Digital Electronics 2018 Semester 2 Full Solution Pdf Download Paper CC-3 (IT) 2018 Group B
Digital Electronics 2018 Semester 2 Full Solution Pdf Download Paper CC-3 (IT) 2018 Group B, Previous Semester Question Paper Solution 2018,
Hello friends, through this post, I am sharing you Semester Two BSc IT Digital Electronics Complete Solution, in which some solution has been uploaded now and some solution which remains will be uploaded very soon, we expect from you. We hope that you will see this solution completely and also share it with your friends and try to get good marks in your exam. Best of luck for your exam.
नमस्कार दोस्तों, इस पोस्ट के माध्यम से मैं आपको सेमेस्टर दो बीएससी आईटी डिजिटल इलेक्ट्रॉनिक्स पूर्ण समाधान साझा कर रहा हूं, जिसमें कुछ समाधान अभी अपलोड किए गए हैं और कुछ समाधान जो शेष हैं, वे बहुत जल्द अपलोड होंगे, हम आपसे उम्मीद करते हैं। हम उम्मीद करते हैं कि आप इस समाधान को पूरा देखेंगे और इसे अपने दोस्तों के साथ भी शेयर करेंगे और अपनी परीक्षा में अच्छे अंक प्राप्त करने का प्रयास करेंगे। आपकी परीक्षा के लिए शुभकामनाएँ।
Friend, I would like to suggest you that whenever you are looking for someone's solution, there is a handwriting note along with the solution, look at it carefully below and do not copy it, rather prepare a new note and write it. Read it will be best for you, by the way, if you want, you can study from here too, there is no problem, if you have any kind of question, then you can share it with us in the comment section so that other students can also be helped. can thank you
मित्र मैं आपको सुझाव देना चाहूंगा कि जब भी आप किसी का समाधान ढूंढ रहे हों तो समाधान के साथ एक हैंडराइटिंग नोट भी हो तो उसे नीचे ध्यान से देखें और उसकी नकल न करें बल्कि एक नया नोट तैयार करके लिखें। पढ़ना आपके लिए सबसे अच्छा रहेगा, वैसे आप चाहें तो यहां से भी पढ़ाई कर सकते हैं, कोई समस्या नहीं है, अगर आपके मन में किसी तरह का कोई सवाल है तो आप उसे हमारे साथ कमेंट सेक्शन में शेयर कर सकते हैं ताकि अन्य छात्रों की भी मदद ली जा सकती है। धन्यवाद कर सकते हैं
GROUP 'A' Solution Available Here
KOLHAN UNIVERSITY PREVIOUS SEMESTER QUESTION
UG Semester 2 2018
BSC.IT CC-3 (DIGITAL ELECTRONICS)
GROUP 'B'
Answer Any Four Question
2.Prove De-Morgan's Law Using Truth Table ?
Ans: De-Morgan's First Law : The first theorem of De Morgan's Law defines that the inverted result from AND Operation is the same as the OR operation of a complement of each variable where the Result equal NAND Operation.
De-Morgan's First Law Formula : (X•Y)' = X' + Y'
De-Morgan's First Law Truth Table
| X | Y | X' | Y' | X•Y | (X•Y)' | X'+Y' |
|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 0 | 0 |
(X•Y)' = X' + Y'
L.h.s = R.h.s
De-Morgan's Second Law : The Second theorem of De Morgan's Law defines that the inverted result from OR Operation is the same as the AND operation of a complement of each variable where the Result equal NOR Operation.
De-Morgan's Second Law Formula : (X+Y)' = X' • Y'
De-Morgan's Second Law Truth Table
| X | Y | X' | Y' | X+Y | (X+Y)' | X'•Y' |
|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 | 0 |
Here Prove The Second De-morgan's Law Formula.
(X+Y)' = X' • Y'
L.h.s = R.h.s
3. Using K-Map, obtain the minimal sum of pro- duct expression of the following function:
F(x,y,z,w)=Σ(1,3,4,5,7,9,11,12,13,15)
4. Difference between sequential and combinational circuit.
Ans : Combinational Circuit: Combinational Circuit is defined as the time independent circuit which does not depend on the previous input to generate any output is called as Combinational circuit.
Sequential Circuit: Sequential Circuit are those which are clock cycle dependent and depend On the current as well as previous input to generate any output
| S.No | Combinational Circuit | Sequential Circuit |
|---|---|---|
| 1. | In this Output depends only on present Input | In this output depends on present as well as past input. |
| 2. | Speed Is Fast | Speed Is Slow |
| 3. | It is designed easy | it is degisigned though as compared to Combinational Circuit |
| 4. | There is no feedback between input and output. | There exists. a feedback path between input and output |
| 5. | This is Time Independent | This is time Dependent |
| 6. | Elementary building blocks: Logic gates | Elementary building blocks: Flip-flops |
xy + x(y + z) + y(y + z).
Ans: x y + x (y + z) + y (y + z)
= xy + xy + xz + y (y + z)
= xy + x z + y ( y + z )
= ху + xz + yy + yz
= xy + xz + y + yz
= xy+ xz + y
= xy + y + x z
= y + x z
6. Explain Half Adder
Ans: A half Adder is a type of adder. an electronic circuit that performs the addition of numbers. The half Adder is able to add two single binary digits and provide, the output plus a carry value. It has two input's, called A and B. and two outputs. S(sum) and C (carry). The common representation uses. a XOR logic gate and am AND logic gate.
Truth Table Of Half Adder
| INPUT | INPUT | OUTPUT | OUTPUT |
|---|---|---|---|
| A | B | S(SUM) | C(CARRY) |
| 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 0 | 0 | 0 | 1 |
7. 8 Bit subtraction of (17)10, (-13)10 using 2'complement.
Ans:
Here A = 00010001, B = 00001101.
Find A - B = ? using 2's complement
First find 2's complement of B = 00001101
Note : 2's complement of a number is 1 added to it's 1's complement number.
Step-1: 1's complement of 00001101 is obtained by subtracting each digit from 1
Step-2: Now add 1 to the 1's complement to obtain the 2's complement :
11110010 + 1 = 11110011
Step-3: Now Add this 2's complement of B to A
The left most bit of the result is 1, called carry and it is ignored.
So answer is 00000100
Find A - B = ? using 2's complement
First find 2's complement of B = 00001101
Note : 2's complement of a number is 1 added to it's 1's complement number.
Step-1: 1's complement of 00001101 is obtained by subtracting each digit from 1
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | ||
| - | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | |
| 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | ||
Step-2: Now add 1 to the 1's complement to obtain the 2's complement :
11110010 + 1 = 11110011
Step-3: Now Add this 2's complement of B to A
| 1 | 1 | 1 | 1 | 1 | 1 | ||||
| 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | ||
| + | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | |
| 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | |
Hints : (Move mouse over the steps for detail calculation highlight)
The left most bit of the result is 1, called carry and it is ignored.
So answer is 00000100
Another Solution uploaded As Soon As Posible
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