SEHH2268 Electronic Circuits - Chapter 1 Frequency Domain Analysis

Overview

Introduction to the transfer function for describing the relationship between a circuit's input and output.
Introduction to Bode plots and their utility for describing a circuit's frequency response.
Coverage of the concept of resonance as it applies to LRC circuits.
Discussion of frequency filters.

Introduction (1)

Electronic engineering utilizes non-linear and active electrical components like electron tubes, semiconductor devices (transistors, diodes, integrated circuits), passive electrical components, and printed circuit boards.

Introduction (2)

Electronic engineering subfields:
- Analog electronics
- Digital electronics
- Consumer electronics
- Embedded systems
- Power electronics
Applications:
- Communication systems
- Control systems
- Computer systems

Revision on phasor representation (1)

AC circuit sources (in time domain) can be represented by phasors (in frequency domain).
- $X(t) = A\cos(\omega t + \theta) \leftrightarrow X(j\omega) = A\angle \theta$
- Where:
  - A = magnitude
  - $\omega$ = angular frequency
  - $\theta$ = phase of the sources
- X can be voltage or current.
For sinusoidal sources with a single frequency $\omega1$ , phasors represent frequency responses measured at $\omega1$ .

Revision on phasor representation (2)

Circuits with multiple frequency sources require the superposition principle for each frequency.
- Phasors at a specific frequency represent frequency responses at that frequency.
- Example: $V(t) = A1 \cos(\omega1 t + \theta1) + A2 \cos(\omega2 t + \theta2)$
- Phasors:
 - At $\omega1$ , $X(j\omega1) = A1 \angle \theta1 V$
 - At $\omega2$ , $X(j\omega2) = A2 \angle \theta2 V$

Revision on phasor representation (3)

Impedance calculations for components:
- Resistance R: $Z_R(\omega) = R \Omega$
- Capacitance C: $Z_C(\omega) = -\frac{j}{\omega C} \Omega$
- Inductance L: $Z_L(\omega) = j\omega L \Omega$
Capacitance and inductance in AC circuits are "frequency dependent resistances."

Sinusoidal frequency response

Sinusoidal frequency response measures load voltage or current variation as a function of excitation signal frequency.
Knowing a circuit's frequency response allows computation of the output signal given the input signal's amplitude, phase, and frequency.

Example: CD player circuit model (1)

Illustrates a physical system (CD Player -> Amplifier -> Speakers) and its representation as a Thévenin equivalent circuit.
$ZT = (Zs + Z1) || Z2$
$VT = VS \frac{Z2}{Zs + Z1 + Z2}$

Example: CD player circuit model (2)

$VL = \frac{ZL}{ZL + ZT} VT = \frac{ZL}{ZL + (ZS + Z1)Z2 / (ZS + Z1 + Z2)} \cdot \frac{Z2}{ZS + Z1 + Z2} VS$
$\frac{VL}{VS}(j\omega) = HV(j\omega) = \frac{ZL Z2}{ZL(ZS + Z1 + Z2) + (ZS + Z1)Z2}$
$HV(j\omega)$ (frequency response) = output voltage $VL(j\omega)$ as a function of source voltage $V_S(j\omega)$ .

The output calculation by frequency response

$\frac{VL(j\omega)}{VS(j\omega)} = HV(j\omega) \rightarrow VL(j\omega) = HV(j\omega) \cdot VS(j\omega)$
$VL e^{j\phiL} = HV e^{j\phiH} \cdot VS e^{j\phiS} \rightarrow VL e^{j\phiL} = HV VS e^{j(\phiH + \phiS)}$
Amplitude response: $VL = HV V_S$
Phase response: $\phiL = \phiH + \phi_S$
Output current can be calculated similarly.

Example

Problem: Compute the frequency response $HV(j\omega) = \frac{VL(j\omega)}{V_S(j\omega)}$ for the given circuit.
Solution: Using equivalent circuit approach (Thévenin equivalent).
$VL = \frac{ZL}{ZT + ZL} VT = \frac{ZL}{\frac{Z1 Z2}{Z1 + Z2} + ZL} VS$
$HV(j\omega) = \frac{VL}{VS} = \frac{ZL Z2}{ZL(Z1 + Z2) + Z1 Z2}$
Given: $Z1 = 10^3 \Omega$ , $Z2 = \frac{1}{j\omega \times 10^{-5}} \Omega$ , $Z_L = 10^4 \Omega$
$H_V(j\omega) = \frac{10^4}{10^3 + \frac{1}{j\omega \times 10^{-5}}} / (10^4 + \frac{10^3}{j\omega \times 10^{-5}}) = \frac{100}{110 + j\omega} = \frac{100}{\sqrt{110^2 + \omega^2}} e^{-j \arctan(\frac{\omega}{110})}$
$|H_V(j\omega)| = \frac{100}{\sqrt{110^2 + \omega^2}}$
$\angle H_V(j\omega) = -\arctan(\frac{\omega}{110})$

