Let's solve the problem step-by-step: ### **Step 1: Input Information** Given input: \[ D(t) = [1, 0, 1, 1, 0, 0] \] ### **Step 2: Differential Encoding** The differential encoding is given by: \[ b(t) = D(t) \oplus b(t-1) \] Let’s assume the initial value: \( b(0) = 1 \) (commonly used as the starting bit) Now, let's compute \( b(t) \) for each \( t \): | \( t \) | \( D(t) \) | \( b(t-1) \) | \( b(t) = D(t) \oplus b(t-1) \) | |---------|------------|-------------|----------------------------------| | 1 | 1 | 1 | 0 | | 2 | 0 | 0 | 0 | | 3 | 1 | 0 | 1 | | 4 | 1 | 1 | 0 | | 5 | 0 | 0 | 0 | | 6 | 0 | 0 | 0 | So, the sequence for \( b(t) \) (including the initial value for reference): \[ b = [1, 0, 0, 1, 0, 0, 0] \] But for the modulation we use \( b(1) \) to \( b(6) \): \[ b = [0, 0, 1, 0, 0, 0] \] --- ### **Step 3: Output Modulated Signal \( s(t) \)** The output modulated signal is: \[ s(t) = \begin{cases} A \sin(\omega_c t) & \text{if } b(t) = 1 \\ -A \sin(\omega_c t) & \text{if } b(t) = 0 \end{cases} \] So, for each bit: | \( t \) | \( b(t) \) | \( s(t) \) | |---------|------------|--------------------| | 1 | 0 | \(-A \sin(\omega_c t)\) | | 2 | 0 | \(-A \sin(\omega_c t)\) | | 3 | 1 | \(A \sin(\omega_c t)\) | | 4 | 0 | \(-A \sin(\omega_c t)\) | | 5 | 0 | \(-A \sin(\omega_c t)\) | | 6 | 0 | \(-A \sin(\omega_c t)\) | --- ### **Step 4: Sketch the Modulated Waveform** #### **DPSK Waveform:** - For \( b(t)=0 \): \( s(t) \) is \( -A\sin(\omega_c t) \) (180° phase shift) - For \( b(t)=1 \): \( s(t) \) is \( +A\sin(\omega_c t) \) (no phase shift) So, the waveform will have a phase reversal (180° shift) wherever \( b(t) \) changes from 0 to 1 or 1 to 0. #### **Waveform Sequence:** 1. **t = 1:** \( b(1) = 0 \) ⇒ \( -A\sin(\omega_c t) \) 2. **t = 2:** \( b(2) = 0 \) ⇒ \( -A\sin(\omega_c t) \) 3. **t = 3:** \( b(3) = 1 \) ⇒ \( +A\sin(\omega_c t) \) (phase reversal here) 4. **t = 4:** \( b(4) = 0 \) ⇒ \( -A\sin(\omega_c t) \) (phase reversal here) 5. **t = 5:** \( b(5) = 0 \) ⇒ \( -A\sin(\omega_c t) \) 6. **t = 6:** \( b(6) = 0 \) ⇒ \( -A\sin(\omega_c t) \) #### **Sketch (ASCII Representation):** ``` t: 1 2 3 4 5 6 ----- ----- +++++ ----- ----- ----- -A sin -A sin +A sin -A sin -A sin -A sin ``` - *"-----"* represents negative sine (180° phase) - *"+++++"* represents positive sine (0° phase) --- ### **Final Answers:** #### **a) Differential Encoded Bits:** \[ b(t) = [0, 0, 1, 0, 0, 0] \] #### **b) Modulated Signal Sequence:** \[ s(t) = [-A\sin(\omega_c t),\ -A\sin(\omega_c t),\ A\sin(\omega_c t),\ -A\sin(\omega_c t),\ -A\sin(\omega_c t),\ -A\sin(\omega_c t)] \] #### **c) DPSK Modulated Waveform:** - The waveform switches phase (180°) when the encoded bit changes from 0 to 1 or 1 to 0. --- **If you want a hand-drawn sketch or a digital plot, let me know!**

