14.15 Write the truth table for the circuits given in Fig. 14.40 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.
14.15 A acts as two inputs of the NOR gate and Y is the output. As shown in the following figure. Hence the output of the circuit is =
The truth table for the same is given as:
A
Y = ()
0
1
1
0
This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.
A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates are and , as shown in the following figure
Above is given the inputs for the last NOR gate.
Hence, the output for the circuit can be written as:
Y = = = A.B
The truth table for the same can b
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14.15 A acts as two inputs of the NOR gate and Y is the output. As shown in the following figure. Hence the output of the circuit is =
The truth table for the same is given as:
A
Y = ()
0
1
1
0
This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.
A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates are and , as shown in the following figure
Above is given the inputs for the last NOR gate.
Hence, the output for the circuit can be written as:
Y = = = A.B
The truth table for the same can be written as:
A
B
Y(=A!B)
0
0
0
0
1
0
1
0
0
1
1
1
This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.
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<p><strong>14.15 </strong>A acts as two inputs of the NOR gate and Y is the output. As shown in the following figure. Hence the output of the circuit is <span contenteditable="false"> <math> <mover accent="true"> <mrow> <mrow> <mi>A</mi> <mo>+</mo> <mi>A</mi> </mrow> </mrow> <mo>?</mo> </mover> </math> </span>=<span contenteditable="false"> <math> <mover accent="true"> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> <mo>?</mo> </mover> </math> </span></p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728034989phpzKodhy_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728034989phpzKodhy.jpeg" alt="" width="181" height="74"></picture></div></div><p>The truth table for the same is given as:</p><table><tbody><tr><td width="30"><p>A</p></td><td width="66"><p>Y = (<span contenteditable="false"> <math> <mover accent="true"> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> <mo>?</mo> </mover> </math> </span>)</p></td></tr><tr><td width="30"><p>0</p></td><td width="66"><p>1</p></td></tr><tr><td width="30"><p>1</p></td><td width="66"><p>0</p></td></tr></tbody></table><p>This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.</p><p>A and B are the inputs and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates are <span contenteditable="false"> <math> <mover accent="true"> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> <mo>?</mo> </mover> </math> </span>and <span contenteditable="false"> <math> <mover accent="true"> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mo>?</mo> </mover> </math> </span> , as shown in the following figure</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728035154php8IiZgh_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728035154php8IiZgh.jpeg" alt="" width="174" height="90"></picture></div></div><p>Above is given the inputs for the last NOR gate.</p><p>Hence, the output for the circuit can be written as:</p><p>Y = <span contenteditable="false"> <math> <mover accent="true"> <mrow> <mrow> <mi>A</mi> <mo>+</mo> <mi>B</mi> </mrow> </mrow> <mo>?</mo> </mover> </math> </span>= <span contenteditable="false"> <math> <mover accent="true"> <mrow> <mrow> <mover accent="true"> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> <mo>?</mo> </mover> <mo>.</mo> <mover accent="true"> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> <mo>?</mo> </mover> <mi> </mi> </mrow> </mrow> <mo>?</mo> </mover> </math> </span> = A.B</p><p>The truth table for the same can be written as:</p><table><tbody><tr><td width="36"><p>A</p></td><td width="28"><p>B</p></td><td width="60"><p>Y(=A!B)</p></td></tr><tr><td width="36"><p>0</p></td><td width="28"><p>0</p></td><td width="60"><p>0</p></td></tr><tr><td width="36"><p>0</p></td><td width="28"><p>1</p></td><td width="60"><p>0</p></td></tr><tr><td width="36"><p>1</p></td><td width="28"><p>0</p></td><td width="60"><p>0</p></td></tr><tr><td width="36"><p>1</p></td><td width="28"><p>1</p></td><td width="60"><p>1</p></td></tr></tbody></table><p>This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.</p>
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