Question

Consider the phases of the received DPSK signal over a noisy channel is [00π0ππ0], while the phases of the transmitted DPSK signal was [0π0π0π0]. Here, the first phase corresponds to the reference bit (encoded) and the reference bit (encoded) is ' 1 '. The encoding scheme of the DPSK is ' 1 ' means a transition.

Draw the block diagram of the transmitter and the receiver with the appropriate logic gate at the transmitter and the decision criteria at the receiver.

Answer 1

The DPSK transmitter uses an XOR gate for encoding, creating a signal with phase shifts to represent binary data. The receiver uses a phase detector to decode this data, with a decision criterion based on detecting phase changes. Noise in the communication channel can cause errors in the received signal.

Step-by-step explanation:

DPSK Transmitter and Receiver Block Diagram

Differential Phase Shift Keying (DPSK) is a common method for digital data transmission where the phase change of the carrier signal is used to represent data. In DPSK, a '1' typically signifies a phase transition from the previous bit, whereas a '0' means no phase change. To visualize the transmission and reception process for a DPSK signal, we can construct block diagrams for both the transmitter and the receiver.

At the transmitter, the binary data stream is processed by a Exclusive-OR (XOR) gate, which compares the current bit to the previous bit (the reference bit is considered to be a '1'). If the bits are the same, the XOR gate outputs a '0', and if they are different, it outputs a '1'. This output is then used to control a phase modulator that shifts the phase of the carrier by π radians for a '1' and maintains the same phase for a '0', generating the transmitted DPSK signal.

The receiver includes a phase detector that compares the received phase with the reference phase from the previous bit. The receiver's decision criterion is based on whether a phase change occurred. If a phase change is detected, the output is '1'; if no phase change is detected, the output is '0'. It's worth noting that noise on the channel can cause errors in the received phases, as in the student's example where the received phases are [00π0ππ0], but the transmitted phases were [0π0π0π0].