In multi-cycle datapath, the first and second cycle is similar for all types of instructions.
The first cycle is Instruction fetch and the second cycle is instruction decode and register fetch.
Jump is performed in three clock cycles. two clock cycles are those I mentioned above. and the third one is left shift 26 LSB of instruction so that we have 28 bit and then add 4 MSB bits of PC to it so we get a 32-bit address and we set it to the pc.
but we can do the third cycle of jump in the second clock cycle, so that jump instruction can be finished in two clock cycles.
Related
If the number of clock cycles it takes to complete an instruction is more than one does that mean program counter gets incremented more than once in the same instruction cycle. I am getting this doubt because from my knowledge registers gets updated on each clock pulse.
Does this mean that if a system is waiting for memory for 3 clock cycles pc will be pc +12?
Each instruction cycle of the example processor contains one to x machine cycles. The download phase consists of as many machine cycles as the number of bytes must be sent from the operating memory to the processor under one instruction. The duration of the execution cycle depends on the type of order downloaded.
I'm not trained in EE.
I'm programming a master-receiver device which controls a MAX11644/MAX11645. The datasheet explains the read cycle, saying:
A read cycle must be initiated to obtain conversion results. Read cycles begin with the bus master issuing a START condition followed by seven address bits and a read bit (R/W = 1). If the address byte is successfully received, the MAX11644/MAX11645 (slave) issues an acknowledge. The master then reads from the slave. The result is transmitted in 2 bytes; first 4 bits of the first byte are high, then MSB through LSB are consecutively clocked out.
All of this I understand, except the very last part: "MSB through LSB are consecutively clocked out". Most significant bit? Isn't this the first bit? We already know the first bit in the first byte is hi. And what does "clocked out" mean?
Most significant bit? Isn't this the first bit?
It may or may not be. There's no unambiguous definition of "first". RS232, for example, outputs the least significant bit first. If you mean the one that happens to be output first, then yes, that's what the next part is saying.
We already know the first bit in the first byte is hi.
Right. But the device outputs it anyway.
And what does "clocked out" mean?
It means that they are produced as output on consecutive clock cycles. That is, each time the clock advances, the next bit (in the order defined there) is placed on the output pin.
This is a very basic question. I can't simulate a PWM file, in system time, from its FPGA VI file.
Details
For a NI cRIO-9067 + LabVIEW 2016 + Windows 8 system, under FPGA Interface Mode, I have the Test VI No.1.vi NI LabVIEW file and the corresponding FPGA Desktop Execution Node block file Test VI No.1 DEN.vi as suggested in the Getting Started information [1] [2].
In both files, the Low Pulse and High Pulse Numeric Controls are filled with the 1000 value. The Loop Timer block is set as "mSec" Counter Unit and "32 Bit" Size of Internal Counter.
The compiled FPGA version of the first file executes a square wave changing each 1 second, as expected, after 7 minutes of local compilation.
Under Simulation (Simulated I/O) as Execution Mode, and for reproducing approximatedly and by trial and error the square wave timing every 1 second, I need to put the value 1750 in the Clock Ticks field, from the FPGA 40MHz Onboard Clock reference clock, shown in the block options.
I dont understand this block, and why i should not put any close divisor of 40,000,000 at the Clock Ticks field, or simply, the value 1. Basically i dont understand how to "time" these FPGA simulations.
The desktop execution node is designed for time based simulation you are definately on the right track.
What you are setting at the top is the number of cycles that are executed each time you call the node. In your case you have 1750 ticks so around 43.75us of simulated time per iteration.
To simulate in real time you need to make sure that you execute the same amount of simulated time as the simulation loop takes to run. In your case, you have no timing in your simulation loop so why 1750 works for you is because that is probably how long that loop takes to execute.
If you put a loop timer in of 1ms and set the clock ticks to 40,000 (1ms simulated time) then I think you will find that it also works.
