
2
0
and from the circuit we see that:
oc
oc
vc (t ) = v −(t ) − vout
(t ) = −vout
(t )
i 2 ( t)
therefore the output voltage is:
v (t ) = −
oc
out
1
C
t
∫ i2(t ′) dt ′
0
vin(t)
+ vc -
R
i 1 ( t)
Jim Stiles
C
The Univ. of Kansas
v-
-
i− = 0
v+
ideal
oc
vout
(t )
+
Dept. of EECS
2/23/2011
The Inverting Integrator lecture
8/8
The same result no matter how we do it!
From KCL, we likewise know that:
i1(t ) = i2(t )
and from Ohm’s Law:
i1(t ) =
vin (t ) − v −(t ) vin (t )
=
R1
R1
i 2 ( t)
C
+ vc -
Therefore:
i2 (t ) =
vin (t )
R1
vin(t)
R
i 1 ( t)
and thus:
v (t ) =
oc
out
=
−1
C
−1
RC
t
∫ i (t ′) dt ′
2
v-
-
i− = 0
v+
ideal
oc
vout
(t )
+
0
t
∫vin (t ′) dt ′
0
The same result as before!
Jim Stiles
The Univ. of Kansas
Dept. of EECS