$5.$ Given PIC$24$H running at Fosc $=20$MHz with one digital input at RA$1$ and one digital output at RB$5.$

$($a$)$ Write a function to configure RA$1$ as an input and generate a change notification interrupt $($CN$).$

$($b$)$ Assume that a clock signal is connected to RA$1.$ The following code decide whether the clock speed is

faster than $100$ kHz or not. If the clock is faster than $100$ kHz then set RB$5$ to $0,$ otherwise set RB$5$ to

$1.$

void $\_$ISR $\_$CNInterrupt$($void$)\{$

CNcounter$++$;

$\_$CNIF $=0$;

$\}$

uint$8$ CNcounter;

void main $()\{$

CNcounter $=0$;

configure$()$;

T$2$CON$=0$x$0000$;

PR$2=99$;

$\_$LATB$5=0$;

$\_$T$2$IF$=0$;

T$2$CONbits.TON$=1$;

while$($CNcounter $<2)$

if $(\_$T$2$IF $==1)\{$

$\_$LATB$5=1$;

$\_$T$2$IF $=0$;

break;

$\}$

$\}$

$\}$

$($c$)$ Compute the percentage of CPU time taken by the above ISR.

Assume the following:

$-$ Each statement consumes one instruction cycle.

$-$ The RA$1$ input is a clock with a frequency equal to $100$ kHz$.$