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I am trying to understand why the calculating the last delay is \$max(t,mt)+2t\$. I suppose thats derived from \$max(max(t,mt), t)+t\$? But how do you extract the \$t\$ from \$max(..., t)\$ out.

UPDATE

I also dont get the next part ...

whats m? and how is \$S_i=max(t,mt)+t\$ and \$C_{i+1}=max(t,mt)+2t\$ simplified to \$2t\$ and \$3t\$ respectively?

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  • \$\begingroup\$ It's about the race between X/Y '0/0' inputs and C 'mt' \$\endgroup\$ – kenny Sep 22 '11 at 11:17
  • \$\begingroup\$ @kenny, what about them? I don't understand ... I also updated my question \$\endgroup\$ – Jiew Meng Sep 22 '11 at 13:01
  • \$\begingroup\$ max(x,y) takes the maximum of x or y and it's algebra \$\endgroup\$ – kenny Sep 22 '11 at 13:16
  • \$\begingroup\$ oh yes, I understand that part. What I dont understand is how I get \$C_{i+1}=max(t,mt)+2t\$. What I got, from my own understanding is \$C_{i+1}=max(max(t,mt)+t, mt)+t\$. How do I simplify down to \$C_{i+1}=max(t,mt)+2t\$? Also I dont get whats \$m\$ in the update. How do I simplify from \$C_{i+1}=max(t,mt)+2t\$ to \$C_5=9t\$ for example? I may know \$i\$ but not \$t\$? \$\endgroup\$ – Jiew Meng Sep 22 '11 at 14:11
  • \$\begingroup\$ @Brian Carlton - I previously removed the homework tag as it's deprecated, so I rolled back to that version. \$\endgroup\$ – stevenvh Sep 22 '11 at 15:50
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The answer to the first question is that if you take max(max(t,mt)+t,mt)+t, you can bring the t in the inner brackets, obtaining max(t+t, mt+t): since the latter is always bigger than mt (unless you have a negative t) you can simplify with max(t+t, mt+t)+t and then take out again the +t and obtain the value shown.

For the second part, m is again the coefficient that multiplies t: let's take your case.

m=7, so max(t, 7t)+2t=9t as shown.

It's easy!

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With the 'mt' for the C input, this just means that the input isn't ready until mt time after the X and Y inputs. Each gate adds propagation delay t, so this just tells you that Cin was produced by m gates.

About those two gates that produce Cout, and the nested max() -

max(max(mt,t), t) reduces to max(mt,t) :

let f(m,t) = max(max(mt,t),t)
let g(m,t) = max(mt,t)

if mt > t, then
f(mt,t) equals max(max(mt,t), t) equals max(mt, t) equals mt
g(mt,t) equals max(mt, t) equals mt

if mt <= t, then
f(m,t) equals max(max(mt,t), t) equals max(t, t), equals t
g(m,t) equals max(mt, t) equals t

So, for any m and t, functions f() and g() are equivalent. I.e., max(max(mt,t),t) reduces to max(mt,t) !

Now, with that out of the way, you should see that the t wasn't extracted from the max() expression; that t delay was incurred by the rightmost AND gate. It was the inputs to that AND gate that were ready at time max(mt, t), and its output was ready t units after that, for max(mt,t) + t. Likewise, the OR that produces the C output adds another t of propagation delay, giving you the max(mt,t) + 2t result.

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You said you have,

Ci+1=max(max(t,mt)+t,mt)+t

Start as far in as possible:

max(t,mt)+t = mt+t

And make your way outward.

max(max(t,mt)+t,mt) => max(mt+t,mt) = mt+t

mt+t +t = mt + 2t

Ci+1 = mt + 2t

Where you could of course retain what you have as the solution:

Ci+1 = max(t,mt) + 2t

But after substitution, max(t, mt) will always evaluate to mt. I have never done this sort of thing before with circuit delays, but I assumed mt was just a constant scaled t. As such, just carry out the algebra where max(a,b) returns the max of the two arguments.

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