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A CPU has only three instructions $I1, I2$ and $I3,$ which use the following signals in time steps $T1-T5$:

$I1 : T1$ : Ain, Bout, Cin
         $T2$ : PCout, Bin
         $T3$ : Zout, Ain
         $T4$ : Bin, Cout
         $T5$ : End

$I2 : T1$ : Cin, Bout, Din
        $T2$ : Aout, Bin
        $T3$ : Zout, Ain
        $T4$ : Bin, Cout
        $T5$ : End

$I3 : T1$ : Din, Aout
        $T2$ : Ain, Bout
        $T3$ : Zout, Ain
        $T4$ : Dout, Ain
        $T5$ : End

Which of the following logic functions will generate the hardwired control for the signal Ain ?

  1. $T1.I1 + T2.I3 + T4.I3 + T3$
  2. $(T1 + T2 + T3).I3 + T1.I1$
  3. $(T1 + T2 ).I1 + (T2 + T4).I3 + T3$
  4. $(T1 + T2 ).I2 + (T1 + T3).I1 + T3$
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We just have to see which all options give $1$ whenever $A_{in}$ is $1$ and $0$ otherwise. 

So, Ain is $1$ in $T3$ of $I1, I2$ and $I3$. Also during $T1$ of $I1$, and $T2$ and $T4$ of $I3$. So, answer will be 

$T1.I1 + T2.I3 + T4.I3 + T3.I1 + T3.I2 + T3.I3$

Since CPU is having only $3$ instructions, $T3.I1 + T3.I2 + T3.I3$ can be replaced with $T3$ (we don't need to see which instruction and $A_{in}$ will be activated in time step $3$ of all the instructions).

So, $T1.I1 + T2.I3 + T4.I3 + T3 $

The answer is Option (A)

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