Advanced Microprocessors
- How many NAND gates are required to make AB + BA?
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Y = AB + AB
This means, half adder requires 4 NAND gates. Hence alternative (C) is the correct answer.Correct Option: C
Y = AB + AB
This means, half adder requires 4 NAND gates. Hence alternative (C) is the correct answer.
- The 2 × 1 MUX having switching delay of 1μ sec connected as shown:
The output is connected to its own select switch. The output will be—
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Here switching delay means after 1 μ sec.,the switch is moved between the point A and B.
Hence, Time period = 2μ sec
∴Frequency = 1 = 0.5 MHz 2 μsec
Hence alternative (B) is the correct answer.Correct Option: B
Here switching delay means after 1 μ sec.,the switch is moved between the point A and B.
Hence, Time period = 2μ sec
∴Frequency = 1 = 0.5 MHz 2 μsec
Hence alternative (B) is the correct answer.
- Write output—
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F1 = (AB)′
F2 = (BA)′
Y = F1F2 = [( AB)′ ( BA)′]′
= [(A + B′) ( B + A)] ′ = AB + BA
Alternative Method:
NAND-NAND circuit can be replaced by AND-OR Hence,
Now, F1 = AB,
F2 = BA
Y=F1 + F2 = AB + BA (this is shortcut method).Correct Option: D
F1 = (AB)′
F2 = (BA)′
Y = F1F2 = [( AB)′ ( BA)′]′
= [(A + B′) ( B + A)] ′ = AB + BA
Alternative Method:
NAND-NAND circuit can be replaced by AND-OR Hence,
Now, F1 = AB,
F2 = BA
Y=F1 + F2 = AB + BA (this is shortcut method).
- The number of bits needed to encode all letters (i.e., 26), 10 symbols and all number (i.e.,10) is—
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⇒ n = 6, as 26 = 64
if we take n = 5, then 25 = 32 which will not encode 46 number, hence n = 6 taken.Correct Option: C
⇒ n = 6, as 26 = 64
if we take n = 5, then 25 = 32 which will not encode 46 number, hence n = 6 taken.
- The switching speed of ECL is very high, because—
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Refer synopsis
Correct Option: C
Refer synopsis