Lecture 13 - School of Information Science and...

Electric Circuits (Fall 2015) Pingqiang Zhou

Lecture 13

- Frequency Response

12/1/2015

Reading: Chapter 14

1Lecture 13


Outline

• Frequency response

• Resonance

• Filters

Lecture 13 2


Frequency Response

• When a linear, time invariant (LTI) circuit is excited by a sinusoid, it’s

output is a sinusoid at the same frequency.

Only the magnitude and phase of the output differ from the input.

• The “Frequency Response” is a characterization of the input-output

response for sinusoidal inputs at all frequencies.

Significant for applications, esp. in communications and control systems.

• Filters play critical roles in blocking or passing specific frequencies or

ranges of frequencies.

Widely used in radio, TV and phone systems.

3Lecture 13[Source: Berkeley]


Magic of Sinusoids

• All real-world signals are sums of sinusoids components

that have proper amplitudes, frequencies and phases.

To be found by Fourier Analysis.

Most natural signals contain thousands of components.

4Lecture 13[Allan, 6th ed.]


Representing a Square Wave as a Sum of Sinusoids

[Allan, 6th ed.]


Frequency Ranges of Selected Signals

• When we listen to music, our ears respond differently to

the various frequency components: some pleasing,

whereas others are not.

Design of signal-processing circuits for audio signals must consider

how the circuit responds to components of different frequencies.

6Lecture 13[Allan, 6th ed.]

The components can be determined by theoretical analysis or

by laboratory measurements (using Spectrum Analyzer).


Transfer Function – Voltage Gain

• One useful way to analyze the frequency response of a

circuit is the concept of the transfer function.

• It is a function of frequency

Complex quantity

Both magnitude and phase are function of frequency

Two Port

networkVin Vout

( )

( )

outout in

in

Vf

V

H f

out

in

VH

V

H(f)

Lecture 13 7


Example

• The transfer function 𝐻 𝑓 is shown below. If the input

signal is 𝑣𝑖𝑛 𝑡 = 2 cos 2000𝜋𝑡 + 40° , find an expression

as a function of time for the output.

8Lecture 13


Example

• The transfer function 𝐻 𝑓 is same as shown on previous

slide. If the input signal is

𝑣𝑖𝑛 𝑡 = 3 + 2 cos 2000𝜋𝑡 + cos 4000𝜋𝑡 − 70° ,

find an expression as a function of time for the output.

9Lecture 13


Example

10Lecture 13


Transfer Function – More General Definition

• The transfer function H(ω) is the frequency-dependent ratio

of a forced function Y(ω) to the forcing function X(ω).

YH

X

Voltage gain

Current gain

Transfer impedance

Transfer admittance

o

i

o

i

o

i

o

i

VH

V

IH

I

VH

I

IH

V

Lecture 13 11


Example

12Lecture 13


Example: Plot

13Lecture 13


Exercise

• Obtain the transfer function 𝐕𝐨/𝐕𝐬 of the RL circuit.

Assuming 𝑣𝑠 = 𝑉𝑚 cos𝜔𝑡.

14Lecture 13


Zeros and Poles

• H(ω) can be expressed as the ratio of numerator N(ω)

and denominator D(ω) polynomials.

Zeros are where the transfer function goes to zero.

Poles are where it goes to infinity.

They can be related to the roots of N(ω) and D(ω).

NH

D

YH

X

Lecture 13 15


Example

• For the circuit below, calculate

the gain 𝐈𝐨 𝜔 /𝐈𝐢 𝜔 and its

poles and zeros.

16Lecture 13


Decibel Scale

• We will soon discuss Bode plots.

• These plots are based on logarithmic scales.

• The transfer function can be seen as an expression of

gain.

Gain expressed in log form is typically expressed in bels, or more

commonly decibels2

10

1

10logdB

PG

P

Lecture 13 17


Bel and Decibel (dB)

• A bel (symbol B) is a unit of measure of ratios of power

levels, i.e. relative power levels.

