Week 11 - Filters

Week 11 - Filters

• Week 11
• Topics: Filters - Quick introduction and passive filters

Filters - Introduction

• Analogy to Car Filters:
  • Filters in cars (oil filter) remove unwanted elements, providing clean output.
• Signal Filters:
  • Separate a mixture of signals into wanted and unwanted signals
  • Function: To filter out unwanted signals. Desired signals are allowed to pass through.

Types of Filters

• Two main types:
  • Passive
  • Active

Passive Filters

• Components Used: Resistors (R), Inductors (L), and Capacitors (C).

Active Filters

• Components Used: Operational Amplifiers (Op-Amps), Resistors (R), Inductors (L), and Capacitors (C).
• Studied in more advanced courses (e.g., electronic applications in the next semester).

Filter Configurations

• Both passive and active filters can be configured as:
  • Low Pass
  • High Pass
  • Band Pass
  • Band Reject (Notch/Stop Band)

Focus

• The lecture focuses on passive filters.
• Active filters are not covered in detail.

Gain Characteristics

Active Filters

• Gain ($\AV$) is greater than 1 ($\AV > 1$).

Passive Filters

• Gain ($\AV$) is approximately equal to 1 ($\AV \approx 1$).
• In reality, the gain is often less than 1.

Ideal Filter Responses

• Ideal responses for different filter types (Low Pass, High Pass, Band Pass, Band Reject) show distinct characteristics.

Low Pass Filter

• Frequency vs. Gain:
  • Constant gain ($\AV$) up to the cutoff frequency ($fc$).
  • Cutoff frequency ($f_c$) separates wanted from rejected frequencies.

High Pass Filter

• Frequency vs. Gain:
  • Good gain ($\AV$) for frequencies higher than $fc$.
  • Frequencies lower than $f_c$ are eliminated.

Band Pass Filter

• Frequency vs. Gain:
  • Only a band of frequencies between $f{low}$ ($f1$) and $f{high}$ ($f2$) is passed.
  • Other frequencies are attenuated.

Band Stop (Band Reject) Filter

• Frequency vs. Gain:
  • A band of frequencies between $f{low}$ and $f{high}$ is rejected.
  • Other frequencies are passed.

Low Pass Filter

• Allows signals with lower frequencies to pass.
• Stops signals at higher frequencies, above a cutoff/center/critical frequency.

RC Low Pass Filter

Circuit

• A resistor (R) in series with a capacitor (C) across a voltage source ($V_{in}$).
• Output voltage ($V_{out}$) is measured across the capacitor (C).
• $V{in}$ is divided into voltages across R and C ($VR$ and $V_C$).

Functionality

• Passes signals at low frequencies.
• Rejects signals at high frequencies.
• Cutoff frequency determines how high or low.

Voltage Divider Rule

• Used to find $V_{out}$:
  • $V{out} = VC = V{in} \cdot \frac{XC}{R + X_C}$
  • $X_C$ is the impedance of the capacitor.

Impedance of Capacitor

• Expressed as:
  • $X_C = \frac{1}{j \omega C}$
  • Where $j$ is the imaginary unit, $\omega$ is the angular frequency, and $C$ is capacitance.

Gain Derivation

• $\frac{V{out}}{V{in}} = \frac{\frac{1}{j \omega C}}{R + \frac{1}{j \omega C}}$
• Simplify the equation:
  • $\frac{V{out}}{V{in}} = \frac{1}{1 + j \omega R C}$

Gain ($\A_V$)

• $\AV = \frac{V{out}}{V_{in}} = \frac{1}{1 + j \omega R C}$
• This is the expression for the gain of the filter.

Maximum Gain

• Maximum gain is 1 because there are no amplifiers.
• Without an amplifier, $V{in}$ will equal $V{out}$ for the preferred range of frequencies.
• Outside this range, signals will be rejected.

Practical Perspective

• Consider what happens to the circuit at low and high frequencies.

Low Frequency

• Capacitor (C) behaves like an open circuit.
• If $V{in}$ is at low frequency, $V{out} \approx V_{in}$.
• $\A_V \approx 1$
• All input voltage makes it to the output.

