White noise is that signal whose frequency spectrum
White noise is that signal whose frequency spectrum is uniform i.e. it has flat spectral density.
The power spectral density (PSD) of white noise is uniform throughout the frequency spectrum as shown:
Concept:
The noise figure is given as:
\(NF= 1 + \frac{{{R_{eq}}}}{{{R_s}}}\)
R_{eq} = Equivalent input resistance of antenna
R_{s} = Noise resistance of the system
Calculation:
Given R_{eq} = R_{s} = 50 Ω
Noise figure will be:
\(= 1 + \frac{{50}}{{50}}\)
= 1 + 1
Noise figure = 2.
Thermal noise:
P_{n} = KTB
Where,
K = Boltz man's constant
T = temperature
B = Bandwidth over which noise is measured
Shot noise:
So, based on the above points all options are true except option 2.
Hence the correct answer will be option 2.
Concept:
Noise figure (NF) and noise factor (F) are measures of degradation of the signal-to-noise ratio (SNR), caused by components in a signal chain.
\(Noise\;Figure = \frac{{{{\left( {SNR} \right)}_{i/p}}}}{{{{\left( {SNR} \right)}_{o/p}}}}\)
(N.F)dB = [(SNR)i/p]dB – [(SNR)o/p]dB
Calculation:
Given:
NF = 2
(SNR)i/p = 37 dB,
N.F(dB) = 10log_{10}(NF)
N.F(dB) = 3 dB
(N.F)dB = [(SNR)i/p]dB – [(SNR)o/p]dB
[(SNR)o/p]dB = 37 dB – 3 dB = 34 dB
Transmission errors detection:
Generally, transmission errors are occurred because of thermal and impulse noise, cross talk, voice amplitude signal compression and jitter, etc.
Transmission errors generally three types namely single bit, multiple bits, and burst type errors.
Error detection:
Cyclic redundancy check:
In shot noise, the RMS value of the shot noise current
\({I_{n,\;rms}} = \sqrt {2{I_{dc}}qB} \)
Where,
I_{dc} = dc current
q = charge of an electron
B = bandwidth over which current is measured.
This relation is given by Schottky.
Shot noise current is directly proportional to the square root of the bandwidth.
So option 3 is correct.
In thermal noise, the RMS value of thermal noise voltage is given by
\({V_{n,\;rms}} = \sqrt {4KTRB}\)
Where,
K = Boltzmann’s constant
T = ambient temperature of resister
R = resistance of the conductor
B = bandwidth over which voltage is measured.
Here,
V_{n,rms} α T, R, and B
Voltage Controlled Oscillator is used for
Depending on the input voltage VCO produces the different frequencies and it is used in the demodulation of Frequency Modulation.
Concept:
Signal to noise ratio:
Calculation:
The initial signal to noise ratio is:
\(\frac{S}{N}\) ---(1)
Now if the signal power increase to 3S and the noise power reduces by half then signal to noise ratio will become:
\(\frac{3S}{\frac{N}{2}}=\frac{6S}{N}\) ---(2)
Dividing equation (1) by (2) we get:
\(\frac{Old \ SNR}{New \ SNR}=\frac{\frac{S}{N}}{\frac{6S}{N}}=\frac{1}{6}\)
Hence option (4)is the correct answer.
Important Points
For practical purposes, the signal to noise ratio for acceptable quality transmission is:
Analog signal: 40 – 60 dB
Digital signal: 10 – 12 dB
A wave has 3 parameters Amplitude, Phase and Frequency. Thus there are 3 types of modulation techniques
Amplitude Modulation:
The amplitude of the carrier is varied according to the amplitude of the message signal.
Frequency Modulation:
The frequency of the carrier is varied according to the amplitude of the message signal.
Phase Modulation:
The Phase of the carrier is varied according to the amplitude of the message signal.
Explanation:-
The effect of noise in FM signal is determined by the extent to which it changes the frequency of the modulated signal.
Noise in FM
PSD of the noise at the detector output is directly proportional to the square of the frequency.
So we can say the effect of noise is more at higher frequencies in FM.
Discussion:-
Figure of Merit (FOM): It is the ratio of output SNR to the input SNR. FOM depends only on the demodulator.
For AM:
In noise performance, the synchronous detector is better than an envelope detector.
In AM noise has nothing to do with frequency, it depends on the amplitude of modulating signal.
\({\rm{FO}}{{\rm{M}}_{{\rm{AM}}}} = \frac{1}{3}\) (for envelope detector).
For PM:
\({\rm{FO}}{{\rm{M}}_{{\rm{PM}}}} = \frac{{{{\rm{\beta }}^2}}}{2}\)
Noise in PM increases exponentially for the entire audio range is a wrong statement.
