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Digital Systems & Basics

Q: Why do we need digital systems rather than analog systems? 1Ans: 1

• High accuracy 1
• Easy storage 1
• Easy processing 1
• Reliable transmission 1
• Error detection and correction 1
• Easy programmability 1
• High reliability 1

Q: When a signal progate there are two things occure 1

1. Attenuation 1
2. phase shift. 1\\gamma \= \\alpha \+ j\\beta $rightarrow$ propagation constant. 1

Q: BJT works like NOT gate. Explain. 1Ans: A BJT can be used as a NOT gate by operating it as a switch. In an NPN transistor, the emitter is connected to the ground, the collector is connected to $V\_cc$ through a resistor, and the output is taken from the collector. When the input $V\_B$ is low (0), the transistor is in the cut-off region (OFF), so no collector current flows and the output becomes high (1). 1 When the V\_B \= 1 or high the transistor enters the saturation region (ON), allowing collector current to flow and making the output low. Thus, the output is always the opposite of the input, so a BJT functions as a NOT gate. 2

• X-OR gate এ odd numbers of input যদি high তাহলে output high. (X-OR gate’s output is high if odd numbers of inputs are high) 2

Q: In buffers gate input = output, than why we used this? 2Ans: We used buffers gate because: 2$rightarrow$ Increase current driving capability 2$rightarrow$ Isolate one circuit 2$rightarrow$ Prevent loading effects 2$rightarrow$ Improve signal strength. 2$rightarrow$ Increase fan-out 2
Q: What means loading effect? 3Ans: Loading effect is the reduction in output voltage or signal strength when a circuit is connected to a load that draws too much current. It occurs because the source cannot supply the required current, causing the output signal to weaken or become distorted. This may lead to improper circuit operation. A buffer gate is used to prevent the loading effect by increasing the current-driving capability while keeping the output logic level the same as the input. 3

Digital Codes

8421 code 3$rightarrow$ The 8421 code is a weighted BCD code in which decimal digit is represented by a 4-bit binary number with weights 8, 4, 2 and 1. 3
Merits: 3$rightarrow$ Simple and easy to understand 3$rightarrow$ Direct conversion between decimal and binary-coded decimal. 3$rightarrow$ Widely used in digital systems such as calculators and digital clocks. 3Demerits: 4$rightarrow$ Requires 4 bits to represent each decimal digit, so it use more memory than pure binary. 4$rightarrow$ Arithmetic operation are more complex than with pure binary numbers. 4
Excess-3 4Excess-3 code is a 4-bit non-weighted, self-complementing binary code in which each decimal digit is represented by the binary equivalent of the decimal digit pluse 3. 4
Advantage: 4$rightarrow$ Self-complementing code 4$rightarrow$ simplifies decimal subtraction using complement arithmetic 4$rightarrow$ No all-0 or all-1 code words. 4$rightarrow$ Useful in digital system for arithmetic operation and decimal calculations. 4$rightarrow$ Each decimal digit has a unique 4-bit code. 4Disadvantage: 5$rightarrow$ Not a weighted code 5$rightarrow$ Requires conversion to/from BCD or binary 5$rightarrow$ Arithmetic is more complex 5$rightarrow$ Needs extra hardware for conversion. 5
Q: Why is excess-3 code called a self-complementary code? 5Ans: Excess-3 is called a self-complementary code because the 1’s complement of the Excess-3 code of a decimal digit gives the Excess-3 code of the 9’s complement of that digit. 5Mathematically,1’s complement of XS-3 (N) = XS-3 (9 - N). 5Now,Take a decimal digit - 2 5$rightarrow$ Excess-3 code of 2 $rightarrow$ 2+3 = 5 = $(0101)\_2$ 5$therefore$ 1’s complement $$\begin{gathered} (0101)\_2 \\ rightarrow(1010)\_2 \end{gathered}$$ 5Now find the 9’s complement of 2 69 \- 2 \= 7 6Excess-3 code of 7: 7+3 \= 10 \= (1010)\_2 6Since the 1’s complement of 0101 is 1010, which is exactly the excess-3 code of 7 (the 9’s complement of 2), Excess-3 is called a self-complementing code. 6
BCD: 6BCD is a weighted code in which each decimal digit (0-9) is represented separately by its 4-bit binary equivalent. 6Advantage: 6$rightarrow$ Easy conversion between decimal and binary 6$rightarrow$ Convenient for displaying decimal numbers on digital displays 6$rightarrow$ Widely used in calculators, clocks. 6$rightarrow$ Each decimal digit is represented separately. 6Disadvantage: 7$rightarrow$ Requires more bits than pure binary representation 7$rightarrow$ Storage is less efficient. 7$rightarrow$ Arithmetic operations are more complex than binary arithmetic. 7$rightarrow$ Extra correction logic is needed in BCD addition. 7
Gray code: 7Gray code is a 4-bit non-weighted binary code in which only one bit changes between two consecutive code words. 7Advantage: 7$rightarrow$ Only one bit changes at a time 7$rightarrow$ Reduces switching errors 7$rightarrow$ Prevents false outputs. 7$rightarrow$ Reliable for position sensing 7$rightarrow$ Improves system stability. 7Disadvantage: 8$rightarrow$ Arithmetic operations are difficult. 8$rightarrow$ Needs conversion. 8$rightarrow$ Not suitable for data storage. 8$rightarrow$ Extra hardware or logic required 8$rightarrow$ Limited use. 8
BCD codes: types. 8
Decimal,8 4 2 1,2 4 2 1,5 4 2 1,8 4 -2 -1
Binary number gula এমনভাবে লিখতে হবে যেন যে position এ binary number থাকবে সেই তার সাথে গুণ করে যোগ করলে যাতে decimal পাওয়া যায়। আর শুরুতে 1 বসানো try করতে হবে। 9

