amplitude modulation early radio ee 442 spring …//. ... fm pm. ... 100 khz carrier modulated by...
TRANSCRIPT
AM Modulation -- Radio 1
http://www.technologyuk.net/telecommunications/telecom_principles/amplitude_modulation.shtml
LO audio baseband mf f f
Amplitude Modulation – Early RadioEE 442 – Spring Semester
Lecture 6
AM Modulation -- Radio 2
Modulation Options
We will study:AMFMPM
AM Modulation -- Radio 3
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.
AM Modulation -- Radio 4
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
AM Modulation -- Radio 5
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
AM Modulation -- Radio 6
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
AM Modulation -- Radio 7
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
AM Modulation -- Radio 8
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)
AM Modulation -- Radio 9
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
AM Modulation -- Radio 12
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.
AM Modulation -- Radio 13
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
AM Modulation -- Radio 14
Amplitude Modulation (DSB with Carrier)
http://hyperphysics.phy-astr.gsu.edu/hbase/Audio/bcast.html
AM Modulation -- Radio 15
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
AM Modulation -- Radio 16
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
AM Modulation -- Radio 17
Phasor Interpretation of AM DSB with Carrier (continued)
https://www.slideshare.net/azizulhoque539/eeng-3810-chapter-4
Tone modulation
AM Modulation -- Radio 18
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
AM Modulation -- Radio 19
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
AM Modulation -- Radio 20
AM Modulation Index Basics – Examples
50% Tone Modulation
100% Tone Modulation
150% Tone Modulation
Overmodulationor
Envelope Distortion
50%
100%
150%
AM Modulation -- Radio 21
AM Overmodulation Envelope Distortion
https://www.physicsforums.com/threads/amplitude-modulation-vs-beats.890811/
AM Modulation -- Radio 22
Power Efficiency of Amplitude Modulation
Agbo & Sadiku; Section 3.2.2; pp. 91 to 92
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
AM Modulation -- Radio 23
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
AM Modulation -- Radio 24
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).
AM Modulation -- Radio 25
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
AM Modulation -- Radio 26
Generation of Amplitude Modulated Signals
Agbo & Sadiku; Section 3.2.3; pp. 93 to 95
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)
AM Modulation -- Radio 27
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)
AM Modulation -- Radio 28
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
AM Modulation -- Radio 29
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)
AM Modulation -- Radio 30
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
AM Modulation -- Radio 31
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
AM Modulation -- Radio 32
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
AM Modulation -- Radio 33
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
AM Modulation -- Radio 35
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 Modulation -- Radio 36
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 Modulation -- Radio 37
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
AM Modulation -- Radio 38
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
AM Modulation -- Radio 39
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
AM Modulation -- Radio 40
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
AM Modulation -- Radio 41
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
AM Modulation -- Radio 42
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
AM Modulation -- Radio 43
( )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)
AM Modulation -- Radio 44
Double-Sideband Suppressed Carrier AM (continued)
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
AM Modulation -- Radio 45
Double-Sideband Suppressed Carrier AM (continued)
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
AM Modulation -- Radio 46
Double-Sideband Suppressed Carrier AM (continued)
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
AM Modulation -- Radio 47
( )DSB SC
C m 0CC m
( )M
0 mm
C m C
C m
2
C mA A
Double-Sideband Suppressed Carrier AM (continued)
Tone modulationshown in example
AM Modulation -- Radio 48
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.
AM Modulation -- Radio 49
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.
AM Modulation -- Radio 50
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.
AM Modulation -- Radio 51
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.
AM Modulation -- Radio 52
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
AM Modulation -- Radio 53
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)
AM Modulation -- Radio 54
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)
AM Modulation -- Radio 55
Double-Balanced Diode Ring Modulator (continued)
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:
AM Modulation -- Radio 56
Double-Balanced Diode Ring Modulator (continued)
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
AM Modulation -- Radio 57
Double-Balanced Diode Ring Modulator (continued)
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.
AM Modulation -- Radio 58
Double-Balanced Diode Ring Modulator (continued)
1 2&D D on3 4&D D on
DSB-SC signal at primary of T2
m(t)
OUTPUT
LO rectangular waveform
t
AM Modulation -- Radio 59
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
AM Modulation -- Radio 60
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).
