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// Clock divider circuit // From 50 MHz to 1 MHz/200 Hz with %50 duty cycle module clk_div(Clk_in, Clk_out); // input ports input Clk_in; // output ports output reg Clk_out; // counter size calculation according to input and output frequencies parameter sys_clk = 50000000; // 50 MHz system clock parameter clk_out = 1000000; // 1 MHz clock output parameter max = sys_clk / (2*clk_out); // max ...
dbc/hz. Because when you are measuring the power of any signal besides a perfect CW tone you have to specify the bandwidth it's measured in. If you were comparing two CW tones it would be dBc. Comparing a tone to noise must be dBc/Hz. Also, it's in dB because it's easier to add than multiply. If you want to know the thermal noise power in a 1 ...
Hz = cycles per second. Bit Rate = 1/ (Tb), Where Tb is the bit duration. In general the minimal clock to transmit at a given Bit Rate, should be at least twice of the bit rate in frequency. In general form, Symbol Rate is the period symbol, for M-ary digital waveform where: k = log2 (M) Where M are the levels to represent k bits in a symbol.
Where do you see UWB spectral power definitions in FCC with GHz bandwidth? There are specifications with 1 kHz, 1 MHz and 50 MHz resolution bandwidth. Also respective ETSI specifications are using similar limit values. Calculating dBm/Ghz spectral density is possible but doesn't seem to make much sense.
England, UK. Activity points. 12,979. Re: MIPS + MHz. Hi, MHz (Megahertz) is the number of clock cycles every second. This is usually determined by a crystal oscillator on the board somewhere. For instance, if your processor or memory runs at 20MHz, it gets 20 million clock cycles every second. What it does with this depends entirely on the chip.
Can anyone provide me code to generate a 1hz frequency clock generator from vhdl with clock cycle of 100Mhz default. please.
The time between one bit and next bit is the baudrate, so if you have your uart setted in 9600bps, the bits are 1/9600 segs long, or 0.104uSegs. So, because you always send bytes of 8 bits each, and 2 aditionally bits (start and stop), the transfer rate of data is 8+2=10, or 1/10 from baudrate. So, you will transfer a maximal data rate of 960 ...
Re: Simple Question. (X bits * Y Mhz) / 8 = Z Mbps. ... Assuming there are 8 bits in a byte: (megabits/second) / 8 = megabytes/second. It looks as though your numbers are not coming out correct because different busses have different number of bits/byte. Some are 8 bits/byte so you would divide by 8, others have 10 or 11 bits per byte.
I want to know the relation between sampling frequency (Hz) and sampling rate (sample per second). For example, a 1 Hz sin wave sampled at 8000 samples per second. Each cycle of the 1Hz tone will span all 8000 samples (since its period is 1 second). Thus sampling period will be 1/ (8000-1) s Or sampling frequency will be 8000-1 ≈ 8000Hz.
8. Activity points. 522. hi every one. i want to build a 0.1 hz to 1mhz sin, triangular, pulse generator with 10vpp adjustable amplitude. for the first design i used a DAC for generating signal and then used a amplifier for signal conditioning. the problem is when generating low frequency signal, the signal is corrupted due to the DAC nature.