Lecture 9:

Natural Response
Prof. Abhishek Dixit, Prof. Lalan Kumar, Prof. S. D. Joshi, Prof. I. N. Kar,
Prof. M. Veerachary
Second order systems C
1 L

𝑖(𝑡)
+ R
V
-

𝑑𝑖 1
𝐿 + න 𝑖𝑑𝑡 + 𝑅𝑖 = 0
𝑑𝑡 𝐶

𝑑2 𝑖 𝑑𝑖 1
𝐿 2+𝑅 + 𝑖 =0
𝑑𝑡 𝑑𝑡 𝐶

𝑑2 𝑖 𝑅 𝑑𝑖 1
2
+ + 𝑖=0
𝑑𝑡 𝐿 𝑑𝑡 𝐿𝐶
Second order systems
𝑑2 𝑖 𝑅 𝑑𝑖 1 [Differential equation]
2
+ + 𝑖=0
𝑑𝑡 𝐿 𝑑𝑡 𝐿𝐶

Let 𝑖 = 𝐴𝑒 𝑠𝑡 𝑅 1
𝑠2 + 𝑠+ =0 [Algebraic equation]
𝐿 𝐿𝐶

2
𝑅 𝑅 2
1 𝑅 𝑅 1
− + −4 − − −4
𝐿 𝐿 𝐿𝐶 𝐿 𝐿 𝐿𝐶
𝑠1 = 𝑠2 =
2 2
Second order systems
𝑑2 𝑖 𝑅 𝑑𝑖 1
2
+ + 𝑖=0
𝑑𝑡 𝐿 𝑑𝑡 𝐿𝐶
2
𝑅 𝑅 2
1 𝑅 𝑅 1
Let 𝑖 = 𝐴𝑒 𝑠𝑡 − + −4 − − −4
𝐿 𝐿 𝐿𝐶 𝐿 𝐿 𝐿𝐶
𝑠1 = 𝑠2 =
2 2

If 𝑖1 = 𝐴1 𝑒 𝑠1 𝑡 or 𝑖2 = 𝐴2 𝑒 𝑠2𝑡 satisfies the linear homogenous equation, then the sum of

the two terms also satisfies the equation.

Hence the solution is 𝑖 = 𝐴1 𝑒 𝑠1 𝑡 + 𝐴2 𝑒 𝑠2𝑡 .

Second order systems
𝑑2 𝑖 𝑅 𝑑𝑖 1
2
+ + 𝑖=0
𝑑𝑡 𝐿 𝑑𝑡 𝐿𝐶
2
𝑅 𝑅 2
1 𝑅 𝑅 1
Let 𝑖 = 𝐴𝑒 𝑠𝑡 − + −4 − − −4
𝐿 𝐿 𝐿𝐶 𝐿 𝐿 𝐿𝐶
𝑠1 = 𝑠2 =
2 2

If 𝑖1 = 𝐴1 𝑒 𝑠1 𝑡 or 𝑖2 = 𝐴2 𝑒 𝑠2𝑡 satisfies the linear homogenous equation, then the sum of

the two terms also satisfies the equation.

Hence the solution is 𝑖 = 𝐴1 𝑒 𝑠1 𝑡 + 𝐴2 𝑒 𝑠2𝑡 .

Note: Though 𝐿, 𝐶, and 𝑅 are all real, 𝑠1 and 𝑠2 can be real, complex, or purely imaginary.
Second order systems
2
𝑅 𝑅 2
1 𝑅 𝑅 1
Let 𝑖 = 𝐴𝑒 𝑠𝑡 − + −4 − − −4
𝐿 𝐿 𝐿𝐶 𝐿 𝐿 𝐿𝐶
𝑠1 = 𝑠2 =
2 2

𝑅 2 1
If >4 ,
𝐿 𝐿𝐶
𝑠1 and 𝑠2 will be real, negative and distinct.

𝑅 2 1
If = 4 ,
𝐿 𝐿𝐶
𝑠1 and 𝑠2 will be real, negative and identical.

