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AM Modulation -- Radio 1

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LO audio baseband mf f f

Amplitude Modulation – Early RadioEE 442 – Spring Semester

Lecture 6

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Modulation Options

We will study:AMFMPM

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Baseband versus Carrier Communication

Baseband communication is the transmission of a message as generated is transmitted.

Carrier communication requires the modulation of the message onto a carrier signal to transmit it over a different frequency band. We usemodulators to do the frequency translation.

(Note: “Pulse modulated” signals, such as PAM, PWM, PPM, PCM and DMare actually baseband digital signal coding (and not the result of frequencyconversion).

Use of Sinusoidal Carrier Signal: Using a sine waveform there are three parameters which we can use to “modulate” a message onto the carrier –they are the amplitude, frequency and phase of the sinusoidal carrier.

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Amplitude modulation (AM) is a modulation technique where the amplitude of a high-frequency sine wave (called a radio frequency) is varied in direct proportion to the modulating signal m(t). The modulating signal contains the intended message or information –sometimes consisting of audio data, as in AM radio broadcasting, or two-way radio communications.

The high-frequency sinusoidal waveform (i.e., carrier) is modulated by combining it with the modulating signal using a multiplier or mixer (mixing is a nonlinear operation because it generates new frequencies).

Amplitude Modulation Definition

Agbo & Sadiku; Section 3.2, pp. 84 to 99


Amplitude Modulation in Pictures

100 kHz carrier modulated by a 5kHz audio tone

100 kHz carrier modulated by an audio signal

(frequencies up to 6 kHz)

5 kHz Audio tone

Tone-modulatedAM signal

Voice-modulatedAM signal

Frequency Domain Time Domain

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Example: Voice Signal – 300 Hz to 3400 Hz Baseband

time

amp

litu

de

m(t)

Symbol m(t) represents the source message signal.

Time Domain Display

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Frequency Domain

Voice Band for Telephone Communication

Voice Channel0 Hz – 4 kHz

Voice Bandwidth300 Hz – 3.4 kHz

f0 Hz 300 Hz 3.4 kHz 4 kHz 7 kHz

PSTN or POTS

Pow

er


3,400 Hz

Speechsignal

Speechspectra

Representative Voice Spectrum for Human Speech

For the telephone AT&T determined many years ago that speech couldbe easily recognized when the lowest frequencies and frequencies above 3.4 kilohertz were cutoff.

Waveform as received froma microphone convertingacoustic energy into electricalenergy.

Fast Fourier transform of the above speech waveformshowing energy over frequency from 0 Hz to 12 kHz.

Time t

Frequency f (in Hz)


A crystal radio receiver, also called a crystal set or cat's whisker receiver, is a very simple radio receiver, popular in the early days of radio. It needs no other power source but that received solely from the power of radio waves received by a wire antenna. It gets its name from its most important component, known as a crystal detector, originally made from a piece of crystalline mineral such as galena. This component is now called a diode.

Early AM Crystal Radio Receiver (Minimalist Radio)

Note: 1N34A is agermanium diode

& operates byrectifying signal

Earphones

LC TunedCircuit

Demodulator

10AM Modulation -- Radio

Foxhole Radio (used in World war I)

http://bizarrelabs.com/foxhole.htm

Ground

Coil – 120 turns of wire

Cold water pipe

Safety pinRazor blade

Earphones

http://bizarrelabs.com/foxhole.htm


Modern Mechanix (December 1952)


Crystal Radio Receiver from 1922

Diagram from 1922 showing the circuit of a crystal radio. This common circuit did not use a tuning capacitor, but used the capacitance of the antenna to form the tuned circuit with the coil.

Galena (lead sulfide) was probably the most

common crystal used in “cat's whisker” detectors.

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Amplitude Modulation (DSB with Carrier)

Amplitude Modulation: The amplitude of a carrier signal is varied linearly with a time-varying message signal.

CNote: Keep & fixed.Carrier signal: ( ) cos( )

Only amplitude is allowed to vary in AM: ( )

( ) ( ) cos cos ( ) cos

C C

C C C

AM C C C C C

c t A t

A A A m t

t A m t t A t m t t

( )m t

cos( )C CA t

( ) cos( )C CA m t t


Amplitude Modulation (DSB with Carrier)

http://hyperphysics.phy-astr.gsu.edu/hbase/Audio/bcast.html

http://hyperphysics.phy-astr.gsu.edu/hbase/Audio/bcast.html


Phasor View of Amplitude Modulation

https://inspirehep.net/record/1093258/plots

Modulated oscillation is a sum of

these three vectors an is given by

the red vector. In the case of

amplitude modulation (AM), the

modulated oscillation vector is

always in phase with the carrier field

while its length oscillates with the

modulation frequency. The time

dependence of its projection onto

the real axis gives the signal

strength as drawn to the right of the

corresponding phasor diagram.

Brown vector Carrier signalRed vector AM modulated signal

Example shows tone modulation

https://inspirehep.net/record/1093258/plots

https://inspirehep.net/record/1093258/files/fig4.png


Phasor View of Amplitude Modulation

( ) Re 12 2

m m

C

j t j tj t

AM

e et e

Ct

mt

mt

Tone signal cos(mt)

Carrier cos(Ct)

C C m C m

Tone modulation


Phasor Interpretation of AM DSB with Carrier (continued)

https://www.slideshare.net/azizulhoque539/eeng-3810-chapter-4

Tone modulation

https://www.slideshare.net/azizulhoque539/eeng-3810-chapter-4


Agbo & Sadiku; Section 3.2.1; pp. 89 to 91

Double-Sideband Amplitude Modulation Spectrum

1 12 2

( ) ( ) cos cos ( ) cos

The spectrum is found from the Fourier transform of ( )

( ) ( ) ( ) ( ) ( )

AM C C C C C

AM AM

AM C C C C

t m t t A t A m t t

t

FT t M M A

CarrierMessage

Baseband

CarrierCarrier

Message

SidebandMessage

https://en.wikipedia.org/wiki/Amplitude_modulation

https://en.wikipedia.org/wiki/File:AM_spectrum.svg

https://en.wikipedia.org/wiki/Amplitude_modulation


AM Modulation Index Basics – Definition

The amplitude modulation (AM) modulation index can be defined as a measure of the

amplitude variation upon a carrier.

