VIPSolutionsIN STEP 1 GIVE THE INTRODUCTION OF THE CONCEPT AND GIVE ANSWER FOR EACH PART OF THE QUESTION IN EACH DIFFERENT STEP WITH CLEAR EXPLANATION AND IN THE FINAL STEP GIVE THE WHOLE FINAL ANSWER IN JUST VERY FEW SENTENCES AND MOREOVER I NEED COMPLETE AND CLEAR ANSWER at last explain what we did in each step in just few sentencesThe ROM shown below must be programmed to perform the multiplication of two 8-bit numbers, X and Y. Determine the number, in decimal, that must stored in the ROM at memory address BSC3H. Xo — A D, x, A D, x, A, D, Xs A, D, X A x, —|a ROM oy X, — A, o. X, A, I d Yo As D, AA A, Dy Y, As Dy Ys Au Dy — Y, A, D,, Ys Ass Dy; Ys Aw Dy Y As Dy Ans = 35,295.
Question:
IN STEP 1 GIVE THE INTRODUCTION OF THE CONCEPT AND GIVE ANSWER FOR EACH PART OF THE QUESTION IN EACH DIFFERENT STEP WITH CLEAR EXPLANATION AND IN THE FINAL STEP GIVE THE WHOLE FINAL ANSWER IN JUST VERY FEW SENTENCES AND MOREOVER I NEED COMPLETE AND CLEAR ANSWER at last explain what we did in each step in just few sentences
The ROM shown below must be programmed to perform the multiplication of two 8-bit numbers, X and Y.
Determine the number, in decimal, that must stored in the ROM at memory address BSC3H.
Xo — A D,
x, A D,
x, A, D,
Xs A, D,
X A
x, —|a ROM oy
X, — A, o.
X, A, I
d
Yo As D,
AA A, Dy
Y, As Dy
Ys Au Dy —
Y, A, D,,
Ys Ass Dy;
Ys Aw Dy
Y As Dy
Ans = 35,295.
This is AI generated Response.
Asked by: IMMANDI KISHORE
Created at: 2025-05-26 11:14:25
Asked by: IMMANDI KISHORE
Created at: 2025-05-26 11:14:25
Final Answer:
Rendered
Raw Text
Did you like this answer?
Let's solve the problem step by step, as you requested:
---
### **Step 1: Introduction to the Concept**
**Concept:**
A ROM (Read Only Memory) can be programmed to perform specific logic functions. In this problem, the ROM is designed to compute the multiplication of two 8-bit numbers, X and Y. The two numbers are supplied as inputs to the ROM through 16 address lines (`A0` to `A15`). The ROM output lines (`D0` to `D15`) give the result of the multiplication as a 16-bit output.
---
### **Step 2: Understanding Memory Address**
The memory address provided in the question is **B5C3H** (hexadecimal).
- In binary:
- B = 1011
- 5 = 0101
- C = 1100
- 3 = 0011
- Combine these:
- B5C3 (hex) = 1011 0101 1100 0011 (binary)
---
### **Step 3: Mapping Address Lines to Inputs X and Y**
- The address bus is 16 bits wide: `A15` to `A0`.
- From the diagram:
- `A0`–`A7` = X (8 bits)
- `A8`–`A15` = Y (8 bits)
So, in address B5C3H (1011 0101 1100 0011):
- Lower 8 bits (`A0`–`A7`): **1100 0011** (C3 in hex, 195 in decimal) → **X**
- Upper 8 bits (`A8`–`A15`): **1011 0101** (B5 in hex, 181 in decimal) → **Y**
---
### **Step 4: Calculating the Output**
The ROM is programmed such that the output at address B5C3H is X × Y.
- X = 195
- Y = 181
- X × Y = 195 × 181 = **35,295**
---
### **Step 5: Final Answer (in few sentences)**
The number that must be stored in the ROM at memory address B5C3H is **35,295** (in decimal).
---
### **Summary of Steps**
1. **Introduction:** Explained that the ROM is used to multiply two 8-bit numbers using their values as address inputs.
2. **Address Decoding:** Converted B5C3H to binary.
3. **Input Mapping:** Mapped the address bits to X and Y.
4. **Calculation:** Multiplied the decoded X and Y values.
5. **Final Answer:** Provided the result and summarized the process.
If you have more questions or need clarification, feel free to ask!
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :