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.
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
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.
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:
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.
The minimum input signal that gives output signal to noise ratio of 10 dB in a system that has an input impedance equal to 50 Ω, a noise figure of 4 dB and bandwidth of 100 KHz is:
[10.log (kT_{o}) = -174 dBm]
Concept:
\(\left( {NF} \right)Noise\;figure = \frac{{\left( {S/N} \right)input\;}}{{\left( {S/N} \right)output}}\)
Input Noise power = KTB = -228.6
+10log T(k) + 10 log B(Hz)
B → Bandwidth
T → temperature
(Si) Input power of the system is given as:
\({S_i} = \frac{{{V^2}}}{{{Z_i}}}\)
Calculation:
Given \({\left( {\frac{S}{N}} \right)_{output}} = 10dB,\) Z_{i} = 50 Ω, NF = 4dB, B = 100 kHz
\({S_i}\left( {Input\;power} \right) = \frac{{{V^2}}}{{{Z_i}}} = \frac{{{{\left( {\frac{{{V_i}}}{2}} \right)}^2}}}{{{Z_i}}}\)
= \(\frac{{V_1^2}}{{4{Z_i}}}\)
= \(\frac{{V_1^2}}{{200}}\)
Input Noise power = KT_{0}B
+10log T(k) + 10 log B(Hz)
= -204(dB Watt) + 10 log (B in Hz)
\(10\log \left( {Noise\;figure} \right) = 10\log {\left( {\frac{S}{N}} \right)_{input}} - 10\log {\left( {\frac{S}{N}} \right)_{\frac{o}{P}}}\)
= \(10\log \frac{{V_1^2}}{{200}} - 10\log \left( {k{T_0} \times {{10}^5}} \right) - 10\)
\(4 = 10\log \frac{{V_1^2}}{{200}} + 204 - 50 - 10\)
After solving V_{i} = 1.414 mV
The following items consist of two statements, one labeled as “Assertion (A)” and the other labeled as the “Reason (R)”. You are to examine the two statements carefully and decide if the Assertion (A) and the Reason (R) are individually true and if so whether the reason is a correct explanation of the assertion. Select your answer to these items using the codes given below and mark your answer accordingly.
Assertion (A): At the output of a two-port network, the total noise is due to the contribution of several, more than two noise source
Reason (R): The message single at the input of the port is accompanied by noise sources, and while passing through the two-port network, additional noise is added to the signal.Explanation:
For a two-port network shown in the figure with the gain(loss) G(L) and NF is the noise figure.
T_{e} = (NF - 1)T_{0}
T_{e}: equivalent noise temperature.
T_{0}: Room temperature equal to 290 K
The white noise power is:
PN_{0} = kT_{0}BW
k: Boltzmann constant
Assuming that the noise temperature of the input port is T_{i}, the input noise power is:
PN_{i} = kT_{i}BW
The noise power of a two-port linear network caused by equivalent noise temperature is:
PN_{a} = kT_{e}BW
The noise figure is expressed by noise power as:
\(NF = 1 + \frac{{P{N_a}}}{{P{N_0}}}\)
The output noise power of the two-port linear network is:
PN_{0} = GPN_{i} + GPN_{a}
NOTE:
T_{i} in the formula is not necessarily equal to T_{0}
Output noise power is not a combination of several sources. So, Statement I is false.
Analysis of the noise characteristics of the passive two-port linear network
L: insertion loss of the passive two-port network
BW: bandwidth
T_{i}: Noise temperature at the input of the passive two-port network
PN_{i}: Noise power at the input of the passive two-port network
PN_{0}: Noise power at the output of the passive two-port network
Let T_{i} = T_{0} + T_{ei} then
PN_{i} = kT_{i}BW = k(T_{0} + T_{ei})BW
The noise power generated by PN_{i} at the network output is:
PN_{0i} = kT_{i}BW/L
The equivalent noise temperature of the passive two-port lossy network converted to the input of the two-port network is:
T_{eL} = (L - 1)T_{0}
The noise power generated by the passive two-port lossy network at the output is:
\(P{N_{0L}} = \left( {1 - \frac{1}{L}} \right)k{T_0}BW\)
The total noise power generated by the passive two-port lossy network at the output is:
PN_{0} = PN_{oi} + PN_{oL}
Conclusion:
At the output of a two-port network, the total noise is due to the contribution of several, more than
two noise sources. So, statement II is true.
