8051 microcontroller Machine cycle time & frequency:To execute an instruction microcontroller takes certain number of cycles these cyles are called machine cycles in 8051 family crystal frequency may vary from 4MHz to 20MHz depending on chip manufacturer. normally 12MHz for general use and 11.0592 MHz when using serial communication to set. LMP execute even though one instruction 32. Find the time delay for the delay subroutine shown to the right, if the system has an 8051 with frequency of 11.0592 MHz the system frouency if the machine cycle 1.2 gs. we machine cycle if the crystal frequency is 18 MHz. e machine cycle if the crystal frequency is 12 MHz. the machine cycle if the. A cycle is, in reality, 12 pulses of the crystal. That is to say, if an instruction takes one machine cycle to execute, it will take 12 pulses of the crystal to execute. With 12 MHz clock frequency how -Amdahls-law--Derive-an-expression-for-CPU-clock-as-a-function-of-instruction-count-clocks-per-instruction-and-clock-cycle-time/1 1. Time of execution of instruction is measured in machine cycle.For example ADD instruction takes 2 machine cycles to execute then in AVR family this instruction executes in 2*clock period seconds while in 8051 family this instruction executes in 2*clock period *12 seconds.So for same crystal frequency the 8051 family takes 12 times greater than. Calculating Machine Cycle, frequency and duty cycle given different Oscillators and asm Hi All, I'm trying to calculate the frequency and duty cycle when using different oscillators and I'm a little confused. a) 4Mhz b) 20Mhz frequency = 1 / Time duty cycle = Time on / Time Waveform: Contains a loop that takes 6 clock cycles to complete 1 loop. bsf(1), bcf(1), nop(1), nop(1), goto(2)
What is the time taken by one machine cycle if crystal frequency is 20MHz? 1.085 micro seconds 0.60 micro seconds 0.75 micro seconds 1 micro seconds. Computer Architecture Objective type Questions and Answers The formula used to calculate the frequency is: f = 1 / T. Symbols. f = Frequency; T = Period; Period Measured. Enter the amount of time it takes to complete one full cycle. Frequency Calculation. This is the number of cycles per unit period of time which corresponds to the entered time period. Help 1-10 ms to H 7. 11.0592 MHz/1 = 11.0592 MHz; machine cycle is 1/11.0592 MHz= 0.0904 µs = 90.4 ns Posted by Satish Kashyap Email ThisBlogThis!Share to TwitterShare to FacebookShare to Pinterest 10 Free Mock Tests for GATE 2018 from GATEpaper.i 8.What is the time duration for one state and one machine cycle if a 6 MHz crystal is connected to 8051? Clock frequency=6 MHz/2=3 MHz One T state= 1/ Clock frequency=1/3 MHz= 0.333 microsec The time taken to execute a machine cycle is 12 clock periods
.0592MHz. Machine Cycle DELAY: MOV R3,#250 1 HERE: NOP 1 NOP 1 NOP 1 NOP 1 DJNZ R3,HERE 2 RET 2 Solution: The time delay inside HERE loop is [250(1+1+1+1+2)]x1.085μs ＝ 1627.5μs. Adding the two. 39(a) 11.0592MHz/12 = 921.6 KHz Machine cycle is 1/921.6 KHz = 1.085µs (microsecond)(b) Oscillator period = 1/16 MHz = 0.0625 µ Machine cycle (MC) = 0.0625 µs ×12 = 0.75 (c) 20 MHz/12 = 1.66 MHzMC = 1/1.66 MHz = 0.60 µ Example 3-13 The following shows crystal frequency for three different 8051- based systems And 11.0592 MHz was the frequency very near the maximum operating frequency which was still a multiple of the standard baud rates. By the way original 8051s executed a single machine cycle every 12 clocks, so at 12 MHz, the instruction execution frequency was 1 million instructions per second. What are the standard baud rates
