Consider a simple DPCM encoder in which a first-order predictor is used for the following message m(t) = A cos(mt ). The sampling interval is Ts such that the k-th sample is given as m[k] = m(kTs). The signal contains an arbitrary phase = 0.5mTs. The first-order predictor is formed by m q[k] = m[k 1], with prediction error (difference) as d[k] = m[k] m q[k] = A cos(mkTs ) cos(mkTs mTs) . (a) Determine the peak value of d[k] and call it dp. Hint 1: Use the following trigonometric identity to simplify d[k] cos() cos() = 2 sin( 2 ) sin( 2 ). (b) Evaluate the amount of SNR improvement in dB that can be achieved by this DPCM over a standard PCM. Hint 2: The SNR for PCM (with peak value mp) and for DPCM (with peak value dp) can be written as follows SNRPCM = 3L 2 S0 m2 p , SNRDPCM = 3L 2 S0 d 2 p , where S0 is the signal power and L is the number of quantized levels. You can assume S0 and L are the same for both