For FM:
\({\rm{FO}}{{\rm{M}}_{{\rm{FM}}}} = \frac{{{\rm{3}}{{\rm{\beta }}^2}}}{2}\)
The figure of merit of FM is far better than the other two. So we can say the noise performance of FM is better than AM and PM.
Important Points
1. To improve noise performance in FM we use the concept of pre-emphasis and de-emphasis.
2. WBFM is preferred over NBFM because the earlier has a better figure of merit.
Two resistors R_{1} and R_{2} (in ohms) at temperatures T_{1}K and T_{2}K respectively, are connected in series.
Their equivalent noise temperature is
Concept:
Thermal noise voltage:
Due to thermal agitation (rise in temp), atoms in the electrical component will gain energy, moves in random motion, collides with each other, and generate heat this heat produced is called thermal noise.
Thermal noise power (N) = KTB
Where,
K = Boltzman constant
T = temperature in °K
B = Bandwidth
If we consider a Noisy Resistor as an ideal Resistor R connected in series with a noise voltage source and connected to match the load.
\(P = \frac{{V_n^2}}{{4R}}\)
Noise Power = KTB
\(KTB = \frac{{V_n^2}}{{4R}}\)
\({V_n} = \sqrt {4\;KTRB} \)
Calculation:
Let equivalent voltage of series combination is \(\overline {{V_{{n_1}}}} \)
\(\overline {V_n^2} = \overline {V_{{n_1}}^2} + \overline {V_{{n_2}}^2} \)
Since noise voltage is AC that’s why we are adding its square value.
The equivalent temperature of the series combination is Te
\(\overline {V_n^2} = 4\;KTe\;BR \)
\(= 4\;kTe\;B\;\left( {{R_1} + {R_2}} \right)\)
\(\overline {V_{{n_1}}^2} = 4\;K{T_1}B{R_1} \to \) Voltage across R_{1}
\(\overline {{V_{{n_2}}}} = 4\;K{T_2}B{R_2} \to \) noise voltage Across R_{2}
4 KTe B (R_{1} + R_{2}) = 4KT_{1} BR_{1} + 4KT_{2} BR_{2}
Te(R_{1} + R_{2}) = T_{1} R_{1} + T_{2} R_{2}
\(Te = \frac{{{R_1}{T_1} + {R_2}{T_2}}}{{{R_1} + {R_2}}}\)
Explanation:
In any communication system, during the transmission of the signal, or while receiving the signal, some unwanted signal gets introduced into the communication, making it unpleasant for the receiver, questioning the quality of the communication. Such a disturbance is called Noise.
What is Noise?
Noise is an unwanted signal which interferes with the original message signal and corrupts the parameters of the message signal. This alteration in the communication process leads to the message getting altered. It is most likely to be entered at the channel or the receiver.
Hence, it is understood that noise is some signal which has no pattern and no constant frequency or amplitude. It is quite random and unpredictable. Measures are usually taken to reduce it, though it can’t be completely eliminated.
The most common examples of noise are −
Hiss sound in radio receivers
Buzz sound amidst of telephone conversations
Flicker in television receivers, etc.
Which of the following statements is correct?
If the channel bandwidth doubles, the S/N ratio becomesThe noise in the system is directly proportional to the bandwidth, i.e.
Noise ∝ Bandwidth
Signal to noise ratio is defined as:
\(SNR=\frac{S}{N}=\frac{Signal~Power}{Noise~Power}\)
Observations:
1) As the bandwidth increases, the noise in the system will increase.
2) As the noise in the system increases, the S/N ratio will decrease.
∴ If Bandwidth is doubled then, S/N ratio will be halved.
Industrial noise:
So option 3 is correct.
Noise in communication system generally divided into two categories-
i) External noise like atmospheric noise, industrial noise solar noise, and cosmic noise.
ii) Internal noise like shot noise, thermal noise, low and high-frequency noise.
Thermal noise:
P_{n} = KTB
Where,
K = Boltz man's constant
T = temperature
B = Bandwidth over which noise is measured
Shot noise:
Explanation:
Noise factor
It is also known as 'Noise Figure'.
It is defined as the ratio of the signal to noise ratio at the input to the signal to the noise ratio at the output.
\(Noise \; factor = \frac{SNR_i}{SNR_o}\)
No units since both are of the same quantity.
\((NF)_{dB} = 10log(SNR_i)-10log(SNR_o)\)
Total Harmonic distortion
It is defined as the ratio of the sum of the powers of all harmonics to the fundamental frequency.
Harmonic distortions
It is defined as the ratio of harmonics to a fundamental when a theoretical sine wave is constructed.
Conclusion:
Option 4 is correct.
Gain and NF of a single-stage amplifier are 10 dB and 3 dB respectively. When two such amplifiers cascaded then gain and NF of the cascaded amplifier will be?
Concept:
The overall noise figure is given by,
\(\rm{F=F_1+\frac{F_2-1}{G_1}+\frac{F_3-1}{G_1G_2}\ldots}\) ---(1)
Where F_{1}, F_{2}, F_{3}...... are noise figures and G_{1}, G_{2}, G_{3}..... are system gains respectively.