A single machine cycle is the minimum amount of time in which a single 8051 instruction can be executed. although many instructions take multiple cycles. A cycle is, in reality, 12 pulses of the crystal. That is to say, if an instruction takes one machine cycle to execute, it will take 12 pulses of the crystal to execute The oscillator formed by the crystal, capacitor and an on-chip inverter generates a pulse train at the frequency of the crystal. The clock frequency f establishes the smallest interval to accomplish any simple instruction. The time taken to complete any instruction is called as machine cycle or instruction cycle. In 8051 one instruction cycle. ONE MACHINE CYCLE FREQUENCY = CRYSTAL FREQUENCY/12 = 11.0592/12 = 921.6ms STEP 3. ONE MACHINE CYCLE TIME PERIOD = 1/FREQUENCY = 1/921.6ms = 1.085us STEP 4 Find the frequency and period used by the timer if the crystal attached to the 8051 has the following values (a) XTAL = 11.0592MHz (b) XTAL = 20MHz (c)XTAL = 24MHz (d) XTAL = 30MHz A). F= (1/12 machine cycles) * XTAL = (1/12) * 11.0592e6 HZ =.92160 MHz Period = T= 1/F = 1/.92160 MHZ = 1.085 microSec B)
In my case I have a 8 MHz external quarts with PLL configured to provide 84 Mhz system clock SYSCLK. That means that one processor cycle is 1.0/84e6 ~ 12 ns. For reference of the how many cycles or SYSCLK one instruction takes you are using the ARM® Cortex®‑M4 Processor Technical Reference Manual. For example the MOV instruction in most of. One ―cycle‖ is the minimum time it takes the CPU to do any work. —The clock cycle time or clock period is just the length of a cycle. —The clock rate, or frequency, is the reciprocal of the cycle time. Generally, a higher frequency is better. Some examples illustrate some typical frequencies
1/frequency (MHz) if frequency increased, the period will. decrease. if period increases, wavelength will. increase. wavelength. the distance (length) is takes for one cycle to occur (m, mm) if wavelength increases, frequency. decreases. wavelength = propagation speed (mm/us)/frequency (MHz) crystal transmits 1% of the time and receives 99%. In this frequencies each 12 cycles take 1 machine cycle ie 11.0952 / 12 = 0.9246, = 926.6KHz this means 926.6 instruction can be execute in second with the 11.952 mhz frequency . we know the Time period , T= 1/F ie T = 1 /926.6 = 1.079us (it is the one machine cycle time ) taking this 1.079 micro second ., 1000mcro second is 1 millisecon 2. Find the timer's clock frequency and its period for various 8051-based systems, with the following crystal frequencies. (3 points) (a) 12MHz 1us (b) 16MHz 0.75us (c) 11.0592MHz 1.085us 3. Find the value for TMOD if we want to program timer 0 in mode 2, use 8051 crystal
A t state is a clock cycle. It is the minimum time unit for the CPU to perform operations. Machine cycle:Generally, the CPU cycle (machine week) is specified by the shortest time for reading a script from the memory.Period), that is, the time required for the CPU to complete a basic operation. Generally, a machine cycle includes 12 clock cycles What is the time taken by one machine cycle if crystal frequency is 20MHz? sushmitahate32 sushmitahate32 03.10.2020 Computer Science Secondary School answered 5. What is the time taken by one machine cycle if crystal frequency is 20MHz? 1 See answer sushmitahate32 is waiting for your help. Add your answer and earn points. sreerenjini563. The amount of time delay depends on the frequency of the 8051 13. one machine cycle lasts 12 oscillator periods Find the period of the machine cycle for 11.0592 MHz crystal frequency Solution: 11.0592/12 = 921.6 kHz; machine cycle is 1/921.6 kHz = 1.085μs.
