Analysis of Amplifiers in Control Systems Assignment

This assignment solution focuses on the analysis of inverting and non-inverting amplifiers within a Control Systems context. The solution begins by defining the components and parameters of the circuit, including resistor values and relationships. It then proceeds to derive the output voltage equations for both inverting and non-inverting amplifier configurations, using the voltage divider rule and other relevant circuit analysis techniques. The solution also provides two methods for relating the input and output voltages, and the impedance circuit and the effective resistance of various components are also considered. The document uses mathematical expressions and substitutions to analyze the behavior of the amplifiers, providing a comprehensive understanding of their functionality within the control system.

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.