𝑅 2 1
If <4 ,
𝐿 𝐿𝐶
𝑠1 and 𝑠2 will be complex conjugate, and purely imaginary if 𝑅 = 0.
Check Yourself C
1 L
2 𝑖(𝑡)
+
V
-

Switch is connected to 2 at 𝑡 = 0. If 𝑅 is very large, then the roots are

1) Real and same 2) Real and distinct

3) Complex conjugate 4) Purely imaginary

Check Yourself C
1 L
2 𝑖(𝑡)
+
V
-

Switch is connected to 2 at 𝑡 = 0. If 𝑅 is very large, then the roots are

1) Real and same 2) Real and distinct

3) Complex conjugate 4) Purely imaginary

Check Yourself
𝑉
• 𝑖 𝑡 = (𝑒 −3𝑡 − 𝑒 −𝑡 )
2𝐿

1.5

0.5 𝑒 −3𝑡

i (current)
0
0 2 Time 4
-0.5

-1 −𝑒 −𝑡
-1.5
Check Yourself C
1 L
2 𝑖(𝑡)
+
V
-

Switch is connected to 2 at 𝑡 = 0. If 𝑠1 = −1, and 𝑠2 = −3, current is

1 2 3 4
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −1, and 𝑠2 = −3

-3 -1
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −1, and 𝑠2 = −10

-10 -1
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −1, and 𝑠2 = −39

-39 -1
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −2, and 𝑠2 = −4

-4 -2
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −3, and 𝑠2 = −5

-5 -3
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −10, and 𝑠2 = −12

-12 -10
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −8, and 𝑠2 = −10

-10 -8
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −6, and 𝑠2 = −10

-10 -6
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −4, and 𝑠2 = −10

-10 -4
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −1, and 𝑠2 = −10

-10 -1
Second order systems (Complex roots case)
2 2
𝑅 𝑅 1 𝑅 𝑅 1
Let 𝑖 = 𝐴𝑒 𝑠𝑡 − + −4 − − −4
𝐿 𝐿 𝐿𝐶 𝐿 𝐿 𝐿𝐶
𝑠1 = 𝑠2 =
2 2

2 2
𝑅 𝑅 1 𝑅 𝑅 1
𝑠1 = − + − 𝑠2 = − − −
2𝐿 2𝐿 𝐿𝐶 2𝐿 2𝐿 𝐿𝐶

𝑅 1 1 𝑅2
Let α = , ω2𝑛 = , and ω2 = − 2 = ω2𝑛 − α2 .
2𝐿 𝐿𝐶 𝐿𝐶 4𝐿

𝑠1 = −α + 𝑗ω, 𝑠2 = −α − 𝑗ω
Second order systems (Complex roots case)
• 𝑖 = 𝐴1 𝑒 𝑠1 𝑡 + 𝐴2 𝑒 𝑠2𝑡
• Let us say that 𝑠1 = −α + 𝑗ω, and 𝑠2 = −α − 𝑗ω
• So, 𝑖 = 𝐴1 exp −α𝑡 + 𝑗ω𝑡 + 𝐴2 exp −α𝑡 − 𝑗ω𝑡
• 𝑖 = exp −α𝑡 𝐴1 exp 𝑗ω𝑡 + 𝐴2 exp −𝑗ω𝑡
• Using Euler’s theorem, 𝑒𝑥𝑝 𝑗ω𝑡 = cos ω𝑡 + 𝑗 sin ω𝑡
• 𝑖 = exp −α𝑡 𝐴1 cos ω𝑡 + jsin ω𝑡 + A2 cos ω𝑡 − jsin ω𝑡
• 𝑖 = exp −α𝑡 cos ω𝑡 𝐴1 + 𝐴2 + jsin ω𝑡 𝐴1 − 𝐴2
• This implies that 𝐴1 + 𝐴2 and 𝑗𝐴1 − 𝑗𝐴2 should be real.
• 𝑖 = exp −α𝑡 cos ω𝑡 𝐵1 + sin ω𝑡 𝐵2 = 𝐴𝑒𝑥𝑝 −α𝑡 sin ω𝑡 + θ
• This is called damped sinusoid, 𝑎 is the damping coefficient, and ω is called the
natural frequency of oscillations.
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −1 + 𝑗, and 𝑠2 = −1 − 𝑗

-1+j

-1-j
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −10 + 𝑗, and 𝑠2 = −10 − 𝑗