The oscillator formed by the crystal , capacitor and an on-chip inverter generates a pulse train at the frequency of the crystal. The clock frequency . f. establishes the smallest interval to accomplish any simple instruction. The time taken to complete any instruction is called as machine cycle or instruction cycle. In 8051 one instruction. The ALE signal goes high at the beginning of each machine cycle indicating the availability of the Clock frequency = crystal frequency / 2 = 6 Mhz/2 =1/(3Χ10 6) = 0.333μsec Time for opcode fetch cycle = 4 Χμs 0.33= 1.332 μsec Time for memory read cycle= 43Χ 0.33=0.999 μsec. 8.Write the operation carried out when 8085 executes DAD. 2. The 8051 has external clock of frequency 11.0592 MHz, which gets divided by 12 to get machine cycle frequency is 921.6 kHz. 3. The 8051 microcontroller serial communication UART circuitry divides the machine cycle frequency of 921.6 kHz by 32, before it is used by Timer 1 to set the baud rate. Therefore, 921.6 kHz divided by 32 gives out.
24MHz respectively. Microcontroller A divides its clock by 4 to give one machine cycle, microcontroller B by 8, and microcontroller C by 12. Microcontrollers A & C take 2 machine cycles to perform an instruction, while microcontroller B takes three cycles. Place the microcontrollers in order of the speed in which they can perform that instruction This example is the same as the previous one, but now we have CLEAR the Zero flag. The instructions that follows the BTFSS instruction will be executed normally! In this case, the BTFSS instruction must be calculated as an 1-Cycle instruction, and the following instruction must be taken into account accordingly. In other words, the conditional branching instructions may need 1 or 2. One machine cycle duration is the 1/12th of the frequency of the crystal attached to the controller. For example, if the frequency of the crystal is 12 MHz, then the frequency for Timer will be 1MHz (1/12 of crystal frequency) and hence the time (T = 1/f) taken by the Timer to count by one is 1µs (1/1MHz)
Despite the trends towards ever-higher frequency processor clocks and data rates, there is still a large need for lower frequency crystal oscillators for timing in extreme low-power applications. For example, the ECS-327MVATX is a miniature surface-mount oscillator operating at a fixed frequency of 32.768 kHz with MultiVolt capability (1.6 to 3. On this occasion, the author would like to discuss the use of Microsoft Excel to simulate the charging of the capacitor (C) through a resistance (R) of the 8051 microcontroller reset circuits . The author would also shows the equation (rule o A single machine cycle is the minimum amount of time in which a single 8051 instruction can be executed. although many instructions take multiple cycles. 8051 has an on-chip oscillator. It needs an external crystal thats decides the operating frequency of the 8051
. What is meant by instruction cycle and clock cycle. How to calculate Instruction cycle time and clock cycle time. Is there any relation between architecture pipe lining and instruction cycle It takes up one statement of high level language program at a time, translates Write a program to generate a delay of 0.5 sec if the crystal frequency is 5 MHz. Ans. The operating frequency is half of te crystal frequency. Machine Cycle: It is the time required by the microprocessor to complete the operation of accessin In this mode, the synchronizing clock signal is the frequency of one machine cycle. In the case of 8051, this is (11.059Mhz/12=921.6Mhz). Just to emphasize, the TXD pin's function can be a little misleading as it is used here for sending the synchronizing clock signal Note. Cases might exist where QueryPerformanceFrequency doesn't return the actual frequency of the hardware tick generator. For example, in many cases, QueryPerformanceFrequency returns the TSC frequency divided by 1024; and on Hyper-V, the performance counter frequency is always 10 MHz when the guest virtual machine runs under a hypervisor that implements the hypervisor version 1.0 interface