In some cases it may be beneficial to execute faster than real time so you would just have to account for that in your maths. For example if you set the clock ticks to 40 (1us simulated time) then you can count the number of iterations and multiply by 1us to get the actual clock time.
I have a high speed clock at 10 MHz going to the processor's TIM4 input capture pin (ch.3). I would like to verify that the clock is running at 10 MHz with the processor's input capture. I coded the processor with the input capture module, and it works fine for lower frequencies (around 1 kHz or so). Once I start to climb the frequency up to the MHz range, the processor starts to miss interrupts and thus gives me an incorrect frequency. I didn't see anywhere in the datasheet that states the maximum frequency that the input capture can read. I have an external clock of 8 MHz, and a core clock of 72 MHz, so I would imagine that I can read a 10 MHz signal. Any ideas?
Take a look at the TIM_ICInitStructure.TIM_ICPrescaler options. Usually you'll have it set to TIM_ICPSC_DIV1 so that interrupts are generated on every valid transition.
Prescaler values of 1,2,4 and 8 are available that will allow you to effictively reduce the rate of interrupt generation by that factor. For example, for a 10MHz signal with a prescaler of 8 you'd expect to count a frequency of 10Mhz/8 = 1.25MHz.
This is still quite tight for a 72MHz HCLK so you'll still need to optimise your IRQ handler carefully.
Looks like you're generating an interrupt request for every rising (or falling) edge of the clock.
If that is indeed the case, then think about this for a second: with a 10 MHz input signal, you're generating an interrupt about every 7 CPU cycles. In these 7 CPU cycles, you need to budget time to save registers to RAM, run the IRQ handler function prolog, run the actual code you wrote for the interrupt handler, run the IRQ handler function epilogue, and restore the registers.
Best case, if you set compiler flags to optimize for speed and you're not doing much processing in the interrupt handler, you're looking at tens of cycles to run all these tasks. Since you only have 7 cycles to run tens of cycles' worth of processing, it's no surprise that you're missing interrupts.
You can't use an interrupt routine at that frequency. You need to feed the 10MHz in as an external trigger to the timer. Then you can use the prescaler and the timer to divide down to a suitable lower interrupt frequency.
I'm currently learning about interrupts but don't understand how you
calculate the data rate for the question below. I have the answers but
I have no idea how you get there. If someone could please explain to
me how it is calculated it would be really appreciated.
Here is the question...
This question concerns the use of interrupts to handle the input and
storage in memory of data arriving at an input interface, and the
consideration of data rates that be achieved using this mechanism. In
this particular question, the arrival of each new data item triggers
an interrupt request to input and store the data item in a queue in
memory.The question is about calculating the maximum data rate
achievable in this scenario.
You are first required to calculate the time to respond to an
interrupt from the interface, run the interrupt service routine (ISR)
and return to the interrupted program.From this and the number of data
bits input on each interrupt, you are required to calculate the
maximum data rate in bits per second, that can be handled. Below you
are given: the number of clock cycles the CPU requires to respond to
the interrupt and switch to the ISR, the number of instructions
executed by the ISR, the average number of clock cycles executed per
instruction in the ISR, the number of bits in the data item input on
each interrupt, and the clock frequency. [You can assume that when the
CPU can be immediately interrupted again as soon as the ISR completes,
but not before this]
clock cycles to respond to interrupt = 15
instructions executed in ISR= 70
average clock cycles per instruction = 5
number of bits per data item = 32
clock frequency = 10MHz
Questions
a) What is the time in microseconds to respond to an interrupt from
the interface, run the interrupt service routine (ISR) and return to
the interrupted program?
b)What is the maximum data rate in Kbits/second?
Answers
a) 36.5 - I understand this
b) 876.7 - ????
Because each ISR takes 36.5 us, the absolute maximum number of ISRs that can happen in a second is 27,397.2603.
In each ISR, 32 bits of data are processed.
Therefore, 27397.2603 * 32 bits = 876.712.33 bits processed per second