The simplest of infinitely many possible solutions, is y = x2
It's simple. Circulator is device which has a peculiar property that each terminal is connected only to the next clockwise terminal. Consider a 3 port circulator with ports 1,2,3 ( arranged in anti clock wise). As there are 3 ports, order of scattering matrix is 3 X 3. [S]ij i= input applied at port i, j= output taken from port j when input is applied at port 1, we get output from port 3 only. i.e,s11=0,s12=0 when input is applied at port 2, we get output from port 1 only. when input is applied at port 3, we get ouput from port 2 only. hence, scattering matrix is 0 0 s13 S = s21 0 0 0 s32 0
rule of 0 is everythign equalls 0
16-16 = 0
2*2 - 12x +16 =0 4 - 12x +16 =0 -12x + 16 +4 =0 -12x + 20 =0 -12x = -20 x = 1.666666667
Assuming the polynomial is written in terms of "x": It means, what value must "x" have, for the polynomial to evaluate to zero? For example: f(x) = x2 - 5x + 6 has zeros for x = 2, and x = 3. That means that if you replace each "x" in the polynomial with 2, for example, the polynomial evaluates to zero.
There are infinitely many possible answers. y = x - 4 y = x*0 y = x2 - 16 y = |sqrt(x)| - 2 that's 4 of them.
A not gate is a logical gate which inverts a digital signal. If the input to a not gate is 1, then the output will be 0. If the input is 0, then the output will be 1.
No, it does not necessarily mean that the system is linear. A linear system will exhibit a constant scaling property, which means that if the input is multiplied by a constant, the output will also be multiplied by the same constant. It is possible for a system to have an output of zero for a zero input, but still be non-linear if it does not exhibit the scaling property.
In AND gate , if through both the terminals minimum values were send then the output will be 0 . if either one of the input is maximum , then the output will be 0 , meanwhile if both the input were 1 then the output will be 1 .In OR gate , if both the input are 0 then the output will be 0, but if either the input is 1 then the output will be 1,when both the inputs are 1 then the output will be 1 .
Because if input A *and* input B is true, then the output is true! Truth table of AND gate: ┌─┬─╥───────┐ │A│B║Q (Output)│ ├─┼─╫───────┤ │0│0║0..............│ ├─┼─╫───────┤ │0│1║0............. │ ├─┼─╫───────┤ │1│0║0............. │ ├─┼─╫───────┤ │1│1║1............. │ └─┴─╨───────┘
Its truth table is: input output 0 1 1 0
Input Output 0 1 1 0
0 1 2
#include<stdio.h> #include<conio.h> #include<string.h> #include<stdlib.h> struct object_code { int locctr; char add[10]; }obcode[300]; void main() { char input[100][16],output[100][16],binary[20],address[20],stloc[10]; int len,bitmask,loc,tlen=0,tloc,textloc,i=0,location,j,k,count=0,start,n,num=0,inc=0; FILE *fp1,*fp2; clrscr(); fp1=fopen("linput.dat","r"); fp2=fopen("reloadout.dat","w"); printf("Enter the location where the program has to be loaded:"); scanf("%s",stloc); start=atoi(stloc); location=start; tloc=start; fscanf(fp1,"%s",input[i]); while(strcmp(input[i],"T")!=0) { strcpy(output[i],input[i]); i++; fscanf(fp1,"%s",input[i]); strcpy(output[i],input[i]); } itoa(start,output[2],10); while(strcmp(input[i],"E")!=0) { strcpy(output[i],input[i]); if(strcmp(input[i],"T")==0) { for(j=0;j<3;j++) { i++; fscanf(fp1,"%s",input[i]); strcpy(output[i],input[i]); } bitmask=atoi(output[i]); itoa(bitmask,binary,2); strcpy(output[i],NULL); textloc=atoi(output[i-2]); textloc=textloc+start; itoa(textloc,output[i-2],10); for(n=0;n<(textloc-(tloc+tlen));n++) { strcpy(obcode[inc].add,"xx"); obcode[inc++].locctr=location++; } tlen=atoi(output[i-1]); tloc=textloc; k=0; } else { if(binary[k]==1) { num=0; len=strlen(output[i]); strcpy(address,NULL); for(j=2;j<len;j++) { address[num]=output[i][j]; output[i][j]='\0'; num++; } loc=atoi(address); loc=loc+start; itoa(loc,address,10); strcat(output[i],address); } k++; len=strlen(output[i]); num=0; for(n=0;n<len;n++) { obcode[inc].add[num++]=output[i][n]; if(num>1) { obcode[inc++].locctr=location++; num=0; } } } i++; fscanf(fp1,"%s",input[i]); } strcpy(output[i],input[i]); i++; fscanf(fp1,"%s",input[i]); loc=atoi(input[i]); loc=loc+start; strcpy(output[i],itoa(loc,address,10)); count=0; i=0; n=0; fprintf(fp2,"%d\t",obcode[n].locctr); for(n=0;n<inc;n++) { fprintf(fp2,"%s",obcode[n].add); i++; if(i>3) { fprintf(fp2,"\t"); i=0; count++; } if(count>3) { fprintf(fp2,"\n%d\t",obcode[n+1].locctr); count=0; } } getch(); }
if G1 is the output the output is 0. If G1 is the input then you must follow the logic to determine the output.
A, B and C are inputs, Q is output.A B C Q0 0 0 10 0 1 00 1 0 00 1 1 11 0 0 01 0 1 11 1 0 11 1 1 0*update*Although the Logism software shows the 3 input XNOR gate output 1 when all 3 are 1, perhaps there is a mistake, and 3 input of 1,1,1 in an XNOR gateoutputs 0. There is a good reason why. In an XOR gate, an even amount of 1 or 0 input will output a 0, and an odd input of 1 (ie: 1, 1, 1) will output a 1. The truth tablefor an XOR gate with 1, 1, 1 input is an output of 1. An XNOR gate will output the opposite of an XOR gate, thus XNOR input of1, 1, 1 should output 0.
It is a data selector. There are 16 digital input lines, 4-bit decoder, strobe and output pin. So you put a 4-bit binary number from 0-15 into ABCD bits and the corresponding input value is found on output at the strobe time.