(A) false, but (R) is true.
Concept:
\(\frac{{{E_b}}}{{{N_0}}} = \frac{{Energy\;received\;per\;information\;bit}}{{noise\;power\;spectral\;density}}\)
Here,
\({E_b} = {P_r} \times {T_b} = \frac{{{P_r}}}{{{R_b}}}\)
Where,
P_{r} = received power (Watt)
R_{b} = data rate or bit rate (bps)
\(\frac{{{E_b}}}{{{N_0}}}\) is a dimensionless parameter and is often expressed in dB.
\(\therefore {\left( {\frac{{{E_b}}}{{{N_0}}}} \right)_{dB}} = 10{\log _{10}}\left( {\frac{{{E_b}}}{{{N_0}}}} \right)\)
Calculation:
Given that,
P_{r} = -151 dBW ; here dBW means decibel-watt
R_{b} = 2400 bPS
T = effective noise temperature of the receiver system = 1500 K
To convert P_{r} in watt from dBW, we use the relation:
P_{r} (dBW) = 10 log_{10} (P_{r}(W)]
-151 = 10 log_{10} [P_{r}(W)]
P_{r}(W) = 10^{-15.1} = 7.94 × 10^{-16} watt.
When noise Temperature is considered in the receiver system than N_{o} will be:
N_{0} = K.T (watt/Hz)
Where,
K = Boltzmann’s constant (J/K)
= 1.38 × 10^{-23} J/K
\(\therefore \frac{{{E_b}}}{{{N_0}}} = \frac{{{P_r}}}{{{R_b} \times \left( {KT} \right)}} = \frac{{7.94 \times {{10}^{ - 16}}}}{{2400 \times 1.38 \times {{10}^{ - 23}} \times 1500}}\)
\(= \frac{{1.598 \times {{10}^{ - 22}}}}{{{{10}^{ - 23}}}}\)
\(= \frac{{1.598}}{{{{10}^{ - 1}}}}\)
= 15.98
\(\therefore {\left( {\frac{{{E_b}}}{{{N_0}}}} \right)_{dB}} = 10{\log _{10}}\left( {\frac{{{E_b}}}{{{N_0}}}} \right) = 10{\log _{10}}\left( {15.98} \right)\)
= 12.03 dB
Important Points:
Power ratio in dB is given by:
\(dB = 10{\log _{10}}\left( {\frac{{{P_1}}}{{{P_2}}}} \right)\)
Ratio’s like the voltage, current, sound, and pressure levels are calculated as the ratio of squares, i.e.
\(\therefore dB = 20{\log _{10}}\left( {\frac{{{V_1}}}{{{V_2}}}} \right)\)
dBm means decibel-milliwatt, and is defined as:
P(dBm) = 10 log_{10} )P(mw)]
dBW means decibel-watt
P(dBW) = 10 log_{10} [P(watt)]
P(dBW) = P(dBm) - 30
The noise level of the spectrum analyzer can be related to Noise Figure (NF) and IF Bandwidth by:
-114 dBm + 20 log (BW/1MHz) + NF
NF (Noise Figure) is given by:
\(NF=\frac{(SNR)_{in}}{(SNR)_{op}}\)
The Noise Figure is usually expressed in dB as:
\(NF=10~log\frac{(SNR)_{in}}{(SNR)_{op}}~dB\)
Concept:
The dynamic range is the ratio between the highest and the lowest-power signal simultaneously present at the input, where both of these signals appear in the display.
TOI (Third-Order Intercept) point is a parameter used to evaluate the linearity of components that are utilized in applications where nonlinear effects can cause distortion—for example, in digitally modulated signals.