Overall Gain of cascaded amplifiers is given by:
G(db) = G_{1}(db) + G_{2}(db) + ..... ---(2)
Calculation:
Given:
G_{1} = G_{2} = 10 db
G1 = G2 = 10log(x) = 10 db
log(x) = 1
x = 10^{1} = 10
Therefore, G1 = G2 = 10
Similarly F_{1} = F_{2} = 3 db = 2
From equation (1) noise figure (F) can be calculated:
\(F=2 \ + \ \frac{(2-1)}{10}\)
F = 2.1
F(db) = 10log(2.1) = 3.2 db
From equation (2) gain can be calculated:
G(db) = 10 db + 10 db = 20 db
Hence option (1) is the correct answer.
Concept:
Junction temperature is calculated by using the thermal resistance as follows.
ΔT [°C] = R_{th}(j-a) [°C/W] × P_{loss} [W]
ΔT: Junction temperature rise
ΔT = T_{j} - T_{a}
R_{th(j-a)} : Thermal resistance
P_{loss}: Power dissipated is a semiconductor device
T_{j }: Junction temperature
T_{a} : Ambient temperature
Given:
T_{a} = 25°C
R_{th(j-a)} = ?
P_{loss} = 100 mW
Calculation:
T_{j }- T_{a} = R_{th(j-a)} × P_{loss}
Rth(j-a) = (Tj - Ta) / Ploss
= [(125 - 25) / 100]
= 1000° C/W
Concept:
Noise Figure: It indicates noise generated within a device. The noise figure relates the noise temperature to a reference temperature.
\({F_N} = \frac{{{{\left( {\frac{S}{N}} \right)}_{input}}}}{{{{\left( {\frac{S}{N}} \right)}_{output}}}}\)
In Cascaded Amplifiers,
If there are three amplifiers with gain G_{1}, G_{2}, and G_{3},
The overall Noise Figure is given by:
\(F=F_1 \ + \ \frac{F_2-1}{G_1} \ + \ \frac{F_3 - 1}{G_1G_2}\)
Calculation:
Given:
For RF Amplifier A:
F_{A} = 3db = 10^{0.3} = 2
_{GA} = 5db = 10^{0.5} = 3.162
For RF Amplifier B:
F_{B} = 3.5db = 10^{0.35 }= 2.24
G_{B} = 15db = 10^{1.5} = 31.6
When RF amplifier A is used as 1st stage:
\(F=F_A \ + \ \frac{F_B-1}{G_A}\)
\(F=2 \ + \ \frac{1.24}{3.16}=2.04\)
When RF amplifier B is used as 1st stage:
\(F=F_B \ + \ \frac{F_A-1}{G_B}\)
\(F=2.24 \ + \ \frac{2-1}{31.6}=2.24\)
We can conclude that when RF amplifier A is at the first stage and amplifier B is at the second stage,
then minimum Noise Figure occurs.
For identical total transmitted power, the ratio of S / N for wideband FM to S / N of DSB-AM (with m_{a} = 1) is:
(Given f_{m} = 15 kHz & frequency deviation = ± 75 kHz)
Concept:
Signal-to-Noise Ratio (SNR):
It is the ratio of the signal power to noise power. The higher the value of SNR, the greater will be the quality of the received output.
\(\left ( SNR \right )= \frac{Average \:\: power \:\:of \:\:modulated \:\:signal}{Average\:\: power \:\:of \:\:noise \:\:in \:\:message \:\:bandwidth}\)
The figure of Merit:
The ratio of (SNR) at the output to the (SNR) at the input is known as the figure of merit.
It is denoted by F.
\(FOM = \frac{{S/N}_{output}}{{S/N}_{input}}\)
For wideband FM
\((F)_{FM}=\frac{3β ^2}{2}\) ----(1)
Where,
β = modulation index.
For DSB - AM,
\((F)_{DSB-AM}=\frac{μ ^2}{2+μ^2}\) ----(2)
Where,
μ = modulation index
Calculation:
Given:
f_{m} = 15 kHz
Δ_{f} = ± 75 kHz
\(\beta =\frac{\Delta_f}{f_m}\)
\(\beta =\frac{75}{15}=5\)
As transmitted power is the same for both wideband FM and DSB-AM
So (S/N)_{input} is the same for both.
From equation (1):
\((FOM)_{FM}=\frac{3 \ \times \ 25}{2}=\frac{75}{2}\) ---(3)
Putting μ = 1 in equation (2):
\((FOM)_{DSB-AM}=\frac{1}{3}\) ----(4)
Dividing equation (3) and (4) we get:
\(\frac{S/N_{FM}}{S/N_{DSB-AM}}=\frac{225}{2}=112.5\)