Frequency Response

Frequency response = variation in a circuit’s behavior with changes in signal frequency.
Important in filter applications.
Filters block or pass specific frequencies/frequency ranges.
Essential for multiple channels of data in radio communications.

Transfer Function

Transfer function $H(\omega)$ analyzes a circuit's frequency response.
It's the frequency-dependent ratio of a forced function $Y(\omega)$ to the forcing function $X(\omega)$ .
$H(\omega) = \frac{Y(\omega)}{X(\omega)}$

Transfer Function

Four possible input/output combinations:
- Voltage gain: $H(\omega) = \frac{Vo(\omega)}{Vi(\omega)}$
- Current gain: $H(\omega) = \frac{Io(\omega)}{Ii(\omega)}$
- Transfer impedance: $H(\omega) = \frac{Vo(\omega)}{Ii(\omega)}$
- Transfer admittance: $H(\omega) = \frac{Io(\omega)}{Vi(\omega)}$

Zeros and Poles

$H(\omega)$ is obtained by converting circuit components to frequency domain equivalents.
$H(\omega)$ can be expressed as the ratio of numerator $N(\omega)$ and denominator $D(\omega)$ polynomials.
Zeros: roots of $N(\omega)$ where the transfer function goes to zero.
Poles: roots of $D(\omega)$ where the transfer function goes to infinity.
$H(\omega) = \frac{N(\omega)}{D(\omega)}$

Example

Obtain the transfer function $\frac{Vo}{Vs}$ for the RC circuit and its frequency response.
By voltage division: $H(\omega) = \frac{Vo}{Vs} = \frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} = \frac{1}{1 + j\omega RC}$
Magnitude: $|H(\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}} = \frac{1}{\sqrt{1 + (\frac{\omega}{\omegao})^2}}$ , where $\omegao = \frac{1}{RC}$
Phase: $\angle H(\omega) = -\tan^{-1}(\frac{\omega}{\omega_o})$

Decibel Scale

Bode plots are based on logarithmic scales.
Transfer function is seen as gain.
Gain in log form is expressed in bels or decibels (1/10 of a bel).
$G{dB} = 10 \log{10} \frac{P2}{P1} = 10 \log{10} \frac{\frac{V2^2}{R2}}{\frac{V1^2}{R1}} = 20 \log{10} \frac{V2}{V1} - 10 \log{10} \frac{R2}{R_1}$
When $R2 = R1$ , $G{dB} = 20 \log{10} \frac{V2}{V1}$

Bode Plots

Transfer function needs to cover a large frequency range.
Semilog plots (x-axis in log form) make it easier.
Bode plots show magnitude (in decibels) or phase (in degrees) as a function of frequency.
Usually drawn on semi-log paper.

Standard Form

Transfer function in terms of factors with real and imaginary parts.
$H(j\omega) = K \frac{(1 + j\omega/z1)(1 + j2\zeta1 \omega/\omegak + (j\omega/\omegak)^2)}{(j\omega)(1 + j\omega/p1)(1 + j2\zeta2 \omega/\omegan + (j\omega/\omegan)^2)}$
Seven standard factors:
- Gain K
- Pole $(j\omega)^{-1}$ or zero $(j\omega)$ at the origin
- Simple pole $\frac{1}{1+j\omega/p1}$ or simple zero $(1+j\omega/z1)$
- Quadratic pole $\frac{1}{[1+j2\zeta2 \omega/ \omegan+ (j\omega/ \omegan)^2]}$ or zero $[1+j2\zeta1 \omega/ \omegak+ (j\omega/ \omegak)^2]$

Bode Plots

Each factor is plotted separately and then added graphically.
Gain, K: magnitude is $20\log_{10}K$ and phase is $0°$ (constant with frequency).

Bode Plots

Pole/zero at the origin: For zero $(j\omega)$ , magnitude slope is $20 \text{ dB/decade}$ and phase is $90°$ . For pole $(j\omega)^{-1}$ , magnitude slope is $-20 \text{ dB/decade}$ and phase is $-90°$ .

