1. Mathematical Questions in CNIM Anurag Tewari MD

3. Given SNR = Signal/Noise Given SNR= 5/20 =1/4=0.25 Required SNR=2 Which is 8 times larger than 0.25 (2/0.25=8) Number of averages required 8X8= 64 Anurag Tewari MD

7. 1 second is equal to 1000000 microsecond (SIX zeros) How many microsecond in 1 minute? 60000000 (SEVEN Zeros) Hertz is Cycle per Second Given Sampling rate is 20µS So it implies that it takes 20µS to get one sample Therefore, 1000000 µS will give us 1000000/20 = 50,000 samples As per Nyquist Theorem Sampling rate should be at least the double of the highest frequency Therefore, the highest frequency that can be resolved is 50,000 divided by two = 25,000 Hz Or 25KHz Anurag Tewari MD

9. Given SNR is 0.5/2 = 0.25 Improvement in SNR = Square root of number of averages Therefore, with 400 averages = 20 Present SNR = 0.25 With 20 times improvement SNR will be 0.25 X 20 = 5 Answer = SNR is 5 after 400 averages Anurag Tewari MD

11. A period is the time it takes for 1 cycle of the frequency to occur It is calculated as 1/frequency Period = seconds /cycle Frequency = cycles/second Thus period = 1/frequency = 1 / (cycles/second) = seconds/cycle The number of periods per second is the same as the number of cycles per second. Period of a waveform whose frequency is 200Hz 1 / (cycles/second) = 1/[(200 periods)/(second)] =.005 seconds/period = 5ms per period Period of a waveform whose frequency is 2,000,000Hz 1 / (cycles/second) = 1 / [(2,000,000 periods)/(second)] =.0000005 seconds/period =.5us per period Anurag Tewari MD

13. FREQUENCY is the number of WAVES that pass through a point in one SECOND Frequency = Waves Per Second (Waves/Second) Given Frequency = 60 Hz Given Time is 250milli seconds i.e. 0.25 seconds (250/1000) Therefore, using formula Frequency = Waves/Second or Waves = Frequency X Second 60 = Waves/0.25 Waves = 60 X 0.25 = 15 Period (T) vs. Frequency (f) Period (T) is Seconds for one cycle (unit S) Frequency (f) is cycles in one second (unit Hz) Therefore Frequency = 1/ Period Anurag Tewari MD

15. FREQUENCY is the number of CYCLES per SECOND Given Interval between each cycle = 0.04mS Frequency = 1/ Period Frequency = 1/0.04mS = 25 Converting milliseconds into Seconds 1/0.04 X 1000 = 25,000 i.e. 25KHz Anurag Tewari MD

17. Given SNR = Signal/Noise = 10/50 = 1/5 = 0.2 Number of averages done = 225 Expected Improvement = Square root of 225 = 15 Given SNR = 1/5 Final SNR = Given SNR X Improvement after averaging Final SNR = 1/5 X 15 = 3 or 3: 1 Anurag Tewari MD

19. Improvement by Averaging = Square root of number of averages done Given number of averages = 576 Square root of 576 = 24 Given Ratio = 1/8 Improvement = 24 times New SNR = Old Ratio X Improvement by averaging New SNR = 1/8 X 24 = 3 or 3/1 = 3 Anurag Tewari MD

21. 50Hz means we have 50 cycles per second So 1 Cycle = 1 second/50Hz 1 Cycle = 1000millisecond/50Hz = 20ms Anurag Tewari MD

23. Given SNR = 2:1 Lets assume it is 2 Required SNR = 20:1 Lets assume = 20 Improvement from 2 to 20 is TEN times So to get TEN times improvement Number of averages would be square of TEN 10 2 = 100 100 averages required to get SNR of 2 Anurag Tewari MD

25. Analysis Period = number of points available for averaging X Dwell Time OR Analysis Period = number of points available for averaging / Sampling Rate Given Sample Points = 120 Given Dwell Time = 10ms Therefore Analysis Period = 120 X 10 =1200 ms Anurag Tewari MD

27. Analysis Period = number of points available for averaging X Dwell Time Analysis Period = 10ms Number of Points available = 1000 Dwell time = Analysis period divided by Number of points available Dwell Time = 10/1000 = 0.01ms Anurag Tewari MD

28. What will be the frequency of a signal if in a 30 ms sweep in the interval between each cycle of a signal is 0.03 ms? A)33KHz B)66KHz C)33Hz D)66Hz Anurag Tewari MD

29. What will be the frequency of a signal if in a 30 ms sweep in the interval between each cycle of a signal is 0.03 ms? A)33KHz B)66KHz C)33Hz D)66Hz 0.03ms = 0.00003 Seconds Frequency = Wave/Second Anurag Tewari MD

31. Frequency resolution is the distance in Hz between two adjacent data points The three relevant quantities here are the total number of samples N, the sampling frequency Fs, and the total duration of the signal in the samples’ interval, T. The three are related by N=T multiplied by Fs and 1/T is the frequency resolution.

32. Anurag Tewari MD