Digital Electronics Test 1

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Digital Electronics Test 1

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Question 1
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A certain microprocessor based system has a memory space of 2²⁰ words where each word has a capacity of 24 bits. The number of memory chips required to construct the given memory space is ______. [Consider each memory chip is having 14 address lines & 8 data lines]

A
192-192

B
1924-1921

C
1927-1926

D
1929-1922

Solution

Memory space = 2²⁰ × 24
Memory chip capacity = 2¹⁴ × 8
No. of memory chips \(= \frac{{{2^{20}} \times 24}}{{{2^{14}} \times 8}} = 192\)
Question 2
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The output voltage of a 5-bit D/A binary ladder has digital input of 11011. Assuming all zeros voltage 0 V and all ‘1’ is voltage as 10 V, the analog output of the DAC is _______ V.

A
8.5-9

B
8.54-91

C
8.57-96

D
8.59-92

Solution

\({V_0} = k\left[ {{2^{n - 1}}{b_{n - 1}} + {2^{n - 2}}{b_{n - 2}} + \ldots + 2{b_1} + {b_0}} \right]\)
b_nb_n-1...b₁b₀ are digits of digital input
V_FS= 10 V, n = 5
⇒ 10 = K [31]
K = 10/31
For Digital input (11011)₂
\({V_0} = \frac{{10}}{{31}}\left[ {27} \right]\)
= 8.71 V
Question 3
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Let us consider the following equation in a 6-bit binary number system X = A + B.
A is given as (001010)₂ in 1's complement binary number system, B is given as (111010)₂ in signed number system. What would be X in the 2’s complement number system?

A
(110000)₂

B
(010000)₂

C
(101011)₂

D
(110101)₂

Solution

A in 1’s complement binary number system = (001010)₂
As the MSB is ‘0’, the number is positive.
Therefore, the number A = (001010)₂ = (10)₁₀
B in signed number system
B = (111010)₂
As the MSB is ‘1’, the number is negative.
Therefore the number B = (-26)₁₀
Now, X = A + B = (10)₁₀ + (-26)₁₀ = (-16)₁₀
The two’s complement representation of (-16)₁₀ is the two’s complement of (+16)₁₀
(+16)₁₀ = 010000
2’s complement = 110000
X in 2’s complement representation = (110000)₂