Bode Plots

Simple pole/zero: For simple zero, magnitude is $20\log{10}|1+j\omega/z1|$ and phase is $\tan^{-1} \omega/z_1$ .
Approximated as flat line and sloped line intersecting at $\omega=z_1$ .
$\omega=z_1$ is the corner or break frequency.

Bode Plots

Phase plotted as straight lines:
- From $\omega=0$ to $\omega \le z_1/10$ , $\phi=0$
- At $\omega=z_1$ , $\phi=45°$
- For $\omega \ge 10z_1$ , $\phi= 90°$
Pole is similar, except corner frequency is at $\omega=p_1$ , and magnitude has negative slope.

Bode Plots

Quadratic pole/zero: Magnitude of quadratic pole $\frac{1}{[1+j2\zeta2 \omega/ \omegan+ (j\omega/ \omegan)^2]}$ is $-20\log{10} |[1+j2\zeta2 \omega/ \omegan+ (j\omega/ \omegan)^2]|$ ([zeta2] is damping factor and $\omega_n$ is corner frequency).
Approximation:
- Two lines: slope zero for \omega<\omegan and slope $-40 \text{dB/decade}$ for $\omega>\omegan$ , with $\omega_n$ as the corner frequency.

Bode Plots

Phase can be expressed as:
- $\phi = \begin{cases} 0, & \omega << \omegan \ -90 \degree, & \omega = \omegan \ -180 \degree, & \omega >> \omega_n \end{cases}$
This will be a straight line with a slope of $-90°/decade$ starting at $\omegan/10$ and ending at $10 \omegan$ .
For the quadratic zero, the plots are inverted.
Exact plots depend on the damping factor $\zeta2$ and corner frequency $\omegan$ .

Bode Plots

Summary of Bode straight-line magnitude and phase plots (Table provided in the slides).

Bode Plots

(Continuation of the Bode plots summary table).

Example 14.3

Constructing Bode plots for the transfer function $H(w) = \frac{200 jw}{(jw + 2)(jw + 10)}$ .

Example 14.5

Drawing Bode plots for $H(s) = \frac{s+1}{s^2 + 12s + 100}$ .

Resonance

Sharp peak in the amplitude characteristics is a prominent feature.
Occurs in systems with complex conjugate pole pairs.
Enables energy storage in oscillations.
Allows frequency discrimination.
Requires at least one capacitor and inductor.

Series Resonance

Series resonant circuit: inductor and capacitor in series.
Resonance when the imaginary part of impedance Z is zero.
Resonant frequency: $\omega0 = \frac{1}{\sqrt{LC}}$ rad/s or $f0 = \frac{1}{2\pi \sqrt{LC}}$ Hz

Series Resonance

At resonance:
- Impedance is purely resistive.
- Voltage $V_s$ and current I are in phase.
- Transfer function magnitude is minimum.
- Inductor and capacitor voltages can be much greater than the source.
Frequency response of the current magnitude: maximum power occurs at $V_m/R$ .

Series Resonance

At half power frequencies $\omega = \omega1, \omega2$ :
Dissipated power is half the maximum value.
$P(\omega) = \frac{1}{2} I^2 R = \frac{1}{2} \frac{Vm^2}{R^2 + (\omega L - 1/\omega C)^2} R = \frac{Vm^2}{4R}$
Relating half-power frequencies with resonant frequency:

Series Resonance

Bandwidth B is the difference between half-power frequencies:
$B = \omega2 - \omega1$
Quality factor Q measures resonance sharpness:
$Q = \frac{\omega0 L}{R} = \frac{1}{\omega0 CR}$
At resonance, reactive energy oscillates between inductor and capacitor.

Quality Factor

$Q = 2\pi \frac{\text{Maximum energy stored}}{\text{Energy dissipated per cycle}}$
Measure of peak energy stored divided by energy dissipated in one period at resonance.
$Q = \frac{\omega_0 L}{R}$

Quality Factor

Ratio of resonant frequency to its bandwidth, B.
$Q = \frac{\omega_0}{B}$
High-Q circuit: $Q \ge 10$ , half-power frequencies are approximated by:

Example14.7

Given: R = 2 $\Omega$ , L = 1 mH, C = 0.4 $\mu$ F, $V(t) = 20 \sin(\omega t)$
Find: Resonant frequency, half-power frequencies, quality factor, bandwidth, and current amplitudes at $\omega0, \omega1, \omega_2$ .

