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 $\omega<em>1$, phasors represent frequency responses measured at $\omega</em>1$.

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: $V(t) = A<em>1 \cos(\omega</em>1 t + \theta<em>1) + A</em>2 \cos(\omega<em>2 t + \theta</em>2)$
  • Phasors:
    • At $\omega<em>1$, $X(j\omega</em>1) = A<em>1 \angle \theta</em>1 V$
    • At $\omega<em>2$, $X(j\omega</em>2) = A<em>2 \angle \theta</em>2 V$

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.
• $Z<em>T = (Z</em>s + Z<em>1) || Z</em>2$
• $V<em>T = V</em>S \frac{Z<em>2}{Z</em>s + Z<em>1 + Z</em>2}$

Example: CD player circuit model (2)

• $V<em>L = \frac{Z</em>L}{Z<em>L + Z</em>T} V<em>T = \frac{Z</em>L}{Z<em>L + (Z</em>S + Z<em>1)Z</em>2 / (Z<em>S + Z</em>1 + Z<em>2)} \cdot \frac{Z</em>2}{Z<em>S + Z</em>1 + Z<em>2} V</em>S$
• $\frac{V<em>L}{V</em>S}(j\omega) = H<em>V(j\omega) = \frac{Z</em>L Z<em>2}{Z</em>L(Z<em>S + Z</em>1 + Z<em>2) + (Z</em>S + Z<em>1)Z</em>2}$
• $H<em>V(j\omega)$ (frequency response) = output voltage $V</em>L(j\omega)$ as a function of source voltage $V_S(j\omega)$.

The output calculation by frequency response

• $\frac{V<em>L(j\omega)}{V</em>S(j\omega)} = H<em>V(j\omega) \rightarrow V</em>L(j\omega) = H<em>V(j\omega) \cdot V</em>S(j\omega)$
• $V<em>L e^{j\phi</em>L} = H<em>V e^{j\phi</em>H} \cdot V<em>S e^{j\phi</em>S} \rightarrow V<em>L e^{j\phi</em>L} = H<em>V V</em>S e^{j(\phi<em>H + \phi</em>S)}$
• Amplitude response: $V<em>L = H</em>V V_S$
• Phase response: $\phi<em>L = \phi</em>H + \phi_S$
• Output current can be calculated similarly.

Example

• Problem: Compute the frequency response $H<em>V(j\omega) = \frac{V</em>L(j\omega)}{V_S(j\omega)}$ for the given circuit.
• Solution: Using equivalent circuit approach (Thévenin equivalent).
• $V<em>L = \frac{Z</em>L}{Z<em>T + Z</em>L} V<em>T = \frac{Z</em>L}{\frac{Z<em>1 Z</em>2}{Z<em>1 + Z</em>2} + Z<em>L} V</em>S$
• $H<em>V(j\omega) = \frac{V</em>L}{V<em>S} = \frac{Z</em>L Z<em>2}{Z</em>L(Z<em>1 + Z</em>2) + Z<em>1 Z</em>2}$
• Given: $Z<em>1 = 10^3 \Omega$, $Z</em>2 = \frac{1}{j\omega \times 10^{-5}} \Omega$, $Z_L = 10^4 \Omega$
• $H_V(j\omega) = \frac{10^4}{10^3 + \frac{1}{j\omega \times 10^{-5}}} / (10^4 + \frac{10^3}{j\omega \times 10^{-5}}) = \frac{100}{110 + j\omega} = \frac{100}{\sqrt{110^2 + \omega^2}} e^{-j \arctan(\frac{\omega}{110})}$
• $|H_V(j\omega)| = \frac{100}{\sqrt{110^2 + \omega^2}}$
• $\angle H_V(j\omega) = -\arctan(\frac{\omega}{110})$

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: $H(\omega) = \frac{V<em>o(\omega)}{V</em>i(\omega)}$
  • Current gain: $H(\omega) = \frac{I<em>o(\omega)}{I</em>i(\omega)}$
  • Transfer impedance: $H(\omega) = \frac{V<em>o(\omega)}{I</em>i(\omega)}$
  • Transfer admittance: $H(\omega) = \frac{I<em>o(\omega)}{V</em>i(\omega)}$

