(Continued)

Sample gains expressed in dBs

Power gain	Decibels (dBs)
$100$	$20$
$10$	$10$
$1$	$0$

Power gain	Decibels (dBs)
$0.5$	$-3$
$0.1$	$-10$
$0.01$	$-20$

Using dBs simplifies calculation in cascaded circuits

Power gain is related to voltage gain

\begin{align*} \text{Power gain (dB)}= 10 \text{log}_{10} \frac{P_2}{P_1} = 10 \text{log}_{10} \frac{V_2^2 / R_2}{V_1^2 / R_1} \end{align*}

If $R _1 = R_2$

\begin{align*} \text{Power gain (dB)}= 10 \text{log}_{10} \frac{V_2^2}{V_1^2} = 20 \text{log}_{10} \frac{V_2}{V_1} \end{align*}
\begin{align*} \text{Power gain (dB)}= 20 \text{log}_{10} \text{(Voltage gain)} \end{align*}

This expression is often used even when $R_1 \mathrel{{=}\llap{/}} R _2$

– see Example 8.4 in the course text

Two-port Networks

A two-port network has two ports:

– an input port

– an output port

We then define voltages and currents at the input and output

Then

\begin{align*} \text{voltage gain }(A_v) =& \frac{V_o}{V_i} \\ \\ \text{current gain }(A_i) =& \frac{I_o}{I_i} \\ \\ \text{voltage gain }(A_p) =& \frac{P_o}{P_i} \end{align*}

The Decibel (dB)

The power gain of modern electronic amplifiers is often very high, gains of $10^6$ or $10^7$ being common

With such large numbers it is often convenient to use a logarithmic expression for gain

This is often done using decibels

The decibel is a dimensionless figure for power gain

\begin{align*} \large \text{Power gain (dB)} = 10 \text{ log}_{10} \frac{P_2}{P_1} \end{align*}

(Continued)

Clearly the transfer function is

\begin{align*} \frac{V_o}{V_i} = \frac{Z_R}{Z_R + Z_C}= \frac{R}{R-j \frac{1}{\large \omega C}} = \frac{1}{1-j \frac{1}{\large\omega CR}} \end{align*}

📷   See as circuit diagram

At high frequencies

– $\omega$ is large, voltage gain ≈ $1$

At low frequencies

– $\omega$ is small, voltage gain → $0$

Since the denominator has real and imaginary parts, the magnitude of the voltage gain is $$ \text{|Voltage gain |} = \frac{1}{\sqrt{1^2 + \bigg(\frac{1}{\omega CR}\bigg) ^2} } $$

📷   See as circuit diagram

When $1 / \omega \;CR = 1$
$$\text{|Voltage gain |} = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} = 0.707$$

This is a halving of power, or a fall in gain of $3$ dB

The half power point is the cut-off frequency of the circuit

– the angular frequency $\omega _C$ at which this occurs is given by
\begin{align*} \frac{1}{\omega _C CR}=& 1 \\ \\ \omega _C = \frac{1}{CR}=& \frac{1}{ \text{T}} \text{rad/s} \end{align*}

– where Τ is the time constant of the $CR$ network. Also
$$ f_C = \frac{\omega_C}{2 \pi} = \frac{1}{2 \pi CR} \text{Hz} $$

Frequency response

Since the characteristics of reactive components change with frequency, the behaviour of circuits using these components will also change

The way in which the gain of a circuit changes with frequency is termed its frequency response

These variations take the form of variations in the magnitude of the gain and in the phase response

We will start by considering very simple circuits

Consider the potential divider shown here

– from our earlier consideration of the circuit
$$ \large {V_o = V_i \times \frac{Z_2}{Z_1 + Z_2}}$$

– rearranging, the gain of the circuit is
$$\large {\frac{V_o}{V_i} = \frac{Z_2}{Z_1 + Z_2}}$$

– this is also called the transfer function of the circuit

A High-Pass RC Network

Consider the following circuit

– which is shown re-drawn in a more usual form

The behaviour in these three regions can be illustrated using phasor diagrams

At low frequencies the gain is linearly related to frequency. It falls at -6dB/octave (-20dB/decade)

Frequency response of the high-pass network

– the gain response has two asymptotes that meet at the cut-off frequency

– figures of this form are called Bode diagrams

(Continued)

Substituting $\omega = 2 \; \pi \; f $ and $CR = 1/2 \; \pi \; f_C$ in the earlier equation gives

\begin{align*} \large {\frac{V_o}{V_i} = \frac{1}{1-j \frac{1}{\omega CR}} = \frac{1}{1-j \Large \frac{1}\Large{(2 \pi f) \bigg( \frac{ 1}{ 2 \pi f_C} \bigg)} } = \frac{1}{1-j \frac{f_C}{f}}} \end{align*}