Let's proceed step-by-step with detailed calculations, based on the provided data, to fully understand the differential encoding and the resulting DPSK waveform. --- ### **Given Data:** - Input data bits: \[ D(t) = [1, 0, 1, 1, 0, 0] \] - Initial differential bit \( b(0) \): **Assumption:** \( b(0) = 1 \) (common initial condition) - Modulated signal \( s(t) \): \[ s(t) = \begin{cases} A \sin(\omega_c t) & \text{if } b(t) = 1 \\ -A \sin(\omega_c t) & \text{if } b(t) = 0 \end{cases} \] --- ### **Step 1: Differential Encoding** The differential encoding rule (XOR operation): \[ b(t) = D(t) \oplus b(t-1) \] Compute \( b(t) \) for each \( t \): | \( t \) | \( D(t) \) | \( b(t-1) \) | Calculation \( b(t) = D(t) \oplus b(t-1) \) | Result \( b(t) \) | |---------|------------|--------------|----------------------------------------------|------------------| | 1 | 1 | 1 | \( 1 \oplus 1 = 0 \) | 0 | | 2 | 0 | 0 | \( 0 \oplus 0 = 0 \) | 0 | | 3 | 1 | 0 | \( 1 \oplus 0 = 1 \) | 1 | | 4 | 1 | 1 | \( 1 \oplus 1 = 0 \) | 0 | | 5 | 0 | 0 | \( 0 \oplus 0 = 0 \) | 0 | | 6 | 0 | 0 | \( 0 \oplus 0 = 0 \) | 0 | **Note:** For clarity, initial \( b(0) = 1 \) was used, and the first \( b(1) \) is obtained as above. **Sequence of \( b(t) \):** \[ b = [b(1), b(2), b(3), b(4), b(5), b(6)] = [0, 0, 1, 0, 0, 0] \] --- ### **Step 2: Modulated Signal \( s(t) \)** Using the rule: \[ s(t) = \begin{cases} A \sin(\omega_c t), & \text{if } b(t) = 1 \\ -A \sin(\omega_c t), & \text{if } b(t) = 0 \end{cases} \] Compute \( s(t) \) for each \( t \): | \( t \) | \( b(t) \) | \( s(t) \) | |---------|------------|-------------------------------------| | 1 | 0 | \(-A \sin(\omega_c t)\) | | 2 | 0 | \(-A \sin(\omega_c t)\) | | 3 | 1 | \(+A \sin(\omega_c t)\) | | 4 | 0 | \(-A \sin(\omega_c t)\) | | 5 | 0 | \(-A \sin(\omega_c t)\) | | 6 | 0 | \(-A \sin(\omega_c t)\) | --- ### **Step 3: Interpretation of the waveform** - The phase of the carrier **flips** (180° phase shift) whenever \( b(t) \) changes from 0 to 1 or 1 to 0. - Since \( b(t) \) stays at 0 for most bits, the waveform remains at the negative phase. - When \( b(t) \) switches to 1 (at \( t=3 \)), the waveform switches to positive phase, indicating a phase change. --- ### **Step 4: Final waveform sketch description** - **At \( t=1 \) and \( t=2 \):** The waveform is in the negative phase: \( -A \sin(\omega_c t) \). - **At \( t=3 \):** The waveform switches to positive phase: \( +A \sin(\omega_c t) \). - **At \( t=4, 5, 6 \):** It reverts back to negative phase. This pattern illustrates the phase shifts characteristic of DPSK modulation—a phase change (180°) whenever the differential bit indicates. --- ### **Final Summary** | Data bit \( D(t) \) | Encoded bit \( b(t) \) | Modulated signal \( s(t) \) | Phase shift indicator | |---------------------|------------------------|------------------------------|------------------------| | 1 | 0 | \(-A \sin(\omega_c t)\) | No shift (phase 0) | | 0 | 0 | \(-A \sin(\omega_c t)\) | No shift | | 1 | 1 | \(+A \sin(\omega_c t)\) | Phase shift (180°) | | 1 | 0 | \(-A \sin(\omega_c t)\) | Phase shift (180°) | | 0 | 0 | \(-A \sin(\omega_c t)\) | No shift | | 0 | 0 | \(-A \sin(\omega_c t)\) | No shift | --- ### **Key points:** - **Differential encoding** determines the phase change. - **DPSK waveform** exhibits phase reversals corresponding to data transitions. - **Waveform sketch** would show continuous sine waves with phase shifts at \( t=3 \). --- **Let me know if you'd like a graphical plot or further elaboration!**