Week 11: Transient Behaviour

Wk 11

Transient Behaviour

Introduction

Charging Capacitors and Energising Inductors

Discharging Capacitors and De-energising Inductors

Response of First-Order Systems

Second-Order Systems

Higher-Order Systems

Introduction

So far we have looked at the behaviour of systems in response to:
– fixed DC signals
– constant AC signals

We now turn our attention to the operation of circuits before they reach steady-state conditions

– this is referred to as the transient response

We will begin by looking at simple $RC$ and $RL$ circuits
Inductor energising

A similar analysis of this circuit

gives
\begin{align*} v =& Ve ^{\Large \frac{Rt}{L}} = Ve ^{\Large \frac{t}{\text{T}}} \\ \\ i =& I( 1 - e ^{\Large \frac{Rt}{L}}) = I (1 - e ^{\Large \frac{t}{\text{T}}}) \end{align*}
where $I = V/R$

– see Section 9.2.2 for this analysis

📷 Thus, again, both the voltage and current have an exponential form - see here

Charging Capacitors and Energising Inductors

Capacitor Charging - Consider the circuit shown here

– Applying Kirchhoff’s voltage law

$ iR + v = V $

– Now, in a capacitor

$ i= C\frac{dv}{dt} $

– which substituting gives

$ CR\frac{dv}{dt} + v = V $

The above is a first-order differential equation with constant coefficients

Assuming $V_C = 0 $ at $t = 0$, this can be solved to give
$$ \large {v = V(1-e ^{\frac{t}{CR}}) = V(1-e ^{\frac{t}{\text{T}}})} $$

– see Section 9.2.1 of the course text for this analysis

Since $i = C \text{d} \; v / \text{d}t$ this gives (assuming $V _C = 0$ at $t = 0$)
$$ \large {i = Ie ^{\frac{t}{CR}} = Ie ^{\frac{t}{\text{T}}}} $$

– where $I = V/R$

📷 Thus both the voltage and current have an exponential form - see here
Inductor de-energising

A similar analysis of this circuit

gives
\begin{align*} v =& - Ve ^{\Large \frac{Rt}{L}} = - Ve ^{\Large \frac{t}{\text{T}}} \\ \\ i =& Ie ^{\Large \frac{Rt}{L}} = I e ^{\Large \frac{t}{\text{T}}} \end{align*}
– where $I = V/R$
– see Section 9.3.1 for this analysis

📷 And once again, both the voltage and the current take the form of decaying exponentials

📷 A comparison of the four circuits

Discharging Capacitors and De-energising Inductors

Capacitor discharging - Consider this circuit for discharging a capacitor

– At $t = 0$, $V_C = V$

– From Kirchhoff’s voltage law
$$\large{iR + v = 0}$$
– giving
$$\large{CR \frac{dv}{dt} + v = 0}$$

Solving this as before gives
\begin{align*} v =& Ve ^{\Large \frac{t}{CR}} = Ve ^{\Large \frac{t}{\text{T}}} \\ \\ i =& - Ie ^{\Large \frac{t}{CR}} = - I e ^{\Large \frac{t}{\text{T}}} \end{align*}
– where $I = V/R$
– see Section 9.3.1 for this analysis

📷 In the above case, both the voltage and the current take the form of decaying exponentials - see here
(Continued)

The nature of exponential curves

📷 Response of first-order systems to a square waveform

– see Section 9.4.3

📷 Response of first-order systems to a square waveform of different frequencies

– see Section 9.4.3

Response of First-Order Systems

Initial and final value formulae
– increasing or decreasing exponential waveforms (for either voltage or current) are given by:
\begin{align*} v =& V_f +(V_i - V_f)e ^{\large - t / \text{T}} \\ \\ i =& I_f +(I_i - I_f)e ^{\large - t / \text{T}} \end{align*}

– where $V_i$ and $I_i$ are the initial values of the voltage and current

– where $V_f$ and $I_f$ are the final values of the voltage and current
– the first term in each case is the steady-state response

– the second term represents the transient response
– the combination gives the total response of the arrangement

Example – see Example 9.3 from course text
The input voltage to the following $CR$ network undergoes a step change from $5$ V to $10$ V at time $t = 0$. Derive an expression for the resulting output voltage

Here the initial value is $5$ V and the final value is $10$ V. The time constant of the circuit equals $CR = 10 \times 10^3 \times 20 \times 10^-6 = 0.2s.$
Therefore, from above, for $t \geq 0 $

\begin{align*} v =& V_f + (V_i - V_f)e ^{\Large -t/\text{T}} \\ \\ =& 10 + (5 - 10)e ^{\Large -t/ 0.2} \\ \\ =& 10 - 5e ^{\Large -t/ 0.2} \text{ volts} \end{align*}
(Continued)

Response of second-order systems

$\zeta = 0$ undamped

$\zeta$ < $1$ under damped

$\zeta = 1 $ critically damped

$\zeta$ > $1$ over damped

Higher-Order Systems

Higher-order systems are those that are described by third-order or higher-order equations

These often have a transient response similar to that of the second-order systems described earlier

Because of the complexity of the mathematics involved, they will not be discussed further here