This is the general form of the gain of the circuit

It is clear that both the magnitude of the gain and the phase angle vary with frequency

Consider the behaviour of the circuit at different frequencies:

When $f$ > > $f_c$

– $ f _c / f $ < < $1$, the voltage gain ≈ $1$

When $f = f_c$

\begin{align*} \frac{V_o}{V_i} = \frac{1}{1-j \frac{\large f_c}{f}} = \frac{1}{1-j} = \frac{1 \times (1+j)}{(1-j) \times (1+j)} = \frac{(1 + j)}{2} = 0.5 + 0.5j \end{align*}

When $f$ < < $f_c$
\begin{align*} \frac{V_o}{V_i} = \frac{1}{1-j \frac{\large f_c}{f}} \text{≈} \frac{1}{- j \frac{\large f_c}{f}} = j \frac{f}{\large f_c} \end{align*}

 

Therefore

– the angular frequency $\omega _C$ at which this occurs is given by
\begin{align*} \omega _C CR =& 1 \\ \\ \omega _C = \frac{1}{CR} =& \frac{1}{\text{T}} \text{rad/s} \end{align*}

– where Τ is the time constant of the $CR$ network, and as before
$$ f _C = \frac{\omega _C}{2 \pi} = \frac{1}{2 \pi CR} \text{Hz} $$

Substituting $\omega = 2 \; \pi \; f \text{ and } CR = 1/2 \; \pi \; f_C$ in the earlier equation gives

\begin{align*} \large{\frac{V_o}{V_i} = \frac{1}{1 + j \omega CR}= \frac{1}{1 + j \frac{\omega }{\omega_C}} = \frac{1}{1 + j \frac{f}{f_C}}} \end{align*}

This is similar, but not the same, as the transfer function for the high-pass network

A Low-Pass RC Network

Transposing the C and R gives

\begin{align*} \frac{V_o}{V_i} = \frac{Z_C}{Z_R + Z_C}= \frac{-j \frac{\large 1}{\large \omega C}}{R-j \frac{\large 1}{\large \omega C}} = \frac{1}{1+j \omega CR} \end{align*}

At high frequencies

– ω is large, voltage gain → 0

At low frequencies

– ω is small, voltage gain ≈ 1

A similar analysis to before gives (see reference diagram above)
$$ \text{|Voltage gain |} = \frac{1}{\sqrt{1 + \omega CR ^2} } $$

Therefore when, when ω $CR$ = 1
$$ \text{|Voltage gain |} = \frac{1}{\sqrt{1 + 1}} = \frac{1}{\sqrt{2}} = 0.707 $$

Which is the cut-off frequency

At high frequencies the gain is linearly related to frequency. It falls at 6dB/octave (20dB/decade)

Frequency response of the low-pass network

– the gain response has two asymptotes that meet at the cut-off frequency

– you might like to compare this with the Bode Diagram for a high-pass network

(Continued)

Consider the behaviour of this circuit at different frequencies:

When $f$ < < $f_c$

– $f / f_c $ < < $1$, the voltage gain ≈ $1$

When $f = f_c$

\begin{align*} \frac{V_o}{V_i} = \frac{1}{1+j \frac{f}{\large f_c}} = \frac{(1-j) (1+j)}{(1+j)} = \frac{(1 - j)}{2} = 0.5 + 0.5j \end{align*}

When $f$ > > $f_c$

\begin{align*} \frac{V_o}{V_i} = \frac{1}{1+j \frac{f}{\large f_c}} \text{≈} \frac{1}{j \frac{f}{\large f_c}} = - j \frac{\large f_c}{f} \end{align*}

The behaviour in these three regions can again be illustrated using phasor diagrams

A Comparison of RC and RL Networks

Circuits using $RC$ and $RL$ techniques have similar characteristics

– see Figure 8.12 in the course text

A Low-Pass RL Network

Low-pass networks can also be produced using $RL$ circuits

– these behave similarly to the corresponding CR circuit

– the voltage gain is

\begin{align*} \frac{V_o}{V_i} = \frac{Z_R}{Z_R + Z_L} = \frac{R}{R + j \omega L} = \frac{1}{1 + j \omega \frac{L}{R}} \end{align*}

– the cut-off frequency is

\begin{align*} \omega _C = \frac{R}{L} = \frac{1}{\text{T}} \text{rad/s} \qquad \qquad f _C = \frac{\omega _C}{2 \pi} = \frac{R}{2 \pi L} \text{Hz} \end{align*}

A High-Pass RL Network

High-pass networks can also be produced using $RL$ circuits

– these behave similarly to the corresponding CR circuit

– the voltage gain is

\begin{align*} \frac{V_o}{V_i} = \frac{Z_L}{Z_R + Z_L} = \frac{j \omega L}{R + j \omega L} = \frac{1}{1 + \frac{R}{j \omega L}} =\frac{1}{1 - j \frac{R}{\omega L}} \end{align*}