clocks per machine cycle from 12 to four, or even one. The frequency for the timer is always 1/12th the frequency of the crystal attached to the 8051, regardless of the 8051 version. We can also use crystal frequency as the clock source. The following are the steps to find TH, TL register values. We can assume XTAL = 11.0592 MHz for 8051 systems Instruction cycle time for the PIC - One instruction cycle consists of four oscillator periods. - To calculate the instruction cycle for the PIC, we take 1/4 of the crystal frequency, then take its inverse. - Unconditional branch takes 2 instruction cycles. - Conditional branch takes 2 instruction cycles if it jumps, and takes 1 when not jumps The time taken to execute a machine cycle is expressed in terms of T-states. In 8085, if the external clock frequency applied through X1 & X2 pins are 6 MHz; then the internal clock frequency is 3 MHz (since, Fint = Fext / 2).Therefore, one T-State equals to T The frequency is the number of oscillations per second and 1 Hertz is one cycle per second. Ultrasound is very high frequency and therefore measured in megahertz. The period of the wave is the time it takes to complete one cycle and is described by the following equation: f = 1 / T where f is the frequency and T is the period for one full cycle
A spark-gap transmitter is an obsolete type of radio transmitter which generates radio waves by means of an electric spark. Spark-gap transmitters were the first type of radio transmitter, and were the main type used during the wireless telegraphy or spark era, the first three decades of radio, from 1887 to the end of World War I. German physicist Heinrich Hertz built the first experimental. Why wase your time running one frequency at a time? And since the GB-4000 is so much more powerful than mose other machines available, it has plenty of horsepower to run those frequencies at the same time. It's like getting 8 frequency generators for the price of one time is just the inverse of the oscillator frequency. We are using an 11.0592 MHz oscillator (this particular frequency makes it easy to use standard RS-232 communication speeds); therefore, each oscillator cycle takes 1/11059200 seconds, or roughly 90 nanoseconds. Thus, one machine cycle (of 12 oscillator cycles) takes about 1.085 microseconds 8051 divides the crystal oscillator frequency by 12 to get the machine cycle frequency. UART divides the machine cycle frequency by 32 and sends it to Timer 1 to set the baud rate. Timer 1, mode 2 (8-bit, auto-reload) Define TH1 to set the baud rate. XTAL = 11.0592 MHz. The system frequency = 11.0592 MHz / 12 = 921.6 kH 8051 microcontrollers completes one machine cycle after every 12 clock cycles. So our Instruction execution frequency is crystal frequency/12. Now if our crystal is of 11.0592 MHz then our effective frequency is 11.0592/12 MHz => 921.6 KHz. 8051 UART further divides this frequency (921.6 KHz) by 32 to generate its baud rate
. The frequency parameters are either 1, 2, or 3 MHz. Frequency is measured in megahertz (MHz) and can range from .8 up to 3 MHz.9 A frequency of 1 MHz can penetrate from 2.5 cm up to 5 cm within the tissue, while 3 MHz reaches up to a 2.5 cm depth.6,8,1 nop ; 1E cycle Dbne X,Loop ; 3E cycles condition not satisfied; 3E cycle condition satisfied - Assume that the HCS12 runs unde r a crystal oscillator with a frequency of 16 MHz, with an E freq. of 8 MHz. The E-clock period is then 125 ns. - Therefore, each repeated loop would take 14*125=1750ns=1.75us to execute
One complete path traveled by the wave is called a cycle. One cycle per second is known as 1 Hz (Hertz). The amplitude of a wave is the maximal excursion in the positive or negative direction from the baseline, and the period is the time it takes for one complete cycle of the wave (Fig. 2.2) For time delay,the timer use the clock source of the crystal frequency of the crystal attached to 8051. The frequency for the timer will be 1/12th of the frequency of the external crystal attached. The time delay of one machine cycle is given below. We use this to generate the delay. TP = 1/ 921583= 1.085 µ sec. For delay of 10ms: Firstly. Frequency range of 0.01 to 20 million hertz. Produces digitally accurate frequencies converted to analog for full analog harmonics. New! Incorporates a groundbreaking new 3.1 MHz RF (Radio Frequency) carrier wave technology to give massive versatility in sideband frequencies when using squarewave audio frequencies the time should not be too long because this increases the current consumption. The time periods are defined with the TSM register by selecting a clock source, either ACLK or the high-frequency clock ESIOSC, and the number of clock cycles of that clock. The ESIOSC source is mainly chosen for precise settle time definition