• 8 4 2 1 (For 5): 0 1 0 1 \\therefore 8\\times0 \+ 4\\times1 \+ 2\\times0 \+ 1\\times1 \= 5 9
• 2 4 2 1 (For 5): 1 0 1 1 \\rightarrow 2\\times1 \+ 4\\times0 \+ 2\\times1 \+ 1\\times1 \= 2+2+1 \= 5 9
• 5 4 2 1 (For 5): 1 0 0 0 \\rightarrow 5\\times1 \+ 4\\times0 \+ 2\\times0 \+ 1\\times0 \= 5 9
• 8 4 -2 -1 (For 5): 1 0 1 1 \\rightarrow 8\\times1 \+ 4\\times0 \+ 1\\times(-2) \+ 1\\times(-1) \= 8-2-1 \= 5 9
• (Note points to 2421 and 84-2-1 indicating:) $rightarrow$ Those are self comple-mantry code. 9

$$\begin{gathered} (XX.XX)\_n \\ rightarrow \\ textDecimal \\ rightarrow \\ textBCD \end{gathered}$$. 9$$\begin{gathered} (XX.XX)\_n \\ rightarrow \\ textDecimal\text{+}3 \\ rightarrow \\ textExcess-3 \end{gathered}$$ 9
Gray code conversion: 100 $rightarrow$ 0000 | 8 $rightarrow$ 11001 $rightarrow$ 0001 | 9 $rightarrow$ 11012 $rightarrow$ 0011 | 10 $rightarrow$ 11113 $rightarrow$ 0010 | 11 $rightarrow$ 11104 $rightarrow$ 0110 | 12 $rightarrow$ 10105 $rightarrow$ 0111 | 13 $rightarrow$ 10116 $rightarrow$ 0101 | 14 $rightarrow$ 10017 $rightarrow$ 0100 | 15 $rightarrow$ 100010
Binary $rightarrow$ Gray 10B\_3 \\ B\_2 \\ B\_1 \\ B\_0$$(3)\_B \= 0011$$G\_3 \= B\_3$$G\_2 \= B\_3 \\oplus B\_2$$G\_1 \= B\_2 \\oplus B\_1$$G\_0 \= B\_1 \\oplus B\_0$$(3)\_G \= G\_3 \\ G\_2 \\ G\_1 \\ G\_0 \= 0 \\ 0 \\ 1 \\ 0 10
Gray $rightarrow$ Binary 11G\_3 \\ G\_2 \\ G\_1 \\ G\_0$$(11)\_G \= 1110$$B\_3 \= G\_3$$B\_2 \= B\_3 \\oplus G\_2$$B\_1 \= B\_2 \\oplus G\_1$$B\_0 \= B\_1 \\oplus G\_0$$(11)\_B \= B\_3 \\ B\_2 \\ B\_1 \\ B\_0 \= 1 \\ 0 \\ 1 \\ 1 11
Q: Represent the decimal number 8620 i) in BCD ii) in excess-3 iii) 2,4,2,1 iv) binary v) Gray. 11Ans:i) 8 6 2 0 $rightarrow$ 1000 0110 0010 0000 = $(1000011000100000)\ast BCD$ 11ii) 8+3 \= 11 \= 1011, 6+3 \= 9 \= 1001, 2+3 \= 5 \= 0101, 0+3 \= 3 \= 0011 11iii) 2,4,2,1: 8 \= 1110, 6 \= 1100, 2 \= 0010, 0 \= 0000 11iv) (8620)*{10} \= (10000110101100)\_2 11v) 8 6 2 0 = $(10000110101100)\ast 2$= 1 1 0 0 0 1 0 1 1 1 0 1 0= $(11100010111010)\ast Gray$ 12(First Gray bit = First binary bit. Next = XOR of the current bit with the previous one). 12
