Dung Trinh, PhD HCMUT / 2019 Chapter 2 du o ng th an co ng .c om Noise, Nonlinear Distortion and System Parameters cu u Trinh Xuan Dung, PhD dung.trinh@hcmut.edu.vn Department of Telecommunications Faculty of Electrical and Electronics Engineering Ho Chi Minh city University of Technology Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 1 Dung Trinh, PhD HCMUT / 2019 an 1. General Considerations co ng .c om Contents ng th 2. Nonlinear Distortion in Microwave System cu u du o 3. Noise in RF/Microwave System Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 2 Dung Trinh, PhD HCMUT / 2019 1. Units in RF Design .c om ļ¶ Units in RF Design: The voltage gain and power gain are expressed in decibels (dB): ššš¢š” š“š£ ššµ = 20ššš ššš ššš¢š” š“š ššµ = 10ššš ššš • These two quantities are equal (in dB) only if the input and output voltages appear across equal impedances. For example, an amplifier having an input resistance of R0 (50ā¦) and driving a load resistance of R0 satisfies the following equations: 2 ššš¢š” ššš¢š” ššš¢š” š 0 š“š ššµ = 10ššš = 10ššš 2 = 20ššš = š“š£ ššµ ššš ššš ššš š 0 • The absolute signal levels are often expressed in dBm rather than watts or volts. The unit dBm refers to “dBs” above “1mW”. To express the signal power, Psig, in dBm, we write: šš šš šš šš ššµš = 10ššš 1 šš Dept. of Telecoms Engineering cu u du o ng th an co ng • CuuDuongThanCong.com https://fb.com/tailieudientucntt 3 Dung Trinh, PhD HCMUT / 2019 1. Units in RF Design du o ng th an co ng ļ¶ Solution: 0dBm is equivalent to 1mW, we have: 2 ššš = 1šš 8š šæ ššš = 632šš. .c om ļ¶ Example 1: An amplifier senses a sinusoidal signal and delivers a power of 0 dBm to a load resistance of 50ā¦. Determine the peak-to-peak voltage swing across the load. cu u ļ¶ Output voltage of the amplifier is of interest in the above example. This may occur if the circuit following the amplifier does not present a 50ā¦ input impedance, and hence the power gain and voltage gain are not equal in dB. ļ¶ Only for a sinusoid can we assume that the rms value is equal to the peak-to-peak value divided by 2 2. Fortunately, for a narrowband 0-dBm signal, it is still possible to approximate the (average) peak-to-peak swing as 632mV. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 4 Dung Trinh, PhD HCMUT / 2019 .c om 1. General Considerations ļ¶ Solution: an • -100dBm is 100 dB below 632 mVpp. co ng ļ¶ Example 2: A GSM receiver senses a narrowband (modulated) signal having a level of -100dBm. If the front-end amplifier provides a voltage gain of 15 dB, calculate the peak-to-peak voltage swing at the output of the amplifier. th • 100 dB for voltage quantities is equivalent to 105. ng • Thus, -100 dBm is equivalent to 6.32šššš . cu u du o • This input level is amplified by 15ššµ ≈ 5.62, resulting in an output swing of 35.5 šššš . Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 5 Dung Trinh, PhD HCMUT / 2019 1. Nonlinearity .c om ļ¶ A system is called “memoryless” or “static” if its output does not depend on the past values of its input. ng ļ¶ For a memoryless linear system, the input/output characteristic is given by: š¦ š” = š¼š„(š”) an co ļ¶ For a memoryless nonlinear system, the input/output characteristic can be approximated with a polynomial: š¦ š” = š¼0 + š¼1 š„ š” + š¼2 š„ 2 š” + +š¼3 š„ 3 š” + āÆ u du o ng th ļ¶ The following figure shows a common-source stage as an example of a memoryless nonlinear circuit (at low frequencies). If M1 operates in the saturation region and can be approximated as a square-law device, then 1 ššš¢š” = šš·š· − š¼š· š š· = šš·š· − šš ššš − ššš» 2 š š· 2 cu ļ¶ The system has “odd symmetry” if y(t) is an odd function of x(t). This occurs if š¼š = 0 for even j. Such system is called “balanced” as exemplified by the differential pair. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 6 Dung Trinh, PhD HCMUT / 2019 1. Nonlinearity .c om ļ¶ Example 3: The MOS transistors operating in saturation, characteristic can be expressed as co ng 1 4š¼šš 2 ššš¢š” = − šš ššš − ššš š š· 2 šš If the differential input is small, approximate the characteristic by a polynomial. 4š¼š·šš š ng ššš¢š” and applying the approximation 1 − š ≈ 1 − š/2, we have šš 2 ≈ − šš š¼š·šš ššš 1 − š š 8š¼š·šš šš š· du o 2 • Assuming ššš āŖ th an ļ¶ Solution: cu u ššš¢š” ≈ − šš š¼š·šš š š· ššš + 3 šš 2 8 š¼š·š 3 š š· ššš • The first term on the right-hand side represents linear operation, revealing the small signal voltage gain of the circuit (−šš š š· ). • Due to symmetry, even-order nonlinear terms are absent. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 7 Dung Trinh, PhD HCMUT / 2019 2. Nonlinear Distortion in Microwave Systems .c om ļ¶ A system is linear if its output can be expressed as a linear combination (superposition) of responses to individual inputs: ng š¦1 š” = š š„1 š” šš¦1 š” + šš¦2 š” = š šš„1 š” + šš„2 š” an th š£š = šš£š ng ļ¶ For a linear system: co š¦2 š” = š š„2 š” cu u du o ļ¶ For a nonlinear system: The input/output characteristic of a memory-less nonlinear system can be approximated with a polynomial: š£š = š0 + š1 š£š + š2 š£š2 + š3 š£š3 + āÆ Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 8 Dung Trinh, PhD HCMUT / 2019 2. Nonlinear Distortion in Microwave Systems .c om ļ¶ The nonlinearity of the system causes the following issues: Harmonic generation (multiples of a fundamental signal) ļ§ Gain Compression (gain reduction in an amplifier) ļ§ Inter-modulation Distortion (products of a two-tone input signal) ļ§ Cross-modulation (modulation transfer from one signal to another) ļ§ AM-PM conversion (amplitude variation causes phase shift) ļ§ Spectral regrowth (intermodulation with many closely spaced signals) cu u du o ng th an co ng ļ§ Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 9 Dung Trinh, PhD HCMUT / 2019 2A. Nonlinear – Harmonic Distortion .c om ļ¶ Input signal of the system: š£š š” = š0 ššš š0 š” 2 + š3 š0 ššš š0 š” an š£0 š” = š0 + š1 š0 ššš š0 š” + š2 š0 ššš š0 š” co ng ļ¶ Output signal of the system: 3 +āÆ u Fundamental cu DC du o ng th 1 3 1 1 = š0 + š2 š02 + š1 š0 + š3 š03 ššš š0 š” + š2 š02 ššš 2š0 š” + š3 š03 ššš 3š0 š” + āÆ 2 4 2 4 Second Harmonic Third Harmonic ļ¶ Even-order harmonics result from αj with even j ļ¶ nth harmonic grows in proportion to An. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 10 Dung Trinh, PhD HCMUT / 2019 .c om 2A. Nonlinear – Harmonic Distortion ļ¶ Problems: More signal loss and distortion. ļ§ Interference to other systems. ļ§ Sometimes can be used to create frequency multipliers. cu u GSM900 Band du o ng th an co ng ļ§ Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 11 Dung Trinh, PhD HCMUT / 2019 2A. Nonlinear – Harmonic Distortion ng .c om ļ¶ Example 4: An analog multiplier “mixes” its two inputs, ideally producing š¦ š” = šš„1 (š”)š„2 (š”) where k is a constant. Assume š„1 š” = š“1 ššš š1 š” and š„2 š” = š“2 ššš š2 š” . co a. If the mixer is ideal, determine the output frequency components. an b. If the input port sensing š„2 (š”) suffers from third-order nonlinearity, determine the output frequency components th ļ¶ Solution: šš“1 š“2 ššš 2 du o = ng š¦ š” = šš“1 ššš š1 š” š“2 ššš š2 š” a. We have: š1 + š2 š” + šš“1 š“2 ššš 2 š1 − š2 š” cu u b. The third harmonic of š„2 š” is š¼3 š“32 /4, we have: š¦ š” = šš“1 ššš š1 š” Dept. of Telecoms Engineering = šš“1 š“2 ššš 2 + šš¼3 š“1 š“32 ššš 8 CuuDuongThanCong.com š¼3 š“32 š“2 ššš š2 š” + ššš 3š2 š” 4 š1 + š2 š” + šš“1 š“2 ššš 2 š1 + 3š2 š” + š1 − š2 š” šš¼3 š“1 š“32 ššš 8 š1 − 3š2 š” https://fb.com/tailieudientucntt 12 Dung Trinh, PhD HCMUT / 2019 .c om 2A. Nonlinear – Harmonic Distortion cu u du o ng th an co ng The mixer now produces two “spurious” component at š1 + 3š2 and š1 − 3š2 , one or both of which often prove problematic. For example, if š1 = 2š × 850šš»š§ and š2 = 2š × 900šš»š§, then š1 − 3š2 = 2š × 1850šš»š§, an undesired component that is difficult to filter out because it lies close to the desired component at š1 + 3š2 = 2š ×1750MHz. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 13 Dung Trinh, PhD HCMUT / 2019 .c om 2B. Nonlinear – Gain Compression GAIN: š® = š šš š½š + š šš š½šš š½š š = šš + šš š½šš š cu u du o ng th an Fundamental co ng 1 3 1 1 š£0 = š0 + š2 š02 + š1 š0 + š3 š03 ššš š0 š” + š2 š02 ššš 2š0 š” + š3 š03 ššš 3š0 š” + āÆ 2 4 2 4 14 Expansive Dept. of Telecoms Engineering Most RF/microwave component/circuits are compressive ! CuuDuongThanCong.com Compressive https://fb.com/tailieudientucntt Dung Trinh, PhD HCMUT / 2019 2B. Nonlinear – Gain Compression .c om ļ¶ P1dB: power at 1-dB compression. ļ¶ Psat: power at saturation. th an co ng ļ¶ If we need very good linearity, we should operate a few dBs below the P1dB point; this is called power back-off. du o ng š ššššš šš + šš š½ššš,šš š© = ššššš šš − š š Dept. of Telecoms Engineering u cu š½šš,šš š© = šš š. ššš šš CuuDuongThanCong.com https://fb.com/tailieudientucntt 15 Dung Trinh, PhD HCMUT / 2019 2B. Nonlinear – Gain Compression ļ§ Amplitude Modulation: ng Frequency Modulation: an co ļ§ .c om ļ¶ Effect on modulated signal: th ļ¶ FM signal carries no information in its amplitude and hence tolerates compression. ng ļ¶ AM contains information in its amplitude, hence distorted by compression. cu u du o ļ¶ Desensitization: the receiver gain is reduced by the large excursions produced by the interferer even though the desired signal itself is small. vi ļ½ V1 cos ļ·1t ļ« V2 cos ļ·2t 3 3 ļ¦ ļ¶ vo ļ½ ... ļ« ļ§ a1V1 ļ« a3V13 ļ« a3V23 ļ· cos ļ·0t ļ« ... 4 2 ļØ ļø Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 16 Dung Trinh, PhD HCMUT / 2019 2B. Nonlinear – Gain Compression co ng .c om ļ¶ Example 5: A 900-MHz GSM transmitter delivers a power of 1W to the antenna. By how much must the second harmonic of the signal be suppressed (filtered) so that it does not desensitize a 1.8-GHz receiver having š1ššµ = −25dBm? Assume the receiver is 1m away and the 1.8-GHz signal is attenuated by 10 dB as it propagates across this distance. an ļ¶ Solution: cu u du o ng th The output power at 900MHz is equal to 30dBm. With an attenuation of 10dB, the second harmonic must not exceed -15dBm at the transmitter antenna so that it is below š1ššµ of the receiver. Thus, the second harmonic must remain at least 45dB below the fundamental at the TX output. In practice, this interference must be another several dB lower to ensure the RX does not compress. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt Dung Trinh, PhD HCMUT / 2019 2C. Nonlinear – Inter-modulation Distortion .c om ļ¶ Single Signal: an co ng Harmonic distortion Desensitization du o ng th ļ¶ Single and one large interferer: cu u ļ¶ Single and two large interferers: Dept. of Telecoms Engineering CuuDuongThanCong.com Intermodulation https://fb.com/tailieudientucntt 18 Dung Trinh, PhD HCMUT / 2019 2C. Nonlinear – Inter-modulation Distortion .c om ļ¶ A received small desired signal along with two large interferers ļ¶ Intermodulation product falls onto the desired channel, corrupts