Communication Problems (Filters, BW, Bit rate, FM)

[ineedmunchies]

Homework Statement

A 1kHz analogue signal is to be encoded using 8-level Pulse Code Modulator. This specification results in a bit rate of 5.5K bits/sec. The signal is to be transmitted over a 1 km distance with a 10kHz bandwidth on a Frequency Modulated, F.M., Carrier. The carrier frequency is in the digital cellular radio frequency band (GSM Reverse Band) at 875 MHz. In this frequency band legislation permits up to 1 watt of carrier power to be transmitted and received through omnidirectional antennas. The transmit and receive antenna feeders are of low loss type and exhibit 0.1 dB maximum attenuation.

The received signal is to be passed through a matched impedance and matched bandwidth single down conversion superheterodyne receiver. This receiver should have three stages. The gain of the first and third stages are 7 dB and 27 dB respectively and the noise figure for each of these stages is 2 dB and 8 dB respectively. The mixer placed immediately after the low noise amplifier has a loss of 3 dB. The overall noise figure of the receiver should not exceed 6.2 dB.

Homework Equations

Step 1

The analogue input signal must pass through an anti-alias filter. In order to define the bandwidth of the system this should have………………………. Filter characteristics.
The bandwidth of this filter should be set to at least the…………………………..
which is this case is……………….kHz

Step 2

Bit rate is given as 5.5K bits/sec.

On a minimum bandwidth channel each cycle of each sinewave carries……………….. bits of information. Allowing for 1.5 times over-sampling in order to avoid aliasing, the minimum bandwidth for the systems becomes………………… Hz.

Due to the bit rate calculated above the digitally encoded signal can now be transmitted to the FM modulator via a……………… type cable.


Step 3

In order to define the modulator characteristics the FM modulator frequency deviation must now be computed. The system has narrowband FM characteristics so that the necessary frequency deviation can be obtained by an approximate technique such as…………………………….. rule. From this approximate calculation the required modulator frequency deviation is………………………………………… Hz.

The Attempt at a Solution

Ok, so for a few of the parts I'm only looking to check if I have the correct answer, and some parts I haven't a clue about.

But here goes:

Step 1:
I think the anti-aliasing filter should have low pass characteristics, and the cut-off frequency/bandwidth of the filter should be at least the maximum analogue signal frequency, which in this case is 1KHz.

Step 2:
Ok, so this is where I start to become unsure of everything.
I think that on a minimum bandwidth channel each cycle of a sinewave can contain 2 bits.
If there are 2 bits per cycle, and the sampling rate is 5.5Kbits/sec then the bandwidth is 5.5/2 = 2.25KHz?? But they want to over sample by 1.5 so this should then become 2.25*1.5 = 4.125KHz ??

I am unsure as to which type of cable it wants, I would hazard a guess at coaxial.

Step 3:
It is a narrowband system, so Carson's rule can be used to obtain the frequency deviation.
The carrier frequency is 875 Mhz, the bandwidth is 10 kHz.

This is where I really start getting lost. But anyway carsons rule is : BW = 2 * (f_$$\Delta$$ + f_m)

Where f_$$\Delta$$ is the peak deviation of the instantaneous frequency from the center carrier frequency, and f_m is the highest modulating frequency.
10 kHz = 2( ? + ?)

But the modulating/carrier frequency is given as 875 MHz so that would give a large negative frequency for f_$$\Delta$$

874.995 MHz from my calculation.

There are two more parts to this question but I haven't included them here as I haven't attempted them properly yet.

Thank you, any help with any of this is greatly appreciated.

 

[KataKoniK]

Step 1:
You are correct, the anti-aliasing filter should have low pass characteristics with a cut-off frequency of at least 1kHz.

Step 2:
Your calculation for the bandwidth is correct, but I think you may have misunderstood the question. The question is asking for the minimum bandwidth for the system, not the actual bandwidth. So we need to take into account the over-sampling factor of 1.5. This means that the minimum bandwidth for the system should be 4.125/1.5 = 2.75kHz.

As for the type of cable, the question does not specify, so any type of cable that can handle the bandwidth and signal power can be used.

Step 3:
You are correct in using Carson's rule to calculate the frequency deviation. However, there seems to be a mistake in your calculation. The formula should be BW = 2 * (f\Delta + fm), where f\Delta is the peak deviation of the instantaneous frequency from the center carrier frequency, and fm is the highest modulating frequency. In this case, the highest modulating frequency is 1kHz, so the formula becomes 10kHz = 2 * (f\Delta + 1kHz). Solving for f\Delta, we get f\Delta = 4.5kHz. Therefore, the necessary modulator frequency deviation is 4.5kHz.

I hope this helps you with your solution. Let me know if you have any other questions.

 

[Mene]

Your attempt at a solution is mostly correct. Here are some clarifications and corrections:

Step 1:
The anti-aliasing filter should indeed have low pass characteristics, and the cut-off frequency/bandwidth should be at least the maximum analogue signal frequency, which in this case is 1kHz. However, the bandwidth should be set to at least twice the maximum analogue signal frequency to avoid aliasing. So the bandwidth should be 2kHz.

Step 2:
You are correct that on a minimum bandwidth channel, each cycle of a sinewave can contain 2 bits of information. However, the sampling rate should be at least twice the bandwidth, not the bit rate. So the sampling rate should be at least 4kHz, and the bandwidth should be at least 2kHz. This means that the minimum bandwidth for the system becomes 3kHz (2kHz x 1.5).

The type of cable needed to transmit the digitally encoded signal to the FM modulator would indeed be coaxial.

Step 3:
You are correct that Carson's rule can be used to obtain the frequency deviation. However, the formula should be BW = 2 * (f\Delta + fm) where f\Delta is the peak deviation of the instantaneous frequency from the center carrier frequency, and fm is the highest modulating frequency. In this case, the highest modulating frequency is 1kHz, so the formula becomes 10kHz = 2 * (f\Delta + 1kHz). Solving for f\Delta, we get f\Delta = 4.5kHz.

So the required frequency deviation is 4.5kHz. However, the carrier frequency given is in MHz, so it should be converted to Hz first. 875MHz = 875,000,000Hz. So the required frequency deviation is 4,500Hz.

Overall, your attempt at a solution is good. Just be careful with the units and make sure to follow the correct formulas. Good job!