BASIC PHYSICAL PRINCIPLES OF ULTRASOUND ©AIU 4 ÜIMPORTANT SUMMARY DIAGRAMS TO REMEMBER CYCLE ONE COMPLETE OSCILLATION OF THE WAVE FREQUENCY NUMBER OF CYCLES PER SECOND OF TIME 1 SECOND WAVELENGTH LENGTH OF ONE COMPLETE CYCLE Frequency and wavelength have an inverse relationship shown in the formula c = fl, where c is the speed of sound in tissue, f is the frequency and l is the wavelength One machine cycle is the 1/12th of the frequency of the crystal attached to the controller. We are here using 6MHz crystal, then a single machine cycle will take. 6/12 = 0.5 MHz. Each count takes 1/0.5 MHz resulting in 2us , so when timer1 counts from 0000H to FFFFH (65535 counts) it takes. Why do we choose 12 MHz? Because in the 8051, a machine cycle consists of 12 clock cycles, so each machine cycle takes one microsecond (µS) and most 8051 instructions take either one or two machine cycles, so these instructions take either one or two µS. Therefore, calculation becomes very convenient
First is the crystal oscillator. This is a good time to point out that a quartz crystal is not an oscillator; rather, a quartz crystal is the central component in a quartz-crystal oscillator. Page 3 of 3 Q-35 Find the time delay generated by the following routine if the XTAL = 22 MHz. HERE : MOV R0,#200 AGAIN : DJNZ R0,AGAIN RET Q-36 Write a program to generate a square wave with frequency of 10KHz and with 50% duty cycle using Timer 0. Assume the external crystal frequency of 12 MHz. Q-37 Timer 0 is used as a counter, and Timer 1 is used as time base of 1 second
The choice of 4.77 MHz instead of 5 MHz only cost 4.5% performance. You can argue that they could have architected the system for 4.77 MHz and then just used a faster crystal in the 5150 PC (planning to downsize the crystal to 4.77 in the future imaginary consumer PC) . In terms of frequency range, power output, and ease of use, the GB4000 is head and shoulders above the rest. Widest frequency range of any machine on the marketSimple to use, but includes advanced features.Comes with both Audio and RF modes.Highest output power available in a.
let if we use 11.0592MHz crystal. u know 8051 takes 12 pulses for 1 machine cycle. then period of one machine cycle would be . 11.0592MHz/12 = 921.6KHz. t = 1/f. so time period of one machine cycle = 1/ 921.6KHz = 1.085us . now when u run loop of 1275.. delay would be 1275×1.085us = 1.3ms. practically this value gives delay of almost 1m A 12 MHZ crystal results in convenient time of 1 microsecond per cycle. An 11.0592 MHZ crystal frequency of 921.6 kloherts, which can be divided evenly by the standard communication baud rates of 19200, 9600, 4800, 2400, 1200 and 300 HZ With 12 MHz clock frequency how many instructions (of 1 machine cycle and 2 machine cycle) it can execute per second? A cycle is, in reality, 12 pulses of the crystal. That is to say, if an instruction takes one machine cycle to execute, it will take 12 pulses of the crystal to execute Let us consider 8051 microcontroller for which an external crystal oscillator circuit of 12MHz is essential, even though (based on model) 8051 microcontroller is capable to run at 40 MHz (max). 8051 requires 12 clock cycles for one machine cycle, such that to give effective cycle rate at 1MHz (considering 12MHz clock) to 3.33MHz (considering. · 8051 divides the crystal frequency by 12 to get the machine cycle frequency · XTAL = 11.0592 MHz, the machine cycle frequency is 921.6 kHz · 8051's UART divides the machine cycle frequency of 921.6 kHz by 32 once more before it is used by Timer 1 to set the baud rat