High Frequency

• Capacitor (C) behaves like a short circuit.
• If $V{in}$ is at high frequency, $V{out} \approx 0V$.

Transfer Function

• If $\AV$ max = 1, $\AV$ is called the transfer function.
• Applicable to both passive and active filters.
• Describes the function performed on the signal while transferring from input to output.
• Transfer function = $\frac{V{out}}{V{in}}$

Circuit Analysis

• At any time:
  • $V{in} = VR + VC = VR + V_{out}$

Low Frequency Behavior

• C is open; therefore, $V_R \approx 0V$.
• $VC \approx V{in}$.

High Frequency Behavior

• C is short; therefore, $VR \approx V{in}$.
• $V_{out} \approx 0V$.

Transfer Function revisited

• $\frac{V{out}}{V{in}} = \frac{1}{1 + j \omega R C}$
• A complex number with magnitude and phase ($\theta$).

Magnitude Calculation

• Magnitude = $\frac{1}{\sqrt{1 + (\omega R C)^2}}$

Phase ($\theta$) Calculation

• $\theta = -\tan^{-1}(\omega R C)$
• Negative sign due to being in the denominator.

Filter Box

• Imagine the circuit inside a box with an input ($V{in}$) and an output ($V{out}$).

Phase Shift

Low Frequency

• $\theta$ (input) = -90 degrees because the capacitor's reactance is much bigger than $R$.
• $\theta$ (output) = -90 degrees (output is always across the capacitor).
• $\theta$ (transfer function) = $\frac{\theta{out}}{\theta{in}} = \frac{-90}{-90} = 0$ degrees.
• Output is identical in magnitude and time to the input.

High Frequency

• $\theta$ (input) = 0 degrees because the capacitor looks like a short circuit.
• $\theta$ (output) = -90 degrees (output is always across the capacitor).
• $\theta$ (transfer function) = $\frac{\theta{out}}{\theta{in}} = \frac{-90}{0} = -90$ degrees.
• $V{out}$ lags $V{in}$ by 90 degrees (lagging filter).

Summary of Transfer Function

• Expression: $\frac{1}{1 + j \omega R C}$
• Magnitude: $\frac{1}{\sqrt{1 + (\omega R C)^2}}$
• $\theta$: $-\tan^{-1}(\frac{\omega R C}{1})$

Cutoff Frequency

• The frequency that separates low and high frequencies.

Condition

• When the real value = the imaginary value (1 = $\omega R C$).
• 1 = $\omega R C = 2 \pi f_c R C$

Calculation

• $f_c = \frac{1}{2 \pi R C}$
• At $f_c$, $\theta = -45$ degrees.
• At the cutoff frequency, the output lags the input by 45 degrees.

Transfer Function in terms of Frequencies

• $\omega = 2 \pi f$ (at any frequency).

Applying to Magnitude

• Starting with Magnitude = $\frac{1}{\sqrt{1 + (\omega R C)^2}}$
• $\R C = \frac{1}{2 \pi f_c}$ (Relationships)
• Therefore, the Transfer function magnitude = $\frac{1}{\sqrt{1 + (\frac{f}{f_c})^2}}$

Transfer Function

• $\theta = -\tan^{-1}(\frac{f}{f_c})$
• These two rules are used for low pass passive RC filters.

Bode Plot

• Approximation of the filter's response.

Frequency vs Transfer Function Magnitude

• Constant magnitude up to $f_c$.
• After $f_c$, response decays at a rate of -20dB/decade (decade is 10 frequencies).

Cutoff Frequency in Reality

• Response starts decaying at $f_c$.
• At $f_c$, the gain is is 0.707 times the maximum.

Gain in dB

• The gain at the cutoff frequency is -3 dB.
• 20 log(0.707) = -3 dB.

Logarithmic Scale

• With filters, use a wide range of frequencies.
• Linear scale does not accurately show megahertz, thus a logarithmic scale is used.
• log(one x 10^0)= Value Zero
• Each division is divided using the log function.

Decades

• The frequency goes in units called decades - from log10^0 to log 10^1

Rate of Decay

• -20 dB per decade.

