Assignment #1
Name
Class
Roll No.
Course
Instructor
Syed Jazib Hasan
2-B
EE-1563
ENA
Dr. Imtiaz Hussain
Table of Contents :
1.1 Introduction : .................................................................................................................................... 3
1.1a Introduction to Second Order Circuits : ........................................................................................... 3
1.1b Response of a Parallel RLC Circuit : ............................................................................................... 3
1.1c Unit Step Function : .......................................................................................................................... 4
1.2 Problem Statement : ........................................................................................................................... 5
1.2a Solution Plan :................................................................................................................................... 5
1.3 Solution .............................................................................................................................................. 6
1.4 GNU Octave ....................................................................................................................................... 9
Plot : ....................................................................................................................................................... 10
Values : (Zoomed View) ........................................................................................................................ 10
1.5 Multisim ........................................................................................................................................... 11
1.1 Introduction :
This section focuses on the fundamentals required to understand the problem statement that
follows. A brief introduction on Second Order Circuits , Response of a Parallel RLC Circuit
and Unit Step Function are included.
1.1a Introduction to Second Order Circuits :
A second-order circuit is an electrical circuit that contains two energy storage elements,
typically capacitors and/or inductors. These circuits are commonly used in electronics and
electrical engineering to model a wide range of physical systems.
Second-order circuits are characterized by a second-order differential equation that describes
the circuit's behavior over time.
In this section we are going to focus on the Parallel RLC Circuit Configuration.
A parallel RLC circuit is an electrical circuit that consists of a resistor (R), an inductor (L),
and a capacitor (C) that are connected in parallel to each other. In this circuit configuration,
each component is connected to the same two nodes, and the voltage across each component
is the same.
1.1b Response of a Parallel RLC Circuit :
The response of a second order circuit is how it behaves when it receives an input signal. A
parallel RLC circuit can be modelled by a second order differential equation of the form :
𝑑2 𝑣
1 𝑑𝑣
1
+
+
𝑣=0
2
𝑑𝑡
𝑅𝐶 𝑑𝑡 𝐿𝐶
This differential equation yields the response 𝑣(𝑡) of the circuit for a general time t.
Now the response of the RLC circuit is mainly of three types and depend upon the
comparison of the neper frequency (α) and the natural frequency (𝜔0 ) of the circuit where :
α =
1
2RC
and 𝜔0 =
1
√𝐿𝐶
In case where α < 𝜔0 , we have a underdamped response which is of the form :
𝑣(𝑡) = 𝑒 −𝛼𝑡 (𝐴𝑐𝑜𝑠(𝜔𝑑 𝑡) + 𝐵𝑠𝑖𝑛(𝜔𝑑 𝑡))
Now if we apply an external voltage or current source to excite the system , we get a Step
Response.
The step response of an RLC circuit is the behavior of the circuit in response to a sudden
change in input voltage, such as a step function. The response depends on the values of the
resistance, inductance, and capacitance in the circuit.
In such a case , the complete response of the circuit becomes :
𝑣(𝑡) = 𝑣𝑡 (𝑡) + 𝑣𝑠𝑠 (𝑡)
Where :
𝑣𝑡 (𝑡) is the transient response and is of the form discussed above.
𝑣𝑠𝑠 (𝑡) is the steady-state response of the circuit.
1.1c Unit Step Function :
A unit step function is a mathematical function that has a value of zero for all negative inputs,
and a value of one for all non-negative inputs. It is denoted as u(t) or Θ(t), where t is the input
variable.
In mathematical notation :
0,
u(𝑡) = {
1,
𝑥<0
𝑥>0
1.2 Problem Statement :
1.2a Solution Plan :
Find the inductor current i and capacitor voltage v at time t = 0− .
Find the inductor current i and capacitor voltage v at time t = 0+ .
𝑑𝑣
Find 𝑑𝑡 at t = 0+ , by using KCL and finding 𝑖𝑐
Calculate α and 𝜔0 for t > 0 and determine 𝑣𝑡 (𝑡).
1.3 Solution
For t = 𝟎− , the circuit is in the state shown :
The circuit is in steady state.
To evalualte 𝑣(0− ) , we find the voltage drop at R2 (𝑣𝑅2 ). Since 𝑣𝑅2 = 𝑣(0− ) ,
𝑣(0− ) =
100 × 103
150 × 103
× 3 = 2𝑉
Finding 𝑖(0− ) :
𝑖(0− ) =
Now for t = 𝟎+ :
5
= 10 𝑚𝐴
500
Since the capacitor voltage can not change abruptly , 𝑣(0− ) = 𝑣(0+ ) = 2𝑉 .