Period of an ultrasound wave is the time that is required to capture one cycle, i.e., the time from the beginning of one cycle till the beginning of the next cycle. The units of period is time and typical values in echo is 0.1 to 0.5 microsecond. Period of ultrasound is determined by the source and cannot be changed by the sonographer frequency Fosc/24. While there are no limitations on the external clock duty cycle, care must be taken to insure that the clock signal has been sampled correctly during the S5P2 time. This is easily accomplished by maintaining the external timer clock in a given state for a minimum of one machine cycle. Interrupt At high frequency, i.e., in the order of more than 10 MHz, as shown in Fig. 9.16, dissolved metal ions were deposited on the microtool. This happens as the cycle time is very small and dissolved metal ions do not get enough time to react with the hydroxyl ion
Quartz Crystal Circuit. The below figure represents the electronic symbol of a piezoelectric crystal resonator and also quartz crystal in an electronic oscillator that consists of resistor, inductor, and capacitors.. Crystal Oscillator Circuit Diagram. The above figure is a 20psc New 16MHz Quartz Crystal Oscillator and it is one kind of crystal oscillators, that works with 16MHz frequency The frequency response of a crystal is as shown below. The graph shows the reactance (X L or X C) versus frequency (f). It is evident that the crystal has two closely spaced resonant frequencies. The first one is the series resonant frequency (f s), which occurs when reactance of the inductance (L) is equal to the reactance of the capacitance C. operates with a crystal frequency 11.04962 MHz The oscillator formed by the crystal, capacitor and an on-chip inverter generates a pulse train at the frequency of the crystal. The clock frequency f establishes the smallest interval to accomplish any simple instruction. The time taken to complete any instruction is called as machine cycle or.
The RF shrinks collagen, speeds up metabolism, and thickens the skin to give help to the skin's surface. The LED colour light rejuvenation of wavelength 625 nm enhances cell activity and provides you a crystal clear and even complexion. Advantages Of Ultrasonic Cavitation Machines Painless and non-invasiv In theory, one standard PSI-5A sheet (1.5 x 2.5 x .0075) used as an extender can do .00035 joules of work on the outside world in a quasi-static cycle (i.e. a slowly executed sinusoidal cycle). When operated just under its first longitudinal resonance of 15 KHz, the theoretically available output power from the sheet would be around 5 watts
the standard unit of frequency, equal to one cycle per second. 1,000,000 is 1 MHz: A mode is_____ also known as amplitude mode. One dimensional image displaying the amplitude strength of the returning echo signals along the vertical axis and the time (therefore the distance from the transducer) along the horizontal axis: Axial is: Depth axis Time is an integer count of some number of clock ticks • One year @ 2 MHz takes about 47 bits to represent as an integer - too big to be useful for most embedded applications • But, most applications don't need time to the nearest 1/2,000,000 second • So, we want time with a bigger granularity than thi The Relationship between the crystal frequency and the baud rate in the 8051 is that the 8051 divides the crystal frequency by 12 to get the machine cycle frequency which is shown in figure1. Here the oscillator is XTAL = 11.0592 MHz, the machine cycle frequency is 921.6 kHz Two implementations of the same instruction set, machine A has a clock cycle time of 1 ns, and a CPI of 2.0, machine B has a clock cycle time of 2 ns and a CPI of 1.2 for the same program. Which machine is faster, and by how much? 1.2 2.0 2.4 CPU time CPU time CPU performance CPU performance CPU time 1.2 2 2.4 CPU time 2.0 1 2.0 A B B A B A. Hz (the frequency range of audible sound is 20 to 20,000 Hz), diagnostic ultrasound uses frequencies in the range of 1-10 million (mega) hertz. The Wavelength The wavelength is the distance traveled by sound in one cycle, or the distance between two identical points in the wave cycle i.e. the distanc