Q: The state of 12-bit register is 100010010111. what is its content if it represents: 12i) Three decimal digits in BCD?** 121000 1001 0111 $rightarrow$ 8 9 7 12ii) Three decimal digits in the excess-3 code? 12$rightarrow$ 8 - 3 = 5$rightarrow$ 9 - 3 = 6$rightarrow$ 7 - 3 = 4Result: 564 12iii) Three decimal digits in the 8 4 -2 -1 code 121000 = 1\\times8 \+ 0\\times4 \+ 0\\times(-2) \+ 0\\times(-1) \= 8 131001 = 1\\times8 \+ 0\\times4 \+ 0\\times(-2) \+ 1\\times(-1) \= 7 130111 = 0\\times8 \+ 4\\times1 \+ 1\\times(-2) \+ 1\\times(-1) \= 4 \- 2 \- 1 \= 1 13$therefore871$ 13iv) A binary number. 12$(100010010111)\ast 2$= $2^{11}\text{+}{2}^7\text{+}{2}^4\text{+}{2}^2\text{+}{2}^1\text{+}{2}^0$= 2048 + 128 + 16 + 4 + 2 + 1= $(2199)\ast 10$ 13

Boolean Algebra & Gates

Q: Prove the duality principle of boolean algebra. 14Ans: Every valid Boolean expression or theorem remains valid if we interchange AND ($cdot$) with OR (+) and interchange 0 with 1, while keeping the variables and their complements unchanged. 14A \+ AB \= AL.H.S = A \+ AB$$= A(1 \+ B)$$= A.1$$= A 14Dual: A \\cdot (A \+ B) \= AL.H.S \\Rightarrow A \\cdot (A \+ B)$$= AA \+ AB$$= A \+ AB$$= A(1 \+ B)$$= A \\cdot 1$$= A 14
Q: Why is it called exclusive? 14Ans: X-OR is called exclusive because it produces an output of only 1 only when odd number of input 1. It excludes the case where even number of inputs 1. 14 Unlike, the ordinary OR gate, which gives an output of 1 when one or both inputs are 1, the XOR gate gives an output of 0 when both inputs are the same. Therefore, XOR is also known as a difference detector, or inequality detector. 15
Application of XOR gate: 15$rightarrow$ Parity bit generation and checking 15$rightarrow$ Parity bit correction 15$rightarrow$ Half adder and full adder 15$rightarrow$ Binary comparator 15$rightarrow$ Data encryption and decryption. 15
Q: Explain XOR is use for parity bit checker, generate and correction. 15Ans: XOR gives 1 when there is an odd numbers of 1-bits and 0 when an even number. This is why it’s perfect for parity. To generate the parity bit, all data bits are XOR-ed together, so that the total number of 1s becomes even. 15 To check for errors, the receiver XORs all the bits, including the parity bit - if the result is 0, there is no error; if it is 1, an error happened. But this method can only catch errors in an odd number of bits, and it can’t tell which bit is wrong. To correct errors, But there are a limitation, if their two bit chang the we will not found the error. 16
A,B,C,X
Using NAND make the other logic gate: 17

