This is the current situation:
802.11a Wireless network with a maximum datarate of 3/4 64QAM.
Calculate the datarate you get with a 2/3 QPSK.
802.11a uses a bandwidth of 54MHz.
Can anyone help me with this calculation?
Related
Assume a MLD (Multi Link Device) with 2SS that operating in 2.4g and 5g.
In 2.4g link, MLD transmit EHT MCS13 and band width is 20M. We suppose theoretical throughput is X.
In 5g link, MLD transmit EHT MCS13 and band width is 320M. we suppose theoretical throughput is Y.
I calculate the throughput of MLD that operating in 2 link.
In theoretical scenario, is the throughput X+Y?
Thx!
A current source is exciting a Loud by an AC current of ±5mA. The voltage through the loud is measured using the NI data acquisition. The resistance of the loud changes with time, the peak-to-peak amplitude of the voltage signal changes accordingly. How to define the relationship between the loud's resistance and the voltage peak-to-peak amplitude?! in other words, how can I plot the graph of signal's peak-to-peak amplitude as a function of time in LabView?
measure with appropriate Sample-Rate, at least 10 times higher than the max. frequency of your signal.
use a DAQ, that has "synchronous sampling"
measure current and voltage (synchronously with high sample rate). You can use either a shunt or a current transducer for current measurement
Sample in "Blocks". This means: let the DAQ device store e.g. 10k Values (at a Sample rate of 100kHz) in it's internal memory, and read that buffer every 100 ms. Go to the Example finder (Help -> Find exampes) and look for "continous analog measurement" examples.
calculate the RMS-Value of both signals of each block and plot that in a graph. If you want it simple, feed both signals into a "Chart".
if the current is constant (which should be a strait line in the graph!), the voltage should rise over time, when the inner resistance of the loudspeaker rises ...
Note: be aware, that with the example numbers above (100kHz SR, Blocksizte 10k) calculating the RMS value will produce wrong results, when your signal main frequency is below 10 Hz!
I'm working with wireless sensor network lead to evaluate its performance in my work. I want to measure the latency and total energy consumption to find the remaining energy in each node. So my problem is that I have some values of tx rx cpu cpu_idle and I don't how to use them to calculate what I need. I found some rules that take the calculation but i don't understand exactly how to apply it in my case.
Energy consumed in communication:
Energy consumed by CPU:
What is the meaning of 32768, and why do we use this number? Is it a standard value?
The powertrace output is printed in timer ticks.
tx - the number of ticks the radio has been in transmit mode (ENERGEST_TYPE_TRANSMIT)
rx - the number of ticks the radio has been in receive mode (ENERGEST_TYPE_LISTEN)
cpu - the number of ticks the CPU has been in active mode (ENERGEST_TYPE_CPU)
cpu_idle - the number of ticks the CPU has been in idle mode (ENERGEST_TYPE_LPM)
The elements of the pair tx and rx are exclusive, as are cpu and idle - the system can never be in both modes at the same time. However, other combinations are possible: it can be in cpu and in tx at the same time, for example. The sum of cpu and idle is the total uptime of the system.
The duration of timer a tick is platform-dependent and defined as the RTIMER_ARCH_SECOND constant. 32768 ticks per second is a typical value of this constant - that's where the number in your equation comes from. For example:
ticks_in_tx_mode = energest_type_time(ENERGEST_TYPE_TRANSMIT);
seconds_in_tx_mode = ticks_in_tx_mode / RTIMER_ARCH_SECOND;
To compute the average current consumption (in milliamperes, mA), multiply each of tx, rx, cpu, cpu_idle with the respective current consumption in that mode in mA (obtain the values from the datasheet of the node), sum them up, and divide by RTIMER_ARCH_SECOND:
current = (tx * current_tx_mode + rx * current_rx_mode + \
cpu * current_cpu + cpu_idle * current_idle) / RTIMER_ARCH_SECOND
To compute the charge (in millicoulumbs, mC), multiply the average current consumption with the duration of the measurement (node's uptime) in seconds:
charge = current * (cpu + cpu_idle) / RTIMER_ARCH_SECOND
To compute the power (in milliwats, mW) multiply the average current consumption with the voltage of the system, for exampe, 3 volts if powered from a pair of AA batteries:
power = current * voltage
Finally, to compute the energy consumption (in millijoules, mJ), multiply the power with the duration in seconds or multiply the charge with the voltage of the system:
energy = charge * voltage
The first formula above computes the energy consumption for communications; the second one: for computation.
This site might be helpful to break down the numbers.
32768 Hz or 32, 768 kHz this is of MSP430F247 Microcontroller frequency, for specifics are Active mode: 32iuA # 3 v / 1 MHz or 1x10 6 Hz and Low Power mode = 1 uA # 3V /32768 Hz
Background:
I've implemented QoS with four queues having strict priorities in wireless device. Queue-0 is having highest priority , Queue-1 is second highest and so on. My wireless device was set as 20Mhz and MCS: -1 which gives throughput around 40-45mbps. I tested this with JDSU having 8 streams of 10mbps that means total JDSU tx rate: 80mbps. In my overnight test, i found frame loss happened in queue-0 and queue-1 which was not expected if we place the device in RF chamber ( Lab environment). However, If i limit the tx rate of JDSU within 45mbps then i don't find any frame loss.Is there any relationship between throughput and frame loss? My topology is like :
jdsu<---->wifi master<------air i/f------>wifi slave >loopback
Just my two cents. Have you considered that the rate you are transmitting at is nearly double the supported data rate of your receiving devices? Wireless nodes broadcast their supported data rate for good reason. This is so other devices on the network can speak at a rate that the other device can understand. So I would say the answer to your question is an emphatic yes. Imagine if I were only capable of comprehending 1000 words per minute but you spoke at a rate of 2500 words per minute. You can safely expect that at some point I am going to be unable to comprehend every word that you are saying.
I intend to use the beaglebone to sample a shaped signal of the order of 1 microsec. I need to fit the signal after and therefore i would like to have a sampling rate of let's 10 MHZ. Something that seems feasible with PRU and libpruio. The point is, looking to the adc specifications it seems there is a limit at 200KHz. Is my reasoning correct?
thanks
You'll need additional hardware for a sampling rate of 10 MHz! libpruio isn't designed to work at that speed, as well as the BBB hardware.
The ADC subsystem in the AM335x CPU is clocked at 24 MHz and needs 15 cycles for a sample (14 in continous mode). This leads to a maximum sample rate of 1.6 (1.74) MSamples/s. See SRM, chapter 12 for details.
The problem is to get the samples in to the host memory. I couldn't get this working faster than ~250 kSamples/s (by CPU access - I didn't try DMA).
As long as you don't need more values than the FIFO can hold, you can sample a single line at maximum 1.7 MHz.
BR