How Can Multiple ALUs be Connected to Create a Layered Structure?

AI Thread Summary
To connect multiple ALUs in a layered structure, start with four 8-bit ALUs that process inputs in parallel, reducing to two, and finally to one. Each ALU's output can serve as the input for the next, allowing for sequential calculations, while unused inputs can be grounded. The design must consider propagation delays, which can be mitigated by incorporating registers and synchronizing operations with a clock. The specific application will dictate the connections and functionality of the ALUs. Ultimately, this setup enables complex calculations by efficiently combining outputs from multiple ALUs.
sjaguar13
Messages
49
Reaction score
0
If I have an 8-bit ALU, how would I hook them up to do the layer thing (I don't know the name)? It's the thing that you start with 4 ALUs, then they go down to 2, then those go down to 1. Are all the inputs for each of the original 4 seperate, or do they share some inputs? How to two go down to one? The outputs of one are the As of the next and the outputs of the other are the Bs of the next? What about the Cout, Zero, and Set? Do they somehow go into the next?
 
Physics news on Phys.org
What exactly are you trying to do here? You want to feed the output of two ALUs to one ALU and feed that to the remaining ALU? What's the spec. for these ALUs?
 
The connections between the ALUs will depend on the application you are using them for. What do you want to do with them?
 
I want it to act in parallel. Looking at the 7 8-bit ALUs, I want it to be able to take 4 sets of numbers and get them down to 2 sets. Then, take those 2 sets and get it down to one. If I had 3 and 4, 3 and 5, 2 and 2, 1 and 2, and I select add, I want the first ALU to add 3 and 4 to get 7. Put the 7 into the next ALU. 3 and 5 get added for 8 and it goes down to the next ALU with the 7 for 15. 15 goes to the last ALU. 2 and 2 is 4 and that goes to the ALU along with 3 from 1 and 2. 4 and 3 is 7 and that goes to the last ALU with the 15 for 22, which is the answer.

If the answer from one becomes the input of the next, what about the Cout, Not B, and basically the other pins do?
 
I see. If you don't need the other inputs, don't use them (set them to ground or something). Simple as that. In real life however, your parallel setup would suffer from propagation delays and whatnot. You can get rid of that however by putting in some registers and have everything synchronized to a clock.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top