Double-Balanced Diode Ring Modulator
( )m t
AM Modulation -- Radio 61
Double-Balanced Diode Ring Modulator
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)
AM Modulation -- Radio 62
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
AM Modulation -- Radio 63
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.
AM Modulation -- Radio 64
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.
AM Modulation -- Radio 65
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.
AM Modulation -- Radio 66
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
AM Modulation -- Radio 67
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
AM Modulation -- Radio 68
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
AM Modulation -- Radio 69
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".
AM Modulation -- Radio 70
Elenco AM/FM Dual-Radio Receiver Kit
AM
FM
Antenna
AM Modulation -- Radio 72
http://www.ni.com/white-paper/13193/en/
AM Modulation -- Radio 73
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
AM Modulation -- Radio 74
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.
AM Modulation -- Radio 75
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
AM Modulation -- Radio 76
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.
AM Modulation -- Radio 77
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
AM Modulation -- Radio 78
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 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 )sin(2 )
We recover ( ) by passing ( ) through a LPF.
C
Q Q
t
m t z t
AM Modulation -- Radio 79
Quadrature-Amplitude Modulation & Demodulation
( ) ( ) cos( ) ( )sin( )QAM I C Q Ct m t t m t t
2 cos( )Ct
2 sin( )Ct
( )Qz tTransmitter Receiver
Channel
( )Iz t
( )QAM tcos( )Ct
sin( )Ct
( )Im t ( )Im t
( )Qm t ( )Qm t
Analogsignals
AM Modulation -- Radio 80
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
AM Modulation -- Radio 81
Review: Spectral Lines for Sine and Cosine Signals
AM Modulation -- Radio 82
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
AM Modulation -- Radio 83
Quadrature-Amplitude Modulation & Demodulation
( ) ( ) cos( ) ( )sin( )QAM I C Q Ct m t t m t t
2 cos( )Ct
2 sin( )Ct
( )Qz tTransmitter Receiver
Channel
( )Iz t
( )QAM tcos( )Ct
sin( )Ct
( )Im t ( )Im t
( )Qm t ( )Qm t
Digitalsignals
Now it becomes a digital communication system
AM Modulation -- Radio 84
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
AM Modulation -- Radio 85
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 t t t m t t m t t
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 t t t m t t 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
AM Modulation -- Radio 86
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)
AM Modulation -- Radio 87
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.
AM Modulation -- Radio 88
Single Sideband (SSB) AM
Multiplication of a USB signal by cos(Ct) shifts the spectrum to left and right.
AM Modulation -- Radio 89
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
AM Modulation -- Radio 90
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.
AM Modulation -- Radio 91
Phase-Shift Method for Receiving SSB Signals
http://www.panoradio-sdr.de/ssb-demodulation/
https://www.dsprelated.com/showarticle/176.php
Reference:
I
Q
AM Modulation -- Radio 92
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
AM Modulation -- Radio 93
Synchronous Demodulation of SSB AM
https://www.dsprelated.com/showarticle/176.php
( ) 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
AM Modulation -- Radio 94
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
AM Modulation -- Radio 95
SSB Mixer and Image Rejection Mixer Comparison
https://commons.wikimedia.org/wiki/File:SSB_and_Image_Rejection_Mixer.svg
AM Modulation -- Radio 96
Pulse Amplitude Modulation (PAM) Digital Signal
How is PAM in digital communication similar to AM in analog communication?
AM Modulation -- Radio 97
Questions
?
AM Modulation -- Radio 98
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.
AM Modulation -- Radio 99
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.
AM Modulation -- Radio 100
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.
AM Modulation -- Radio 101
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.
AM Modulation -- Radio 102
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
AM Modulation -- Radio 103
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)
AM Modulation -- Radio 104
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.
AM Modulation -- Radio 105
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.
AM Modulation -- Radio 106
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.
AM Modulation -- Radio 107
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.
AM Modulation -- Radio 108
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
AM Modulation -- Radio 109
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?
AM Modulation -- Radio 110
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
AM Modulation -- Radio 111
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
AM Modulation -- Radio 112
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.
AM Modulation -- Radio 113
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)?
AM Modulation -- Radio 114
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
AM Modulation -- Radio 115
More Questions
24. What is the one question you think we should have asked?
Answer: The one that is troubling you.