Parallel Resonance

Parallel RLC circuit is the dual of the series circuit.
Resonance when the imaginary part of the admittance is zero.
Same resonant frequency as in the series circuit: $\omega_0 = \frac{1}{\sqrt{LC}}$

Parallel Resonance

Relevant equations for parallel resonant circuit:
- $B = \omega2 - \omega1 = \frac{1}{RC}$
- $Q = \frac{\omega0}{B} = \omega0 RC = \frac{R}{\omega_0 L}$
For high-Q circuit $(Q \ge 10)$

Example

Given: R = 8 k $\Omega$ , L = 0.2 mH, C = 8 $\mu$ F, $V(t) = 10 \sin(\omega t)$
Calculate: $\omega0$ , Q, B, $\omega1$ , $\omega2$ , and power dissipated at $\omega0$ , $\omega1$ , and $\omega2$ .

Time domain issues for the complex frequency variable

Time domain representation of s is the derivative operator $d/dt$ .
For capacitance, $i = C \frac{d}{dt}v = sCv$
For inductance, $v = L \frac{d}{dt}i = sLi$
Time domain solution by inverse Laplace transform: $v{out}(t) = \mathcal{L}^{-1}{v{out}(s)} = \mathcal{L}^{-1}{H(s)v_{in}(s)}$

The characteristic equation of a circuit (1)

In general, $s^n = \frac{d^n}{dt^n}$ , converts differential equation to characteristic equation.
\frac{d^n}{dt^n}x(t) + a1 \frac{d^{n-1}}{dt^{n-1}}x(t) + … + a{n-1} \frac{d}{dt}x(t) + an x(t) = 0 \rightarrow s^n + a1 s^{n-1} + … + a{n-1}s + an = 0
Characteristic equation gives free-oscillating response.
Roots affect system stability.

The characteristic equation of a circuit (2)

Output stimulus, Y, obtained by multiplying the input signal, X, to the transfer function, $Y = H(s)X$ .
Transfer function is a rational function, $H(s) = \frac{N(s)}{D(s)}$ .
$D(s) = 0$ is the characteristic equation of the circuit. (Same as the one obtained in the corresponding differential equation)

The characteristic equation of a circuit (3)

Proof: $\frac{Y}{X} = H(s) = \frac{N(s)}{D(s)} = \frac{s^m + a1 s^{m-1} + …}{s^n + b1 s^{n-1} + …}$
$(s^n + b1 s^{n-1} + …)Y = (s^m + a1 s^{m-1} + …)X$
Free oscillating response when $X(s) = 0$ :
$(s^n + b_1 s^{n-1} + …)Y = 0$

Complex frequency concept (1)

Substitution $s = j\omega$ :
$H(s) = \frac{1}{R+ \frac{1}{sC}} = \frac{1}{1+sRC}$ (Frequency response becomes a rational function).
s: complex frequency variable.
Impedance in terms of s:
- $Z_C = \frac{1}{j\omega C} = \frac{1}{sC}$
- $Z_L = j\omega L = sL$

Complex frequency concept (2)

Transfer function calculated in terms of s.
Specifies output to input relationship.
Important tool for circuit/system design, determining frequency response and stability.

Example

Transfer function in terms of complex frequency variable s:
$\frac{V2}{V1} = \frac{R || \frac{1}{sC}}{sL + R || \frac{1}{sC}} = \frac{1}{1 + s \frac{L}{R} + s^2 CL}$
Characteristic equation: $1 + s \frac{L}{R} + s^2 CL = 0$

Pole and zero in transfer function

By factorization:
$H(s) = \frac{(s-zm)(s-z{m-1})…(s-z1)}{(s-pn)(s-p{n-1})…(s-p1)}$
$z_i$ : zero of transfer function (amplitude response becomes zero).
$p_j$ : pole of transfer function (amplitude response goes to infinity).

Example

Amplitude response for one pole transfer function.
The amplitude response looks like a surface function on top of the s plane.

Mapping of the s plane to the frequency domain

Frequency response obtained by projecting the amplitude response to the imaginary axis of the s-plane.
Setting $s = j\omega$ in the transfer function.