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 $\frac{V<em>o}{V</em>s}$ for the RC circuit and its frequency response.
• By voltage division: $H(\omega) = \frac{V<em>o}{V</em>s} = \frac{\frac{1}{j\omega C}}{R + \frac{1}{j\omega C}} = \frac{1}{1 + j\omega RC}$
• Magnitude: $|H(\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}} = \frac{1}{\sqrt{1 + (\frac{\omega}{\omega<em>o})^2}}$, where $\omega</em>o = \frac{1}{RC}$
• 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).
• $G<em>{dB} = 10 \log</em>{10} \frac{P<em>2}{P</em>1} = 10 \log<em>{10} \frac{\frac{V</em>2^2}{R<em>2}}{\frac{V</em>1^2}{R<em>1}} = 20 \log</em>{10} \frac{V<em>2}{V</em>1} - 10 \log<em>{10} \frac{R</em>2}{R_1}$
• When $R<em>2 = R</em>1$, $G<em>{dB} = 20 \log</em>{10} \frac{V<em>2}{V</em>1}$

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.
• $H(j\omega) = K \frac{(1 + j\omega/z<em>1)(1 + j2\zeta</em>1 \omega/\omega<em>k + (j\omega/\omega</em>k)^2)}{(j\omega)(1 + j\omega/p<em>1)(1 + j2\zeta</em>2 \omega/\omega<em>n + (j\omega/\omega</em>n)^2)}$
• Seven standard factors:
  • Gain K
  • Pole $(j\omega)^{-1}$ or zero $(j\omega)$ at the origin
  • Simple pole $\frac{1}{1+j\omega/p<em>1}$ or simple zero $(1+j\omega/z</em>1)$
  • Quadratic pole $\frac{1}{[1+j2\zeta<em>2 \omega/ \omega</em>n+ (j\omega/ \omega<em>n)^2]}$ or zero $[1+j2\zeta</em>1 \omega/ \omega<em>k+ (j\omega/ \omega</em>k)^2]$

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

Bode Plots

• Phase can be expressed as:
  • $\phi = \begin{cases} 0, & \omega << \omega<em>n \ -90 \degree, & \omega = \omega</em>n \ -180 \degree, & \omega >> \omega_n \end{cases}$
• This will be a straight line with a slope of $-90°/decade$ starting at $\omega<em>n/10$ and ending at $10 \omega</em>n$.
• For the quadratic zero, the plots are inverted.
• Exact plots depend on the damping factor $\zeta<em>2$ and corner frequency $\omega</em>n$.

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: $\omega<em>0 = \frac{1}{\sqrt{LC}}$ rad/s or $f</em>0 = \frac{1}{2\pi \sqrt{LC}}$ 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 $\omega = \omega<em>1, \omega</em>2$:
• Dissipated power is half the maximum value.
• $P(\omega) = \frac{1}{2} I^2 R = \frac{1}{2} \frac{V<em>m^2}{R^2 + (\omega L - 1/\omega C)^2} R = \frac{V</em>m^2}{4R}$
• Relating half-power frequencies with resonant frequency:

Series Resonance

• Bandwidth B is the difference between half-power frequencies:
• $B = \omega<em>2 - \omega</em>1$
• Quality factor Q measures resonance sharpness:
• $Q = \frac{\omega<em>0 L}{R} = \frac{1}{\omega</em>0 CR}$
• 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 $\omega<em>0, \omega</em>1, \omega_2$.

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:
  • $B = \omega<em>2 - \omega</em>1 = \frac{1}{RC}$
  • $Q = \frac{\omega<em>0}{B} = \omega</em>0 RC = \frac{R}{\omega_0 L}$
• 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: $\omega<em>0$, Q, B, $\omega</em>1$, $\omega<em>2$, and power dissipated at $\omega</em>0$, $\omega<em>1$, and $\omega</em>2$.

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: $v<em>{out}(t) = \mathcal{L}^{-1}{v</em>{out}(s)} = \mathcal{L}^{-1}{H(s)v_{in}(s)}$

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: $\frac{Y}{X} = H(s) = \frac{N(s)}{D(s)} = \frac{s^m + a<em>1 s^{m-1} + …}{s^n + b</em>1 s^{n-1} + …}$
• $(s^n + b<em>1 s^{n-1} + …)Y = (s^m + a</em>1 s^{m-1} + …)X$
• 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:
  $\frac{V<em>2}{V</em>1} = \frac{R || \frac{1}{sC}}{sL + R || \frac{1}{sC}} = \frac{1}{1 + s \frac{L}{R} + s^2 CL}$
• Characteristic equation: $1 + s \frac{L}{R} + s^2 CL = 0$

Pole and zero in transfer function

• By factorization:
• $H(s) = \frac{(s-z<em>m)(s-z</em>{m-1})…(s-z<em>1)}{(s-p</em>n)(s-p<em>{n-1})…(s-p</em>1)}$
• $z_i$: 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

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

• 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 $\omega_c$.