The frequencies are in KHz. We use the ETDFL group's list and it is the most current, effective and up to date list available. The machine comes with everything you need including an instruction/frequency manual. If you wanted to use a Hz frequency, divide it by 1000 and you will get the KHz value 3.2.1 System Frequency and Clock Division One of the first decisions that users migrating designs from AT89S52 to AT89LP52 in Fast mode must make is at what frequency to run the AT89LP52. By default the AT89LP52 in Fast mode is at least 6 times faster than the AT89S52, meaning it fetches bytes from memory in one-sixth the time The wavelength is the distance traveled by sound in one cycle, or the distance between two identical points in the wave cycle i.e. the distance from a point of peak compression to the next point of peak compression. It is inversely proportional to the frequency. Wavelength is one of the main factors affecting axial resolution of an ultrasound. Period is the time for a sound wave to complete one cycle; the period unit of measure is the microsecond (µs).Wavelength is the length of space over which one cycle occurs; it is equal to the travel distance from the beginning to the end of one cycle.Frequency is the number of cycles repeated per second and measured in hertz (Hz).Acoustic velocity is the speed at which a sound wave travels. The original 1950's equipment did not use an RF carrier frequency, so they could only output one audio frequency at a time with only 1/5th of one watt of power (0.20 of one watt). This power level worked well with only one audio frequency at a time
8051 microcontrollers completes one machine cycle after every 12 clock cycles. So our Instruction execution frequency is crystal frequency/12. Now if our crystal is of 11.0592 MHz then our effective frequency is 11.0592/12 MHz => 921.6 KHz. 8051 UART further divides this frequency (921.6 KHz) by 32 to generate its baud rate In this way, a 400 MHz output of the voltage controlled oscillator (VCO) will yield a 16 MHz clock at the other input of the phase/frequency detector. If the 400 MHz clock is too slow, the up signal will add to the charge pump, increasing the input to the VCO, leading to an increase in the 400 MHz frequency First the machine applies a voltage to the crystal to expand it and transmit. The ultrasound machine then very quickly switches to a listening mode by monitoring the voltage across the piezo electric crystal. This transmit and receive cycle is repeated very rapidly. The above examples show only one crystal for clarity
Upgrading your desktop or laptop to a solid-storage solution—whether that's a traditional 2.5-inch drive or a cutting-edge M.2 one—is a quick, often inexpensive way of adding some much-needed. Note that this is four times the 3.579545 MHz frequency of the NTSC color standard, which was called a color clock. They divided that crystal frequency to derive their actual clock frequency. For instance . Apple II was 1.023 MHz (1/14 of crystal, 3.5 color clocks per CPU cycle) and used a 1 MHz rated CPU A rubidium standard, or rubidium atomic clock, is a high accuracy frequency and time standard, usually accurate to within a few parts in 10 11.This is still several orders of magnitude less than. The time taken in decoding and execution of an instruction is one clock cycle. In some situations, an execute cycle may involve one or more read or write cycles or both. Read Cycle: If an instruction contains data or operand address which are in the memory, the CPU has to perform some read operations to get the desired data A period is the time it takes for a single cycle to occur. From the start of one cycle to the start of the next cycle is called the period. While period is measured in time, wavelength is measured in length. A period is determined by the sound source, not the medium through which the ultrasound signal travels
Multiplying a very stable low-frequency reference signal can still produce signals with better Phase Noise than producing them directly in the microwave frequency range. • For example typical Phase Noise of a 10 MHz Crystal Oscillator is: -170 dBc/Hz @ 100 kHz offset You can't measure exact delay_cycles time using such approach because for(;;) cycle handling and also port I/O consume CPU cycles so definitely results will be more than 1us high or low time. Also you don't use DCO calibration constants, most probably your clock is not 1MHz but somewhere in 800..1000 kHz range Q can seem mysterious, but is a measure of energy loss with Q = Energy stored during one cycle / Energy lost during one cycle. For a component, Q is the ratio of reactance to resistance. For example, if an inductor has 500 Ω of reactance and 5 Ω of loss resistance, it's Q = X L / R = 500 / 5 = 100 — a typical value for inductors.