The name was coined in the early 20th century in honor of

Alexander Graham Bell, a telecommunications pioneer.

The bel is a logarithmic measure. The number of bels for a given

ratio of power levels is calculated by taking the logarithm, to the

base 10, of the ratio.

– one bel corresponds to a ratio of 10:1.

– B = log10(P1/P2) where P1 and P2 are power levels.

• The bel is too large for everyday use, so the decibel (dB),

equal to 0.1B, is more commonly used.

dB = 10 log10(P1/P2)

dB are used to measure

– Electric power, Gain or loss of amplifiers, Insertion loss of filters.

Lecture 13 18


Logarithmic Measure for Power

• To express a power in terms of decibels, one starts by

choosing a reference power, Preference, and writing

• Exercise:

Express a power of 50 mW in decibels relative to 1 watt.

P (dB) =10 log10 (50 x 10-3) = - 13 dB

Express a power of 50 mW in decibels relative to 1 mW.

P (dB) =10 log10 (50) = 17 dB

• dBm to express absolute values of power relative to a

milliwatt.

100 mW = 20 dBm

10 mW = 10 dBm

dBm = 10 log10 (power in milliwatts / 1 milliwatt)

Power P in decibels = 10 log10(P/Preference)

[Source: Berkeley] Lecture 13 19


• From the expression for power ratios in decibels, we can

readily derive the corresponding expressions for voltage or

current ratios.

• Suppose that the voltage V (or current I) appears across (or

flows in) a resistor whose resistance is R. The corresponding

power dissipated, P, is V2/R (or I2R). We can similarly relate

the reference voltage or current to the reference power, as

Hence,

Voltage, V in decibels = 20log10(V/Vreference)

Current, I, in decibels = 20log10(I/Ireference)

Logarithmic Measures for Voltage or Current

Preference = (Vreference)2/R or Preference= (Ireference)

2R

[Source: Berkeley]


• Note that the voltage and current expressions are just like the

power expression except that they have 20 as the multiplier instead

of 10 because power is proportional to the square of the voltage or

current.

• The input voltage (current, or power) is taken as the reference value

of voltage (current, or power) in the decibel defining expression:

Logarithmic Measures for Voltage or Current

Voltage gain in dB = 20 log10(Voutput/Vinput)

Current gain in dB = 20log10(Ioutput/Iinput

Power gain in dB = 10log10(Poutput/Pinput)

Exercise: How many decibels larger is the voltage of a 9-volt transistor

battery than that of a 1.5-volt AA battery? Let Vreference = 1.5.

Example: The voltage gain of an amplifier whose input is 0.2 mV and

whose output is 0.5 V is 20log10(0.5/0.2x10-3) = 68 dB.

20 log10(9/1.5) = 20 log10(6) = 16 dB.

[Source: Berkeley]


Bode Plots

• One problem with the transfer function is that it needs to

cover a large range in frequency. So we can

Plotting the frequency response on a semilog plot

– X is in log scale

– Y scale in dB

These plots are referred to as Bode plots.

– Show magnitude (in decibels) or phase (in degrees) as a function of

frequency.

• Log Frequency Scale

Decade Ratio of higher to lower frequency = 10

Octave Ratio of higher to lower frequency = 2

Lecture 13 22


Bode Plot of First-Order Lowpass Filter

23Lecture 13

𝐻 𝑓 =1

1 + 𝑗 𝑓/𝑓𝐵𝑓𝐵 =

1

2𝜋𝑅𝐶


Magnitude Plot

Convert magnitude to decibels, we have

𝐻 𝑓 𝑑𝐵 = 20 log 𝐻 𝑓 = −10 log 1 +𝑓

𝑓𝐵

2

• When 𝑓 ≪ 𝑓𝐵, i.e., low frequency, 𝐻 𝑓 𝑑𝐵 ≅ 0;

• When 𝑓 ≫ 𝑓𝐵, 𝐻 𝑓 𝑑𝐵 ≅ −20 log𝑓

𝑓𝐵.