When expressed as a percentage it is the same as the depth of modulation. In other words

it can be expressed as:

where AC is the carrier amplitude, and

mp is the modulation amplitude (peak change in the RF amplitude relative

to its un-modulated value.

Example: An AM modulation index of 0.5 means the signal increases by a factor of 0.5,

and decreases to 0.5, centered around its unmodulated level. See drawings below.

( ) ( ) cos( )AM C Ct A m t t

Modulation Indexp

C

m

AAgbo & Sadiku

Page 85-86

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AM Modulation Index Basics – Examples

50% Tone Modulation

100% Tone Modulation

150% Tone Modulation

Overmodulationor

Envelope Distortion

50%

100%

150%

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AM Overmodulation Envelope Distortion

https://www.physicsforums.com/threads/amplitude-modulation-vs-beats.890811/

https://www.physicsforums.com/threads/amplitude-modulation-vs-beats.890811/


Power Efficiency of Amplitude Modulation


2

/2 /2

2 2 2

/2 /2

Given the AM signal: ( ) cos( ) ( ) cos( )

The power in the carrier is 2

The power in the sidebands (modulated message) is

1 1 1lim ( )cos ( ) lim ( ) 1 cos(

2

AM c c c

cc

T T

S cT T

T T

t A t m t t

AP

P m t t dt m tT T

/2

2

/2

/2

2

/2

2 )

But ( ) cos(2 ) 0, and

1 1 1we are left with lim ( )

2 2

That is, is one-half the total message power .

In AM the power in the message (useful power) is the power

c

T

c

T

T

S mT

T

S m

t dt

m t t dt

P m t dt PT

P P

in the

sidebands. Now we can finally define .power efficiency


Power Efficiency in Amplitude Modulation (continued)

The power efficiency of a modulated signal is the ratio of the power inthe message part of the signal relative to the total power of themodulated signal.

212

12

2

messsage power sideband powerPower efficiency = =

total power total power

In symbols, = , and2

=

mS CC

C S C m

m

C m

PP AP

P P P P

P

A P


Power Efficiency in Amplitude Modulation (continued)

In general, the form of Pm is complicated and not known precisely. However,we can study AM power efficiency using tone modulation.

22 2 2

2 2 2

For tone modulation ( ) cos( ) cos( ) ,

and2 2 2 2

m C C C

Cm mm

C m

m t A t A t

AA AP

A A

0.25 0.0303 or 3.03 %

0.5 0.111 or 11.1 %

1.0 0.333 or 33.3 %

Examples:

2

22

Modulation index

Conclusion: AM power efficiency is very low (highly undesirable).


Preview: Categories of Amplitude Modulation

Baseband spectrum (message spectrum)

Conventional AM (Double-SideBand With Carrier)

Double-Sideband-Suppressed Carrier (DSB-SC)

Single-Sideband /Upper Sideband SSB/USB

Single-Sideband /Lower Sideband SSB/LSB

f

f

f

f

fAlso Vestigial Sideband and Amplitude Companded SSB

Special cases of AM:

DSB-w/C


Generation of Amplitude Modulated Signals


Agbo & Sadiku present two methods for AM generation:

1. Nonlinear AM modulatorAlmost any nonlinearity will work, but a very inexpensive but strongly nonlinear device is the diode. Transistors are also nonlinear and work well as modulators.

2. Switching AM modulatorSwitching is an easily attained function with diodes and transistors in electronic circuits.

There is also a third method:

3. Electronic multipliers (such as Gilbert cells)


Diode Operation Applied to AM Modulators & Demodulators

1. As nonlinear circuit components (primarily the “square law” part )

2. As “on-off” switches (they have to be driven hard to do this)

Cu

rre

nt

(mA

)

Voltage (V)


Using Nonlinearity For Modulation (i.e., AM Generation)

R

+

_y(t)

m(t)+

_

Accos(Ct)+

_

Diode

BPFFilter(c)

+

_x(t)

The diode is the nonlinear component (it has an exponentialcharacteristic). Using a Taylor’s series we can express the diodecurrent iD as (only first two terms of Taylor’s series),

2

1 2

2 2

1 2 1 2

( ) ( ) ( ); ( ) is diode voltage.

The voltage across resistor R is given by

( ) ( ) ( ) ( ) ( ) ( )

D D D D

D D D D D

i t b v t b v t v t

x t i t R b Rv t b Rv t a v t a v t

“Square Law”behavior

Di


Using Nonlinearity For Modulation (continued)

2

1 2

22 2

1 2

21

1

We now can evaluate voltage ( )

( ) ( ) cos ( ) ( ) cos ( )

( ) ( ) ( ) 1 cos (2 )2

2 ( )1 cos ( )

Applying the bandpass filter about , the output voltage (

C C C C

CC

C C

C

x t

x t a m t A t a m t A t

a Ax t a m t a m t t

a m ta A t

a

y

21

1

) is

2 ( )( ) ( ) 1 cos ( )AM C C

t

a m ty t t a A t

a

Note: For to be less than unity, we must demand 2

1

2 ( )1.

a m t

a

(Eq. 3.23)


Using Nonlinearity For Modulation (continued)

Comments:

1. Can use a general nonlinear element (not just a “square law” device)2. The filter can be as simple as a LC resonator3. This is a about the simplest of all modulators (it is unbalanced)

R

m(t)+

_

Accos(Ct)+

_

GeneralNonlinearElement

+

_y(t)C

Tuned to radianfrequency C

1C

LC


Using General Nonlinearity For Modulation

fRF = 0.8fLO = 1.0

LO

RF

LO - RF LO + RF

Vout()

2LO

-R

F

3RF 3LO

2LO

+ R

F

2RF 2LO

1 2 30

2RF

–LO

2RF

+ L

O

vout RF

and LO

(linear)

v2out (

RF+

LO ), (

LO -

RF ), 2

RFand 2

LO(square law)

v3out (2

RF +

LO), (2

RF -

LO), (2

LO+

RF), (2

LO -

RF), 3

RF & 3

LO

Conclusion: Nonlinearity generates new frequencies.

2 3

out DC in in inv v Gv Av Bv

Input signals: cos( ) cos( )in RF RF LO LOv A t B t

IF

Taylor’s series

tone carrier


Switching Amplitude Modulator – The Switch

1 2 1 1( ) cos( ) cos(3 ) cos(5 )

2 3 5C C Cp t t t t

Ron = 0

Roff is infinite

No Capacitance

(t)

By driving a diode with sufficient AC voltage it acts like a switch:

Forward bias Reverse bias

No current flowSwitch open

Current flowSwitch closed


1 2 1 1( ) cos( ) cos(3 ) cos(5 )

2 3 5p t t t t

FourierSeriesrepresentation

Switching Amplitude Modulator – Pulse Spectrum Generated

1

p(t)

02

T

2

T

2

T

2

time

Infinitely long pulse train

P()

1

T

2

T

1

T

3

T0

Fourier series ofthe pulse trainof period T

1

1

Duty CycleT

Shown for aduty cycle of 1/4

2


Switching Modulator – Generating m(t)cos(Ct)

Pulse trainp(t)

m(t)p(t)


A “hopelessly unsophisticated” mixer.

Tom Lee (Stanford University)

The unbalanced single-diode mixer has no isolation and no conversion gain.

Single-diode mixers have been used in many applications --

(1) Detectors for radar in WW II(2) Early UHF Television tuners(3) Crystal radio detectors(4) mm-wave & sub-mm-wave

receivers

Diode Mixer For Modulation and Demodulation

IFRF +

LO

Filter

Diode


AM Demodulation

Section 3.2.4 (pp. 95 to 99)

Coherent (i.e., synchronous) demodulation (or detection) is a methodto recover the message signal from the received modulated signal thatrequires a carrier at the receiver. This carrier signal must match infrequency and phase to the received signal.

But . . . Amplitude Modulation has the advantage of not requiring coherent detection methods. Non-coherent methods can be used which are much simpler to implement.

1. AM Envelope Detector

2. AM Rectifier Detector


AM Envelope Detector Circuit

Two conditions must be met for an envelope detector to work:

(1) Narrowband [meaning fc >> bandwidth of m(t)](2) AC + m(t) 0

Incoming AMmodulated signal

Rectified AMmodulated signal

Capacitor stores energyfrom the peaks of the

rectified signal

Envelope Detection requires the an RC network with time constant = RC

Key idea: Capacitor captures the voltage peaks of rectified waveform


Choosing the RC Time Constant in Envelope Detector

Time constant = RC too short.t

t Time constant = RC too long.

1

Design criteria is 2 2 CB fRC

How the envelope is constructed


The user has a choice of changing the filter to meet their needs.

Practical Demodulation of an AM Signal

V V

ModulatedRF signal

inputIF output

• • •

•

•

•

IF

output

RF

AM input

Diode

DC blockingEnvelopedetection

FrequencySelectivity

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AM Rectifier Detection

1 13 5

( ) ( ( ))cos( ) ( )

1 2( ( ))cos( ) (cos( ) (cos(3 ) (cos(5 ) ....

2

1( ) other terms.

Note: Multiplication with ( ) allows

= dc term + baseband term

rect C C

C C C C C

C

V t A m t t p t

A m t t t t t

A m t

p t rectifier detection to act essentially as a

synchronous detection without a carrier being generated at the receiver.

DC componentis removed by

capacitor C

LPF


Double-Sideband Suppressed Carrier AM

Conventional AM transmits both the message and carrier signals.

Hence, the its power efficiency is low,

If (AC)2 approaches zero, then approaches 100%.

2100%m

C m

P

A P

( ) ( ) cos( )

For DSB-SC (double sideband -- suppressed carrier) we have

( ) ( ) cos( ) with a FT pair: ( ) ( )

1( )cos( ) ( ) ( ) ( )

2

AM C C

DSB SC C

C DSB SC C C

t A m t t

t m t t m t M

FT m t t M M


Double-Sideband Suppressed Carrier AM

( )m t

( )m t

( )m t

( ) cos( )Cm t t

t

t

Note phasereversal

( )m t( ) cos( )Cm t t

cos( )Ct

(Modulator)

Figure 3.11 in Agbo & Sadiku

Modulation:

DSB-SC Output


( )DSB SC

C m 0CC m C m CC m

TransmittedDSB-SC Signal

USBLSB

DSB-SC has USB and LSB spectra but not carrier impulses at C.

( )M

0 m m

( ) ( )m t M

Double-Sideband Suppressed Carrier AM (continued)

( )m t( ) cos( )Cm t t

cos( )Ct

(Modulator)



Demodulation:

( )m t

( ) cos( )Cm t t

2cos( )Ct

(Modulator)

( )x t

Low-passfilter

2

2

( ) 2 ( ) cos ( ) ( ) ( ) cos(2 )

1Use identity: cos 1 cos(2 )

2

The Fourier transform of ( ) is

1( ) ( ) ( 2 ) ( 2 )

2

C C

C C

x t m t t m t m t t

x t

X M M M



Let us examine the spectrum of the demodulated DSB-SC signal.

The message sidebands are shifted from being centered at C

back to the about the origin ( = 0) and 2C. This is illustrated

below.

Note that we now needed coherent (synchronous) detection!

( )X

02 C 2 C

Low-pass filterSelects baseband

m m

We filter out the signals centered at and 2C.

Message

recovered



Example 3.5 (on page 101):

We are given a carrier signal of Ac cos(Ct) and a tone message signal of

m(t) = Am cos(m

t)

Therefore, the AM signal is

( ) cos( ) cos( ) cos( ) cos( )2

1This comes from the identity: cos( ) cos( ) = cos( ) cos( )

2

Next we take the Fourier transform,

( ) ( ) (2

C mDSB SC C m C m C m C m

C mDSB SC C m C m

A At A A t t t t

A A

) ( ) ( )C m C m


( )DSB SC

C m 0CC m

( )M

0 mm

C m C

C m

2

C mA A


Tone modulationshown in example


Analog Product Modulator

https://en.wikibooks.org/wiki/Electronics/Analog_multipliers

Log

Log

Anti-Log

BufferOutput

X

Y

outV kX Y

DSB-SC is used primarily today for point-to-point communicationswhere a small number of receivers is involved.

