Transcript Chap05--Amplitude Mo..

Chapter V. Amplitude Modulations
 An unmodulated sinusoidal carrier signal can be described as
ec(t) = Ec cos2pfct
(5-1)
where Ec is the peak continuous-wave (CW) amplitude
and fc is the carrier frequency in hertz.
 Figure 5-3 illustrates the result of amplitude modulation of
the carrier by a squarewave and a sinusoid.
 The sinusoidal modulating signal of Figure 5-3c can be described by
em(t) = Em cos2pfmt,
(5-2)
where Em is the peak voltage of the modulation signal of frequency fm.
 The sinusoidally modulated AM signal is shown in Figure 5-4.
For an arbitrary information signal em(t), the AM signal is
e(t) = (Ec + em(t)) cos2pfct
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
(5-3)
Chapter V. Amplitude Modulations
Figure 5-3. Amplitude modulation.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Chapter V. Amplitude Modulations
Figure 5-4. (a) Amplitude-modulated signal.
(b) Information signal to be transmitted by AM.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Modulation Index
 AM modulation index is defined by ma = Em/Ec. Hence, the AM signal
can be written for sinusoidal modulation as
e(t) = Ec(1+ ma cos2pfmt) cos2pfct.
 A convenient way to measure the AM index is to use an oscilloscope:
simply display the AM waveform as in Figure 5-4, and measure the
maximum excursion A and the minimum excursion B of the amplitude
"envelope" (the information is in the envelope).
 The AM index is computed from Figure 5-4 as
A = 2(Ec + Em)
(5-4a)
B = 2(Ec - Em)
(5-4b)
and, solving for Ec and Em in terms of A and B will then yield
(5-4c)
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Chapter V. Amplitude Modulations
 It should be clear that the peak measurements A/2 and
B/2 will yields ma also. The numerical value of ma is
always in the range of 0 (no modulation) to 1.0 (full
modulation) and is usually expressed as a percentage
of full modulation.
 If more than one sinusoid, such as a musical chord (that
is, a triad, 3 tones), modulates the carrier, then we get
the resultant AM, index by RMS-averaging the indices
that each sine wave would produce.

Thus, in general,
ma = (m12 +m22 +m32 +… +mn2)1/2
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
(5-5)
Chapter V. Amplitude Modulations
Figure 5-6. AM represented as the vector
sum of sidebands and carrier.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Chapter V. Amplitude Modulations
Figure 5-7.
(a) ma= 1.0(100% AM).
(b) The result of over-modulation
that corresponds to the
spectrum in (c).
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Chapter V. Amplitude Modulations
 In Fig. 5-6 the AM signal is shown to be the instantaneous
phasor sum of the carrier fc, the lower-side frequency fc–fm,
and the upper-side frequency fc+fm.
 The phasor addition is shown for six different instants,
illustrating how the instantaneous amplitude of the AM
signal can be constructed by phasor addition.

Notice how the USB (fc+fm), which is a higher frequency
than fc, is steadily gaining on the carrier, while the LSB
(fc–fm), a lower frequency, is steadily falling behind.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Chapter V. Amplitude Modulations
 The last phasor sketch in Figure 5-6 shows the phasor
relationship of sidebands to carrier at the instant
corresponding to the minimum amplitude of the AM
signal.
 You can see that if the amplitude of each sideband is
equal to one-half of the carrier amplitude, then the
AM envelope goes to zero.
 This corresponds to the maximum allowable value of
Em; that is, Em =Ec and ma = 1.0 or 100% modulation.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Chapter V. Amplitude Modulations
 As illustrated in Figures 5-7b and c, an excessive
modulation voltage will result peak clipping and
harmonic distortion, which means that additional
sidebands are generated.
 Not only does over-modulation distortion result in the
reception of distorted information, but also the
additional sidebands generated usually exceed the
maximum bandwidth allowed.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Chapter V. Amplitude Modulations
Figure 2-2. The AM
signal sisplayed in
the frequency
domain where fc is
the carrier,A is
magnitude, the
modulating
frequency is fixed,
and m% is the
variable.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Chapter V. Amplitude Modulations
Figure 2-3. The signal
displayed in the
frequency domain
where fc is the carrier,
A is magnitude, m is
constant, and the
modulation frequency
fm is the variable.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AM SPECTRUM AND BANDWIDTH
 Let us analyze the mathematical expression for the AM
signal
e(t) = (Ec + Em cos2pfmt) cos2pfct
= Ec cos2pfct + Em cos2pfmt cos2pfct

