ECE424_HW8_Solution
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ECE 424 (524) - Radio System Design Homework 8 Department of Electrical and Computer Engineering NC State University Your solution should be submitted to moodle 1.
Below is a frequency spectrum illustrating the power of a 0.1mV sinusoidal signal of 162MHz applied to the input of a receiver. The receiver has a 50-ohm load and operates within a bandwidth of 300MHz. a)
What is the power of the signal (dBm) at 162MHz? 𝑃?𝑖𝑔 =
?
2
2?
=
(0.1?)
2
2 ∗ 50
= 1 ∗ ?
−10
? = −70?𝐵?
b)
If the SNR of the signal is 15dB, how much is the noise power? ?𝑁? =
𝑃𝑠𝑖𝑔?𝑎?
𝑃??𝑖𝑠?
,
𝑃??𝑖𝑠? = −70 − 15 = −85?𝐵? = 3.16?
−12
??
c)
What is the equivalent temperature of this receiver? 𝑃??𝑖𝑠? = 𝐾?𝐵,
? =
𝑃??𝑖𝑠?
𝐾 ∗ 𝐵
=
3.16?
−12
1.38?
−23
∗ 300?
6
= 763.8𝐾 2.
Below is a receiver system featuring a bandpass filter (BPF) with a bandwidth of 150 MHz. Considering the system is at room temperature. Answer the following questions.
Amplifier 1
G=10 dB
NF=2dB
Amplifier 2
G=15 dB
NF=2dB
Loss=1.5 dB
BPF
0
50
100
150
200
250
300
350
Frequency (MHz)
ECE 424 (524) - Radio System Design Homework 8 Department of Electrical and Computer Engineering NC State University a)
What is the noise figure of the overall system? NF1=1.41, NF2=1.58, NF3=1.58 G1=0.7, G2=10, G3=31.6 𝑁?
𝑡𝑜𝑡𝑎𝑙
= 𝑁?1 +
𝑁?2 − 1
?1
+
𝑁?3 − 1
?1?2
= 2.32
b)
What is the output SNR if the input signal level is -85 dBm? Output noise level is = KTBxNF=-88.4dBm Output SNR = 3.4dB 3.
A receiver system with 50dB gain is tested with a two-tone signal at its input, and the intermodulation products at the output are 65dB lower than main tones (which is at 0dBm). Calculate the input referred IIP3 of this receiver. OIP3=Pout_Main + Delta/2=0dBm+(65/2) =32.5 IIP3=OIP3-Gain=32.5-50=-17.5 4.
A receiver system consists of two components (G1=15dB, IIP31=0dBm) and (G2=0dB, IIP32=0dBm) a.
Calculate the OIP3in dB? OIP3_1 = IIP3_1 + G1 = 15dBm = 31.6mW
ECE 424 (524) - Radio System Design Homework 8 Department of Electrical and Computer Engineering NC State University OIP3_2 = IIP3_2 + G2 = 0dBm = 1mW OIP3_Total = ((1/1) + (1/(31.6*1)))^-1 = 0.969mW = -0.135 dBm b.
The gain G2 is increased by 7dB. Calculate the new OIP3? OIP3_1 = IIP3_1 + G1 = 15dBm = 31.6mW OIP3_2 = IIP3_2 + G2 = 7dBm = 5mW OIP3_Total = ((1/5) + (1/(31.6*5)))^-1 = 4.84 mW = 6.85 dBm c.
How much has the IP3 increased with the new G2 gain? 6.85dBm –
(-0.135dBm) ≈ 7dBm
5.
You are designing an RX system that has a noise figure of 10dB and 33dB gain. The initial system has a channel bandwidth of 100MHz, temperature of 300K and SNR of 10dB at the output is required. Answer the following questions. a.
What is the minimum detectable input signal power? (in dBm) KTB is in Watts, it is multiplied by 1000 to convert to mW, So Noise level = KTB*NF*1000 Noise level = KT*B*1000xNF=4.14e-21*100e6*1e3*10^1= 4.14E-9 or-83.83 dBm Pmin=Noise+SNR Pmin=-83.83+10=-73.83 dBm b.
You are assigned to decrease minimum detectable input signal power and you have two options: 1) System cooling on the receiver to have the temperature be 100K or 2) Decrease the operation bandwidth to 1Mhz. What would be the minimum detectable input signal for these changes? 1)
Noise level = K (100) *B*1000xNF=1.38e-21*100e6*1e3*10^1=4.14E-9 or-88.6 dBm Pmin=Noise+SNR Pmin=-88.6 +10=-78.6 dBm 2)
Noise level = KT*B*1000xNF=4.14e-21*1e6*1e3*10^1= 4.14E-11 or -103.8 dBm Pmin=Noise+SNR Pmin=-103.8 +10=-93.83 dBm
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ECE 424 (524) - Radio System Design Homework 8 Department of Electrical and Computer Engineering NC State University c.
What is the option that you would pick to meet the customers’ requirements?
The best decision would be to drop the bandwidth of the receiver. This will meet the customer requirements and can be achieved by modifying the circuit opposed to adding additional components to provide the cooling. Adding in the cooling will also increase the cost of operation of the receiver.
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