Question on active low and active high logic

[BlueCerealBox]

Hi ,

I have a homework question which asks me to find the boolean equation of X.L in the image attached , the solutions are also attached. But I'm rather confused with all the active low and active high signals.

Why does it say that /Z1 = /A/B/C/D ? Shouldn't not /A/B/C/D be Z1 instead ? and /Z1 be /(/A/B/C/D) ?

I am aware that X.L = /X . H

I am seriously confused. Hoping to seek clarification.

Thanks!

 

[Harald Kapp]

I do understand your confusion - I'm confused, too.

It should be either
/Z0 = /A*/B*/C*/D
or
/Z1 = /A*/B*C*/D - this is more likely because the equation also states /Z1=/A2*/A1*A0*E3*/E2*/E1

THe notation /Z only signals that Z is active low, meaning it is low if the condition of the equation is fulfilled, high otherwise.

The notation X.L or X.H I do not recognize.

 

[BlueCerealBox]

I went and thought about it for a bit and came up with this :

Assume you connect signal B.H to a NOT gate , you would get /B.H at the output of the NOT gate. Condensing this simple connection would give you a signal B.L

That is what they're probably doing at the output. The NOT gate connected to Z1 could be condensed into Z1.L , which would explain why X.L = Z1.Z3.Z4.Z7 , if you were to expand the signal X.L at the output of the NAND gate , the 2 NOT gates would cancel out , leaving you with a NAND gate.

I'm unsure of the validity of this theory though.

 

[Harald Kapp]

I cannot comment on this as I do not recognize the notation X.L or X.H. Sorry. Maybe someone else can?

 

[BlueCerealBox]

X.L stands for active low signal output X and X.H stands for active high output X.

 

[Harald Kapp]

I'd be using /X for a low active signal and X for a high active signal. For me this evokes considerably less confusion since I can directly insert this notation into Boolean equations.

In the light of your last reply, this statement or yours

Assume you connect signal B.H to a NOT gate , you would get /B.H at the output of the NOT gate. Condensing this simple connection would give you a signal B.L

seems completely logical.