The second term of this expression can be expanded by
the trigonometric identity
cosA.cosB = (1/2)[cos(A-B) + cos(A+B)],
so that
e(t) = Ec cos2pfct
carrier
+ (Em/2) cos2p(fc-fm)t lower sideband, LSB
+ (Em/2) cos2p(fc+fm)t upper sideband, USB
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AM SPECTRUM AND BANDWIDTH
Figure 5-5. Frequency spectrum of the AM signal shown
in Figure 5-4.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AM SPECTRUM AND BANDWIDTH
Figure 5-11. Amplitude-modulated signal. (a) Generating the
AM signal. (b) The AM signal (time domain). (c) The “envelope.”
(d) One-sided spectrum of the AM signal (frequency domain).
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
POWER in an AM SIGNAL


Consider Equation (5-6), if the voltage signal is present on
an antenna of effective real impedance R, then the power
of each component will be determined from the peak
voltages of each sinusoid.
For the carrier, Pc = Ec2/2R, and for each of the two
sideband components,
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
POWER in an AM SIGNAL
 Therefore,
P1sb = ma2Pc/4
(5-7)
where P1sb denotes the power in one sideband only.
 The total power in the AM signal will be the sum of
these powers:
Ptotal = Pc + PLSB + PUSB
= Pc + (m2/4)Pc + (m2/4)Pc
= Pc(1+ m2/2) = Pt
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
(5-8)
POWER in an AM SIGNAL
Relative AM Signal Energy versus Modulation Index
%m
0
10
20
30
40
50
60
70
80
90
100
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Energy (Pt/Pc)
1
1.005
1.02
1.045
1.08
1.125
1.18
1.245
1.32
1.405
1.5
POWER in an AM SIGNAL
 From the tabulation we see that the energy contribution
to the carrier is 1/2 that of the carrier itself at 100%
modulation. Each of the two sidebands contributes 1/2 of
this value or 1/4 of the energy.
 As an example, a 100 watt carrier 100% AM by a sine
wave, will have an average of 50 watts of sideband power,
composed of 25 watts from each sideband.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
NONSINUSOIDAL MODULATION SIGNALS
 The information signal in modulated systems, as
illustrated by Figure 5-8a, is often referred to as
the baseband signal,

and the spectrum of Figure 5-8c is the (one-sided)
baseband spectrum, where only positive frequencies
are shown.
 The modulated signal spectrum (one-sided) of Fig.
5-8d consists of the upper and lower sidebands on
either side of the carrier.

Figure 5-8d clearly shows that the information
bandwidth of the AM signal is 2fm(max).
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
NONSINUSOIDAL MODULATION SIGNALS
Figure 5-8. (a) Information signal (modulation). (b) AM output (time
domain). (c) One-sided frequency spectrum of m(t). (d) AM output
frequency spectrum (one sided).
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
NONSINUSOIDAL MODULATION SIGNALS
 The mathematically formal method of determining the
frequency spectrum of a time-varying signal is to
employ the Fourier transform.
vAM(t) = Ec.cos2pfct + m(t).cos2pfct
(5-9)
VAM(f) = (Ec/2)[d(f - fc) + d(f + fc)]
+ (1/2)[M(f - fc) + M(f + fc)]
(5-10)
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
NONSINUSOIDAL MODULATION SIGNALS
 For the two-sided idealized audio frequency spectrum of
Figure 5-9a, the plot of Fourier transform of vAM(t),
VAM(f), is illustrate in Figure 5-9b.
Figure 5-9. (a) Idealized audio time-domain signal and baseband
Fourier transform spectrum. (b) Fourier transform spectrum of m(t)
amplitude modulated on a (cosine) carrier signal of frequency fc.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AMPLIIUDE SHIFT KEYING (ASK) or
ON/OFF KEYING (OOK)
 ASK or OOK consists of a carrier which is turned on,
for by a mark and off by a space. The carrier takes on
the form of an interrupted carrier, such as in telegraphy,
only the data is encoded differently. The signal takes on
the form shown in Figure 2-5.
 Detection of such a signal may be non-coherent or
coherent, with the latter being the better performer,
although more difficult to achieve.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AMPLIIUDE SHIFT KEYING (ASK) or
ON/OFF KEYING (OOK)
Fig. 2-5. The ASK signal in the time domain.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AMPLIIUDE SHIFT KEYING (ASK) or
ON/OFF KEYING (OOK)

