Question:

Please convert the below content to rendered equations without any errors.Alright, let’s walk through it **cleanly and step-by-step**, exactly the way this kind of digital communications problem is expected to be solved in exams ? --- ## Given data * **Channel bandwidth** ( B = 50 \text{ kHz} ) * **Symbol rate** ( R_s = 4800 \text{ symbols/sec} ) * **Desired data rate** ( R_b = 19200 \text{ bits/sec} ) We are asked to: 1. **Select an appropriate FSK signal constellation** 2. **Find the roll-off factor ( \alpha )** of a **raised-cosine pulse** such that the entire frequency band is utilized. --- ## Step 1: Determine bits per symbol [ \text{Bits per symbol} = \frac{R_b}{R_s} ] [ = \frac{19200}{4800} = 4 \text{ bits/symbol} ] --- ## Step 2: Determine FSK order ( M ) For FSK: [ \text{Bits per symbol} = \log_2 M ] [ \log_2 M = 4 \Rightarrow M = 2^4 = 16 ] ### ✅ **Selected modulation** [ \boxed{\text{16-FSK}} ] --- ## Step 3: Bandwidth of raised-cosine pulse For a **raised-cosine shaped signal**, the **occupied bandwidth** is: [ B = (1 + \alpha) R_s ] where * ( \alpha ) = roll-off factor * ( 0 \le \alpha \le 1 ) --- ## Step 4: Use entire available bandwidth We are told to **utilize the full 50 kHz bandwidth**, so: [ (1 + \alpha) R_s = 50{,}000 ] Substitute ( R_s = 4800 ): [ (1 + \alpha)(4800) = 50{,}000 ] [ 1 + \alpha = \frac{50{,}000}{4800} ] [ 1 + \alpha = 10.4167 ] [ \alpha = 9.4167 ] --- ## Step 5: Check physical feasibility A **raised-cosine filter** must satisfy: [ 0 \le \alpha \le 1 ] But we obtained: [ \alpha = 9.4167 \quad \text{❌ (not physically possible)} ] --- ## Step 6: Interpretation & final choice Since: * Required roll-off factor exceeds allowable limit * Even **maximum roll-off (( \alpha = 1 ))** cannot fill the full 50 kHz The **best practical choice** is: [ \boxed{\alpha = 1} ] This **maximizes bandwidth usage** while remaining realizable. --- ## ✅ Final Answer * **FSK constellation:** [ \boxed{16\text{-FSK}} ] * **Raised-cosine roll-off factor:** [ \boxed{\alpha = 1} ] This configuration achieves the required **19200 bps data rate**, uses the **4800 symbol/sec rate**, and **occupies the maximum feasible portion** of the 50 kHz band. --- If you want, I can also: * Sketch the **16-FSK constellation** * Compare with **QAM/PSK efficiency** * Convert this into **exam-ready final statements** Just tell me ?