The two equations given below are used to calculate the intermodulation free dynamic range:
For 2^{nd} Order Intercept, the dynamic range in dB is given by:
\(D{R_2} = \frac{1}{2}\left( {I{P_2} - NL} \right)\)
IP_{2} = Second-Order intercept point
NL = Noise Level
For 3^{rd} Order Intercept, the dynamic range in dB is given by:
\(D{R_3} = \frac{2}{3}\left( {I{P_3} - NL} \right)\)
Calculation:
Given IP_{3} = + 26dBm
NL = - 85 dBm
The Dynamic Range for the given third-order intercept will be:
\(D{R_3} = \frac{2}{3}\left( {26 - \left( { - 85} \right)} \right)dB\)
\(D{R_3} = \frac{2}{3}\left( {111} \right) = 74\;dB\)
The noise figure of an amplifier is 3 dB. Its noise temperature will be about
Noise temperature is \({T_0} = {T_{eq}}\left( {F - 1} \right)\)
\({T_0} = 290\left( {{{10}^{.3}} - 1} \right) = 290\left( {1.995 - 1} \right)K = 288\;K\)
Noise level in dB = 10Log_{10} (Noise in watts)
10 dB = 10Log_{10} (Noise in watts)
Noise in watts for 10 dB = 10^{1} = 10 Watts …(1)
1dB = 10 Log_{10} (Noise in watts)
Noise in watts for 1 dB = 10^{0.1} = 1.26….(2)
P_{1}/P_{2} = 10/1.26 = 7.94 W
Best suitable option is option bAntenna’s noise temperature \({{\rm{T}}_{{\rm{AN}}}} = 50{\rm{K}}\)
Ambient temperature \({{\rm{T}}_{\rm{A}}} = 290{\rm{K}}\)
Preamplifier noise figure \({{\rm{F}}_{\rm{P}}} = 2{\rm{dB}}\)
Amplifier gain \({{\rm{G}}_{\rm{P}}} = 40{\rm{dB}}\)
Now, effective noise temperature of amplifier input \({{\rm{T}}_{\rm{e}}} = \left( {{\rm{F}} - 1} \right){{\rm{T}}_{\rm{A}}}\)
Noise figure of antenna \({\rm{F}} = {10^{\frac{2}{{10}}}} = {10^{\frac{1}{5}}} = 1.585\)
Thus, \({{\rm{T}}_{\rm{e}}} = 290\left( {1.585 - 1} \right) = 290 \times 0.585 = 169.7{\rm{\;K}}\)
Now, noise power at the preamplifier output, \({{\rm{P}}_{{\rm{ao}}}} = {\rm{GK}}{{\rm{T}}_{{\rm{IN}}}}{\rm{B}}\)
Gain , \({\rm{G}} = {10^{\frac{{40}}{{10}}}} = {10^4}\) and \({{\rm{T}}_{{\rm{IN}}}}\) is the input noise temperature at amplifier input
\(\begin{array}{l} {{\rm{T}}_{{\rm{IN}}}} = {{\rm{T}}_{{\rm{AN}}}} + {{\rm{T}}_{\rm{e}}} = 50{\rm{K}} + 169.6{\rm{K}}\\ \Rightarrow {{\rm{T}}_{{\rm{IN}}}} = 219.6{\rm{K}} \end{array}\)
Amplifier output noise, \({{\rm{P}}_{{\rm{ao}}}} = {\rm{GK}}{{\rm{T}}_{{\rm{IN}}}}{\rm{B}}\)
\(\begin{array}{l} {{\rm{P}}_{{\rm{ao}}}} = {10^4} \times \left( {1.38 \times {{10}^{ - 23}}} \right) \times \left( {219.6} \right) \times \left( {12 \times {{10}^6}} \right)\\ {{\rm{P}}_{{\rm{ao}}}} = 3.63 \times {10^{ - 10}}{\rm{W}} \end{array}\)
Thus, option a is correct.