If the carrier burst is defined as
(5-11)
then the Fourier transform may be expressed as
(5-12)

where
d = t/T
(5-13)
is the duty cycle of the bursts, and T is the pulse
repetition period (inverse of the pulse repetition
frequency, PRF).
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AMPLIIUDE SHIFT KEYING (ASK) or
ON/OFF KEYING (OOK)
Figure 5-10. 25% duty cycle on-off key (OOK) signal and
one-sided frequency spectrum.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AMPLIIUDE SHIFT KEYING (ASK) or
ON/OFF KEYING (OOK)
Figure 2-6. Non-coherent detection of ASK.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Non-Coherent-Detection of ASK
 Non-coherent detector in its simplest form consists of
envelope detector followed by decision circuitry, as
shown in Fig. 2-6.
 The decision threshold grossly affects the error
probability (Pe) for mark and space independently,
and they are therefore not equally probable.
 The “mark” or “carrier on” signal consists of carrier
plus noise whereas the “space” or “carrier off” signal
is noise alone. It has been shown that Pe(mark) can be
made equal to Pe(space) for a given C/N.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Non-Coherent-Detection of ASK
 Minimum probability of error results when a thresholds
of roughly
(pulse-amplitude/2).(1+2Eb/No)1/2
(2-8)
is used, where
Eb is the pulse energy
No is the noise density per reference bandwidth
 In general, ASK is a poor performer although it is used in
non-critical applications.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Non-Coherent-Detection of ASK
 For Eb/No >> 1 and a decision threshold of half the pulse
amplitude, the probability of error for space is
Pe(space) = exp[-(Eb/2No)]
(2-9)
and for mark
Pe(mark) = exp[-(Eb/2No)]/(2pEb/No)1/2
(2-10)
 From this, it is seen that the majority of errors are spaces
converted to marks.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Coherent Detection of ASK
 Coherent detection requires a product detector with
a reference signal which is phase coherent with the
incoming signal carrier (see Figure 2-7).
 The product detector is followed by an integrator
and a decision circuit timed to function at the end
of bit time t.
Fig. 2-7. Coherent detection of ASK using synchronous detection.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Coherent Detection of ASK
 An equivalent performer is the matched-filter detector
shown in Figure 2-8.
 Here, the output of the matched filter is the convolution
of the pulse and the impulse response of the matched
filter.
 The resulting output is ideally diamond shaped and of
duration 2t, with maximum signal energy at a time of
A2/2, where A is the signal amplitude and t is the pulse
duration.
 To complete the system, the decision circuitry is timed to
function at time t for optimum performance.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Coherent Detection of ASK
Fig. 2-8. Matched filter detection of ASK signals.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Coherent Detection of ASK
 The probability of error for coherent ASK signaling is:
(2-11)
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AM DEMODULATION
Figure 5-13. Peak amplitude detector.
 The diode shown in Figure 5-13 conducts whenever vin exceeds
the diode cut-in voltage of about 0.2V for germanium. Hence,
with no capacitor, the detector output is just the positive peaks
of the input AM signal.
 The value of vo will rise and fall at the same rate as the
information -- 5 kHz in this case. All that is required is some
filtering to smooth out the recovered information.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AM DEMODULATION
 If a capacitor is added to the circuit, as shown in Fig. 5-14,
not only is filtering provided but also the average value of
the demodulated signal is increased.
 The capacitor charged up to the positive peak value of the
carrier pulse while the diode is conducting.
 The capacitor is allowed to discharge just slowly enough
through the resistor that the very next carrier peak will
exceed vo, thereby allowing the diode to conduct and charge
the capacitor up to the new peak value.
 The result is that the output voltage will follow the input
AM peaks with a loss of only the voltage dropped across the
diode.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AM DEMODULATION
Figure 5-14. AM demodulation.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Diagonal Clipping Distortion
 The values of R and C at the output of the AM detector must
be chosen to optimize the demodulation process. As illustrated
in Figure 5-15, if the capacitor is too large, it will not be able
to discharge fast enough for vo to follow the fast variations of
the AM envelope.
 The result will be that much of the information will be lost