More on Low pass filters

• Transfer function: $H(j\omega) = \frac{V<em>0}{V</em>i} = \frac{1}{1+j\omega CR}$
• Corner frequency: $\omega_c = \frac{1}{RC}$
• Magnitude: $|H(j\omega)| = \frac{1}{\sqrt{1+ (\omega CR)^2}} = \frac{1}{\sqrt{1+ (\frac{\omega}{\omega_c})^2}}$
• Phase: $\angle H(j\omega) = -\tan^{-1}(\omega CR) = -\tan^{-1}(\frac{\omega}{\omega_c})$
• when $\omega = 0$, $|H(j\omega)| = 1$, implying $V<em>0=V</em>i$.
• But at when $\omega \rightarrow \infty$, $|H(j\omega)| = 0$.
• At $\omega = \omega<em>c = \frac{1}{RC}, |H(j\omega</em>c)| = \frac{1}{\sqrt{2}} = 0.707$ or $-3dB$ (20log(1/sqrt(2))

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.

More on High Pass Filter

• Transfer function: $H(j\omega) = \frac{V<em>0}{V</em>i} = \frac{j\omega CR}{1+j\omega CR}$
• Corner frequency: $\omega_c = \frac{1}{RC}$
• Magnitude: $|H(j\omega)| = \frac{\omega CR}{\sqrt{1+ (\omega CR)^2}} = \frac{\frac{\omega}{\omega<em>c}}{\sqrt{1+ (\frac{\omega}{\omega</em>c})^2}}$
• Phase: $\angle H(j\omega) = \frac{\pi}{2} - \tan^{-1}(\omega CR) = \frac{\pi}{2} - \tan^{-1}(\frac{\omega}{\omega_c})$
• When $\omega -> \infty$, $|H(j\omega)| = 1, -> V<em>0 = V</em>i$
• At $\omega = 0, H(j\omega) = 0$
• At $\omega = \omega<em>c, H(j\omega</em>c) = \frac{1}{\sqrt{2}} = 0.707 (-3db)$

#

• Note: the low pass and high pass filters can be constructed by using inductors as well.

Bandpass Filter

• RLC series resonant circuit with output taken off the resistor.
• Center frequency: $\omega_0 = \frac{1}{\sqrt{LC}}$
• Filter passes frequencies from $\omega<em>1$ to $\omega</em>2$.
• Made by feeding the output from a lowpass to a highpass filter.

More on Bandpass Filter

• Transfer function: $H(j\omega) = \frac{V<em>0}{V</em>i} = \frac{j\omega CR}{1+j\omega CR+(j\omega)^2LC}$
• $H(j\omega) = \frac{jA\omega}{(\frac{j\omega}{\omega<em>1} +1)(\frac{j\omega}{\omega</em>2} +1)}$
• Magnitude: $|H(j\omega)| = \frac{A\omega}{\sqrt{[1+ (\frac{\omega}{\omega<em>1})^2][1+ (\frac{\omega}{\omega</em>2})^2]}}$
• Phase: $\angle H(j\omega) = \frac{\pi}{2} - \tan^{-1}(\frac{\omega}{\omega<em>1}) - \tan^{-1}(\frac{\omega}{\omega</em>2})$
• At $\omega -> \infty, |H(j\omega)| = 0$
• At $\omega = 0, |H(j\omega)| = 0$
• $H( \omega_0 ) = H(\frac{1}{\sqrt{LC}}) = A$

#

• Note: the bandpass filter acts as a combination of a high- pass and a low pass 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.

More on Bandstop Filter

• Transfer function: $H(j\omega) = \frac{V<em>0}{V</em>i} = \frac{1+(j\omega)^2LC}{1+j\omega CR+(j\omega)^2LC}$
• $H(j\omega) = A \frac{1+ (\frac{j\omega}{\omega<em>0})^2}{(\frac{j\omega}{\omega</em>1} +1)(\frac{j\omega}{\omega_2} +1)}$
• Magnitude: $|H(j\omega)| = A \frac{\sqrt{1- (\frac{\omega}{\omega<em>0})^2}}{\sqrt{[1+ (\frac{\omega}{\omega</em>1})^2][1+ (\frac{\omega}{\omega_2})^2]}}$
• Phase: $\angle H(j\omega) = -\tan^{-1}(\frac{\omega}{\omega<em>1}) - \tan^{-1}(\frac{\omega}{\omega</em>2})$
• At $\omega -> \infty, |H(j\omega)| = A$
• At$\omega = 0, |H(j\omega)| = A$
• $H( \omega_0 ) = H(\frac{1}{\sqrt{LC}}) = 0$

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.