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?
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!
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.
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.
homeworktag as it's deprecated, so I rolled back to that version. – stevenvh Sep 22 '11 at 15:50