Could you please solve #1 parts A & B. I am not really sure what I am doing for

Question

Could you please solve #1 parts A & B. I am not really sure what I am doing for this.

Safari File Edit View History Bookmarks Window Help , 879 Tue 9:31 AM a 0 myasucourses.asu.edu Blackboard Learn https://myasucourses.asu.edu/bbcswebdav/pid-1... Blackboard Learn CSE100 The objective of this homework is to practice your knowledge on some simple Verilog coding. 1. Consider the following Verilog Module 3 module prbl (a,b,c,d,e,f,); 5 input [3:0] a, b, c output [7:0] d, e, f; 7 8 wire [3:0] g; 9 wire [1:0] h; 10 b); assign g = ~(a assign h= b[2:1] 12 13 14 15 16 17 18 assign d= assign e assign f= ((2(h)),c); a & (b | c); {g2,c

Explanation / Answer

a)

first erroe is at line 3 there is a comma after f but it should not be there.

error free line is : module prb1(a,b,c,d,e,f);

second error is at line no. 12 here semicolon is missing at the end of line

error free line : assign h = b[2:1];

b)

a = 0111

b = 0110

c = 0101

d = {{2{h}},c}

h = b[2:1] = 2a nd 1 bit of b = 11

so h = 11

2{h} is replication operator which is 2 times h = 1111

{{2{h}},c} is conctenation operator which is conctenation of 2{h} and c = 11110101

so,d = 11110101

e = a&(b|c)

b|c is nothing but bitwise or of b and c

which is equal to 0110|0101 = 0111

a&(b|c) is bitwise and of (bitwise or if b and c)

which is equal to 0111 & 0111 = 0111

so, e = 0111

f = {g<<3,b>>2,c<<1}

g = ~(a^b)

a^b is bitwise xor of 0111 and 0110 = 0001

so g = 0001

<< is shift left operator.

g<<3 euqals to 3 times left shift of g

we have g = 0001

first shift g = 0010

simillarly after 3 shift g = 1000

b>>2,

>> is right shift

so after first shift b = 1100 and after second shift b = 1000

c<<1 is left shift of c

c<<1 = 1010

and,

f{g<<3,b>>2,c<<1}

is concatenation of shifted variables.

but g,b and c is 4 bit variable and f is 8 bit so it may give concatenation error.

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