Digital Electronics Test 1

Concept:

A quadratic equation with factor α and β can be written as:

x² – (α + β) x + (αβ) = 0

Application:

Given the quadratic equation:

x² – 11x + 22 = 0         ---(1)

Also, the solutions to the given quadratic equation are 3 and 6.

∴ We can write the quadratic equation as:

x² – (6 + 3) x + (6 × 3) = 0       ---(2)

Comparing the two-equations, we can write:

(6)_b + (3)_b = (11)_b i.e. for some base ‘b’, this equation must hold true.

Converting both the RHS and LHS to their respective decimal equivalent, we can write:

(6 × b⁰) + (3 × b⁰) = (1 × b¹ + 1 × b⁰)

6 + 3 = b + 1

b = 8

∴ The base of the number is 8 that satisfies the above quadratic equation.

Concept:

Δt = NT_c

Where,

Δt is the total delay required

N is number of flip flops

T_c is the clock time cycle

Calculation:

Given that,

Δt = 0.32 ms

Clock frequency, f_c = 1.25 MHz

\({T_c} = \frac{1}{{{f_c}}} = \frac{1}{{1.25\; \times \;{{10}^6}}} = 0.8\;\mu sec\) 

\(N = \frac{{{\rm{\Delta }}t}}{{{T_c}}} = \frac{{0.32\; \times \;{{10}^{ - 3}}}}{{0.8\; \times\; {{10}^{ - 6}}}} = 400\)

We know that,

V_in T₁ = V_ref T₂

⇒ 120 × 10^-3 × 50 × 10^-3 = 100 × 10^-3 × T₂

⇒ T₂ = 60 ms

Concept:

1’s complement representation of a binary number is obtained by toggling all the bits, i.e. replacing 1 with 0, and 0 with 1.

2’s complement representation of a binary number is obtained by adding 1 to the 1’s complement representation.

Application:

(127)₁₀ = (01111111)₂

1’s complement representation will be:

1’s complement = 10000000

Number of 1’s is the 1’s complement is, n = 1

Now, the two (2’s) complement representation will be:

2’s complement = 10000000 + 1

= 10000001

Number of 1’s in 2’s complement is, m = 2

∴ The required ratio is m : n = 2 : 1

\(\begin{array}{l}\left( {\bar A + \bar B} \right)\left[ {\overline {A\left( {B + C} \right)} } \right] + A\left( {\bar B + \bar C} \right)\\= \left( {\bar A + \bar B} \right)\left[ {\bar A + \left( {\overline {B + C} } \right)} \right] + A\left( {\bar B + \bar C} \right)\end{array}\)

\(\begin{array}{l}= \left( {\bar A + \bar B} \right)\left[ {\bar A + \bar B\bar C} \right] + A\bar B + A\bar C\\= \bar A + \bar A\bar B\bar C + \bar A\bar B + \bar B\bar C + A\bar B + A\bar C\end{array}\)

\(\begin{array}{l}= \bar A\left( {1 + \bar B\bar C + \bar A\bar B} \right) + \bar B\bar C + A\bar B + A\bar C\\= \bar A + \bar B\bar C + A\bar B + A\bar C\end{array}\)

\(\begin{array}{l}= \bar A + A\bar B + A\bar C\\= \bar A + A\left( {\bar B + \bar C} \right)\end{array}\)

\(\begin{array}{l}= \bar A + \bar B + \bar C\\= \overline {ABC} \end{array}\)

The given Boolean expression represents NAND gate.

Given, contents A = AA H

CPI 99 H:

Compares the data 99 H with contents of A, the compression is made by subtracting 99 H from contents of A (i.e. AA H). the contents of A remains unaffected but status of result is reflected by Flags.

\(\begin{array}{*{20}{c}}{A;}&{10101010}\\{99\;H;}&{\begin{array}{*{20}{c}}{\underline {10011001} }\\{000010001}\end{array}}\end{array}\) 

After comparison both zero flag and carry flag are reset.

XRA A:

The instruction XRA A performs ExOR operation on contents of A with contents of A only and stores the result in A. The instruction clears the contents of A. So, the zero flag is set by the instruction.

