= 36 M symbols/second

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1 Tutorial (3) Solution Problem 1: Suppose a CATV system uses coaxial cable to carry 100 channels, each of 6 MHz bandwidth. Suppose that QAM modulation is used. What is the symbol rate/channel if a four-point constellation is used? Eight-point? Given a 8 level quantizer. Given 100 channel each has: W = Hz 8 level quantizer 4-QAM m Quantizer = log 2 (M quantizer ) = log 2 8 = 3 m QAM = log 2 (M QAM ) = log 2 4 = 2 R = 2 w m quantizer 2 m QAM = QAM m Quantizer = log 2 (M quantizer ) = log 2 8 = 3 m QAM = log 2 (M QAM ) = log 2 8 = 3 R = 2 w m quantizer 2 m QAM = = 36 M symbols/second = 24 M symbols/second

Problem 2: Suppose that the receiver in a QAM system is not perfectly synchronized to the carrier of the received signal; that is, it multiplies the received signal by 2cos(2πf c t + φ) and by

2 Problem 2: Suppose that the receiver in a QAM system is not perfectly synchronized to the carrier of the received signal; that is, it multiplies the received signal by 2cos(2πf c t + φ) and by 2sin(2πf c t + φ) where φ is a small phase error. What is the output of the demodulator? However due to the imperfect synchronization the Received signal is instead multiplied by 2cos(2πfc t + φ) and by 2sin(2πfc t + φ). The transmitted signal in QAM is y(t) = Ak cos(2πfc t) + Bk sin(2πfc t ) y(t) 2cos(2πfc t + φ) = 2cos(2πfc t + φ) Ak cos(2πfc t) + 2cos(2πfc t + φ) Bk sin(2πfc t ) = Ak {cos(φ) + cos(4πfc t + φ) } + Bk {-sin(φ) + sin(4πfc t + φ) } The low pass filter removes the double-frequency component, so the output of the upper demodulator circuit is: Ak cos(φ) - Bk sin(φ). When the phase error φ is small, then cos φ is approximately 1, and sin φ is approximately 0, so the phase error causes a small error in the demodulator output. As the phase error increases however, the desired signal Ak becomes harder to discern because the cosine term decreases and the sine term increases. It can be similarly shown that the output of the lower demodulator is: Bk cos(φ) + Ak sin(φ).

3 Problem 3: A phase modulation system transmits the modulated signal Acos(2πf c t + φ) where the phase φ is determined by the 2 information bits that are accepted every T-second interval: for 00, φ = 0 ; for 01, φ = π/2 ; for 10, φ = π; for 11, φ = 3π/2. a. Plot the signal constellation for this modulation scheme. The transmitted signals corresponding to the phase values are as follows: for 00 φ = 0, so x(t) = Acos(2πfc t) for 01 φ = π/2, so x(t) = Acos(2πfc t + π/2) = -Asin(2πfc t) for 10 φ = π, so x(t) = Acos(2πfc t + π) = - Acos(2πfc t) for 11 φ = 3π/2, so x(t) = Acos(2πfc t - 3π/2) = Asin(2πfc t) The signal constellation is shown below: sin(2πf c t) cos(2πf c t) cos(2πf c t) 01 sin(2πf c t)

4 b. Explain how an eight-point phase modulation scheme would operate. The generalization to an eight-point constellation is straightforward. In the above figure we can see that the four constellation points are placed at equidistant points in a circle about the origin. The figure below shows how eight points can be placed in a circle with angle π/4 between them. sin(2πf c t) cos(2πf c t) cos(2πf c t) sin(2πf c t) Problem 4: a. Use Figure 2 below to draw the signal constellation for the modulation system described in Figure 1 Assuming that A k takes the values -1 and 1 and B k takes the values -2, - 1,1 and 2 ( please define the axis you use on Figure 2 below) Figure 1

5 Figure 2 A K = [ 1, 1] B K = [ 2, 1, 1, 2] The transmitted signals corresponding to: A K, B K Possible combinations: [-1, -2], [-1, -1], [-1. 1], [-1, 2], [1, -2], [1, -1], [1. 1], [1, 2] b. How many bits per symbol can the modulator in part a (of this question) support? 8 points M = 8 m = log 2 8 = 3 bit/symbol

6 Problem 4: Assume a digital broadcast system that supports 50 stereo music channels. The stereo music from each Channel is digitized using a 12 bit per sample quantizer. The stereo music signal has a bandwidth of 22KHz. The system is operating in an environment that has noise level equal to -50 db a. What is the sampling rate per stereo music channel Sampling rate = 2 Ws = = 22 KHz b. Calculate the bandwidth required to support the system assuming that the system uses a 16 QAM Modulator m QAM = log 2 M = log 2 16 = 4 Rate 1 channel = 2W m quantizer 2 m QAM = 2 22K = 264 K symbol/sec Rate 50 channels=system = 264K 50 = K symbol/sec c. What minimum power needed to support the system C System = W System log 2 (1 + SNR) C System = Rate System = 13200K symbol/sec W System = Bandwidth BandPass one channel 50 channels = Bandwidth Baseband one channel 2 50 channels = 22K 2 50 = 2200 KHz 13200K = 2200K log 2 (1 + SNR) 13200K 2200K = log 2(1 + SNR) = 1 + SNR SNR = SNR = P rx 1 = 63 Noise in ratio P rx in ratio = SNR Noise in ratio = = 315

Outline. EECS 3213 Fall Sebastian Magierowski York University. Review Passband Modulation. Constellations ASK, FSK, PSK.

Outline. EECS 3213 Fall Sebastian Magierowski York University. Review Passband Modulation. Constellations ASK, FSK, PSK. EECS 3213 Fall 2014 L12: Modulation Sebastian Magierowski York University 1 Outline Review Passband Modulation ASK, FSK, PSK Constellations 2 1 Underlying Idea Attempting to send a sequence of digits through