Electronic Analog Computer
by
Dr. Amin Danial Asham
References
Modern Control Engineering
Katsuhiko Ogata
III. Solving Linear Differential Equations
 The mathematical operation amplifiers can by used to
solve various mathematical equations and results can be
displayed on an oscilloscope.
 In this sections, examples of using the different types of
amplifiers to solve mathematical problems will be
introduced.
III. Solving Linear Equations (continue)
Example 1:
𝑉𝑜 = 4 𝑉𝑖
𝑉𝑖
4
𝑉1
1
Solution
Since
𝑅
𝑉1 = − 𝑅 𝑉𝑖 = −4𝑉𝑖
4
and
𝑅
𝑅
𝑉𝑜 = − 𝑉1 = −𝑉1
∴ 𝑉𝑜 = 4𝑉𝑖
𝑹 = 𝟏 𝑴𝜴
𝑉𝑜
III. Solving Linear Equations (continue)
Example 2:
𝑑2𝑦
𝑑𝑡 2
=𝑔
Solution :
• To solve this problem, the input signal 𝑉𝑖 = 𝑔 has to be integrated twice to get the
output 𝑉𝑜 = 𝑦.
𝑉1 =
1
−
𝑅𝐶
𝑉𝑖 𝑑𝑡 =
1
𝑅𝐶
𝑉1 𝑑𝑡 =
𝑉𝑜 = −
1 𝑑𝑦
−
𝑅𝐶 𝑑𝑡
1 2
𝑅𝐶
𝑉𝑖
𝑦
If the value of 𝑅 = 1 𝑀Ω and 𝐶 = 1𝜇𝐹
∴ 𝑅𝐶 = 1
1
2
∴ 𝑉𝑜 = 𝑦 = 𝑡 2 𝑔
1
𝑅𝐶
𝑉1
0
𝑉𝑜
1
𝑅𝐶
0
III. Solving Linear Equations (continue)
Example 2 (continue):
Simulating this example with 𝑔 = 2, 𝑅 = 1𝑀Ω, and 𝐶 = 1𝜇𝐹. We get the following solution.
III. Solving Linear Equations (continue)
Example 3: Solve the following differential equation:
𝑥 + 10𝑥 + 16𝑥 = 0,
𝑥 0 = 0,
𝑥 0 =8
Solution:
From the differential equation, we get:
𝑥 = −10𝑥 − 16𝑥
Therefore, we have to integrate 𝑥 twice to get 𝑥
First Integrator integrates 𝑥 to get −𝑥
Therefore, initial condition −𝑥(0) = −8
Second integrator integrates −𝑥 to 𝑥
𝒙
10
The analog computer
diagram for solving
the
Differential
Equation
−𝒙
16
0
-8
−𝒙
𝒙
1
1
III. Solving Linear Equations (continue)
Example 3: (Continue) - Circuit Diagram of the analog computer
These contacts
change their
state after a
short time at the
start of the
circuit
III.
Solving Linear Equations (continue)
Example 3: (Continue) – The result from analog computer
III. Solving Linear Equations (continue)
Example 3: (Continue) – Numerical Solution using Matlab on a digital computer.
III. Solving Linear Equations (continue)
Example 3: (Continue)
 For the differential equation:
𝑥 + 10𝑥 + 16𝑥 = 0,
𝑥 0 = 0,
𝑥 0 =8
• If the scale factors are 𝑘1 for 𝑥 and 𝑘2 for 𝑥 then
10
16
𝑥+
𝑘 𝑥 +
𝑘 𝑥 =0
𝑘2 2
𝑘1 1
Therefore
10
16
𝑥=−
𝑘2 𝑥 −
𝑘1 𝑥
𝑘2
𝑘1
𝟏𝟎
𝒌𝟐
𝟏𝟔
𝒌𝟏
−𝒙
1
0
-8 𝒌𝟐
−𝒌𝟏 𝒙
𝒌𝟏 𝒙
𝒌𝟏
𝒌𝟐
−𝒌𝟐 𝒙
𝒙 𝒌
𝟐
1
III. Solving Linear Equations (continue)
Example 3: (Continue)
• The diagram can be simplified for the minimum number of OpAmp’s
as follows:
𝟏𝟎
𝟏𝟔𝒌𝟐
𝒌𝟏
0
-8 𝒌𝟐
−𝒌𝟏 𝒙
𝒌𝟏 𝒙
𝒌𝟏
𝒌𝟐
−𝒌𝟐 𝒙
1
III. Solving Linear Equations (continue)
Example 3: (Continue)
•For 𝑘1 = 4 𝑎𝑛𝑑 𝑘2 = 2
𝟏𝟎
−𝟐𝒙
4𝒙
𝟐
𝟖
0
-8 𝒌𝟐
−𝟒𝒙
1
III. Solving Linear Equations (continue)
Example 3: (Continue)
III. Solving Linear Equations (continue)
Example 3: (Continue)
III. Solving Linear Equations (continue)
Time scale factor:
Time scale factor is used to slow down or speed up the time of the simulation
compared to the real time to make the analysis easier.
Example 4:
𝑑2 𝑦
𝑑𝑦
𝑑𝑦
+ 0.25
+ 𝑦 = 1 𝑦 0 = 1,
0 =2
2
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝜏
Let 𝜏 = 𝜆𝑡, hence for 𝑦 𝑡 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 𝑦( ) (𝑡 is the real time and 𝜏is the
𝜆
simulation time)
In other words each 𝒕 is replaced with
𝝉
𝝀
Therefore:
𝑑𝑦 𝑑𝑦 𝑑𝜏
𝑑𝑦
=
.
=𝜆
𝑑𝑡 𝑑𝜏 𝑑𝑡
𝑑𝜏
And
2𝑦
𝑑2𝑦
𝑑 𝑑𝑦
𝑑
𝑑𝑦 𝑑𝜏
𝑑
2
=
=
𝜆
.
=
𝜆
𝑑𝑡 2 𝑑𝑡 𝑑𝑡
𝑑𝜏
𝑑𝜏 𝑑𝑡
𝑑𝜏 2
III. Solving Linear Equations (continue)
Example 4: (continue)
The Differential equation becomes:
2
𝑑
𝑦
𝑑𝑦
2
𝜆
+ 0.25𝜆
+𝑦 =1
2
𝑑𝜏
𝑑𝜏
Therefore
𝑑 2 𝑦 0.25 𝑑𝑦 𝑦
1
+
+ 2= 2
2
𝑑𝜏
𝜆 𝑑𝜏 𝜆
𝜆
Consequently since 𝜏 = 0 𝑤ℎ𝑒𝑛 𝑡 = 0
𝑑𝑦
𝑑𝑦
0 =2=𝜆
0
𝑑𝑡
𝑑𝜏
Therefore
𝑑𝑦
0
𝑑𝑦
𝑑𝑡
0 =
= 2/𝜆
𝑑𝜏
λ
III. Solving Linear Equations (continue)
Example 4: (continue)
1
𝟎. 𝟐𝟓
𝝀
𝟏
𝝀𝟐
𝟏
𝝀𝟐
−𝒚
−
−𝒚
𝒚
𝟏
𝟐
𝝀
1
1
III. Solving Linear Equations (continue)-Example 4: (continue)
Thanks