b) Find the transmitted Phase sequence for

### Reasons and Explanations

– Reason 1: **QPSK Basics:** QPSK encodes two bits per symbol by varying the phase of the carrier signal. Common phase mappings exist, and we need to define one to translate the binary sequence into phase shifts.
– Reason 2: **Binary to Dibit Conversion:** The input binary sequence needs to be grouped into dibits (pairs of bits) because QPSK transmits two bits at a time.
– Reason 3: **Phase Mapping:** Each dibit is mapped to a specific phase shift (0°, 90°, 180°, or 270°). The choice of mapping affects the resulting phase sequence.
– Reason 4: **Waveform Sketching:** Each phase shift is represented by a sinusoidal segment (cosine or sine wave) with the corresponding phase. The duration of each segment is the symbol duration.

### Summary

The question asks for the transmitted phase sequence and a sketch of the waveform for a given binary sequence using QPSK. The process involves converting the binary sequence to dibits, mapping each dibit to a phase shift, and then representing these phase shifts as segments of a sinusoidal waveform.

### Core Answer

**Transmitted Phase Sequence:** {0°, 270°, 180°, 90°}

**Transmitted Waveform:**

* **0° (00):** $A\cos(2\pi f_c t)$ for $0 \le t < T$
* **270° (11):** $-A\sin(2\pi f_c t)$ for $T \le t < 2T$
* **180° (10):** $-A\cos(2\pi f_c t)$ for $2T \le t < 3T$
* **90° (01):** $A\sin(2\pi f_c t)$ for $3T \le t < 4T$

Where A is the amplitude, $f_c$ is the carrier frequency, and T is the symbol duration.