One can buy commercial ICs that perform this function.

https://en.wikibooks.org/wiki/Electronics/Analog_multipliers


Non-Linear DSB-SC Modulator

22 2 2

1 1 2 1

1

22 2 2

2 1 2 1

1

1 2 1 2

2

2 ( )( ) ( ) ( ) 1 cos(2 ) 1 cos( )

2

2 ( )( ) ( ) ( ) 1 cos(2 ) 1 cos( )

2

( ) ( ) ( ) 2 ( ) 4 ( ) cos( )

( ) 4

CC C C

CC C C

C C

DSB SC C

a A a m ts t a m t a m t t a A t

a

a A a m ts t a m t a m t t a A t

a

s t s t s t a m t a A m t t

t a A m

( ) cos( ) BPF selected thisCt t

BPF

( )DSB SC t

Refer to slide 29 for equations.


Non-Linear DSB-SC Modulator (continued)

2( ) 4 ( ) cos( )DSB SC C Ct a A m t t

Note that this expression contains no carrier signal. Why?

Answer: The modulator is a balanced configuration and this results

in the carrier signal being cancelled (that assumes perfect balance

of course).

Definition: A balanced modulator does not output either a carrier

component or the message component. When both are missing we

say it is double balanced.


Switching DSB-SC Modulators

Agbo & Sadiku present three switching modulators:

1. Series-bridge modulator

2. Shunt-bridge modulator, and

3. Ring modulator

We are only going to discuss the ring modulator because it is the

most widely used and contains the fundamental principle of

Operation of all of them. It is important you understand how it

works.


Double-Balanced Diode Ring Modulator

1 1

3 5

4cos( ) cos(3 cos(5C C Ct t t

cos( )C CA t

( )cos( )Ck m t t( )m t ( )x t

( )p t

( )m t ( ) cos( )i Cv p t t

The LO is driven hard enoughto operate the diodes

as on/off switches.

T1 T2

BPF

D1

D2

D4

D3

a b

Bipolar square wave:

1:1 1:1


Assume the diodes act as perfect switches (either “on” or “off”) and are controlled by the RF carrier signal (requires large amplitude).

T1T2

D1

D2

D3

D4

RF carrier signal

AC cos(ct)

m(t)

DSB-SC:

( ) cos( )Cm t t

Double-Balanced Diode Ring Modulator (continued)



Operation in the positive half-cycle of the carrier signal

T1T2

D1

D2

m(t) = 0

+

currents

These currents cancel in the primary, thus,

no output.

Diodes D3 &D4 are Off

Positive Half-Cycle:

AC cos(ct)



AC cos(ct)

Operation in the negative half-cycle of the carrier signal

T1T2

D3

D4

m(t) = 0

+

currents

These currents cancel in the

primary sono output.

Diodes D1 &D2 are Off

Negative Half-Cycle:



AC cos(ct)

Operation in the positive half-cycle of the carrier signal passes message signal m(t) to output.

Diodes D1 and D2 are “on” and the secondary of T1 is applied directly to T2.

T1T2

D1

D2

+_

+_

+

_

+_

+_

+

m(t)

+

_

+ m(t)output



AC cos(ct)

Diodes D3 and D4 are “on” and the secondary of T1 is applied directly to T2.

T1T2

D3

D4

+_

+_

+_

+_

+

_

+

m(t)

+

_

- m(t)output

Operation in the negative half-cycle of the carrier signal inverts message signal m(t) at the output.



1 2&D D on3 4&D D on

DSB-SC signal at primary of T2

m(t)

OUTPUT

LO rectangular waveform

t


Double-Balanced Diode Ring Modulator Waveforms

( )m t

cos(2 )C CA f t

( )cos( )Ck m t t

( ) ( )bipolarp t m t

D1 & D2 are “on”

D3 & D4 are “on”

After filtering


Requirements:

1. The carrier signal is higher in amplitude than the modulating signal m(t).

2. The carrier signal must be of sufficient amplitude to fully switch the diodes between “on” and “off” states.

3. The carrier signal switches the diodes on and off at a rate higher than the highest frequency contained in m(t).

4. The message signal m(t) is chopped into segments, alternating between two amplitudes; +m(t) and –m(t).


( )m t



1( ) 2 ( ) 2 ( ) 1

2

1 2 1 1( ) 2 cos cos(3 ) cos(5 )

2 3 5

4 1 1( ) cos cos(3 ) cos(5 )

3 5

Therefore, the ou

This is bipolar square wave train.bipolar

bipolar

bipolar

C C C

C C C

p t p t p t

p t t t t

p t t t t

tput is found by the product,

4 ( ) ( )( ) ( ) ( ) ( ) cos cos(3 ) cos(5 )

3 5

Upon passing through the BPF,

4( ) ( ) cos

bipolar C C C

DSB SC C

m t m tx t m t p t m t t t t

t m t t

The mathematics behind the Diode Ring Modulator:

t

+1

-1

pbipolar(t)


Double-Balanced Diode Ring Modulator/Mixer

LO RF

IF

It is inexpensive and easy to build a ring mixer.

AB

C

D

GG

Trifilar-Wound Toroid

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwignayV_vrKAhUS-2MKHVoTBbkQjRwIBw&url=http://www.robkalmeijer.nl/techniek/electronica/radiotechniek/hambladen/qst/1993/12/page32/&psig=AFQjCNGdlvdrV0wTXTsxs8EgGft77-iwng&ust=1455666390555567


Commercial Diode Ring Mixer (Mini-Circuits)

Mixer in surface-mount package

Ring of FET devicesoperated as nonlinear

resistances

It is even easierto buy a ring mixercomponent.

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwjf99mNg_vKAhVD5GMKHfd-DwsQjRwIBw&url=http://mwrf.com/components/gauge-wideband-lo-noise-passive-txrx-mixers&psig=AFQjCNGdlvdrV0wTXTsxs8EgGft77-iwng&ust=1455666390555567


Mixers

Frequency mixing frequency conversion heterodyning

( )IF t( )RF t

cos( )LOt

(Mixer)

( )x t

Band-passfilter

RF

LO

IF

A mixer translates the modulation around one carrier frequency to another frequency. In a receiver, this is usually from a higher RF frequency to a lower IF frequency. In a transmitter, it’s the inverse.