Similarly since the inductor current can not change abruptly , 𝑖(0− ) = 𝑖(0+ ) = 10 𝑚𝐴
To find
𝑑𝑣(0+ )
𝑑𝑡
,
𝑑𝑣(0+ ) 𝑖𝑐
=
𝑑𝑡
𝐶
To find 𝑖𝑐 , we apply KCL at the blue node :
−𝑖1 − 𝑖3 − 𝑖𝑐 − 𝑖(0+ ) = 0
𝑖𝑐 = −𝑖3 − 𝑖1 − 𝑖(0+ )
𝑖𝑐 = −
2
3
−
− 0.01
500 150 × 103
𝑖𝑐 = −0.014 𝐴
Hence ,
𝑑𝑣(0+ )
−0.014
=
= −𝟏𝟒𝟎𝟎 𝑽/𝒔
𝑑𝑡
10 × 10−6
To find the natural response of the Parallel RLC Circuit , we calculate α and ω0 .
α =
1
= 100 𝑛𝑝/𝑠
2 × 500 × 10 × 10−6
𝜔0 =
1
√2 × 10 × 10−6
= 223.6 𝑟𝑎𝑑/𝑠
Since α < ω0 ,we have an under-damped case here , the solution to which is of the form
discussed in the introductory section of this document.
The general solution is :
𝑣(𝑡) = 𝑒 −𝛼𝑡 (𝐴𝑐𝑜𝑠(𝜔𝑑 𝑡) + 𝐵𝑠𝑖𝑛(𝜔𝑑 𝑡)) 𝑉
Since 𝜔𝑑 = √𝜔0 2 − 𝛼 2 , 𝜔𝑑 = √223.62 − 1002 = 200 𝑟𝑎𝑑/𝑠
Substituting this and the value of α in our general solution we have :
𝑣(𝑡) = 𝑒 −100𝑡 (𝐴𝑐𝑜𝑠(200𝑡) + 𝐵𝑠𝑖𝑛(200𝑡)) 𝑉
To find the particular solution we use our initial conditions we worked out previously.
Using 𝑣(0+ ) = 2𝑉 :
𝑒 −100(0) (𝐴𝑐𝑜𝑠(200(0)) + 𝐵𝑠𝑖𝑛(200(0))) = 2
A = 2
Using
𝑑𝑣(0+ )
𝑑𝑡
= 1400 𝑉/𝑠 :
𝑑𝑣(𝑡)
= 𝑒 −100𝑡 [−400𝑠𝑖𝑛(200𝑡) + 200𝐵𝑐𝑜𝑠(200𝑡)] + 𝑒 −100𝑡 [−200𝑐𝑜𝑠(200𝑡) − 100𝐵𝑠𝑖𝑛(200𝑡)]
𝑑𝑡
𝑑𝑣(0+ )
= 200𝐵 − 200 = −1400
𝑑𝑡
B = −6
The particular solution is :
𝒗(𝒕) = 𝒆−𝟏𝟎𝟎𝒕 (𝟐𝒄𝒐𝒔(𝟐𝟎𝟎𝒕) − 𝟔𝒔𝒊𝒏(𝟐𝟎𝟎𝒕)) 𝑽
𝒗(𝒕) at t= 1ms :
𝑣(0.001) = 𝑒 −100(10
−3 )
(2𝑐𝑜𝑠(200 × 10−3 ) − 6𝑠𝑖𝑛(200 × 10−3 ))
𝐯(𝟎. 𝟎𝟎𝟏) = 𝟎. 𝟔𝟗𝟓 𝐕
t at 𝒗(𝒕) = 0 :
2𝑐𝑜𝑠(200𝑡) − 6𝑠𝑖𝑛(200𝑡) = 0
tan(200t) =
𝐭𝟎 =
1
3
𝟏
𝟏
𝐭𝐚𝐧−𝟏( ) = 𝟏. 𝟔𝟏 𝐦𝐬
𝟐𝟎𝟎
𝟑
1.4 GNU Octave
1.4a Code
t = linspace(0, 0.05, 10000000);
v = exp(-100*t) .* (2*cos(200*t) - 6*sin(200*t));
plot(t, v);
xlabel('Time (s)');
ylabel('Voltage (V)');
hold on
xval = 1e-3;
yval = 0.6950211577;
plot(xval, yval, 'ro')
text(xval, yval, sprintf('(%.0f ms, %.9f V)', xval*1e3, yval), 'HorizontalAlignment', 'right')
xval = 1.608752772e-3;
yval = 0;
plot(xval, yval, 'go')
text(xval, yval, sprintf('(%.9f s, %.1f V)', xval, yval), 'HorizontalAlignment', 'right')
hold off
1.4b Explanation
1- A one dimensional vector t has been created using the linspace() function with start value =
0 , end value = 0.05 and spacing = 10000000
2- v = exp(-100*t) .* (2*cos(200*t) - 6*sin(200*t)) is the voltage function.
3- plot() function plots the voltage-time function.
4- hold on keyword holds the plot so that further plots are added.
5- Two points are further plotted using the plot() function and specifying the xval and yval of
the points.
6- hold off keyword releases the plot so that no further plots are added.
1.4c Simulation Results
Plot :
Values : (Zoomed View)
v(t) at t = 1 ms
𝐭𝟎
1.5 Multisim
1.4a Circuit Schematics
1.4a Result