prof. jen-fa huang, fiber-optic communications lab. national cheng kung university, taiwan. chapter...

Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.Prof. Jen-Fa Huang, Fiber-Optic Communications Lab.National Cheng Kung University, Taiwan.National Cheng Kung University, Taiwan.

Chapter V. Amplitude Modulations

An unmodulated sinusoidal carrier signal can be described as ec(t) = Ec cos2fct (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 cos2fmt, (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)) cos2fct (5-3)



Figure 5-3. Amplitude modulation.



Figure 5-4. (a) Amplitude-modulated signal. (b) Information signal to be transmitted by AM.


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 cos2fmt) cos2fct.

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)



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 = (m1

2 +m22 +m3

2 +… +mn2)1/2 (5-5)



Figure 5-6. AM represented as the vector sum of sidebands and carrier.



Figure 5-7. (a) ma= 1.0(100% AM).

(b) The result of over-modulation that corresponds to the spectrum in (c).



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.



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.



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.



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.



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 f

m is the variable.


AM SPECTRUM AND BANDWIDTH

Let us analyze the mathematical expression for the AM signal e(t) = (Ec + Em cos2fmt) cos2fct

= Ec cos2fct + Em cos2fmt cos2fct

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 cos2fct carrier + (Em/2) cos2(fcfm)t lower sideband, LSB

+ (Em/2) cos2(fcfm)t upper sideband, USB



Figure 5-5. Frequency spectrum of the AM signal shown in Figure 5-4.



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).


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 = Ec

2/2R, and for each of the two

sideband components,



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 (5-8)



Relative AM Signal Energy versus Modulation Index

%m Energy (Pt/Pc) 0 1 10 1.005 20 1.02 30 1.045 40 1.08 50 1.125 60 1.18 70 1.245 80 1.32 90 1.405 100 1.5



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.


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).



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).



The mathematically formal method of determining the frequency spectrum of a time-varying signal is to employ the Fourier transform.

vAM(t) = Ec.cos2fct + m(t).cos2fct (5-9)

VAM(f) = (Ec/2)[(f - fc) + (f + fc)] + (1/2)[M(f - fc) + M(f + fc)] (5-10)



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.


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.



Fig. 2-5. The ASK signal in the time domain.



If the carrier burst is defined as

(5-11) then the Fourier transform may be expressed as (5-12)

where d = /T (5-13) is the duty cycle of the bursts, and T is the pulse repetition period (inverse of the pulse repetition frequency, PRF).



Figure 5-10. 25% duty cycle on-off key (OOK) signal and one-sided frequency spectrum.



Figure 2-6. Non-coherent detection of ASK.


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.



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.



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)]/(2Eb/No)1/2 (2-10)

From this, it is seen that the majority of errors are spaces converted to marks.


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 .

Fig. 2-7. Coherent detection of ASK using synchronous detection.



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 2, with maximum signal energy at a time of A2/2, where A is the signal amplitude and is the pulse duration.

To complete the system, the decision circuitry is timed to function at time for optimum performance.



Fig. 2-8. Matched filter detection of ASK signals.



The probability of error for coherent ASK signaling is:

(2-11)


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.


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.


AM DEMODULATION

Figure 5-14. AM demodulation.


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.



Figure 5-15. Diagonal clipping.



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)



Figure 5-16. A higher-index AM requires a shorter RC time constant.



Figure 5-17. Complete AM detector and volume (loudness) control.



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.


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.


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


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.



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.


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!



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)



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 = Vc

2/2R (where Vc is peak volts) and a typical equivalent detector impedance is 1 k.

The result is that 2mW of carrier power is typically required at the detector.



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.


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)



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.


Example 5-4: A receiver antenna has an output voltage of 10 mV (carrie

r only) when connected to a 50 receiver. 1. Determine the power level in dBW and dBm. 2. The receiver has one RF amplifier with 10 dB of gain, mixe

r 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.




Solution: 1. P = (10 x 10-6 V)2 / 50 = 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.



Figure 5-21. Signal-level distribution in superhet receiver with 10-V (-87-dBm) input. The gain of IF5 has been adjusted to provide 1-mW at the detector.


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.



Figure 5-24. Effect of clipping due to excessive signal in IF.



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.


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.


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.

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.

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.

Delayed AGC


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.

Noise Figure Equation Derivation

Figure 5-26. Amplifier with Noise.


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)



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). (5-32)



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.



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.



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) (5-38a) = 1+ (NR1 - 1) + (NR2 - 1)/G1 (5-38b)

where Equ. (5-37b) has been substituted in Equ. (5-38a) to get Equ. (5-38b).



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.



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).

Bandwidth Improvement


Figure 5-28. A filter bandwidth narrower than the information bandwidth, with resulting reduced information power.



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)



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.



Fig. 5-29. Superheterodyne receiver block diagram referred with Example 5-5.



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.



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.



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. chapter...

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