; Replace with your application code
start:
ldi r20, 0b00000100 ; load to r20 the mask to set pin 2 as output
out DDRB, r20 ; Set PIN 2 on port B to output and all others to input
loop:
ldi r20, 0b00000100 ; load to r20 the mask to set pin 2 to 1
out PORTB, r20 ; Set pin 2 to 1 on port B
call delay ; wait for 250ms
ldi r20, 0b00000000 ; load to r20 the mask to set pin 2 to 0
out PORTB, r20 ; Set pin 2 to 0 on port B
call delay ; wait for 250ms
rjmp loop ; repeat the loop

delay:
; By default on Xplained mini is extenal clock
; On 5V it is at 16MHz
; On 3.3V it is 8MHz
; We have to do a 250ms delay
; On 16MHz 1 clock is 1/16MHz = 62.5ns
; It takes 16 clocks for 1us, if there is a loop which will take 16 clocks (1us) executed 250 times it is 250us.
; Doing this a 250 times, it will elapse 62.5ms
; And doing this 4 times, we will have our 250ms of delay
; There will be some inaccurancy due to setting the values after end of loops.

ldi r16, 4 ; load 4 to r16 - 1 clock
cycle1: ; 4 * 250 * 250us loop = 250ms
ldi r17, 250 ; load 250 to r17 - 1 clock
cycle2: ; 250 * 250us loop = 62.5ms
ldi r18, 250 ; load 250 to r18 - 1 clock
cycle3: ; 250us loop
dec r18 ; decement r18 - 1clock
nop ; no operation, just spend a clock cycle - 1clock
nop ; 1clock
nop ; 1clock
nop ; 1clock
nop ; 1clock
nop ; 1clock
nop ; 1clock
nop ; 1clock
nop ; 1clock
nop ; 1clock
nop ; 1clock
nop ; 1clock
nop ; 1clock
brne cycle3 ; branch if r18 not 0 - 2 clocks, 1 if no branching
; in total 16 clocks or 1 us
; it will execute 250 times, so it will spend 250us
dec r17 ; decrement r17 - 1 clock
brne cycle2 ; branch if r17 not 0 - 2 clocks, 1 if no branching
; it will execute 250 times, so it will spend 250 * 250us = 62.5ms

Problem 2-1 (6 points) Consider the following periodic input curren...