• NOT gate: $A$ connected to both inputs of NAND $$\begin{gathered} rightarrow \\ barA \end{gathered}$$ 17
• OR/NOR gate: $A$ through NAND $$\begin{gathered} rightarrow \\ barA \end{gathered}$$, $B$ through NAND $$\begin{gathered} rightarrow \\ barB \end{gathered}$$, both into NAND $rightarrowA+B$ (OR). Into another NAND $$\begin{gathered} rightarrow \\ overlineA+B \end{gathered}$$ (NOR). 17
• AND gate: $A$, $B$ into NAND $$\begin{gathered} rightarrow \\ overlineAB \end{gathered}$$, then both inputs to another NAND $rightarrowAB$ (AND). 17
• XOR/XNOR: Network of 4 NAND gates for $$\begin{gathered} A \\ oplusB \end{gathered}$$ (XOR), into a final NAND for $$\begin{gathered} overlineA \\ oplusB \end{gathered}$$ (XNOR). 17

Using NOR make the other logic gate: 18

• NOT gate: $A$ connected to both inputs of NOR $$\begin{gathered} rightarrow \\ barA \end{gathered}$$ 18
• OR/NOR: $A$, $B$ into NOR $$\begin{gathered} rightarrow \\ overlineA+B \end{gathered}$$ (NOR). Into another NOR $rightarrowA+B$ (OR). 18
• AND gate: $A$ through NOR $$\begin{gathered} rightarrow \\ barA \end{gathered}$$, $B$ through NOR $$\begin{gathered} rightarrow \\ barB \end{gathered}$$, both into NOR $rightarrowAB$ (AND). 18
• XOR/XNOR gate: Network of 4 NOR gates for $$\begin{gathered} overlineA \\ oplusB \end{gathered}$$ (XNOR), into a final NOR for $$\begin{gathered} A \\ oplusB \end{gathered}$$ (XOR). 18

Logic Families & Boolean Functions

Significance of boolean algebra. 19

• Boolean algebra is a branch of algebra in which variables can hold only two values - 0 and 1. 19Significance:
• Mathematical foundation of digital logic systems and digital electronics. 19
• Basis of all logic gates used in processors, memory and digital devices. 19
• Helps simplify complex logical expression using boolean law 19
• Used with K-Map to minimize circuits, reducing cost and power consumption and increasing speed. 19
• Essential for designing combinational and sequential circuits. 19
• Basis of logical operators and conditional statements in computer programming. 19

Q: Why we simplify boolean algebra. 19Ans:

• To reduce the numbers of logic gates required to implement a circuit. 19
• Fewer gates mean lower hardware cost and less circuit complexity. 20
• Reduces power consumption of digital circuit. 20
• Reduces propagation delay, so the circuit works faster. 20
• Simplified circuits occupy less space. 20
• Improves reliability of the overall system since fewer components mean fewer chances of failure. 20

Fan-out: Fan-out is the maximum number of standard logic gate inputs that can be connected to the output of a logic gate without affecting its proper operation. 20Propagation delay: Propagation delay is the time taken for a change in the input of a logic gate to produce the corresponding change at its output. It is usually measured in nanoseconds. 20Power Dissipation: Power dissipation is the amount of electrical power consumed by a logic gate during its operation. It is expressed in milliwatts. 20Nois margin: Nois margin is the maximum amount of unwanted noise voltage that a digital circuit can tolerate without causing an incorrect logic output. A higher noise margin indicates better noise immunity. It’s depends on the material properties. 21
SOP (sum of products) 21$rightarrow$ Sum of products is a boolean expression in which two or more AND terms are combined using the OR operation. 21Ex: F(A,B,C) \= AB' \+ AB' \+ ABC. 21POS (Products of sum) 21POS is a boolean expression in which two or more OR terms are combined using the AND operation. 21F(A,B,C) \= (A+B')(A'+C)(B+C). 21
Canonical form 22