24Lecture 13

Corner (break) frequency

where two straight-line

asymptotes intersect


Phase Plot

• When 𝑓 < 𝑓𝐵/10, i.e., low frequency, ∠𝐻 𝑓 ≅ 0;

• When 𝑓 > 10𝑓𝐵, 𝐻 𝑓 𝑑𝐵 ≅ −90°.

25Lecture 13

The actual phase curve departs

from these straight-line

approximations by less than 6°.


Standard Form of General Transfer Function

• The transfer function may be written in terms of factors

with real and imaginary parts. For example:

• This standard form may include the following seven

factors in various combinations:

Lecture 13 26


Bode Plots

Lecture 13 27


Bode Plots

Lecture 13 28


Example

29Lecture 13


Example - Magnitude

30Lecture 13


Example - Phase

31Lecture 13


Resonance

• The most prominent feature of the

frequency response of a circuit may

be the sharp peak in the amplitude

characteristics.

• Resonance occurs in any system

that has a complex conjugate pair

of poles.

It enables energy storage in the form of

oscillations

It allows frequency discrimination.

It requires at least one capacitor and

inductor. [Source: Georgia State U]

Lecture 13 32


Series Resonance

• A series resonant circuit consists of an

inductor and capacitor in series.

Resonance occurs when the imaginary part

of Z is zero.

The value of ω that satisfies this is called the

resonant frequency.

0

1rad/s

LC

Lecture 13 33


Series Resonance

• At resonance:

The impedance is purely resistive

The voltage Vs and the current I are in phase

The magnitude of the transfer function is minimum.

The inductor and capacitor voltages can be much more than the

source.

Lecture 13 34


Half-Power Frequencies

• The frequency response of the current magnitude:

0

1rad/s

LC

Lecture 13 35


Bandwidth

• Bandwidth: the difference

between the two half-

power frequencies

Lecture 13 36

𝐵 = 𝜔2 −𝜔1 =𝑅

𝐿


Quality Factor

• Quality factor Q: measure the

“sharpness” of the resonance.

[Source: Georgia State U]

Lecture 13 37


Q and B

38Lecture 13

The higher the Q,

the smaller the B.


Approximation of Half-Power Frequencies

• For high-Q (𝑄 ≥ 10) circuits,

half-power frequencies can

be approximated as

39Lecture 13


Example

In the circuit, 𝑅 = 2Ω, 𝐿 =1mH and 𝐶 = 0.4𝜇F• Find resonant frequency

• Find half-power frequencies

• Calculate the quality factor and

bandwidth

• Determine the amplitude of the

current at 𝜔0,𝜔1 and 𝜔2

40Lecture 13


Parallel Resonance

• The parallel RLC circuit shown here is

the dual of the series circuit.

• Resonance occurs when the imaginary

part of the admittance is zero.

This results in the same resonant frequency

as in the series circuit.

Lecture 13 41


𝝎𝟏, 𝝎𝟐, 𝑸 𝐚𝐧𝐝 𝑩

42Lecture 13


Example

In the circuit, 𝑅 = 8kΩ, 𝐿 =0.2mH and 𝐶 = 8𝜇F• Find Q, B, 𝜔1 and 𝜔2

• Find half-power frequencies

• Determine the power dissipated at

𝜔0,𝜔1 and 𝜔2

43Lecture 13


Summary

44Lecture 13


Filters

• Circuit designed to retain a certain

frequency range but discard or

attenuate others

Low-pass: pass low frequencies and reject

high frequencies

High-pass: pass high frequencies and

reject low frequencies

Band-pass: pass some particular range of

frequencies, reject other frequencies

outside that band

Notch: reject a range of frequencies and

pass all other frequencies

Lecture 13 45


Common Filter Transfer Function vs. Freq

(Magnitude Plots shown)

( )H f

Frequency

High

Pass

( )H f

Frequency

Low

Pass

( )H f

Frequency

Band Pass

Frequency

Band

Reject

( )H f


Passive Filters

• A filter is passive if it consists only of passive elements, R,

L, and C.

• LC filters have been used in practical applications for

more than eight decades.

• They are very important circuits in that many technological

advances would not have been possible without the

development of filters.

LC filter technology feeds related areas such as equalizers,

impedance-matching networks, transformers, shaping networks,

power dividers, attenuators, and directional couplers …

Lecture 13 47


First-Order Lowpass Filter

• A typical lowpass filter is formed when

the output of a RC circuit is taken off

the capacitor.

• The half power frequency is:

Also referred as the cutoff frequency.

Filter is designed to pass from DC up to ωc.

1c

RC

Lecture 13 48


First-Order Highpass Filter

• A highpass filter is also made of a

RC circuit, with the output taken off

the resistor.

• The cutoff frequency will be the same

as the lowpass filter.

• The difference being that the

frequencies passed go from ωc to

infinity.Lecture 13 49


First-Order Filter Circuits

L+

–VS

C

R

Low Pass

High

Pass

HR = R / (R + jL)

HL = jL / (R + jL)

+

–VS

R

High Pass

Low

Pass

HR = R / (R + 1/jC)

HC = (1/jC) / (R + 1/jC)


Bandpass Filter

• The RLC series resonant circuit

provides a bandpass filter when

the output is taken off the resistor.

• The center frequency is:

• The filter will pass frequencies

from ω1 to ω2.

• It can also be made by feeding

the output from a lowpass to a

highpass filter.

0

1

LC

Lecture 13 51


Bandstop Filter

• A bandstop filter can be created from a RLC circuit by

taking the output from the LC series combination.

• The range of blocked frequencies will be the same as the

range of passed frequencies for the bandpass filter.

Lecture 13 52


Second-Order Filter Circuits

53Lecture 13[Source: Berkeley]


Example

• Determine what type of filter

is shown below. Calculate the

corner or cutoff frequency.

𝑅 = 2kΩ, 𝐿 = 2H and 𝐶 = 2𝜇F.

54Lecture 13


Active Filters

• Passive filters have a few drawbacks.

They cannot create gain greater than 1.

They do not work well for frequencies below

the audio range (300 Hz < f < 3,000 Hz).

They require inductors, which tend to be

bulky and more expensive than other

components.

• It is possible, using op-amps, together

with resistors and capacitors, to create

all the common filters.

• Their ability to isolate input and output

also makes them very desirable.

Lecture 13 55


First Order Lowpass

• If the input and feedback elements in

an inverting amplifier are selectively

replaced with capacitors, the

amplifier can act as a filter.

If the feedback resistor is replaced with a

parallel RC element, the amplifier

becomes a lowpass filter.

Corner frequency

1c

f fR C

Lecture 13 56


First Order Highpass

• Placing a series RC combination in

place of the input resistor yields a

highpass filter.

• The corner frequency of the filter

will be:1

c

i iR C

Lecture 13 57


Bandpass

Lecture 13 58


Constructing the Bode Plot of a Bandpass

59Lecture 13[Nilsson, 10th ed.]


Bandpass

• To avoid the use of an inductor,

it is possible to use a cascaded

series of lowpass active filter

into a highpass active filter.

• To prevent unwanted signals

passing, their gains are set to

unity, with a final stage for

amplification.

Lecture 13 60


Active Bandpass Filter

61Lecture 13


Constructing the Bode Plot of a Bandreject

62Lecture 13[Nilsson, 10th ed.]


Bandreject

• Creating a bandstop filter requires using a lowpass and

highpass filter in parallel.

Both output are fed into a summing amplifier.

It will function by amplifying the desired signals compared to the

signal to be rejected.

Lecture 13 63


Active Bandreject Filter

64Lecture 13


Further Reading

• Refer to Ch. 14.9 for Scaling

Magnitude scaling

Frequency scaling

• Read Ch. 14.11 on how to use MATLAB to do Bode Plot

• Refer to Ch. 14.12 for applications of resonant circuits and

filters

Radio receiver

Touch-tone telephone

• Refer to [Allan, Ch. 6.10] for DSP.

65Lecture 13

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