LXI B, 0007H:

Loads BC pair with 16-bit data 0007 H. So, contents of BC pair become 0007 H.

LOOP: DCX B:

Decrements the contents of BC pair by one.

JNZ LOOP:

Jumps the execution of a program to LOOP if zero flag is not set.

In the above program zero flag is set when XRA A is executed and the status of zero flag remains unaffected by DCX B instruction as it does not affect the flags. So, the execution comes out of LOOP during first instance itself as Z flag is already set. So, the loop runs only once.

The truth table is shown for the given information.

Decimal	Binary				Gray				Gray [(n + 1) mod 16]
					h3	h2	h1	h0	g3	g2	g1	g0
0	0	0	0	0	0	0	0	0	0	0	0	1
1	0	0	0	1	0	0	0	1	0	0	1	1
2	0	0	1	0	0	0	1	1	0	0	1	0
3	0	0	1	1	0	0	1	0	0	1	1	0
4	0	1	0	0	0	1	1	0	0	1	1	1
5	0	1	0	1	0	1	1	1	0	1	0	1
6	0	1	1	0	0	1	0	1	0	1	0	0
7	0	1	1	1	0	1	0	0	1	1	0	0
8	1	0	0	0	1	1	0	0	1	1	0	1
9	1	0	0	1	1	1	0	1	1	1	1	1
10	1	Question 9 2 / -0.33 A certain 8-bit DAC has a full scale output of 2mA and full scale error of ± 0.5 % full scale reading. The maximum possible output current for an input of 10000000 is__________ μA. A 1012-1016 B 10124-10161 C 10127-10166 D 10129-10162 Solution Step size \(= \frac{{2mA}}{{{2^8} - 1}} = \frac{{2mA}}{{255}}\) = 7.84 μA. 10000000 = 12810 Ideal output = 128 × 7.84 μA = 1004 μA Error = ± 0.5% × 2mA = ± 10 μA Range of output = 1004μA ± 10 μA Maximum possible output : 1014 μA User Question Analysis Correct - Wrong - Skipped - 1 2 3 4 5 6 7 8 9 My Perfomance Score - out of - Rank - out of - Re-Attempt Solution Weekly Quiz Competition OK Login Start Test Resend Email Already Verified Get latest Exam Updates & Study Material Alerts! No, Thanks Allow Click on Allow to receive notifications Allow Notification To enable notifications follow this 2 steps: First Click on Secure Icon Second click on the toggle icon Allow Notification Get latest Exam Updates & FREE Study Material Alerts! × Open Now Sign up to your account Log in to your account "Log in, submit answer and win prizes" Sign Up with Google Log in with Google New to Selfstudy? Sign Up Already have an account? Log in By clicking Log in/Sign Up, you agree to our Terms of Use & our Privacy Policy. Get personalized exam alerts & new study material made just for you "Enter your details to claim your prize if you win!" Select Class* Select Stream Verify Send OTP Processing, please wait... Select Board Select State Select City I agree to the Terms of Use & Privacy Policy Save Save Sign up to your account Log in to your account Varify Mobile Number Unlimited Access to PDFs: Provide prime members with unlimited access to all study materials in PDF format. Multiple Attempts in MCQ Tests: Allow prime members to attempt MCQ tests multiple times to enhance their learning and understanding. Exclusive Premium Content Access. Provide prime users with access to exclusive PDF study materials that are not available to regular users. Verify your mobile number to become FREE PRIME MEMBER. Send OTP Already have an account Log in New to Selfstudy? Sign Up By Sending OTP, you agree to our Terms of Use & our Privacy Policy. Varify Mobile Number Please enter verification code sent to Didn't Recieve the code? Resend Verify By clicking Verify, you agree to our Terms of Use & our Privacy Policy. Your Answer Has Been Successfully Submitted! Thank you for participating in Quiz Time! Results Announcement on: coming Monday Stay Tuned: We’ll update you on your result via email or SMS. Good luck, and stay tuned for exciting rewards! Continue Just to Confirm! Your Selected Class: Your Selected Stream: Edit Save