topics

 The Rise of 6G: How It Will Change Connectivity Forever

The Future of Meta verse: How Mixed Reality is Changing Our World Humanoid Robots Are Evolving Fast: Are We Ready for What’s Coming? Transformative power of quantum computing and its impact on cyber security. Futuristic Gadgets You Won’t Believe Exist: 10 Mind-Blowing Inventions You’ll Wish You Owned! Quantum Computing Is Changing Everything: Will It Break the Internet’s Security? Dark Web Secrets: What Really Happens in the Digital Underworld?

Comments

Post a Comment

Popular posts from this blog

Chapter no 1

  2. Advantages of a Client-Server System Using a LAN Compared to a large centralized timesharing system, a client-server model using a LAN offers the following advantages: Improved Performance Each client has its own CPU and can perform many tasks locally. This reduces the load on the central server and results in faster response times for users. Scalability and Flexibility It is easier to upgrade or add new clients without affecting the overall system. Services can be distributed across multiple servers (e.g., one for printing, another for files), which simplifies system management and increases reliability. 4. Additional QoS Parameter Needed Based on Traffic Type Besides bandwidth and latency , different applications require specific Quality of Service (QoS) parameters: (i) Digitized Voice Traffic – Jitter Jitter is the variation in arrival time of packets. High jitter can cause voice distortion or gaps during a conversation, affecting call quality. (ii) Vi...
Read more

chapter 2

  1: Compute the Fourier coefficients for the function f(t) = t (0 ≤ t ≤ 1) Solution: To compute the Fourier series of the function f ( t ) = t f(t) = t f ( t ) = t on the interval 0 ≤ t ≤ 1 0 \leq t \leq 1 0 ≤ t ≤ 1 , we first need to express it as a periodic function of period 1. The Fourier series of a function f ( t ) f(t) f ( t ) on the interval [ 0 , 1 ] [0,1] [ 0 , 1 ] is given by: f ( t ) = a 0 + ∑ n = 1 ∞ ( a n cos ⁡ ( 2 π n t ) + b n sin ⁡ ( 2 π n t ) ) f(t) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(2\pi nt) + b_n \sin(2\pi nt) \right) f ( t ) = a 0 ​ + n = 1 ∑ ∞ ​ ( a n ​ cos ( 2 πn t ) + b n ​ sin ( 2 πn t ) ) Where: a 0 = ∫ 0 1 f ( t )   d t = ∫ 0 1 t   d t = 1 2 a_0 = \int_0^1 f(t) \, dt = \int_0^1 t \, dt = \frac{1}{2} a 0 ​ = ∫ 0 1 ​ f ( t ) d t = ∫ 0 1 ​ t d t = 2 1 ​ a n = 2 ∫ 0 1 t cos ⁡ ( 2 π n t )   d t = 0 ( since this is an odd function about  t = 0.5 ) a_n = 2 \int_0^1 t \cos(2\pi nt) \, dt = 0 \quad (\text{since...
Read more

chapter no 3

  1. The message is split into 10 frames, each with an 80% chance of arriving undamaged. What is the expected number of times the message must be sent to get the entire message through? Answer: Let the probability of a frame being transmitted successfully be 0.8, and the probability of a frame being damaged (and requiring retransmission) be 0.2. For one frame to arrive undamaged, the expected number of transmissions is the reciprocal of the probability of success: E [ transmissions per frame ] = 1 0.8 = 1.25 E[\text{transmissions per frame}] = \frac{1}{0.8} = 1.25 E [ transmissions per frame ] = 0.8 1 ​ = 1.25 Since there are 10 frames, the total expected number of transmissions is: E [ total transmissions ] = 10 × 1.25 = 12.5 E[\text{total transmissions}] = 10 \times 1.25 = 12.5 E [ total transmissions ] = 10 × 1.25 = 12.5 Thus, on average, the message will need to be sent  12.5 times  to get the entire message through. 2. Character Encoding...
Read more