We want to use a ROM memory to implement a digital circuit that receives two 2-bit numbers X and Y (X=x1x0, Y=y1y0) and detects whether X is greater (G=1), lower (L=1) or equal (E=1) to Y. What information must be stored in the memory address words 0101 and 1100?
Note: The most significant bit of the address bus is x1.
We want to use a ROM memory to implement a digital circuit that receives two 2-bit numbers X and Y (X=x1x0, Y=y1y0) and detects whether X is greater (G=1), lower (L=1) or equal (E=1) to Y. What information must be stored in the memory address words 1001 and 1111?
What information must be stored in the memory address words 0001 and 1100
Please also provide explanation. Will be grateful for that. :)
Oh, so we're diving into the world of digital circuits, huh? Well, let's see...
In this case, since we're dealing with two 2-bit numbers, X and Y, the memory address words 0101 and 1100 would likely store the output values for the three conditions: greater, lower, and equal.
So, in address word 0101, we would probably want to store the output value for when X is greater than Y. Let's call it "G" for greater. And in address word 1100, we would want to store the output value for when X is equal to Y. We'll call that "E" for equal.
Now, the trick here is that you didn't mention what should be stored in the memory address word when X is lower than Y. Maybe we should put a little note there saying "Sorry, not enough bits in this ROM for that condition. Yikes!"
But hey, don't worry, it's all part of the fun of working with limited resources. It keeps life interesting, right? So, go ahead and pop those values into the memory address words and let's see what happens when we feed some inputs into this digital circus... I mean, circuit!
To determine whether X is greater, lower, or equal to Y using a ROM memory, we need to store the corresponding results in the memory address words 0101 and 1100.
Let's break down the problem step by step:
1. Determine the number of bits required to represent all possible values of X and Y. In this case, both X and Y are 2-bit numbers, so we need 2 bits to represent each of them.
2. Understand how the address bus is connected. The most significant bit of the address bus is x1; this means that the address bus has two bits, x1 and x0.
3. Determine the number of possible address combinations. Since we have two bits in the address bus, there are four possible combinations: 00, 01, 10, and 11.
4. Define the output for each address combination. In this case, we want to detect whether X is greater (G=1), lower (L=1), or equal (E=1) to Y. We need to store these results in the memory for each address combination.
Now, let's determine the information that must be stored in the memory address words 0101 and 1100.
For address word 0101:
- The most significant bit (x1) of the address is 0, indicating that X is less than Y.
- The least significant bit (x0) of the address is 1, which means X has a value of 1.
Since X is less than Y, the output (G=1, L=0, E=0) should be (0, 1, 0). Therefore, the information stored in the memory address word 0101 is (0, 1, 0).
For address word 1100:
- The most significant bit (x1) of the address is 1, indicating that X is greater than Y.
- The least significant bit (x0) of the address is 0, which means X has a value of 2.
Since X is greater than Y, the output (G=1, L=0, E=0) should be (1, 0, 0). Therefore, the information stored in the memory address word 1100 is (1, 0, 0).
To summarize:
- Memory address word 0101: (0, 1, 0)
- Memory address word 1100: (1, 0, 0)