Ecet3110 Unit 1 Question Paper
Homework 2_1
ECET310
NOISE
1. Assume that a receiver has a first amplifier stage with a 5K input resistance, a gain of 300, an input audio signal of 20 V, and an operating temperature of 27oC. When the amplifier is operating with a bandwidth first of 10 MHz, find
(a) the rms input noise levels
(b) the audio output levels
(c) the rms output noise levels
PN = noise power in watts
Tk = 27°C + 273 = 300°K
a) Noise level Pn(in) = KT(k)BW = (1.38x10-23)(300°K)(10MHz) = 41.4 x 10^-15 W
b) Ps(out) = Gain * Ps(in) = 60mW
c) Pn(out) = Gain * Pn(in) = 12.42pW
2. We have an amplifier with a 1 MHz bandwidth and a 20 V input signal. The input noise = 1 V, output signal = 6.0 mV, and output noise = 1 mV.
Find
(a) SNR at …show more content…
A 300 resistor is connected across the 300 antenna input of a television receiver. The bandwidth of the receiver is 6 MHz, and the resistor is at room temperature (20oC). Find the noise power and noise voltage applied to the receiver input.
Pn = KTBW = 24.26 x 10^-15 W
Vn = = (24.26x10^-15 W) * (300Ω) = 2.7µV
5. The signal at the input of an amplifier has an SNR of 42 dB. If the amplifier has a noise figure of 6 db, what is the SNR at the output (in dB’s) ?
SNRo = SNRi / NF = 7dB
6.
A three-stage amplifier is shown. The operating bandwidth is 500 KHz at a temperature of 25o C. The output load is 1000.
(a) Calculate the total power gain in dB’s and as a number
(b) Calculate the total NF as a number
(c) Calculate the output noise power
(d) Calculate the output noise voltage
Ap1=10, Ap2=31.62, Ap3=3.16
NF1=1.58, NF2=3.16, NF3=2.51
a) Ap(T) = 10 + 15 + 5 = 30dB
Ap(T) = 1000
b) NFT = NF1 + = 1.8
c) Pn(in) = KTBW = 2.06x10^-15 W
Pn(o) = NFT * Ap(T) * Pn(in) = 3.71pW
d) Vn = =
ECET310
NOISE
1. Assume that a receiver has a first amplifier stage with a 5K input resistance, a gain of 300, an input audio signal of 20 V, and an operating temperature of 27oC. When the amplifier is operating with a bandwidth first of 10 MHz, find
(a) the rms input noise levels
(b) the audio output levels
(c) the rms output noise levels
PN = noise power in watts
Tk = 27°C + 273 = 300°K
a) Noise level Pn(in) = KT(k)BW = (1.38x10-23)(300°K)(10MHz) = 41.4 x 10^-15 W
b) Ps(out) = Gain * Ps(in) = 60mW
c) Pn(out) = Gain * Pn(in) = 12.42pW
2. We have an amplifier with a 1 MHz bandwidth and a 20 V input signal. The input noise = 1 V, output signal = 6.0 mV, and output noise = 1 mV.
Find
(a) SNR at …show more content…
A 300 resistor is connected across the 300 antenna input of a television receiver. The bandwidth of the receiver is 6 MHz, and the resistor is at room temperature (20oC). Find the noise power and noise voltage applied to the receiver input.
Pn = KTBW = 24.26 x 10^-15 W
Vn = = (24.26x10^-15 W) * (300Ω) = 2.7µV
5. The signal at the input of an amplifier has an SNR of 42 dB. If the amplifier has a noise figure of 6 db, what is the SNR at the output (in dB’s) ?
SNRo = SNRi / NF = 7dB
6.
A three-stage amplifier is shown. The operating bandwidth is 500 KHz at a temperature of 25o C. The output load is 1000.
(a) Calculate the total power gain in dB’s and as a number
(b) Calculate the total NF as a number
(c) Calculate the output noise power
(d) Calculate the output noise voltage
Ap1=10, Ap2=31.62, Ap3=3.16
NF1=1.58, NF2=3.16, NF3=2.51
a) Ap(T) = 10 + 15 + 5 = 30dB
Ap(T) = 1000
b) NFT = NF1 + = 1.8
c) Pn(in) = KTBW = 2.06x10^-15 W
Pn(o) = NFT * Ap(T) * Pn(in) = 3.71pW
d) Vn = =