• Canonical form is a boolean expression in which every term contains all the variables exactly once, either in complemented or uncomplemented form. It provides a unique representation of a boolean function. 22Ex: CSOP: F(A,B,C) \= A'BC \+ ABC \+ ABC 22CPOS: F(A,B,C) \= (A+B+C)(A+B'+C)(A'+B+C') 22

Standard form 22

• A standard form is a boolean expression written in either sum of products or product of sums form. In standard form, each term does not necessarily contain all the variables. 22Ex: SSOP: F(A,B,C) \= AB \+ BC \+ A'C 22SPOS: F(A,B,C) \= (A+B)(B+C)(A'+C). 22

Q: What are the differences and similarities between SOP and POS? 23Ans: * Similarities between SOP and POS.

1. Both are boolean algebra expression. 23
2. Both represent the same boolean function. 23
3. Both are used to design and simplify digital logic circuits. 23
4. Both can be simplified using boolean algebra or k-map. 23
5. Both can be expressed in standard/canonical form. 23

• Differences between SOP and POS. 23

SOP,POS
Q: F(A,B,C) = $Sigma(0,3,4,6)$ convert this into sop form, and pos form. 24Ans: 24
A,B,C,X,Terms,Minterms
The standard SOP form isY \= \\bar{A}\\bar{B}\\bar{C} \+ \\bar{A}BC \+ A\\bar{B}\\bar{C} \+ AB\\bar{C} 24The standard POS form isZ \= (A+B+\\bar{C})(A+\\bar{B}+C)(\\bar{A}+B+\\bar{C})(\\bar{A}+\\bar{B}+\\bar{C}) 24
Y \= \\bar{A}\\bar{B}\\bar{C} \+ \\bar{A}BC \+ A\\bar{B}\\bar{C} \+ AB\\bar{C} 25$$\begin{gathered} = \\ barB \\ barC(A+ \\ barA)\text{+} \\ barABC\text{+}AB \\ barC \end{gathered}$$ 25$$\begin{gathered} = \\ barB \\ barC\text{+}B( \\ barAC\text{+}A \\ barC) \end{gathered}$$ 25$$\begin{gathered} = \\ barB \\ barC\text{+}B(A \\ oplusC) \end{gathered}$$ $leftarrow$ SOP simplified. 25
$rightarrow$ To make standard SOP $rightarrow$ Multiply by $(x+x^{\prime })$ term of missing variable. (0+1=1) 25Ex: \\bar{A}+B \= \\bar{A}(B+\\bar{B}) \+ B(A+\\bar{A})$$= \\bar{A}B \+ \\bar{A}\\bar{B} \+ AB \+ \\bar{A}B$$= \\bar{A}B \+ \\bar{A}\\bar{B} \+ AB$$= A \\oplus B \+ \\bar{A}\\bar{B} \\rightarrow simplified version. 25
$rightarrow$ To make standard POS $rightarrow$ add $(x{x}^{\prime })$ by term of missing variable. (0.1=0) 25Ex: \\bar{A}(B+C) \= (\\bar{A} \+ B\\bar{B} \+ C\\bar{C})(B+C \+ A\\bar{A})$$= (\\bar{A}+C+B)(\\bar{A}+B+\\bar{C})(\\bar{A}+\\bar{B}+C) \\cdot (B+C+A)(B+C+\\bar{A}) 25