– the cut-off frequency is

\begin{align*} \omega _C = \frac{R}{L} = \frac{1}{\text{T}} \text{rad/s} \qquad \qquad f _C = \frac{\omega _C}{2 \pi} = \frac{R}{2 \pi L} \text{Hz} \end{align*}

Combining the Effects of Several Stages

📷   The effects of several stages ‘add’ in bode diagrams - see here

Multiple high- and low-pass elements may also be combined

– this is illustrated in Figure 8.16 in the course text

Bode Diagrams

🎥  Watch a video on Bode diagrams - Adobe Flash video

Straight-line approximations

Creating more detailed Bode diagrams

The series $RLC$ circuit is an acceptor circuit

– the narrowness of bandwidth is determined by the $Q$
\begin{align*} \text{Quality factor } Q = \frac{\small \text{Resonant frequency}}{Bandwidth} = \frac{f_o}{B} \end{align*}

– combining this equation with the earlier one gives
\begin{align*} B = \frac{R}{2 \pi L} \text{Hz} \end{align*}

Parallel $RLC$ circuits

– as before
$$ \omega _o = \frac{1}{\sqrt{LC}} \qquad f _o = \frac{1}{2 \pi \sqrt{LC}} $$

The parallel arrangement is a rejector circuit

– in the parallel resonant circuit , the impedance is at a maximum at resonance

– the current is at a minimum at resonance

– in this circuit
$$ Q = R \sqrt{\bigg(\frac{C}{L}\bigg)} \qquad B = \frac{1}{2 \pi RC} \text{Hz} $$

RLC Circuits and Resonance

Series $RLC$ circuits

– the impedance is given by

\begin{align*} Z = R + j \omega L + \frac{1}{j \omega C} = R + j (\omega L -\frac{1}{\omega C }) \end{align*}

– if the magnitude of the reactance of the inductor and capacitor are equal, the imaginary part is zero, and the impedance is simply $R$

– this occurs when

\begin{align*} \omega L = \frac{1}{ \omega C} \qquad \omega ^2 = \frac{1}{LC} \qquad \omega = \frac{1}{\sqrt{LC}} \end{align*}

This situation is referred to as resonance

– the frequency at which is occurs is the resonant frequency
$$ \omega _o = \frac{1}{\sqrt{LC}} \qquad f _o = \frac{1}{2 \pi \sqrt{LC}} $$

– in the series resonant circuit , the impedance is at a minimum at resonance

– the current is at a maximum at resonance

The resonant effect can be quantified by the quality factor, $Q$

– this is the ratio of the energy dissipated to the energy stored in each cycle

– it can be shown that
\begin{align*} \text{Quality factor } Q = \frac{X_L}{R} = \frac{X_C}{R} \end{align*}

– and
\begin{align*} Q = \frac{1}{R} \sqrt{\bigg(\frac{L}{C}\bigg)} \end{align*}

(Continued)

Common forms include:

Butterworth

– optimised for a flat response

Chebyshev

– optimised for a sharp ‘knee’

Bessel

– optimised for its phase response see Section 8.13.3 of the course text for more information on these

Stray Capacitance and Inductance

All circuits have stray capacitance and stray inductance

– these unintended elements can dramatically affect circuit operation

– for example:

(a) $C_s$ adds an unintended low-pass filter

(b) $L_s$ adds an unintended low-pass filter

(c) $C_s$ produces an unintended resonant circuit and can produce instability

Filters

$RC$ Filters

The $RC$ networks considered earlier are first-order or single-pole filters

– these have a maximum roll-off of 6 dB/octave
– they also produce a maximum of 90° phase shift

Combining multiple stages can produce filters with a greater ultimate roll-off rates (12 dB, 18 dB, etc.) but such filters have a very soft ‘knee’

An ideal filter would have constant gain and zero phase shift for frequencies within its pass band , and zero gain for frequencies outside this range (its stop band )

Real filters do not have these idealised characteristics

$LC$ Filters

Simple $LC$ filters can be produced using series or parallel tuned circuits

– these produce narrow- band filters with a centre frequency $f_o$

Active filters

– combining an op-amp with suitable resistors and capacitors can produce a range of filter characteristics

– these are termed active filters

Further Study

🎥  Watch a video on the design of a radio frequency filter - Adobe Flash video

The Further Study section at the end of Chapter 8 looks at the filter needed to select a single radio station from the multitude being transmitted.

You are required to design a simple $RLC$ filter to select a given station.

When you have finished your design watch the video to see why this is not as simple as it perhaps appears.