signal. co ng interferers ššš − šš = šš cu u du o ng th an desired Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 19 Dung Trinh, PhD HCMUT / 2019 2C. Nonlinear – Inter-modulation Distortion .c om ļ¶ The two interferers at frequency š1 and š2 go into the system: š£š š” = š0 ššš š1 š” + ššš š2 š” co ng ļ¶ The signal at the output of the system: 2 + š3 š03 ššš š1 š” + ššš š2 š” 3 +āÆ an š£0 š” = š0 + š1 š0 ššš š1 š” + ššš š2 š” + š2 š02 ššš š1 š” + ššš š2 š” ng th 1 1 = š0 + š1 š0 ššš š1 š” + š1 š0 ššš š2 š” + š2 š02 1 + ššš 2š1 š” + š2 š02 1 + ššš 2š2 š” 2 2 + š3 š03 + š3 š03 Dept. of Telecoms Engineering u 3 1 3 1 ššš š1 š” + ššš 3š1 š” + š3 š03 ššš š2 š” + ššš 3š2 š” 4 4 4 4 cu + š3 š03 du o + š2 š02 ššš š1 − š2 š” + š2 š02 ššš š1 + š2 š” 3 3 3 ššš š1 š” + ššš ššš − šš š” + ššš 2š1 + š2 š” 2 4 4 3 3 3 ššš š2 š” + ššš ššš − šš š” + ššš 2š2 + š1 š” + āÆ 2 4 4 CuuDuongThanCong.com https://fb.com/tailieudientucntt 20 Dung Trinh, PhD HCMUT / 2019 .c om 2C. Nonlinear – Inter-modulation Distortion Harmonics Formulation ng š0 + š2 š02 DC co 3 3 3 š1 š0 + š3 š0 ššš š1 š” + š1 š0 + š3 š03 ššš š2 š” 4 4 an Fundamental š2 š02 ššš š1 − š2 š” + š2 š02 ššš š1 + š2 š” 2nd harmonic 1 1 2 š2 š0 ššš 2š1 š” + š2 š02 ššš 2š2 š” 2 2 du o ng th 2nd order IMD 3 3 3 š3 š0 ššš 2š1 − š2 š” + š3 š03 ššš 2š2 − š1 š” 4 4 3rd order IMD (IM3) 3 3 š3 š03 ššš 2š1 + š2 š” + š3 š03 ššš 2š2 + š1 š” 4 4 cu u 3rd order IMD (IM3) Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 21 Dung Trinh, PhD HCMUT / 2019 2C. Nonlinear – Inter-modulation Distortion cu u du o ng th an co ng .c om Example 6: Suppose four Bluetooth users operate in a room as shown in figure below. User 4 is in the receive mode and attempts to sense a weak signal transmitted by User 1 at 2.410 GHz. At the same time, Users 2 and 3 transmit at 2.420 GHz and 2.430 GHz, respectively. Explain what happens. ļ¶ Since the frequencies transmitted by Users 1, 2, and 3 happen to be equally spaced, the intermodulation in the LNA of RX4 corrupts the desired signal at 2.410 GHz. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 22 Dung Trinh, PhD HCMUT / 2019 2C. Nonlinear – Inter-modulation Distortion .c om ļ¶ IM3 products degrades system performance because they are very close to the passband of the system th du o ng 3 š3 š½šš ššš 2š1 − š2 š” 4 an co ng ļ¶ The power of IM3 products increases more rapidly than the fundamental output when the input power is increased cu u 3 š3 š½šš ššš 2š2 − š1 š” 4 Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 23 Dung Trinh, PhD HCMUT / 2019 2C. Nonlinear – Inter-modulation Distortion .c om ļ¶ Third-order intercept point designates the point where the third-order intermodulation product has the same power as the fundamental output power. The input power at the IP3 point is call input-referred IP3: IIP3. ļ§ The output power at the IP3 point is call output-referred IP3: OIP3. an th ng ļ¶ In practice, the amplifier never works close to the IP3 point because it would have saturated way before that. co ng ļ§ cu u du o ļ¶ IP3 is extrapolated from measurements of the fundamental output and IM3 products in the linear region of the amplifier. ļ§ IP3 is really a measure of linearity, rather than power handling. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 24 Dung Trinh, PhD HCMUT / 2019 2C. Nonlinear – Inter-modulation Distortion .c om ļ¶ Third Intercept Point: is the point where the output power at š1 equals to output power at 2š1 − š2 . th an co ng 3 š1 šš¼š¼š3 = š3 š½šš°š°š·š → šš¼š¼š3 = 4 š1 š → š¼š¼š3 = 0.145 š1ššµ š3 4 ā 9.6ššµ 0.435 cu u du o ng š1ššµ = 4 š1 3 š3 Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 25 Dung Trinh, PhD HCMUT / 2019 2C. Nonlinear – Inter-modulation Distortion .c om ļ¶ Example 7: A Bluetooth receiver employs a low-noise amplifier having a gain of 10 and an input impedance of 50ā¦. The LNA senses a desired signal level of -80dBm at 2.410 GHz and two interferers of equal levels at 2.420 GHz and 2.430 GHz. For simplicity, assume the LNA drives a 50ā¦ load. ng a. Determine the value of š¼3 that yields a š1ššµ of -30dBm. an co b. If each interferer is 10 dB below š1ššµ , determine the corruption experienced by the desired signal at the LNA output. th ļ¶ Solution: ng a. Note that −30ššµš = 20šššš = 10ššš . We have: du o š1ššµ = š1 0.145 = 10ššš š3 cu u • Since š¼1 = 10 then š¼3 = 14.500 š −2 b. Each interferer has a level of -40dBm (6.32 mššš ). Setting š“1 = š“2 = 6.32 mššš , we determine the amplitude of the IM product 3š¼3 š“12 š“2 = 0.343ššš = −59.3ššµš 4 • The desired signal is amplified by 0 (20dB), produce an output at level -60dBm. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 26 Dung Trinh, PhD HCMUT / 2019 cu u du o ng th an co ng ļ¶ Input IP is the point where the output power at ļ·1 equals to output power at (2ļ·1 - ļ·2 ) .c om 2C. Nonlinear – Inter-modulation Distortion Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 27 Dung Trinh, PhD HCMUT / 2019 2C. Nonlinear – Inter-modulation Distortion ng .c om Example 9: A low-noise amplifier senses a -80dBm signal at 2.410 GHz and two 20dBm interferers at 2.420 GHz and 2.430 GHz. What IIP3 is required if the IM products must remain 20 dB below the signal? For simplicity, assume 50Ω interfaces at the input and output. co Solution: an ļ¶ At the LNA output: ng th 20ššš10 š1 šš šš − 20 = 20ššš10 du o ļ¶ Then: 30 š3 š½šššš 4 ļ¶ Thus: cu u š1 šš šš = 3 š3 š½šššš 4 š¼š¼š3 = Dept. of Telecoms Engineering CuuDuongThanCong.com 4 š1 = 3.65šš = 15.2(ššµš) 30 š3 https://fb.com/tailieudientucntt 28 Dung Trinh, PhD HCMUT / 2019 2D. Nonlinear – Cross Modulation .c om ļ¶ Suppose that the interferer is an amplitude-modulated signal: ng š£ššš”ššššššš š” = š0 1 + šššš šš š” ššš š2 š” co ļ¶ Then: cu u du o ng th an š£š š” = š0 1 + šššš šš š” ššš š2 š” 3 š1 š0 + š3 š03 ššš š0 š” 4 Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 29 Dung Trinh, PhD HCMUT / 2019 3. Noise in RF/Microwave System .c om ļ¶ The effect of noise is critical to the performance of most RF and microwave communications, radar, and remote sensing systems. cu u du o ng th an co ng ļ¶ Noise ultimately determines the threshold for the minimum signal that can be reliably detected by a receiver. ļ¶ Noise power in a receiver will be introduced from: ļ§ The external environment through the receiving antenna – External Noise. ļ§ Internal circuits of receivers – Internal Noise. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 30 Dung Trinh, PhD HCMUT / 2019 3. Noise in RF/Microwave System Thermal noise from the ground ļ§ Cosmic background noise from the sky ļ§ Noise from stars (including the sun) ļ§ Lightning ļ§ Gas discharge lamps ļ§ Radio, TV, and cellular stations ļ§ Wireless devices ļ§ Microwave ovens ļ§ Deliberate jamming devices cu u du o ng th an co ļ§ ng .c om ļ¶ External noise may be introduced into a system either by a receiving antenna or by electromagnetic coupling. Some sources of external RF noise include the following: Noise vs. Interferer Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 31 Dung Trinh, PhD HCMUT / 2019 3. Noise in RF/Microwave System .c om ļ¶ Internal Noise generated in a device or component is usually caused by random motions of charges or charge carriers in devices and materials. Such motions may be due to any of several mechanisms, leading to various types of noise: Thermal noise is the most basic type of noise, being caused by thermal vibration of bound charges. It is also known as Johnson or Nyquist noise. ļ§ Shot noise is due to random fluctuations of charge carriers in an electron tube or solid-state device. ļ§ Flicker noise occurs in solid-state components and vacuum tubes. Flicker noise power varies inversely with frequency, and so is often called 1/ f -noise. cu u du o ng th an co ng ļ§ Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 32 Dung Trinh, PhD HCMUT / 2019 3. Thermal Noise ļ§ .c om ļ¶ A resistor not connected to anything can have a voltage across it. The average voltage is of course 0 and its amplitude is random and very small ng ļ¶ Random motion of charge carriers produce a random variation of voltage (or current) with respect to time ļ noise voltage and/or noise current. co ļ¶ A conductor (equivalently a resistor) has lots of charge carriers. cu u du o ng th an ļ¶ The higher the temperature, the more random motion, and therefore the higher the noise. Dept. of Telecoms Engineering CuuDuongThanCong.com Higher temperature https://fb.com/tailieudientucntt 33 Dung Trinh, PhD HCMUT / 2019 3. Thermal Noise āš š šš ng −1 Noise voltage: š½š = ššš»š©š¹ h = 6.626 x 10-34 Jās, Planck’s constant k = 1.38 x 10-23 J/K, Boltzmann’s constant B, bandwidth in Hz āš : šš f, center frequency of the bandwidth R, resistance in Ω T, temperature in Kelvin š½š Maximum Noise Power: š·š = šš¹ š š¹ = šš»š© cu u ļ§ −1ā du o ļ§ āš šš ng ļ¶ At low frequencies (f < 5 THz), š th an ļ¶ It can be shown that *: 4āššµš co šš2 = .c om ļ¶ The noisy resistor can be replaced with a Thevenin equivalent circuit of a noiseless resistor and a random noise voltage generator Vn. š·š š š©š = ššššššš š. šš × šš−šš š»š© = ššššššš š»š© − ššš. š When: š = 290š¾, šµ = 1š»š§ š = 290š¾, šµ = 1š¾š»š§ š = 290š¾, šµ = 1šš»š§ → š·š = −šššš š©š → š·š = −šššš š©š → š·š = −šššš š©š * For derivation, refer to: http://www.pas.rochester.edu/~dmw/ast203/Lectures/Lect_20.pdf Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 34 Dung Trinh, PhD HCMUT / 2019 3. Thermal Noise .c om ļ¶ At most RF and microwave frequencies, the thermal noise from a resistor appear uniform with respect to frequency, as is evident from the absence of a frequency variable in the Vn expression. cu u du o ng th an co ng ļ¶ We call this type of frequency independent noise “white noise”. ļ¶ We also note that the noise power is independent of the resistance! Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 35 Dung Trinh, PhD HCMUT / 2019 3. Noise Figure š šššššš šššššš ššššš ššš šššššš ššššš ššššš du o ng th an co ng Signal to Noise Ratio (SNR): ššš = .c om ļ¶ Noise Figure: measures the degradation in the signal to noise ratio between the input and output of the component. u ļ¶ It is important to understand that while amplifiers are great at making your signal stronger, they also make noise stronger! cu ļ¶ Noise Factor (linear scale): š¹= ļ¶ Noise Figure (dB scale): Dept. of Telecoms Engineering CuuDuongThanCong.com šŗšµš¹šš šŗšµš¹ššš šš¹ = ššššššš š https://fb.com/tailieudientucntt 36 Dung Trinh, PhD HCMUT / 2019 3. Noise Figure .c om Example 10: Consider a device of bandwidth šµ = 20šš»š§ and having a gain šŗ = 60ššµ with noise figure of šš¹ = 5ššµ. What is the noise power seen at the output of the device? co ng Solution: cu u du o ng th an Nout = -35.82dBm. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 37 Dung Trinh, PhD HCMUT / 2019 3. Noise Figure .c om ļ¶ The noise contributions from subsequent stages are suppressed by the preceding stages. It follows that the first stage in a cascade system needs to have low noise and high gain. ng ļ¶ Amplifiers specifically designed to do this are called low noise amplifiers (LNA). LNAs are often designed for the lowest noise, sacrificing efficiency, linearity, and even gain. ļ§ To a large extent, LNAs determine the system sensitivity. th an co ļ§ cu u du o ng š1 = šŗ1 šš0 šµ + šŗ1 ššš1 šµ š0 = šŗ2 š1 + šŗ2 ššš2 šµ = šŗ1 šŗ2 ššµ š0 + šš1 + šš¶š“š = šš1 + 1 š šŗ1 š2 šš2 šš3 + +āÆ šŗ1 šŗ1 šŗ2 šš = (š¹ − 1)š0 š¹2 − 1 š¹3 − 1 š¹š¶š“š = š¹1 + + +āÆ šŗ1 šŗ1 šŗ2 Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 38 Dung Trinh, PhD HCMUT / 2019 3. Noise Figure an co ng .c om Example 11: The block diagram of a wireless receiver front-end is shown in following figure. Compute the overall noise figure of this subsystem. If the input noise power from a feeding antenna is šš = ššš“ šµ, where šš“ = 150š¾, find the output noise power in dBm. If we require a minimum signal-to-noise ratio (SNR) of 20ššµ at the output of the receiver, what is the minimum signal voltage that should be applied at the receiver input? Assume the system is at temperature šš , with a characteristic impedance of 50Ω and an IF bandwidth of 10MHz. cu u du o ng th Solution: Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 39 Dung Trinh, PhD HCMUT / 2019 3. Noise Figure .c om ļ¶ We have: šŗš = −1ššµ = 0.79 š¹š = 1ššµ = 1.26 ng šŗš = 10ššµ = 10 š¹š = 2ššµ = 1.58 š¹š −1 šŗš + š¹š −1 šŗš šŗš = 1.58 + 1.26−1 2.51−1 + 10 10×0.79 = 1.8 = 2.55ššµ an š¹ = š¹š + co ļ¶ The overall noise figure of the system: šŗš = −3ššµ = 0.5 š¹š = 4ššµ = 2.51 th ļ¶ The equivalent noise temperature of the overall system: du o ļ¶ The output noise power: ng šš = š¹ − 1 š0 = 1.8 − 1 × 290 = 232š¾ šš = š šš“ + šš šµšŗ = 1.38 × 10−23 150 + 232 × 107 × 3.95 = −96.8ššµš šš = šš šŗ = cu u ļ¶ For an output SNR of 20dB, the input signal power must be: šš šš šš šŗ = 2.08×10−13 100 3.95 = −82.8ššµš ļ¶ For a 50Ω system impedance, this corresponds to an input signal voltage of šš = š0 šš = 1.62 × 10−5 š = 16.2šš Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 40 Dung Trinh, PhD HCMUT / 2019 3. Noise Figure of a Lossy Element ng .c om ļ¶ Let’s assume that a resistor at the same physical temperature is connected to the input of the network. Since they in thermal equilibrium, the resistor network can be treated as one noise source at T, the output noise power should be co š0 = šššµ ng th an ļ¶ We can also treat the input resistor as the signal source, then the output noise power should be: 1 1 š0 = šššµ + ššš šµ š“ š“ du o ļ¶ Equating the two equation gives: šš = š“ − 1 š cu u ļ¶ Also we have: šš = š¹ − 1 š. Finally, š¹ = š“. Dept. of Telecoms Engineering CuuDuongThanCong.com A = attenuation, defined as Pin/Pout B, bandwidth in Hz T, physical temperature (in Kelvin) of the network https://fb.com/tailieudientucntt 41 Dung Trinh, PhD HCMUT / 2019 3. Noise Figure Measurement .c om ļ¶ Cold source method: a termination impedance at room temperature, also called “cold source”, is placed at the DUT input. co ng ļ¶ If the gain of the amplifier is well known over the frequency of interest it can be subtracted from the output power noise resulting in the amplifier’s excess noise factor F. an ššš¢š” = ššš × šŗ × š¹ cu u du o ng th ļ¶ The gain G of the amplifier can be accurately measured over a bandwidth B by a calibrated two-port measurement with a VNA. https://www.youtube.com/watch?v=IgXKS0yJ1-E Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 42 Dung Trinh, PhD HCMUT / 2019 3. Noise Figure Measurement ng co š1 = šŗšš1 šµ + šŗššš šµ .c om ļ¶ Y factor method: the idea is simple: measure the noise from two known noise sources and compare the noise