We know that a LTI circuit can’t perform frequency translation. Mixers can be realized with either time-varying circuits or non-linear circuits.


Digression: What is Heterodyning?

Heterodyning is a signal processing technique invented in 1901 by Canadian inventor-engineer Reginald Fessenden that creates new frequencies by combining or mixing two frequencies using a nonlinear device.

Using an electronic circuit to combine an input radio frequency signal(RF) with another signal that is locally generated (LO) to produce new frequencies (IF): one being the sum of the two frequencies and the other being the difference of the two frequencies.

http://www.yourdictionary.com/heterodyning#SgeI5zTvo7VV96o7.99

Applications of heterodyning:1. Used in communications to generate new frequencies. 2. Move modulated signals from one frequency channel to another.3. Used in the superheterodyne radio receivers able to select from

multiple communication channels.

http://www.yourdictionary.com/heterodyning#SgeI5zTvo7VV96o7.99


Mixers Perform Frequency Translation

Let ( ) ( ) cos( ) and ( ) ( ) cos( )

The local oscillator (LO) is proportional to cos( )

The mixer (or multiplier) output ( ) is given by

( ) 2 ( ) cos( ) cos( )

Choos. ing

RF C IF IF

LO

C LO

LO C IF

t m t t t m t t

t

x t

x t m t t t

A

, we have

( ) ( ) cos ( ) cos ( )

( ) ( ) cos ( ) cos (2 )

Note: Used even property of cosines [ . ., cos( ) cos( )]

Choosing , then we have

( ) ( ) co (

.

s

C IF C C IF C

IF C IF

LO C IF

C IF

x t m t t t

x t m t t m t t

i e

x t m t

B

) cos ( )

( ) ( ) cos ( ) cos (2 )

C C IF C

IF C IF

t t

x t m t t m t t


Frequency Conversion From C to IF

With a Mixer

Multiplying a modulated signal by a sinusoidal moves the frequency band to sum and difference frequencies.

Note: Super-heterodyning: c + IF ; Sub-heterodyning: c - IF

Example: We want to convert from frequency C to frequency IF .

( )x t ( ) ( ) cos( )RF ct m t t ( ) ( ) cos( )IF IFt m t t

2 cos ( )c IF t

2 f

Input frequency c Output frequency fIF

X()Filtered out

BPF response

IF 2 C IF 2 C IF2 C

Negative not shown


Mixer Example (Page 110)

Example: Derive the relationship between LO and C so that centering the bandpass filter of the mixer is at LO - C and also ensure that IF is less than (i.e., below) C.

Answer:

We know that LO = C IF in general. We must meet two conditions:

(1) IF = LO - C and (2) IF < C

Start by assuming LO = C + IF that meets the first condition; then thesecond condition, IF < C , implies that

LO - C < C LO < 2C


Superheterodyne Receiver is Widely Used

AM radio receiver:

A superheterodyne receiver, often called superhet, is a type of radio

receiver using frequency mixing to convert a received signal to a fixed

intermediate frequency (IF) which can be more conveniently processed

than the original carrier frequency. It was invented by US engineer Edwin

Armstrong in 1918 during World War I. Virtually all modern radio

receivers use the superheterodyne principle.

The word heterodyne is derived from

the Greek roots hetero- "different", and -dyne "power".


Elenco AM/FM Dual-Radio Receiver Kit

AM

FM

Antenna

https://www.fcc.gov/media/radio/broadcast-radio-links



http://www.ni.com/white-paper/13193/en/

http://www.ni.com/white-paper/13193/en/


Another Mixer Example (Page 110)

For a frequency converter the carrier frequency of the output signal is 425 kHz and the carrier frequency of the AM input signal ranges from 500 kHz to 1500 kHz. Find the tuning ratio of the local oscillator

If the frequency of the local oscillator is given by (a) IF = LO - C and(b) IF = C + LO .

Answer: (a) IF = LO - C LO = C + IF superheterodyning

(b) IF = C + LO LO = C - IF sub-heterodyning

,max

,min

,LO

LO

,max ,max

,min ,min

1500 4252.081

500 425

LO C IF

LO C IF

,max ,max

,min ,min

1500 42514.33

500 425

LO C IF

LO C IF


Quadrature Amplitude Modulation (QAM)

Fact: Both conventional AM and DSB-SC AM are wasteful of bandwidth.

One way to improve of bandwidth efficiency is with quadrature amplitudemodulation (QAM). It involves two data streams: the I-channel and theQ-channel.

Bandwidth efficiency is improved by allowing two signals to share the same bandwidth of a channel. But this can only be done if the twomodulated signals are orthogonal to each other. Let’s see how this can beaccomplished.

A better name for this might be quadrature-carrier multiplexing.


Quadrature-Carrier Multiplexing

Quadrature-carrier multiplexing allows for transmitting two message signals on the same carrier frequency.

(1) Two quadrature carriers are multiplexed together,

(2) Signal mI(t) modulates the carrier cos(Ct), and Signal mQ(t) modulates the carrier sin(Ct).

(3) The two modulated signals are added together & transmitted over the channel as

( ) ( ) cos( ) ( ) sin( )QAM I C Q Ct m t t m t t


Quadrature Amplitude Modulation Features

1. QAM transmits two DSB-SC signals in the bandwidth of one DSB-SC signal.

2. Interference between the two modulated signals of the same frequencyis prevented by using two carriers in phase quadrature. This is becausethey are orthogonal to each other.

3. The In-phase (I-phase) channel modulates the cos(Ct) signal and the Quadrature-phase (Q-phase) channel modulates the sin (Ct) signal.

4. The carriers used in the transmitter and receiver are synchronous witheach other. In fact, they must be almost exactly in quadrature with each other, otherwise they experience cochannel interference.