during the discharge time. This effect is called diagonal
clipping because of the diagonal appearance of the discharge
curve.
 The distortion that results, however, is not just poor sound
quality (fidelity) as for peak clipping; it can also result in a
considerable loss of information.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Diagonal Clipping Distortion
Figure 5-15. Diagonal clipping.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Diagonal Clipping Distortion
 The optimum time constant is determined by analyzing the
diagonal clipping problem. Compare the RC discharge rate
required for the low modulation index illustrated in Figure
5-16a with that required for the same modulating signal but
higher index seen in b.
 Clearly, the modulation index is an important parameter,
and the appropriate RC time constant depends not only on
the highest modulating frequency fm(max), but also on the
depth or percentage of modulation ma. In fact, the maximum
value of C is determined from Equation (5-16).
(5-16)
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Diagonal Clipping Distortion
Figure 5-16. A higher-index AM requires a
shorter RC time constant.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Diagonal Clipping Distortion
Figure 5-17. Complete AM detector and volume (loudness) control.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Diagonal Clipping Distortion
 In 5-17, the demodulated information signal is ac coupled by
capacitor Cc to the audio amplifiers. Coupling capacitor Cc
is made large enough to pass the lowest audio frequencies
while blocking the dc bias of the audio amplifier and the
average (dc) value of vo.
 The audio amplifier input impedance ZA should be much
greater than the output impedance of the detector R to
avoid peak clipping distortion that occurs when the peak ac
current required by ZA is greater than the average current
available.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
SUPERHETERODYNE RECEIVERS
 The standard AM broadcast band in North America extends
from 535 to 1605 kHz, with transmitted carrier frequencies
every 10 kHz from 540 to 1600 kHz (20Hz tolerance).
 The 10 kHz of separation AM stations allows for a maximum
modulation frequency of 5 kHz.
Figure 5-19. Superheterodyne receiver.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
SUPERHETERODYNE RECEIVERS
 Indeed, many use a second down conversion and second
IF amplifier system following the first IF. Such a superheterodyne receiver is called a double-conversion receiver.
 The LO frequency is almost always higher than the RF
carrier frequency, a characteristic referred to as high-side
injection to the mixer.
 For example, to receive the AM station whose RF carrier is
fRF = 560 kHz, the LO must be tuned to
fLO = fRF + fIF
= 560 kHz + 455 kHz
= 1015 kHz
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Choice of IF Frequency and Image Response
 Receiver selectivity, tuning in one station while rejecting
interference from all others, is determined by filtering at
the receiver RF input and in the IF.
Figure 5-20. IF filter at mixer output.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Choice of IF Frequency and Image Response
 Adjacent channels are rejected primarily by IF filtering.
Filter out adjacent channel transmissions right at the RF
input are difficult.
 The first consideration is that the RF input circuit may be
required to tune over a relatively wide frequency range.
Maintaining a high Q and constant bandwidth in such a
circuit is very difficult.
 The second consideration is that multipole networks are
employed. Tuning a multiple filter from station to station
over even a moderate tuning range is not practical.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Choice of IF Frequency and Image Response
 The image frequency is that frequency which is exactly one
IF frequency above the LO when high-side injection is
used. That is,
fimage = fLO + fIF
= fRF + 2fIF.
 Image response rejection is achieved by filtering before
the mixer.
 An AM receiver (IF = 455kHz) is tuned to receive a station
whose carrier frequency is fRF = 1MHz. The LO is fLO =
1.455 MHz and the interfering signal is at 1.910 MHz.
 Consequently the difference frequency is 1.910 MHz –
1.455 MHz = 455 kHz, exactly our IF center frequency!
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
RECEIVER GAIN and SENSITIVITY
 Suppose a 90% AM signal is received, the receiver must
amplify this until it is large enough to cause the diode to
conduct. Indeed to prevent negative peak distortion at the
detector, the minimum positive peak Vmin must cause
conduction.
 A conservative figure for Vmin when using a germanium
detector diode is about 0.2V, including junction potential
and I2R losses in the diode and detector circuitry.