-10+j

-10-j
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −0.5 + 𝑗, and 𝑠2 = −0.5 − 𝑗

-0.5+j

-0.5-j
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −0.5 + 2𝑗, and 𝑠2 = −0.5 − 2𝑗

-0.5+2j S-plane/complex
frequency plane

-0.5-2j
Roots (Real and Equal)
• 𝑖 = 𝐴1 𝑒 𝑠1 𝑡 + 𝐴2 𝑒 𝑠2𝑡
• E.g., 𝑠1 = 𝑠2
• 𝑖 = 𝐴1 + 𝐴2 𝑒 𝑠1𝑡 = 𝐴3 𝑒 𝑠1 𝑡
• There is only one known 𝐴3 ,and there exists two initial conditions to solve:
• 𝑖 0 {e.g., in the previous example, capacitors forces it to be 0}
𝑑𝑖(0) −𝑉
• {e.g., in the previous example, it is }
𝑑𝑡 𝐿
• Thus, we need two unknowns.
• Thus, 𝑖 = 𝐴1 + 𝑡𝐴2 𝑒 𝑠1 𝑡 .
′ 𝑉 𝑉 𝑉
• For 𝑖 0 = 0, and 𝑖 0 = − . 𝐴1 = 0, and 𝑡𝐴2 𝑠1 + 𝐴2 = − → 𝐴2 = − .
𝐿 𝐿 𝐿
Check Yourself
1 L
2 𝑖(𝑡)
+
V
-

𝑠1 = −0.5, and 𝑠2 = −0.5

S-plane/complex
frequency plane

2
-0.5
Root locus as a function of resistance
2
𝑅 𝑅 1
𝑅=0 𝑠1 = − + −
2𝐿 2𝐿 𝐿𝐶
𝐿, 𝐶 = 1
2
𝑅 𝑅 1
𝑠2 = − − −
2𝐿 2𝐿 𝐿𝐶

𝑅 1
Let α = , ω2𝑛 = , and ω2 =
2𝐿 𝐿𝐶
1 𝑅2
− 2 = ω2𝑛 − α2 .
𝐿𝐶 4𝐿

𝑠1 = −α + 𝑗ω, 𝑠2 = −α − 𝑗ω

𝑅=0 1
𝑅 = 0, 𝑠1 , 𝑠2 = ±𝑗ω𝑛 = ±𝑗
𝐿𝐶
Root locus as a function of resistance
2
𝑅 𝑅 1
𝑅=0 𝑠1 = − + −
2𝐿 2𝐿 𝐿𝐶
𝐿, 𝐶 = 1
2
𝑅 𝑅 1
4𝐿 𝑠2 = − − −
𝑅= 2𝐿 2𝐿 𝐿𝐶
𝐶

𝑅 1
Let α = , ω2𝑛 = , and ω2 =
2𝐿 𝐿𝐶
1 𝑅2
− 2 = ω2𝑛 − α2 .
𝐿𝐶 4𝐿

𝑠1 = −α + 𝑗ω, 𝑠2 = −α − 𝑗ω

𝑅=0 1
𝑅 = 0, 𝑠1 , 𝑠2 = ±𝑗ω𝑛 = ±𝑗
𝐿𝐶
Root locus as a function of resistance
𝑅 1
Let α = , ω2𝑛 = , and ω2
𝑅=0 2𝐿 𝐿𝐶
1 𝑅 2
= − = ω 2 − α2 .
𝐿𝐶 4𝐿2 𝑛
𝐿, 𝐶 = 1
𝑠1 = −α + 𝑗ω, 𝑠2 = −α − 𝑗ω
4𝐿
𝑅=
𝐶
α2 + ω2 = ω2𝑛

𝑅=0
Root locus as a function of resistance
2
𝑅 𝑅 1
𝑅=0 𝑠1 = − + −
2𝐿 2𝐿 𝐿𝐶
𝐿, 𝐶 = 1
2
𝑅 𝑅 1
4𝐿 𝑠2 = − − −
𝑅= 2𝐿 2𝐿 𝐿𝐶
𝐶

𝑅=∞ 𝑅=∞

𝑅=0
Check Yourself
• Root locus as a function of capacitance (after fixing resistance and inductance)
is?
Check Yourself
• Root locus as a function of capacitance (after fixing resistance and inductance)
is?