Key Points

The reactance of capacitors and inductors is dependent on frequency

Single $RC$ or $RL$ networks can produce an arrangement with a single upper or lower cut-off frequency

In each case the angular cut-off frequency $\omega _o$ is given by the reciprocal of the time constant $\text{Τ}$

For an $RC$ circuit $\text{Τ} = CR$, for an $RL$ circuit $\text{Τ} = L/R$

Resonance occurs when the reactance of the capacitive element cancels that of the inductive element

Simple $RC$ or $RL$ networks represent single-pole filters

Active filters produce high performance without inductors

Stray capacitance and inductance are found in all circuits

8.14 Determine the frequencies that correspond to:

(a) an octave below $30$ Hz;

(b) two octaves above $25$ kHz;

(c) three octaves above $1$ kHz;

(d) a decade above $1$ MHz;

(e) two decades below $300$ Hz;

(f ) three decades above $50$ Hz.

8.15 Calculate the time constant $\text{T}$ , the angular cut-off frequency $ \omega _c$ and the cyclic cut-off frequency $ f _c$ of the following arrangement. Is this a high- or a low- frequency cut-off ?

8.16 A parallel $RL$ circuit is formed from a resistor of $150 \; \omega$ and an inductor of $30$ mH. What is the time constant of this circuit?

8.17 Calculate the time constant $\text{T}$ , the angular cut-off frequency $ \omega _c$ and the cyclic cut-off frequency $ f _c$ of the following arrangement. Is this a high- or a low- frequency cut-off ?

⬇Download chapter 8 tutorial

Exercises

8.1 What is meant by a ‘two‐port network’, and what are the two ports?

8.2 Derive expressions for the voltage gain, current gain and power gain of a two‐port network in terms of the input and output voltages, and the input and output currents.

8.3 Determine the voltage gain, current gain and power gain of the following arrangement.

8.4 Calculate the overall power gain of the following arrangement if the power gain of each stage is as shown in the diagram.

8.5 For the arrangement shown in Exercise 8.4, determine the gain of each stage in decibels, and then compute the gain of the overall arrangement in decibels.

8.6 A circuit has a gain of $25$ dB. What is its power gain (expressed as a simple ratio)?

8.7 A circuit has a gain of $25$ dB. What is its voltage gain?

8.8 Calculate the reactance of a $1 \mu F$ capacitor at a frequency of $10$ kHz, and the reactance of a 20 mH inductor at a frequency of $100$ rad/s. In each case include the units in your answer.

8.9 Express an angular frequency of $250$ rad/s as a cyclic frequency (in Hz).

8.10 Express a cyclic frequency of $250$ Hz as an angular frequency (in rad/s).

8.11 Determine the transfer function of the following circuit.

8.12 A series $RC$ circuit is formed from a resistor of $33 \; k \; \omega$ and a capacitor or $15 \; nF$. What is the time constant of this circuit?

8.13 Calculate the time constant $\text{T}$ , the angular cut-off frequency $\omega _c$ and the cyclic cut-off frequency $f _c$ of the following arrangement. Is this a high- or a low- frequency cut-off ?

8.24 Calculate the resonant frequency $ f _o$, the quality factor $Q$ and the bandwidth B of a parallel circuit with a resistor of $1 \;k \Omega$, an inductor of 50 mH and a capacitor of $22 \; \mu F$.

8.25 Why is it more common to construct first order filters using combinations of resistors and capacitors, rather than resistors and inductors.

8.26 Explain the difference between a passive and an active filter.

8.27 Why are inductors often avoided in the construction of filters?

8.28What form of active filter is optimised to produce a flat response within its pass band?

8.29 What form of active filter is optimised to produce a sharp transition from the pass band to the stop band?

8.30 What form of filter is optimised for a linear phase response?

8.31 Explain why stray capacitance and stray inductance affect the frequency response of electronic circuits.

⬇ Tutorial Solutions

Exercises (cont.)

8.18 Calculate the time constant T , the angular cut-off frequency $ \omega _c$ and the cyclic cut-off frequency $ f _c$ of the following arrangement. Is this a high- or a low- frequency cut-off ?

8.19 Sketch a straight-line approximation to the Bode diagram of the circuit of Exercise 8.14. Use this approximation to produce a more realistic plot of the gain and phase responses of the circuit.

8.20 A circuit contains three high-frequency cut-offs and two low-frequency cut-offs. What are the rates of change of gain of this circuit at very high and very low frequencies?

8.21In the arrangement described in Exercise 8.20, what phase shift is produced at very high and very low frequencies?

8.22 Explain what is meant by the term ‘resonance’.

8.23 Calculate the resonant frequency $ f _o$ , the quality factor $Q$ and the bandwidth $B$ of the following circuit.