Second-Order Systems

Circuits containing both capacitance and inductance are normally described by second-order differential equations. These are termed second-order systems
– for example, this circuit is described by the equation below it

$$ LC \frac{d^2v_C}{dt^2} + RC \frac{dv_C}{dt} +v_C = V $$

When a step input is applied to a second-order system, the form of the resultant transient depends on the relative magnitudes of the coefficients of its differential equation. The general form of the response is
$$ \frac{1}{\omega_n \; ^2} \frac{d^2 y}{dt^2} + \frac{2 \zeta}{\omega_n} \frac{d y}{dt} + y = x $$
– where $\omega_n $ is the undamped natural frequency in rad/s and $\zeta$ (Greek Zeta) is the damping factor
Further Study

🎥 Watch a video on determination of a circuit's time constant - Adobe Flash video

The Further Study section at the end of Chapter 9 considers the problem of determining the time constant of a circuit, so that the initial and final value theorems can be applied.

Two sample circuits are given so that you can test your understanding.

Calculate the time constants of the circuits and then check your results by looking at the video.

Key Points

The charging or discharging of a capacitor, and the energising and de-energising of an inductor, are each associated with exponential voltage and current waveforms

Circuits that contain resistance, and either capacitance or inductance, are termed first-order systems

The increasing or decreasing exponential waveforms of first-order systems can be described by the initial and final value formulae

Circuits that contain both capacitance and inductance are usually second-order systems. These are characterised by their undamped natural frequency and their damping factor
9.9 The switch in the following circuit is closed at $t = 0$. Derive an expression for the current in the circuit and hence calculate the current in the inductor at $t = 20$ ms.

9.10 In the circuit of Exercise 9.9, what is the final or steady‐state voltage across the inductor?

9.11 In the circuit of Exercise 9.9, what is the final or steady‐state voltage across the resistor?

9.12 A capacitor of $25 \;\mu F$ is initially charged to a voltage of $50$ V. At time $t = 0$, a resistance of $1 \; k \Omega$ is connected directly across its terminals. Derive an expression for the voltage across the capacitor as it is discharged and hence determine the time taken for its voltage to drop to $10$ V.

9.13 An inductor of $25$ mH is passing a current of $1$ A. At $t = 0$, the circuit supplying the current is instantly replaced by a resistor of $100 \Omega$ connected directly across the inductor. Derive an expression for the current in the inductor as a function of time and hence determine the time taken for the current to drop to $100$ mA.

9.14 What is meant by a ‘first-order system’, and what kind of circuits fall within this category?

9.15 Explain how the equation for an increasing or decreasing exponential waveform may be found using the initial and final values of the waveform.

9.16 The input voltage to the following $RC$ network undergoes a step change from $20$ V to $10$ V at time $t = 0$. Derive an expression for the resulting output voltage.

⬇Download chapter 9 tutorial

Exercises

9.1 Explain the meanings of the terms ‘steady-state response’ and ‘transient response’.

9.2 When a voltage is suddenly applied across a series combination of a resistor and an uncharged capacitor, what is the initial current in the circuit? What is the final, or steady-state, current in the circuit?

9.3 The switch in the following circuit is closed at $t = 0 s$. Derive an expression for the current in the circuit after this time and hence calculate the current in the circuit at $t = 4s$.

9.4 The switch in the following circuit is closed at $t = 0 s$. Derive an expression for the voltage v and hence calculate the voltage across the capacitor at $t = 5 s$

9.5 In the circuit of Exercise 9.4 what is the final or steady‐state voltage across the capacitor?

9.6 In the circuit of Exercise 9.4 what is the final or steady‐state voltage across the resistor?

9.7 When a voltage is suddenly applied across a series combination of a resistor and an inductor, what is the initial current in the circuit? What is the final, or steady-state, current in the circuit?

9.8 The switch in the following circuit is closed at $t = 0$. Deduce an expression for the output voltage of the circuit and hence calculate the time at which the output voltage will be equal to $8$ V.
9.22 Under what circumstances does the behaviour of a first-order high-pass filter resemble that of a differentiator?

9.23 Under what circumstances does the behaviour of a first-order low-pass filter resemble that of an integrator?

9.24 What is meant by a ‘second-order system’, and what kind of circuits fall within this category?

9.25 Derive an expression for the current in the circuit of Figure 9.8.

9.26 Explain what is meant by the terms ‘undamped natural frequency’ and ‘damping factor’ as they apply to second-order systems.

9.27 What is meant by ‘critical damping’ and what value of the damping factor corresponds to this situation?

⬇ Tutorial Solutions

Exercises (cont.)

9.17 The input voltage to the following $RL$ network undergoes a step change from $20$ V to $10$ V at time $t = 0$. Derive an expression for the resulting output voltage.

9.18 What is meant by a saturating exponential waveform?

9.19 What is meant by a decaying exponential waveform?

9.20 Sketch the exponential waveform $v = 5e ^{\large − t /10}$.

9.21 For each of the following circuit arrangements, sketch the form of the output voltage when the period of the square-wave input voltage is:
(a) much greater than the time constant of the circuit;
(b) equal to the time constant of the circuit;
(c) much less than the time constant of the circuit.