Referring to Figure 5-4, Vc is the average value between A
and B; that is Vc = (A+B)/2. Notice also that B = Vmin, so we
solve for B in terms of A in m = (A–B)/(A+B) which gives
A = [(1+m)/(1-m)]B
(5-20)
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
RECEIVER GAIN and SENSITIVITY

Now substitute this into
(5-21)

For m = 0.90, Vc = (1/2)(1.9/(0.1+1))Vmin = 10Vmin.
Hence, we find that Vc = 10Vmin = 2Vpk.P = Vc2/2R
(where Vc is peak volts) and a typical equivalent detector
impedance is 1 kW.

The result is that 2mW of carrier power is typically
required at the detector.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
RECEIVER GAIN and SENSITIVITY
 The other piece of information needed to determine the
required receiver gain is the desired receiver sensitivity.
When we talk about the sensitivity of an electronic device,
we are talking about the weakest input signal required for
a specified output.
 An important parameter in determining the quality of the
received message is the signal-to-noise ratio (SNR). This
would lead to a definition of receiver sensitivity as “the
minimum input signal, when modulated at 90% AM,
required to produce a specified SNR at the audio output.”
 Receivers of digital data are specified in terms of bit error
rate (BER) of the output data. The BER is directly related
to the SNR at the receiver output.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
POWER LEVEL in dBm and dBW
 A widely used and very useful way to express power
levels is to put them in decibels (dB) relative to some
reference power level.
 Two of the most frequently used reference levels in
communications are dBm with a reference level of 1 mW
and dBW with a reference level of 1 W.
 The power level P is converted to dBm using
(5-22)
and to dBW by
(5-23)
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
POWER LEVEL in dBm and dBW
 A The point of giving the power expressed in dBm is that
circuit gains and losses are usually expressed in dB and
we can operate on the power levels using very simple
arithmetic once the conversion has been calculated.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
POWER LEVEL in dBm and dBW
Example 5-4:
 A receiver antenna has an output voltage of 10 mV (carrier
only) when connected to a 50W receiver.
1. Determine the power level in dBW and dBm.
2. The receiver has one RF amplifier with 10 dB of gain,
mixer with 6dB of conversion loss, followed by a multipole
filter with 1 dB of insertion loss.
If available IF amplifiers gave 20dB of gain each,
determine the number of IF amplifiers necessary to
provide at least 0 dBm (1 mW) to the detector.
3. Sketch a block diagram of a superheterodyne AM receiver
showing the power level, in dBm, at each block.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
POWER LEVEL in dBm and dBW
Solution:
1. P = (10 x 10-6 V)2 / 50W = 2 x 10-12 W = 2 pW.
P(dBW) = 10log(2x10-12 W/1W) = -117 dBW.
P(dBm) = 10log(2x10-12 W/10-3 W) = -87 dBm.
 2. With a 10-dB gain, the RF amplifier output will be –87dBm
+ 10dB = 77dBm. Following 6dB of loss due to the mixer
and 1dB of filter insertion loss in the passband, the IF input
power will be
P(dBm) = -77dBm + (-7dB) = -84 dBm.
 The IF system must provide an overall gain of Po/Pi or
Po(dBm) - Pi(dBm) = 0 dBm - (-84 dBm) = 84 dB.
At 20 dB/stage, we need five IF amplifiers, one of which
requires only 4 dB of gain.
 3. The completed block diagram is shown in Figure 5-21.

Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
POWER LEVEL in dBm and dBW
Figure 5-21. Signal-level distribution in superhet receiver with 10mV (-87-dBm) input. The gain of IF5 has been adjusted to provide
1-mW at the detector.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AGC and DYNAMIC RANGE
Fig. 5-22. Effect of clipping due to excessive signal in IF.
Fig. 5-23. Information loss due to severe clipping of
the AM signal.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AGC and DYNAMIC RANGE
Figure 5-24. Effect of clipping due to excessive signal in IF.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
AGC and DYNAMIC RANGE
 The usual scheme for AGC is as shown in Figure 5-24,
where the gain of the input stage is controlled by
controlling the bias of the amplifying device.
 Notice that the output of the detector Vo consists of the
audio signal (information desired) and a dc voltage that
will be proportional to the IF output carrier signal
strength.
 The audio variations are smoothed out with the low-pass
filter RC, providing a voltage to control the bias and
consequently the gain of Q1.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Delayed AGC
 Despite the use of AGC with numerous IF stages,
sufficiently strong input signals to the receiver can
overload the mixer (and even the RF amplifier) and
distort the AM signal. The cure for this is to provide
an AGC voltage to the RF amplifier.
 The receiver noise figure (and added noise) can be
kept low if the RF amplifier gain is kept high.
This means that if we AGC the RF amplifier on weak
signals and its gain is reduced, then the signal-to-noise
ratio (S/N) of the information will be spoiled.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Delayed AGC
 The way to a solution is to look at the problem this way:
What we want is to have maximum RF amplifier gain
for weak signals but, for very strong signals where S/N is
no problem anyway, reduce the gain, thereby avoiding
overload distortion in the mixer and RF amp.
 The solution, then, is to put a voltage-level delay circuit
after the regular AGC line to keep the RF amplifier gain
high until a sufficiently strong input signal causes the
AGC to exceed a set threshold voltage.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Delayed AGC
 The voltage-level delay circuit is illustrated in Figure 5-24
by D2 and R1 with R2 connected to -Vcc.
 R1 and R2 place a negative bias of a few tenths of a volt on
the cathode of the silicon diode D2.
 D2 is reverse biased until the AGC voltage rises to about –
0.1 V (Q1 cuts off at about -0.2V).
 If the AGC voltage rises above -0.1V, D2 becomes forward
biased and conducts to reduce the gain of the RF amplifier.
 Thus, the delayed-AGC threshold can be set by R2.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Delayed AGC
 For input signals strong enough to cause the AGC to
exceed the threshold value, delayed-AGC controls the
RF amplifier gain, thereby eliminating high-signal-level
distortion while maintaining good S/N for weak signals.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Noise Figure Equation Derivation
 Referring to Figure 5-26, let the power gain of the amplifier
be Ga and the noise power ratio be NRa, and let Na be the
internal amplifier noise power that would be measured at the
output if absolutely no noise could present at the input.
 Of course, there is always thermal noise
Nth = kTB
(5-26)
present at the input to the amplifier by virtue of the fact that
conductor at temperature T have Nth watts of power as
measured in a bandwidth B.
Figure 5-26. Amplifier with Noise.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Noise Figure Equation Derivation

The total noise power at the amplifier output will be
No = NthGa + Na
(5-27)
where the input thermal noise power is amplified by
power Ga.
 Hence, NRa = (No/Ga)/Nth
(5-30)
That is, the ratio of the total output noise power referred
to the amplifier input that is, divided by amplifier power
gain to input thermal noise power yields the amplifier
noise ratio (noise figure).

At the output, Equ. (5-30) may be written as
NRa = (NthGa+Na)/NthGa
(5-31)
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Noise Figure Equation Derivation
 That is, total noise power out, divided by the output
thermal noise power level is the ratio by which the
amplifier degrades input signal-to-noise ratio.
 Equation (5-31) is a more functional definition of noise
ratio (noise figure) than Equ. (5-30) is.
 Notice that the right-hand side of Equ. (5-31) may be
written as the sum of equations
NRa = (NthGa + Na)/NthGa
= 1 + (Na/NthGa).
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
(5-32)
Noise Figure Equation Derivation
 That is,
NRa - 1 = Na/NthGa ,
(5-33)
from which we see that the noise power added to any input
noise can be written in terms of amplifier noise ratio as
Na = NthGa(NRa - 1)
(5-34)
 If multiplied through, Equ. (5-34) expresses the original
notion that
Na = NthGaNRa - NthGa
(5-35)
That is, input noise times gain, subtracted from total
output noise, equals noise added by the amplifier.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Noise Figure Equation Derivation
 The noise figure of the two-amplifier cascade of Fig. 5-27
is derived as follows: The total noise power at the output
of amplifier 1, N01, is, from Equations (5-27) and (5-35),
N01= NthG1 + [NthG1(NR1-1)]
(5-36)
Figure 5-27. Two noisy amplifiers cascaded.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Noise Figure Equation Derivation
 Noise power N01 is now amplified by G2 and added to the
amount of noise, Na2= NthG2 (NR2-1), added by amplifier 2.
That is,
N02 = NthG2(NR2-1) +G2{NthG1 +[NthG1(NR1-1)]} (5-37a)
= NthG1G2 +NthG1G2(NR1-1) +NthG2(NR2-1) (5-37b)