Q: Use de morgan’s theorem to simplify the expression $(A+B+C{)}^{\prime }$ using both SOP and POS forms. 26Sol${}^n$:(A+B+C)' \= A' \\cdot B' \\cdot C' \= \\Sigma m(0) 26$$\begin{gathered} therefore \\ PiM(1,2,3,4,5,6,7) \end{gathered}$$ 26$=(A+B+C^{\prime })(A+B^{\prime }+C)(A+B^{\prime }+C^{\prime })(A^{\prime }+B+C)(A^{\prime }+B+C^{\prime })(A^{\prime }+B^{\prime }+C)(A^{\prime }+B^{\prime }+C^{\prime })$ 26
(Note: এ রাফ এ করতে হবে। Final ans এ লিখা যাবে না। Ans সব 1 বানতে হবে for SOP, 0 হবে for POS) 26
A,B,C,SOP
0,0,0,1 / $A^{\prime }{B}^{\prime }{C}^{\prime }$ (0)
0,0,1,0 / $A^{\prime }{B}^{\prime }C$ (1)
0,1,0,0 / $A^{\prime }B{C}^{\prime }$ (2)
0,1,1,0 / $A^{\prime }BC$ (3)
1,0,0,0 / $A{B}^{\prime }{C}^{\prime }$ (4)
1,0,1,0 / $A{B}^{\prime }C$ (5)
1,1,0,0 / $AB{C}^{\prime }$ (6)
1,1,1,0 / $ABC$ (7)
26 (0 হবে for POS)
Q: Convert each of the following expression int POS and SOP form. i) (AB+C)(B+C’D) ii) $x^{\prime }\text{+}x(x+y^{\prime })(y+z^{\prime })$ 26Sol${}^n$:i) $(AB+C)(B+C^{\prime }D)$ 27$=(A+C)(B+C)(B+C^{\prime })(B+D)$ 27$=(A+C+B{B}^{\prime }+D{D}^{\prime })(B+C+A{A}^{\prime }+D{D}^{\prime })(B+C^{\prime }+A{A}^{\prime }+D{D}^{\prime })(B+D+A{A}^{\prime }+C{C}^{\prime })$ 27$=(A+B+C+D)(A+B+C+D^{\prime })(A+B^{\prime }+C+D)(A+B^{\prime }+C+D^{\prime })(A^{\prime }+B+C+D)(A^{\prime }+B+C+D^{\prime })$ (same aro 8 ta term). 27$$\begin{gathered} =(A+B+C+D)(A+B+C+ \\ barD)(A+ \\ barB+C+D)(A+ \\ barB+C+ \\ barD)( \\ barA+B+C+D)( \\ barA+B+C+ \\ barD)( \\ barA+ \\ barB+C+D)( \\ barA+ \\ barB+C+ \\ barD) \end{gathered}$$ 27= \\Pi M(0,1,2,3,4,5,8,9,10,11)$$\\therefore \\Sigma m(6,7,12,13,14,15) 27
This is already in SOP form. So we can simplify this. 28(AB+C)(B+C'D)$$= ABB \+ ABC'D \+ BC \+ CC'D | B.B=B$$= AB \+ ABC'D \+ 0 \+ BC$$\\Rightarrow AB(1+C'D) \+ BC$$\\Rightarrow AB+BC. 28
ii) This $x^{\prime }\text{+}x(x+y^{\prime })(y+z^{\prime })$. This is already in SOP form and it could not expression in POS form. But if we simplify then it could be expressed in POS form. 28x' \+ x(x+y')(y+z')$$= x' \+ (xx+xy')(y+z')$$= x' \+ xy \+ xz' \+ xy.y' \+ xy'z' 28$=x^{\prime }\text{+}xy\text{+}x{z}^{\prime }\text{+}x{y}^{\prime }{z}^{\prime }$ 29$=x^{\prime }+y\text{+}x{z}^{\prime }(1+y^{\prime })$ | x'+xy \= x'+y, x'+xy' \= x'+y' 29$=x^{\prime }+y\text{+}x{z}^{\prime }.1$ | x'+xz' \= x'+z' 29$=(x^{\prime }+y+z^{\prime })$ $swarrow$ 29This in a pos form if we consider it in one term. 29