Sketch the output $Y$ from a $NAND$ gate having inputs $A$ and $B$ as shown in the figure.
(N/A) The truth table for a $NAND$ gate is:
If $A=0, B=0$,then $Y=1$.
If $A=0, B=1$,then $Y=1$.
If $A=1, B=0$,then $Y=1$.
If $A=1, B=1$,then $Y=0$.
Based on the input waveforms:
For $t < t_{1}$; $A=1, B=1$; hence $Y=0$.
For $t_{1}$ to $t_{2}$; $A=0, B=0$; hence $Y=1$.
For $t_{2}$ to $t_{3}$; $A=0, B=1$; hence $Y=1$.
For $t_{3}$ to $t_{4}$; $A=1, B=0$; hence $Y=1$.
For $t_{4}$ to $t_{5}$; $A=1, B=1$; hence $Y=0$.
For $t_{5}$ to $t_{6}$; $A=0, B=0$; hence $Y=1$.
For $t > t_{6}$; $A=1, B=1$; hence $Y=0$.
If $A=0, B=0$,then $Y=1$.
If $A=0, B=1$,then $Y=1$.
If $A=1, B=0$,then $Y=1$.
If $A=1, B=1$,then $Y=0$.
Based on the input waveforms:
For $t < t_{1}$; $A=1, B=1$; hence $Y=0$.
For $t_{1}$ to $t_{2}$; $A=0, B=0$; hence $Y=1$.
For $t_{2}$ to $t_{3}$; $A=0, B=1$; hence $Y=1$.
For $t_{3}$ to $t_{4}$; $A=1, B=0$; hence $Y=1$.
For $t_{4}$ to $t_{5}$; $A=1, B=1$; hence $Y=0$.
For $t_{5}$ to $t_{6}$; $A=0, B=0$; hence $Y=1$.
For $t > t_{6}$; $A=1, B=1$; hence $Y=0$.
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