5. Low-pass filters are used to extract the baseband signals mI(t) and mQ(t) in the receiver.


Quadrature Amplitude Modulation and Demodulation

2 cos( )Ct

2 sin( )Ct

( )Qz tTransmitter Receiver

Channel

( )Iz t

( )QAM tcos( )Ct

sin( )Ct

Note: cos( 90 ) sin( )C Ct t

( )Im t

( )Qm t

( )Im t

( )Qm t


Quadrature Amplitude Demodulation (QAM)

2

( ) ( )cos( ) ( )sin( )

( ) 2 cos( ) ( ) 2 cos( ) ( )cos( ) ( )sin( )

( ) 2 ( ) cos ( ) 2 ( ) cos( ) sin( )

( ) ( ) ( ) cos(2 ) ( )sin(2 )

We rec

QAM I C Q C

I C QAM C I C Q C

I I C Q C C

I I I C Q C

t m t t m t t

z t t t t m t t m t t

z t m t t m t t t

z t m t m t t m t t

2

over ( ) by passing ( ) through a LPF.

and

( ) 2 sin( ) ( ) 2 sin( ) ( )cos( ) ( )sin( )

( ) 2 ( ) sin ( ) 2 ( ) sin( ) sin( )

( ) ( ) ( ) cos(2 ) (

I I

Q C QAM C I C Q C

Q Q C I C C

Q Q Q C I

m t z t


z t m t t m t t t

z t m t m t t m t )sin(2 )

We recover ( ) by passing ( ) through a LPF.

C

Q Q

t

m t z t


Quadrature-Amplitude Modulation & Demodulation

( ) ( ) cos( ) ( )sin( )QAM I C Q Ct m t t m t t

2 cos( )Ct

2 sin( )Ct


Channel

( )Iz t

( )QAM tcos( )Ct

sin( )Ct

( )Im t ( )Im t

( )Qm t ( )Qm t

Analogsignals


Quadrature-Amplitude Demodulation

-/2

cos(ct)

sin(ct)

mI(t)

LORF

mQ(t)

Quadrature Downconverter

+C-C

+LO-LO

½ ½

+C-C

+LO

-LO

+½j

-½j

+IF-IF

+IF

-IF -½j

+½j

Re

Im

Im

Re mI(t)

mQ(t)

cos(ct)

sin(ct)

j


Review: Spectral Lines for Sine and Cosine Signals

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwjr4si3pIrLAhVJ42MKHSibA5QQjRwIBw&url=http://www.dsprelated.com/showarticle/192.php&psig=AFQjCNFlO-QBIwHeSwTMWZTg_vrkawoc7w&ust=1456191828896979


Re

Im

Re

Im

mI(t)

mQ(t)

Quadrature-Amplitude Demodulation (continued)

This shows the orthogonality of the two modulated signals.

In-phase signal occupies thereal axis-frequency plane

Quadrature signal occupies theimaginary axis-frequency plane


Quadrature-Amplitude Modulation & Demodulation

( ) ( ) cos( ) ( )sin( )QAM I C Q Ct m t t m t t

2 cos( )Ct

2 sin( )Ct


Channel

( )Iz t

( )QAM tcos( )Ct

sin( )Ct

( )Im t ( )Im t

( )Qm t ( )Qm t

Digitalsignals

Now it becomes a digital communication system


QAM: Phase Error in Synchronous Detection

The local carrier in a DSB-SC receiver and a QAM receiver is 2cos(Ct + ),while the signal carrier at the input of each receiver is cos(Ct). Thatmeans the signal carrier and local carrier in the receivers are phase shifted relative to each other. Derive expressions for the demodulatedoutput signals for both receivers. Compare your results for DSB-SC and QAM.

Solution: (Example 3.9 on pp. 112-113)

For the DSB-SC receiver:

( )=2cos( ) ( ) 2 ( ) cos( ) cos( )

( ) ( ) cos( ) ( ) cos ( )

Therefore, low-pass filering gives ( ) ( ) cos( )

C DSB SC C C

C

x t t t m t t t

x t m t m t t

y t m t


QAM: Phase Error in Synchronous Detection (continued)

1

For the QAM receiver:

( ) 2cos( ) ( ) 2cos( ) ( ) cos( ) ( ) sin( )

( ) 2 ( ) cos( ) cos( ) 2 ( ) sin( ) cos( )

( ) ( ) cos( ) ( ) cos(2 ) ( ) sin( )

I C QAM C I C Q C

I I C C Q C C

I I C Q


z t m t t t m t t t

z t m t m t t m t

1

( ) sin(2 )

and

( ) 2sin( ) ( ) 2sin( ) ( ) cos( ) ( ) sin( )

( ) 2 ( ) cos( ) sin( ) 2 ( ) sin( ) sin( )

( ) ( ) sin( ) ( ) sin(2 ) ( ) cos( )

Q C

Q C QAM C I C Q C

Q I C C Q C C

Q I C Q

m t t


z t m t t t m t t t

z t m t m t t m t m

( ) cos(2 )

The low-pass filters suppress the terms centered at 2

Therefore,

( ) ( ) cos( ) ( ) sin( )

( ) ( ) cos( ) ( ) sin( )

Q C

C

I I Q

Q Q I

t t

y t m t m t

y t m t m t

Co-channel interference


QAM: Frequency Error in Synchronous Detection

Compare the effect of a small frequency error in the local carrier for a DSB-SC receiver and a QAM receiver. The carrier at the transmitter is cos(Ct) and the carrier at the receiver is 2cos(C+ )t.

Answer: (Example 3.10 on pp. 113-114)

( ) ( ) cos( ) ( )sin( )

( ) ( ) cos( ) ( )sin( )

I I Q

Q Q I

y t m t t m t t

y t m t t m t t

Note the similarity to the answer to Example 3.9.You should now be able to guess the answer to aquestion involving both phase error and frequencyerror . (Practice Problem 3.10 on page 114)


Single Sideband (SSB) AM

Why single sideband? DSB-SC is spectrally inefficient because it uses twice the bandwidth of the message. SSB addresses that issue.

The signal can be reconstructed from either the upper sideband (USB) orthe lower sideband (LSB).

SSB transmits a bandpass filtered version of the modulated signal.


Single Sideband (SSB) AM

Multiplication of a USB signal by cos(Ct) shifts the spectrum to left and right.


Phase-Shift Method to Generate SSB AM

( ) ( )cos( ) ( )sin( )

where minus sign applies to USB

and plus sign applies to the LSB.

( ) is ( ) phase delayed by - /2

SSB C h C

h

t m t t m t t

m t m t


Phase-Shift Method to Generate SSB AM

• The phasing method uses two balanced mixers to eliminate the carrier.

• The phasing method for SSB generation uses a phase-shift to cancel one of the sidebands.

