There are a few questions in CNIM exam that would require you to use your knowledge of simple mathematics to derive to an answer. Here are a few representative questions. Please do read more and practice as many questions as you can.

1. Mathematical Questions
in CNIM
Anurag Tewari MD
Anurag Tewari MD

2. 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

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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

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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

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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

12. 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

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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

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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

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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

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

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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
102 = 100
100 averages required to get SNR of 2
Anurag Tewari MD

24. 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

26. 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

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31. Anurag Tewari MD
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.

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