2
𝑅 𝑅 1
𝑠1 = − + −
2𝐿 2𝐿 𝐿𝐶

2
𝑅 𝑅 1
𝑠2 = − − −
2𝐿 2𝐿 𝐿𝐶
Types of responses
• If roots are real and distinct = Overdamped
• If roots are real and equal = Critically damped
• If roots are complex conjugate = Underdamped
• If roots are purely imaginary = Oscillatory
Impedance function
• Impedance = Ratio of Voltage to current when input is exponential function.
• Input an exponential is not a major restriction.
• 𝑒 𝑠𝑡 can represent sinusoids, and when 𝑠 = 0, they can represent constants.
• In future, you may see that sum of exponentials can represent any useful signal in the
world (this idea is known as Fourier series/transforms).
• Key property of exponential function is that its time derivate is another
exponential.
Impedance functions
• Resistor:
𝑣 𝑅𝑖
• 𝑍𝑅 = = = 𝑅 in ohms.
𝑖 𝑖

• Inductance: 𝑣 = 𝐿 𝑑𝑖/𝑑𝑡 = 𝐿𝑠𝑖

𝑣 𝐿𝑠𝑖
• 𝑍𝐿 = = = 𝐿𝑠 in ohms
𝑖 𝑖

𝑑𝑣
• Capacitance: 𝑖 = 𝐶 = 𝐶𝑠𝑣
𝑑𝑡
𝑣 𝑣 1
• 𝑍𝑐 = = = in ohms
𝑖 𝑠𝐶𝑣 𝑠𝐶
Impedance functions

𝑖
𝑖 = 𝑖1 + 𝑖2
𝑖1 𝑖2
𝑣 𝑣 𝑣
R C = +
𝑍𝑒𝑞 𝑅 𝑍𝐶

𝑍𝑒𝑞 = 𝑍𝐶 ||𝑅

Impedances can be treated similar to resistances.

Check Yourself
Find the impedance in the figure below, if voltage applied is 6𝑒 −2𝑡 V.

𝑖 𝑅1 = 2Ω

𝑅 = 4Ω 𝐶=
0.25𝐹.

1) 2Ω 2) -2Ω 3) 4Ω 4) -4Ω
Check Yourself

𝑍𝑅 𝑍𝐶 𝑅 1/𝑠𝐶
𝑅1 = 2Ω • 𝑍 = 𝑍𝑅 + = 𝑅1 +
𝑖 𝑍𝑅 +𝑍𝐶 𝑅+ 1/𝑠𝐶

𝑅 4
𝑅 = 4Ω 𝐶=
= 𝑅1 + =2+
𝑠𝐶𝑅 + 1 −2 1/4 4 + 1
0.25𝐹.

= 2 − 4 = −2 Ω
Check Yourself
Find the impedance in the figure below, if voltage applied is 6𝑒 −2𝑡 V.

𝑖 𝑅1 = 2Ω

𝑅 = 4Ω 𝐶=
0.25𝐹.

1) 2Ω 2) -2Ω 3) 4Ω 4) -4Ω
Check Yourself
Find the impedance function in the figure below.

𝑖 𝑅1 = 2Ω

𝑅 = 4Ω 𝐶=
0.25𝐹.

2𝑠+6 2𝑠+6 𝑠+3 𝑠+6

1) 2) 3) 4)
𝑠+2 𝑠+1 𝑠+1 𝑠+1
Check Yourself
Find the impedance function in the figure below.
𝑍𝑅 𝑍𝐶 𝑅 1/𝑠𝐶
𝑖 𝑅1 = 2Ω • 𝑍 = 𝑍𝑅 + = 𝑅1 +
𝑍𝑅 +𝑍𝐶 𝑅+ 1/𝑠𝐶

𝑅 4
𝑅 = 4Ω 𝐶= = 𝑅1 + =2+
0.25𝐹. 𝑠𝐶𝑅 + 1 𝑠 1/4 4 + 1

4 2𝑠+6
=2 + =
𝑠+1 𝑠+1
Check Yourself
Find the impedance function in the figure below.