• The carrier oscillator is applied to the upper balanced modulator along with the modulating signal.

•

• The carrier and modulating signals are both shifted by 90 degrees and applied to another modulator.

• Phase-shifting causes one sideband to cancel when the two modulator outputs are summed together.


Phase-Shift Method for Receiving SSB Signals

http://www.panoradio-sdr.de/ssb-demodulation/

https://www.dsprelated.com/showarticle/176.php

Reference:

I

Q

http://www.panoradio-sdr.de/ssb-demodulation/



Reference Note: Quadrature Phase-Shifts

sin( )tcos( )t

2

For a + 90° (or /2) phase shift:

sin( ) cos( )

cos( ) sin( )

t t

t t

+j

-j

( ) sgn( )H j

=

sin( )t

+j/2

-j/2

-1/2 -1/2

cos( )t

Hilbert transform

For a - 90° (or -/2) phase shift:

sin( ) cos( )

cos( ) sin( )

t t

t t

https://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwi0uc_In4rLAhUBw2MKHeFxAekQjRwIBw&url=https://en.wikipedia.org/wiki/Phase_(waves)&psig=AFQjCNFlO-QBIwHeSwTMWZTg_vrkawoc7w&ust=1456191828896979


Synchronous Demodulation of SSB AM


( ) 2 ( ) cos( ) ( ) cos( ) ( ) sin( ) 2cos( )

( ) ( ) ( ) cos(2 ) ( ) sin(2 )

and upon low-pass filtering we have

( ) ( )

SSB C C h C C

C h C

x t t t m t t m t t t

x t m t m t t m t t

x t m t



Hartley Image-Rejection Architecture

cos(Ct)

sin(Ct)

IF-IF

LO-LO

-IF

-IF

IF-IF

IFIF

90

RF

IF

Antenna

http://www.microwavejournal.com/articles/3226-on-the-direct-conversion-receiver-a-tutorial

http://www.microwavejournal.com/articles/3226-on-the-direct-conversion-receiver-a-tutorial


SSB Mixer and Image Rejection Mixer Comparison

https://commons.wikimedia.org/wiki/File:SSB_and_Image_Rejection_Mixer.svg

https://commons.wikimedia.org/wiki/File:SSB_and_Image_Rejection_Mixer.svg


Pulse Amplitude Modulation (PAM) Digital Signal

How is PAM in digital communication similar to AM in analog communication?


Questions

?

https://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwjLw6LT8P3KAhUN72MKHa99B-gQjRwIBw&url=https://commons.wikimedia.org/wiki/File:KBRC_AM_radio_antenna_tower.JPG&psig=AFQjCNHzTRNFwYQOJHap6VIxaD8wv1Yp1g&ust=1455766920352627


Questions

1. What is the point of creating a rectified output when using a diode forAM modulation?

It combines the carrier signal with the message signal m(t). See slides 32, 33 & 34 for illustration of this.

2. More about how mixers work. (Three questions asked about mixers.)

Two principles are used in mixers to create new frequencies: (1) Nonlinearitythe I-V characteristics of a nonlinear device do this, and (2) time-varying switching will create new frequencies. We provided examples of the use ofboth in slides 27 through 34.

3. Explain the idea of images in mixers again.

See the next four slides.


Image Signals in Mixers (1)

RF IF

LO

frequency

Signal band

LORF

frequency

IF band

IF

RF

spec

trum

Desireddown-conversion

IF = LO - RF

This converts thespectrum at the RFcarrier frequency

down to the spectrumcentered at the IF

frequency.

There is no signal inthis part of spectrum.


Image Signals in Mixers (2) – Now an Image Signal Appears

Now both the spectrum at the RF

carrier frequency andthe undesired imagespectrum are down

converted to the spectrum centered at

the IF frequency.

RF IF

LO

frequency

Signal band

Image band

LORF image

frequency

IF band

IF

RF

spec

trum

IF = LO - RF

IF = image - LO

and

Now both signalsappear in the IF band.

The image spectrum is not wanted.


Image Signals in Mixers (3)

RF IF

LO

frequency

Signal band

LORF

frequency

IF band

IF

RF

spec

trum

Again, the desireddown-conversion

IF = RF - LO

This converts thespectrum at the RFcarrier frequency

down to the spectrumcentered at the IF

frequency.

Suppose the LO frequency isBelow the RF frequency.


Image Signals in Mixers (4) – With Image Signal

As before both the spectrum at the RF

carrier frequency andthe undesired imagespectrum are down

converted to the spectrum centered at

the IF frequency.

RF IF

LO

frequency

Signal band

LORF

frequency

IF band

IF

RF

spec

trum

IF = RF - LO

The LO frequency isbelow the RF frequency.

image

Again both signalsappear in the IF band.

IF = LO - image

and


More Questions

4. Why does FM take so much more bandwidth than AM?

It is because we force FM (and PM) signals to have constant amplitude. We show this in detail when we cover Angle Modulation.

5. Is there a point where having too many mixers can impact your frequencynegatively?

Issues with using multiple mixers:a. Adds complexity (such as many mixing products come into play)b. Most mixers are lossy and need power gain to continue to

process signals (and mixers add noise of their own to signals)c. Cost (not only of mixers but for LO oscillators and amplifiers)


More Questions

6. In an antenna why does the length have to be /4?

It actually does not have to be /4, but making it longer does not have An advantage in maximizing its efficiency. Making it shorter does decrease the signal strength received. Also, wewant the antenna to resonate at it operating frequency to increase theefficiency of the antenna.

7. Also, if an AM signal has a wavelength of hundreds of feet, how can itsantenna be so small?

The coil of wire picks up the time-variation of the electromagnetic wave’smagnetic field and induces a current in the coil which becomes the signal at the input of the AM receiver.


More Questions

8. When you have no carrier signal, are you sending the message signal?If so, is there even any modulation of the amplitude or justthe original signal?

In the absence of a carrier signal, then only the message signal can be sentat the frequency band of the message signal. We say the message signal “modulates” the carrier signal.

9. Does the local oscillator change its frequency depending upon the RFfrequency you want to select?

I assume you are referring to the superheterodyne receiver. Yes, the local oscillator frequency is changed to select the RF frequency you want to select. We require that frequency difference between theRF and LO frequencies is a constant.


