DSD and HDL Simulation, Assignment Questions

Problem Give implementation of 3 bit UP counter using ACT3 LMs Solution:

PS NS

𝑨 𝑩 𝑪 𝑨 𝑩 𝑪

0 0 0 0 0 1

0 0 1 0 1 0

0 1 0 0 1 1

0 1 1 1 0 0

1 0 0 1 0 1

1 0 1 1 1 0

1 1 0 1 1 1

1 1 1 0 0 0

Design equations are as follows:

𝐴 = 𝐴 𝐵 𝐶 + 𝐴 𝐶 + 𝐴 𝐵

𝐵 = 𝐵 𝐶 + 𝐵 𝐶

𝐶 = 𝐶

Consider the first equation

𝐴 = 𝐴 𝐵 𝐶 + 𝐴 𝐶 + 𝐴 𝐵

Using Shannon’s expansion theorem

𝐴 = 𝐴 𝐵 (0) + 𝐴 𝐵 (𝐶 ) + 𝐴 𝐵 (1) + 𝐴 𝐵 (𝐶 )

Consider the second equation

𝐵 = 𝐵 𝐶 + 𝐵 𝐶 Using Shannon’s expansion theorem

𝐵 = 𝐵 𝐶 (0) + 𝐵 𝐶 (1) + 𝐵 𝐶 (1) + 𝐵 𝐶 (0)

Similarly, implement the third equation

-------------------------------------------------------------------------------------------------------------------------------

Problem Give implementations of 3 bit Johnson (or twisted ring) counter using ACT2 LM Solution:

PS NS

𝑨 𝑩 𝑪 𝑨 𝑩 𝑪

0 0 0 1 0 0

1 0 0 1 1 0

1 1 0 1 1 1

1 1 1 0 1 1

0 1 1 0 0 1

0 0 1 0 0 0

Design equations are

𝑨 = 𝐶 ; 𝑩 = 𝐴 ; 𝑪 = 𝐵

3 ACT2- S LMs are required

Similarly, implement Bn+1, Cn+1

Problem Give implementation of SR latch using ACT1 series Solution: Consider the TT

S R Operation

X 1 0 Set 1

X 0 1 Reset 0

0 0 0

Last state 0

1 0 0 1 X 1 1 Forbidden X

X-Don’t care

Reduced implicant table is shown to be

S R

X 1 X 1

1 X 0 1

= +

Using Shannon’s expansion theorem

= ( ) + ( )

= ( ) + ( )

Problem Give implementation of 3 bit binary- to-gray code converter using ACT1 series Solution:

Binary input Gray code output

𝑩

MSB

𝑩 𝑩 𝑮

MSB

𝑮 𝑮

0 0 0 0 0 0

0 0 1 0 0 1

0 1 0 0 1 1

0 1 1 0 1 0

1 0 0 1 1 0

1 0 1 1 1 1

1 1 0 1 0 1

1 1 1 1 0 0

Note that gray code satisfies unit distance property, reflective and cyclic property.

Considering K-map, it can be shown that

= 𝐵

= 𝐵 𝐵

= 𝐵 𝐵

2 ACT1 LMs are sufficient to implement 3 bit binary- to- gray code converter

Problem Give implementation of 3 bit gray- to-binary code converter using ACT1 series Solution:

Gray code input Binary output

𝑮

MSB

𝑮 𝑮 𝑩

MSB

𝑩 𝑩

0 0 0 0 0 0

0 0 1 0 0 1

0 1 1 0 1 0

0 1 0 0 1 1

1 1 0 1 0 0

1 1 1 1 0 1

1 0 1 1 1 0

1 0 0 1 1 1

Considering K-map, it can be shown that 𝐵 = 𝐵 = 𝐵 = These equations can be further simplified as given below (so as to reduce the number of LMs required): 𝐵 = 𝐵 = 𝐵 𝐵 = 𝐵 2 ACT1 LMs are sufficient to implement 3 bit gray- to-binary code converter

Transparent low latch

If C = 0, Q = D then it is known as transparent low latch as shown.

Transparent high latch

If C = 1, Q = D then it is known as transparent high latch.

Dr. D. V. Kamath Professor, Dept. of E&C Engg., MIT