Decibels (dB)

Derivation

• Starts with P, V, I gain
• Power Gain ($AP$): $\frac{P{out}}{P_{in}}$ (ratio, no unit).

Logarithmic response

• Our ears respond to volume in a logarithmic way

Power in Bels

• Log base 10 of that ratio = Bels.
• Sound unit is in Bels

Decibels

• Decibel = 1/10 of a bell.

Calculations with Voltage

• P=IV since P = \frac{V^2}{R} using the same equation. Use that to change 10log\10 to what you have instead
• 10 log base 10 and instead of P out = \frac{V^2}{R}
• 10logBase 10 (vout^2/rout)/ (vin^2/rin). Input power MUST EQUAL output, since they must have the same impedance
• The ratio will cancel out giving --> 10logbase 10 vOut/ vin to the squared
• Use the LOGARITHMIC RULE TO MAKE IT 20. The outcome from there: 20 LOG base 10 vOut/vin.
• from there -> aV in db
• If aV, which is the transfer function = 1, then Av AND dB has to be 20 log 1 ,which is ZERO
• Maximum = Zero db
• At the cutoff frequency, The output Voltage is equal to 0.7 or seven in which: aV = point = transfer fx, which is point 707 .
• AT dB point equals 7 or seven. It is half PowerPoint SHOW. So there's like 30 log .5 . Can you find a point is equal = minus three DB point. 7 or seven?
  So a b at f c equals five. This is just a ratio B = over being equal 2.5. A B out is half of the power at the cut off frequency at the cutoff frequency at F C

Expressed

• AV in dB is equal to 20 log Av
• If AV, which is the transfer function = 1, = for low frequencies = AND dB has to be 20 log 1 which is ZERO dB.

20 dB per decade

• Reduction and set by step reduction in the voltage gain.
• This specifically applies to filters. passive filters0 DB hundre .
  • a the cutoff frequency. The power is half. A B. Not not that. to a B F C equals 5.
  • this is just the ratio B about equals over being equal 2.5

Magnitude is at 0 dB

• Minus dB gets a - +20dB
• Decades of frequency.
• fC was like 100, The frequency will go at the point.707
  So in this case, if FC can be thousands, so these the TEN FC.

Angle

• Minus45, one over f = minus40 -5dB. Get at that point
  At point one 1 FC This one is 10 FC point one fc is approximately from the degree and the equation angle, a one f see And if you get a negative to point is get a negative approximately zero. From minus approximately zero.
• F equal point equals 10 one point point and one FC so - 10 inverse point one 1 f= minus 10 verse Point.1

If F equals to point zero

• Equation if M is equal to Z - 10 inverse.1 get approximately seven degrees this point, one f see or minus three 10 in verse 10 at C minus seven. This equal that approximate 1 point zero is approximately zero.
• full body part digram for a low passed filter with Examples the the lecture. On Friday and Monday next week will will get there.

Low Pass Filter RL

• thinking about the low pass everything will stay the same, keeping the mentality alive inductors and capacitors are opposite from each other
• IF we continue to think about a passive component we make a change that includes L and R. To maintain the same mentality, what does that indicate?, switching output voltages. this is because its output is a vL.
• This can now will not work as a low pass. If we change the position, excuse me,

Input side with a low frequency

• inductor = behaves like a 0V circuit / short circle. At low frequency the output is always VL short 0 ohm -r
• All the output voltage will be in the zero volts to 0V
  It is rejecting at high frequency.

To Calculate transferred to function:

• Function, but to function, where v Out over V in, we will, use it to calculate vO = v In multiply . The expression and transfer is with our or are plus day day make the and make it go to one after we do the divided by R. Is the is the transferred function
• Magnitude is where it = the route, transfer = to is equal that is equal equals r over to l with that equals 54 = the function

To Derive Magnitude to F

• equation FC 1. = Meg L L R. So if two points is not to r is one
• OMEGA ALWAYS WILL = to FI F
  from here : L r to equal equal of FC all omega equal with this two of you
  Back with this we will cancel out two pi will and it has to be, both of they re both the best formula of what we re supposed to find out
  And, for. They that two of this we can to get the equation. Here that all these together