More Questions

10. How is a mixer similar/different from amplitude modulation?

The mathematics of the AM says we use a multiplier to multiply the carriersignal with the message signal. A mixer performs signal multiplication as required in amplitude modulation.

11. Why do we use sub-heterodyning? What are the common applicationsof sub-heterodyning?

When analyzing a system we focus upon the RF frequencies involved andPossible local oscillator frequencies. The focus is upon frequency conversion.Considerations: Do we want to work with higher or lower frequencies? What frequency bands are to be avoided?

12. Which modulation is the most useful today? AM, FM or PM Why?

FM is the most widely used. It has better noise immunity.


More Questions

13. How do you build a mixer?

One example is slide 62 showing the diode ring mixer.

14. How would you handle having your oscillator being 1% off?

A 1% offset in frequency is very large. You might lock it to a referencefrequency (such as a precision crystal oscillator). You might use a phase-lock loop to slave the local oscillator to the correct frequency. You mightuse a pilot signal broadcast with the modulated carrier. There are manyother possible solutions.

15. From our homework assignments what would you consider the mostimportant problems?

All of them. During the review session before the first midterm I will give abetter answer.


More Questions

16. What happens to an over-modulated signal? Can the signal still be used?

Over-modulation leads to distortion in the message. It can still be used in voice communication if the voice over-modulation is not too severe. The point is that voice can still be understood even with moderate distortion.

17. Why do RLC resonator circuits have a -3 dB frequency corresponding toa bandwidth of B = 1/RC?

2 22 2

1 2

2 1

1 1 1 1&

2 2 2 2

1Bandwidth

LC LCRC RC RC RC

BRC

2( )

( ) ( )Ri j L

Hi t R j L j RCL


More Questions

R

+

_y(t)

m(t)+

_

Accos(Ct)+

_

Diode

BPFFilter(c)

+

_x(t)

The diode is the nonlinear component (it has an exponentialcharacteristic). Using a Taylor’s series we can express the diodecurrent iD as (only first two terms of Taylor’s series),

2

1 2

2 2

1 2 1 2

( ) ( ) ( ); ( ) is diode voltage.

The voltage across resistor R is given by

( ) ( ) ( ) ( ) ( ) ( )

D D D D

D D D D D

i t b v t b v t v t

x t i t R b Rv t b Rv t a v t a v t

“Square Law”behavior

( )Di t

18. How does x(t) =iD(t)R come about? Why is it not KVL?


More Questions

19. Can you explain using nonlinearity for modulation much like Problem 3 in Homework #3?

2

2

2 2 2

( ) 4 , but 1

Substituting for ( ( ) cos( )) ,

( ) 4 ( ) cos( ) ( ) cos( )

( ) 4 ( ) cos( ) ( ) 2 ( ) cos( ) cos ( )

But we k

D D D

D C C

C C C C

C C C C C C

x t i R v v R R

v m t A t

x t m t A t m t A t

x t m t A t m t A m t t A t

2

2 2now cos ( ) 1 cos(2 )2

CC C C

AA t t

Continued next slide

Modulator

Given relationship


2 2 2

2 22

( ) 4 ( ) cos( ) ( ) 2 ( ) cos( ) cos ( )

( ) 4 ( ( )) ( ) 2 ( ) cos( ) cos(2 )2 2

The band-pass filter (BPF) passes only terms of cos( )

C C C C C C

C CC C C

C

x t m t A t m t A m t t A t

A Ax t m t m t A m t t t

t

, thus ( ) is

( ) 4 cos( ) 2 ( ( )) cos( )

( ) 4 2 cos( ) ( ) cos( )

( ) cos( ) ( ) cos( )

C C C C

C C C C

C C

y t

y t A t A m t t

y t A A t m t t

y t K t m t t

Problem 3 in Homework #3 continued:

More Questions


More Questions

20. What is the primary form of noise production in AM systems?

The greatest noise problem in AM channels is interference, noise pickup,& fading in wireless transmission, all of which distort the amplitude of thetransmitted signal.

21. QAM (Several asked about QAM so we need to cover it again)

I will review QAM again after questions are answered.


2

( ) ( )y t A B g t

22( ) ( ) cos( ) 1 cos(2 )

2

( ) cos(2 )2 2

By t A B g t A B t A t

B By t A t

2 3

( ) ( ) ( ) ( ) .y t A Bg t C g t D g t other terms

22. Go over Problem 2 in Homework #2

Problem 2 Square Law Device (20 points)

Answer: The square-law device generates a frequency that is the double of the single tone frequency f. To show this we make use of the trigonometric identity:

Answer: The cubic term in the series generates a frequency that is triple of the frequency of g(t), that is,

frequency 3f. This comes from using the trigonometric identity of cos3(x) = ¼[3cos(x) + cos(3x)].

Thus, the cos3(2ft) term gives us both a cos(2ft) term (not so interesting) and a cos(3·2ft) term

(which is a new frequency being introduced).

More Questions

You are given a square-law component with an input to output relationship of

(a) To explore the behavior of this device we let the input signal g(t) be a sinusoidal

tone, that is, g(t) = cos(t).

(b) What frequencies does the cubic term (that is, D[g(t)]3) generate when driven by g(t) = cos(t)?


More Questions

23. Why does milliwatts relate to dBm rather than just use milliwatts?

We express milliwatts (mW) in decibels by

We use this because logarithms add rather than multiply in calculations.Example:

Suppose a signal of 3 dBm power drives an amplifier of gain = 13 dB. What is theoutput power of the amplifier.

Answer: Output power (in dB) = 3 dBm + 13 dB = 16 dBm,

rather than 2 mW (= 3 dBm) multiplied by gain of 20 (= 13 dB) = 40 mW

10Power in dBm 10 log dBm Note: logarithm of a ratio1 mW

mWP


More Questions

24. What is the one question you think we should have asked?

Answer: The one that is troubling you.


A Typical Superheterodyne Receiver


Generating m(t)cos(Ct) using Convolution

Fold, Shift & Multiply

From Convolution Theorem: (t) g(t) Cn( ) G( )

F

F

G( )

Cn( ) (t)

g(t)

Output Spectrum

LO

RF

( ) cos( )Cm t t