...DeVry University College of Engineering and Information Science Pomona, California MODULATION IN THE BIOMEDICAL FIELD By JeanPaul Gagner Aceflyte99@gmail.com Submitted in Partial Fulfillment of the Course Requirements for Communications Systems w/Lab ECET-310 Professor: Mohammad Muqri April 20, 2014 Since the 1950s, wireless medical implant devices have become increasingly useful in monitoring, diagnosing, and adapting a person’s physiological inability to perform mundane tasks. While there are many advantages to these devices, the engineering behind them reveals many issues with the communicating ability of these wireless implants. With patient safety and comfort in mind, overall size and power requirements became an issue. Since the most sensitive parts of the implant are inside the patient’s body, surgery for repair and maintenance can present a risk to the patient and device. Cost is always an issue as well, so creating the most efficient, reliable, and cost-effective device is of the utmost priority. Possibly the most concerning aspect of the wireless design process is conserving power and reducing interference. Since most internal devices are self-controlled, they can limit their own operation according to the needs of the patient, such as a pacemaker. The power to operate these types of devices is supplied by external battery, or created by radio-frequency signals running through inductive coupling links. This creates...
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...Homework 1_1 ECET310 1. Add: (a) 60 dB + 25 dB= 85 dB (b) 20 dBW + 5 dBW= 25 dBW (c) 22 dB - 4 dB= 18 dB 2. Convert: (a) 65 W into dB= 18.1 dB (b) 10 dB into watts= 10/10 = log x => 10W (c) – 5 dB into watts= 5 = 10 log x x=10 ^(-5/10) = 0.316W (d) –30 dBm into dBrn= 30 dBrun (e) 15 dBrn into dBm= 10 log (p/1mW) 15/10= log (p/1mW) 1.5 =log (p/mW) = -15dB 2. A three-stage amplifier is shown with power gains. Calculate total power gain in decibels, and as a number. Ap(tot)=(Ap1)(Ap2)(Ap3)=7500 Ap(tot) (dB) = 10 log(10) +10 log (25) +10 log(30) = 38.75dB 3. (a) Determine the overall power gain. (b) Find the output power if Pin = 15 mW a) Ap(t) = (10)(10)(20) = 2000 b) Po = (Pin)(Ap(t)) = (15mW)(2000) = 30W 4. If Pin = -20 dBm, determine the output power in dBm and watts. Ap(t) = 13+16-6 = 23dB Ap(t) = log^-1 (23/10) = 199.526 Pin = -20dBm = (1mW)(log^-1(-20/10)) = 1mW* 10^-2 = 10 uW Po = (Pin)(Ap(t)) = 10u*199.526 = 1.995 mW Po(dBm) = 10 log (Po)/1mW = 10 log (1.995mW/1mW) = - 27dBm 5. Determine: (a) Net gain (b) Net gain in dBs (c) Power output (a) Net...
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...Homework 1_1 ECET310 1. Add: (a) 60 dB + 25 dB = 85 dB (b) 20 dBW + 5 dBW = 25 dBW (c) 22 dB - 4 dB = 18dB 2. Convert: (d) 65 W into dB P(dB) = 10 log(65W/1W) = 18.1 dB (e) 10 dB into watts P(watt) = log^-1(P(10)/10) = 10 W (f) – 5 dB into watts P(watt) = log^-1(P(-5)/10) = 316.228 mW (g) –30 dBm into dBrn dBrn = dBm + 90 = -30dBm+90 = 60 dBrn (h) 15 dBrn into dBm AP3 = 30 AP2 = 25 AP1 = 10 2. A three-stage amplifier is shown with power gains. Calculate total power gain in decibels, and as a number. Ap(tot) = (10)(25)(30) = 7500 Ap(tot)(dB) = 10 log(10) + 10 log (25) + 10 log (30) = 38.75 dB AP3 = 20 AP2 = 10 AP1 = 10 3. (a) Determine the overall power gain. Ap(tot) = (10)(10)(20) = 2000 (b) Find the output power if Pin = 15 mW Po = (Pin)(AP(T)) = (15mW)(2000) = 30 W AP3 = -6 dB AP2 = 16 dB AP1 = 13 dB 4. If Pin = -20 dBm, determine the output power in dBm and watts. Ap(tot) = 13dB + 16dB+ -6dB = 23 dB Ap(tot) = log-1(23/10) = 199.526 Pin = log-1(-20 dBm/10) = 0.01 mW Po = (Pin)(AP(T)) = (0.01mW)(199.526) = 1.995 mW Po(dBm) = 10 log (Po)/1mW) = 10 log (1.995mW/1mW) = 2.999 dBm PS = 0.1W AP1 AP2 AP1 = 400 L = 2000 AP2 = 500 RL 0 1 5. Determine: (a) Net gain (b) Net gain in dBs (c) Power output AP(T) = (AP1)(1/L)(AP2) = (400)(1/2000)(500) = 100 AP(T)(dB) = 10log(100)...
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