By extending Equ. (5-30) to a system of two amplifiers in
cascade, it follows that
NRsys = [N02/(G1G2)]/Nth = N02/(NthG1G2)
= 1+ (NR1 - 1) + (NR2 - 1)/G1
(5-38a)
(5-38b)
where Equ. (5-37b) has been substituted in Equ. (5-38a) to
get Equ. (5-38b).
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Noise Figure Equation Derivation
 Hence, the final result for two amplifiers in cascade is
NRsys = NR1 + (NR2 - 1)/G1
(5-39)
 By extension of this procedure to n amplifiers, the family
system noise figure,
(5-40)
is obtained.
 The equivalence between noise temperatures and noise
figure is given by
Teq = To(NR-1)
(5-41)
where To = 290°K.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Bandwidth Improvement
 In the receiver, there are practical and system-requirement
limitations to the improvement achievable by narrowing
the bandwidth. The practical problem is in the difficulty
of building stable, narrowband filters.
 For example, if we are receiving a signal in the standard
AM band at 1.5 MHz, building a front-end filter of 10-kHz
bandwidth requires an equivalent circuit quality factor of
Q = 1.5MHz/10kHz = 150. This is barely achievable.
 However, it is easily accomplished at the IF amplifier:
for a standard fIF = 455 kHz, Q = 455kHz/10kHz = 45.5.
 The system-requirement constraint is that the circuit bandwidth must exceed the information one or the information
power and spectrum will be reduced (Fig. 5-28).
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Bandwidth Improvement
Figure 5-28. A filter bandwidth narrower than the information
bandwidth, with resulting reduced information power.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Bandwidth Improvement
 The simplest method for getting the correct predetection
S/N is to use the IF bandwidth as the noise bandwidth in
the calculation of Nth = kTB.
 The other method is to determine the bandwidth exchange,
or noise bandwidth improvement factor, BI.
 The bandwidth improvement factor is the ratio by which
noise power is reduced by a reduction in bandwidth.
 As an example, suppose a receiver has an RF bandwidth
of 5MHz and an IF bandwidth of 200 kHz. The noise
bandwidth improvement is
BI(dB) = 10log(BRF/BIF)
(5-42)
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Bandwidth Improvement
 For the simple case where all single-tuned amplifiers have
equal bandwidths, the overall 3-dB bandwidth is computed
from
(5-43)
where BW1 = bandwidth of each stage
n = number of stages with bandwidth of BW1.
 For three amplifiers, each of which has BW1 = 10 kHz,
the overall 3-dB bandwidth is
BW =
= 5.09 kHz.
This formula is for single-tuned, synchronously tuned
amplifier systems only.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Bandwidth Improvement
Fig. 5-29. Superheterodyne receiver block diagram
referred with Example 5-5.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Bandwidth Improvement
Example 5-5:
Given that
NF1 = 2dB, NR = 1.6; NF2,3= 6 dB, NR = 4.0;
NF4,5 = 18 dB, NR = 63.1; Ap1 = 8 dB, G1 = 6.3;
Ap2 = 12 dB, G2 = 15.8; Ap3 = -6 dB, G3 = 0.25;
Ap4,5 = 20 dB, G4,5 = 100:
1. Calculate the system NF(dB).
2. If the RF bandwidth is 5 MHz and the IF bandwidth is
200 kHz, determine the predetection S/N (dB) for a
receiver input signal of -80 dBm.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Bandwidth Improvement
Solution:
1. From Equ. (5-40),
NF (dB) = 10 log 4.6 = 6.7 dB.
2. The received signal is -80 dBm. The thermal noise power
at the input is, from Equ. (5-24),
Nth = -174 dBm + 10 log5×106 = -107 dBm.
Consequently, SNR(dB) = -80 dBm - (-107 dBm) = + 27 dB.
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.
Bandwidth Improvement
 The input SNR is reduced because of system noise (NF =
6.7 dB) but is increased by the effect of reduced noise in
a narrow IF bandwidth (BI).
 The bandwidth narrowed noise improvement will be, from
Equ. (5-42),
BI(dB) = 10.log(5000/200) = 14 dB.
The final result is
Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.
National Cheng Kung University, Taiwan.