Example

Frequency response of a simple RC circuit

Frequency response of first order circuit

$G(s) = \frac{1}{1+sCR}$
$G(\omega) = \frac{1}{1+ \omega CR}$
pole $= - \frac{1}{CR}$
When $s \rightarrow - \frac{1}{CR}$ , $G(s) \rightarrow \infty$
When $s \rightarrow \infty$ , $G(s) \rightarrow 0$
Magnitude response drops to $\frac{1}{\sqrt{2}}$ of the highest value is the 3dB corner frequency.
Corner frequency: $\omega_C = \frac{1}{CR}$

Phase shift of the first order circuit

$\phi(\omega) = -\tan^{-1}(\omega CR)$
At $\omega = 0$ , $\phi(\omega) = 0°$
At corner frequency, $\phi(\omega) = -45°$
When $\omega \rightarrow \infty$ , $\phi(\omega) = -90°$

Example

Find the transfer function, magnitude response, and phase response for the given circuit.
Find the magnitude of the transfer function at $\omega = 1 \text{rad/s}, 100 \text{rad/s}, 100k \text{rad/s}$ .

Solution

$\frac{V{out}}{V{in}} = G(s) = \frac{4}{28+65s}$

So, $|G(\omega)| = \frac{4}{\sqrt{28^2 + 65^2\omega^2}}$
And $\phi(\omega) = -\tan^{-1} \frac{65\omega}{28}$
At $\omega = 1 \text{rad/s}, |G(1)| = 0.0565$
At $\omega = 100 \text{rad/s}, |G(100)| = 6.154 \times 10^{-4}$
At $\omega = 100k \text{rad/s}, |G(100k)| = 6.154 \times 10^{-7}$

Passive Filters

A filter passes signals with desired frequencies and rejects others.
Passive filter: consists only of passive elements (R, L, C).
Important circuits for technological advances.

Passive Filters

Four types of filters:
- Lowpass: passes low frequencies, blocks high frequencies.
- Highpass: passes high frequencies, blocks low frequencies.
- Bandpass: allows a range of frequencies to pass through.
- Bandstop: blocks a range of frequencies.

Lowpass Filter

Output of RC circuit taken off the capacitor.
Half power frequency: $\omega_c = \frac{1}{RC}$
Also referred to as the cutoff frequency.
Filter passes from DC up to $\omega_c$ .

3 db point Or cut off point

Note: The frequency axis has been scaled logarithmically as it enables viewing a very board range of frequencies on the same plot without excessively compressing the low frequency end of the plot.

Highpass Filter

RC circuit with output taken off the resistor.
Cutoff frequency: Same as lowpass filter $\omega_c = \frac{1}{RC}$ .
Frequencies passed go from $\omega_c$ to infinity.

Bandpass Filter

RLC series resonant circuit with output taken off the resistor.
Center frequency: $\omega_0 = \frac{1}{\sqrt{LC}}$
Filter passes frequencies from $\omega1$ to $\omega2$ .
Made by feeding the output from a lowpass to a highpass filter.

Bandstop Filter

RLC circuit with output from the LC series combination.
Range of blocked frequencies matches the range of passed frequencies for the bandpass filter.

Example

Determine the type of filter shown. Calculate the corner or cutoff frequency. Take R = 2 k $\Omega$ , L = 2 H, and C = 2 $\mu$ F.

SEHH2268 Electronic Circuits - Chapter 1 Frequency Domain Analysis

Overview

Introduction (1)

Introduction (2)

Revision on phasor representation (1)

Revision on phasor representation (2)

Revision on phasor representation (3)

Sinusoidal frequency response

Example: CD player circuit model (1)

Example: CD player circuit model (2)

The output calculation by frequency response

Example

Frequency Response

Transfer Function

Transfer Function

Zeros and Poles

Example

Decibel Scale

Bode Plots

Standard Form

Bode Plots

Bode Plots

Bode Plots

Bode Plots

Bode Plots

Bode Plots

Bode Plots

Bode Plots

Example 14.3

Example 14.5

Resonance

Series Resonance

Series Resonance

Series Resonance

Series Resonance

Quality Factor

Quality Factor

Example14.7

Parallel Resonance

Parallel Resonance

Example

Time domain issues for the complex frequency variable

The characteristic equation of a circuit (1)

The characteristic equation of a circuit (2)

The characteristic equation of a circuit (3)

Complex frequency concept (1)

Complex frequency concept (2)

Example

Pole and zero in transfer function

Example

Mapping of the s plane to the frequency domain

Example

Frequency response of first order circuit

Phase shift of the first order circuit

Example

Solution

Passive Filters

Passive Filters

Lowpass Filter

More on Low pass filters

3 db point Or cut off point

Highpass Filter

More on High Pass Filter

Bandpass Filter

More on Bandpass Filter

Bandstop Filter

More on Bandstop Filter

Example