𝑖 𝑅1 = 2Ω

𝑅 = 4Ω 𝐶=
0.25𝐹.

2𝑠+6 2𝑠+6 𝑠+3 𝑠+6

1) 2) 3) 4)
𝑠+2 𝑠+1 𝑠+1 𝑠+1
Check Yourself
What is impedance for DC if the impedance function is given below?

𝑖 𝑅1 = 2Ω
2𝑠 + 6
𝑍 𝑠 =
𝑠+1
𝑅 = 4Ω 𝐶=
0.25𝐹.
Check Yourself
What is impedance for DC if the impedance function is given below?

𝑖 𝑅1 = 2Ω
2𝑠 + 6
𝑍 𝑠 =
𝑠+1
𝑅 = 4Ω 𝐶=
𝑍 0 =6
0.25𝐹.
General Impedance Function

• In all cases, the impedance can be expressed as a rational function i.e., a

function which is a ratio of two polynomials

Thursday, September 5, 2024

General Impedance Function

• In all cases, the impedance can be expressed as a rational function i.e., a

function which is a ratio of two polynomials

• If the numerator and denominators are factored then

Thursday, September 5, 2024

General Impedance Function

• In all cases, the impedance can be expressed as a rational function i.e., a

function which is a ratio of two polynomials

• If the numerator and denominators are factored then

• Here, the zi’s are called zeros as 𝑍 𝑧𝑖 = 0,

• The pi’s are called poles.

Thursday, September 5, 2024

Physical Interpretation of Poles and Zeros
• Let us look at 𝑣 = 𝑍𝑖
• What is the meaning of “zero impedance”?
• If the impedance is small, the current can exist with small voltage applied.
• Taking this idea to limit, current can exist with 0 voltage applied.
• But current for 0 voltage applied is natural response.
• Each zero 𝑠 = 𝑠1 corresponds to a possible component 𝐼𝑒 𝑠1 𝑡 of the natural
response.
• Similarly, each pole corresponds to natural response of the voltage for zero
current.
Check Yourself
S L

𝑖
+ 𝑅
V 𝑅
-

𝑍 = 𝑅 + 𝑠𝐿 + 𝑅 = 2𝑅 + 𝑠𝐿

Zero: 𝑧 = −2𝑅/𝐿

𝑖(𝑡) = 𝐼0 𝑒 −2𝑅𝑡/𝐿
Check Yourself 𝐶
S
a

+
V L
- 𝑅

Open circuit voltage at terminal 𝑎 − 𝑏?

Check Yourself 𝐶
S
a

+
V L
- 𝑅

1 𝑅𝑠𝐿 𝑅 + 𝑠𝐿 + 𝑠 2 𝑅𝐿𝐶 1
𝑍𝑎𝑏 𝑠 = + = =
𝑠𝐶 𝑅 + 𝑠𝐿 𝑠𝐶(𝑅 + 𝑠𝐿) 𝑌(𝑠)
Check Yourself 𝐶
S
a

+
V L
- 𝑅

1 𝑅𝑠𝐿 𝑅 + 𝑠𝐿 + 𝑠 2 𝑅𝐿𝐶 𝑅𝐿𝐶 𝑠 2 + 1/𝑅𝐶 𝑠 + 1/𝐿𝐶 1

𝑍𝑎𝑏 𝑠 = + = = =
𝑠𝐶 𝑅 + 𝑠𝐿 𝑠𝐶(𝑅 + 𝑠𝐿) 𝑠𝐿𝐶(𝑠 + 𝑅/𝐿) 𝑌(𝑠)

1 𝑅𝑠𝐿 1 𝑠 − 0 𝑠 − −𝑅/𝐿
𝑌 𝑠 = + =
𝑠𝐶 𝑅 + 𝑠𝐿 𝑅 (𝑠 − 𝑠1 ) 𝑠 − 𝑠2

𝑣 = 𝑉1 𝑒 0 + 𝑉2 𝑒 −𝑅𝑡/𝐿
Check Yourself 𝐶
S
a

+
V L
- 𝑅

𝑣 = 𝑉1 𝑒 0 + 𝑉2 𝑒 −𝑅𝑡/𝐿

𝑣=𝑉
Thanks for your attention &
Questions ?

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