Number of solutions on IF/RF reception & system noise
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1. A receiver tunes the HF band (3 to 30 MHz), utilizes up-conversion with an intermediate frequency of 40.525 MHz, and uses high-side injection. Calculate the required range of local oscillator frequencies.
2. An AM broadcast receiver's pre-selector has a total effective Q of 90 to a received signal at 1180 kHz and uses an IF of 455 kHz. Calculate the image frequency and its dB of suppression.
3. We want to operate a receiver with NF = 8 dB at S/N = 15 dB over a 200-kHz bandwidth at ambient temperature. Calculate the receiver's sensitivity.
4. A receiver with a 10-MHz bandwidth has an S/N of 5 dB and a sensitivity of -96 dBm. Find the required NF.
5. The receiver in problem 4 has a dynamic range of 78.7 dB. If it has a 6 dB NF preamp (gain = 20 dB) added to its input, find its sensitivity and dynamic range.
6. An μ-law companding system with μ=100 is used to compand a 0 to 10 V signal. Calculate the system output for inputs of 0, 0.1, and 1 V.
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Solution Summary
A number of questions and solutions concerning AM, HF, RF reception, intermodulation, heterodyning, noise and receiver SNR and noise figure
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1. The IF frequency f(IF) is deduced from the difference between the local oscillator frequency f(LO) in effect and the tuned frequency f as below.
f(IF) = f(LO) - f (1)
At the low end of the tuning range we put f = 3 MHz into (1) and the fixed IF frequency of f(IF) = 40.525 MHz to get the required local oscillator frequency f(LO MIN).
40.525 = f(LO MIN) - 3 (2A)
f(LO MIN) = 40.525 + 3 = 43.525 MHz
Similarly for the high end tuning frequency we put f = 30 MHz into (1) and the fixed IF frequency of f(IF) = 40.525 MHz to get the required local oscillator frequency f(LO MAX).
40.525 = f(LO MAX) - 30 (2B)
f(LO MAX) = 40.525 + 30 = 70.525 MHz
Therefore the required range of local oscillator frequencies range from 43.525 MHz to 70.525 MHz.
2. Image frequency f(IM) for an AM broadcast occurs when the difference frequency f(IM) - f(LO) is equal to the IF frequency f(IF).
f(IF) = f(IM) - f(LO) (3)
But the IF frequency is also derived from the difference between the same local oscillator frequency and the wanted tuned frequency according to (4).
f(IF) = f(LO) - f (4)
Re-arranging (4) in terms of the unknown local oscillator frequency f(LO) we get
f(LO) = f(IF) + f (5)
Substitution of f(LO) as given by (5) in (3) gives
f(IF) = f(IM) - {fIF + f} (6)
Re-arrangement of (6) in terms of the image frequency gives
f(IM) = 2*f(IF) + f (7)
We are given the IF frequency as f(IF) = 455 kHz; we are told that the tuned frequency is f = 1180 kHz, so in (7) we deduce the image frequency as
f(IM) = 2*455 + 1180 kHz
Image frequency is f(IM) = 2090 kHz or 2.09 MHz
To calculate the dB of suppression for the image frequency we use (8) which gives the Q value in terms of the dB suppression of the image, tuned frequency f and the separation between upper and lower sidebands Δf (effectively separation ...
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