power difference at the output. th an š1 š1 + šš š= = š2 š2 + šš ng š2 = šŗšš2 šµ + šŗššš šµ š1 − šš2 šš = š−1 du o ļ¶ It seems that taking one measurement is enough to calculate Te but: That would be less accurate; you would have to know GB precisely. ļ§ Taking two measurements allows you to get rid of measurement errors that are common to both measurements. ļ§ You need two or more measurement points to https://www.youtube.com/watch?v=bbdTRX4_2DE accurately extrapolate the zero crossing value. cu u ļ§ Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 43 Dung Trinh, PhD HCMUT / 2019 3. Noise Figure .c om Example 12: An X-band amplifier has a gain of 20 dB and a 1 GHz bandwidth. Its equivalent noise temperature is to be measured via the Y-factor method. The following data are obtained: š1 = −62ššµš For š2 = 77š¾, š2 = −64.7ššµš co ng For š1 = 290š¾, ng th an Determine the equivalent noise temperature of the amplifier. If the amplifier is used with a source having an equivalent noise temperature of šš = 450š¾, what is the output noise power from the amplifier, in dBm? ļ¶ The Y factor in dB is: du o Solution: cu u š = š1 − š2 = −62 + 64.7 = 2.7ššµ = 1.86 ļ¶ This gives the equivalent noise temperature as šš = š1 −šš2 š−1 = 270−1.86×77 1.86−1 = 170š¾ ļ¶ If šš = 450š¾, the output noise power from the amplifier is šš = šŗššš šµ + šŗššš šµ = 100 × 1.38 × 10−23 × 109 × 450 + 170 = −60.7ššµš Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 44 Dung Trinh, PhD HCMUT / 2019 .c om 3. Noise Figure Measurement co th an šššš = ššš¶ šµšŗš¹ ng ļ¶ In the OFF or cold state, the noise source is switched off, corresponding to a temperature TC and the noise power at the output is then: u du o ng ļ¶ With the noise source ON or in hot state, the excess noise PN of the noise source is added to the output noise power: cu ššš = ššš¶ šµšŗš¹ + šš šŗšµ = ššā šµšŗš¹ ļ¶ The Excess Noise Ratio of the used noise figure is calculated as: šā − šš šøšš = šš Dept. of Telecoms Engineering CuuDuongThanCong.com šøšš š¹= š−1 https://fb.com/tailieudientucntt 45 Dung Trinh, PhD HCMUT / 2019 3. Noise Figure Measurement .c om ļ¶ Would it help if we take more than two measurements? Good accurate noise sources can be expensive ļ§ Getting two noise sources with drastically different noise temperature can also be difficult ļ§ Remember that temperature is specified in Kelvin; 0°C and 100°C are only about 30% different. u du o ng th an co ļ§ ng ļ¶ That would require more noise sources: cu ļ¶ Example: Calculate the Noise Figure NF of your Spectrum Analyzer: 1. Connect a 50Ω termination at the analyzer’s input 2. Select a resolution bandwidth RBW, e.g. RBW=10KHz 3. Place a marker at the center frequency and read out the power level šš in dBm, e.g. šš = −110ššµš. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 46 Dung Trinh, PhD HCMUT / 2019 3. Sensitivity .c om Sensitivity: Minimum detectable signal by the receiver according to a fixed SNR determined by the BER. ļ§ Path loss. ļ§ Receiver sensitivity – govern by standards and applications. ļ§ Required SNR – depends on BER requirement and modulation scheme. du o ng th Path Loss Receiver Sensitivity Input Noise Floor (No/G) Required SNRo u Noise floor – thermal noise or circuit noise limited depending on the modulation schemes. cu ļ§ co Transmit Power – FCC regulated. Transmit Power an ļ§ ng ļ¶ To find Receiver NF: Noise Figure Input Noise (Ni) ļ¶ Example: Mobile network: -100dBm; GPS: -127.5dBm. Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 47 HCMUT / 2019 co ng .c om Dung Trinh, PhD cu u du o ng th an Q&A Dept. of Telecoms Engineering CuuDuongThanCong.com https://fb.com/tailieudientucntt 48