Control Systems (a) Inverting amplifier (b) Non-inverting

Control Systems
(a)
Inverting amplifier
(b)
Non-inverting amplifier
(c)
Where: RA =∑ R4 R5 R6 R7=11.01 M Ω
RB=∑ R8 R9 =20 k Ω

(d)
For the inverting amplifier:
V y=−V x
R5
R4
V y
V x
=−R5
R4
Substituting for values:
V y
V x
=−10 M
1 M =−10
(e)
For the non-inverting amplifier:
Assigning an arbitrary value, V y
' at the positive input of the second amplifier due to the
voltage divider
V y
' =V y
R7
R6 + R7
...(i)
The output, V out is expressed as:
V out
V y
' = R9
R8
+1
Making V out the subject:
V out =V y
'
( R9
R8
+1 ) ....(ii)
Substituting (i) into (ii)

V out =V y ( R7
R6 +R7 )( R9
R8
+1 )
V out
V y
=¿ ( R7
R6 + R7 )( R9
R8
+1 )
Substituting for values:
V out
V y
=¿ ( 5 k
10 k )( 10 k
10 k +1)=1
(f)
Method 1
Involves relating the input of the inverting op-amp to the output which is in turn an input to
the second amplifier.
V out =V y ( R7
R6 +R7 )( R9
R8
+1 )
Substituting for V y ∈terms of V x :
V out =−V x
R5
R4 ( R7
R6 + R7 )( R9
R8
+1 )
Rearranging:
V out
V x
=−R5
R4 ( R7
R6 + R7 )( R9
R8
+1 )
Substituting for values:
V out
V x
=−10 M
1 M ( 5 k
10 K )(10 k
10 k +1 )=−10

Method 2
From the impedance circuit, the relationship between V out and V x can be derived:
The effective resistance of R3and XC 3is: R3 XC 3
R3 + XC 3
= A
A is in series with XC 2and their effective resistance is: (A+ XC 2) = B
B is in parallel with R2and their effective resistance is: R2 B
R2+ B = C
C is in series with XC 1and their effective resistance is: C+ XC 1=D
D is in parallel with R2 and the effective resistance is: R1 D
R1+ D = E

Simplifying the impedance diagram:
Using the voltage divider rule:
V out = RB
RB + E V x
Hence
V out
V x
= RB
RB + E
Where RB=∑ R8 R9 and the effective resistance E is shown above.