Nyquist Frequency in Hz
Consider sampling input cosine functions, with a sample rate of 100Hz.
(a) What is Nyquist Frequency in Hz?
(b) What is the period associated with Nyquist Frequency?
(c) During one period, how many samples are taken at the Nyquist frequency signal?
(d) Give the values at the sample times when you start sampling at time zero.
(e) Give the values at the sample times if you are sampling a sine function that is oscillating at
Nyquist frequency.
(f) Give the values at the sample times if you are sampling a cos (𝜔'𝑡 −
*
+) at times 𝑡 = 𝑘𝑇
where T is the sample time interval for Nyquist frequency (the
*
+ means that the cosine is
shifted 45 degrees back in time compared to the original cosine, i.e. a phase delay from input to
output).
(g) We commented in class that if a cosine is sampled such that there are 4 or more samples in
each period, then the control action based on the sampled values might be able to make some
effective corrective action. If Nyquist frequency is 100Hz, what sample rate is needed for this
condition.
(h) Now consider sampling a cosine whose frequency is 75 Hz. Use the Slide 15 and Slide 21 to
give all other frequencies that will have the same values at the sample times.
Sample Solution
(a) The Nyquist Frequency in Hz is defined as half of the sample rate. In this case, at a sample rate of 100Hz, the Nyquist frequency would be 50Hz.
(b) The period associated with the Nyquist Frequency is 1/50 seconds or 0.02 seconds.
(c) During one period at the Nyquist frequency signal, there will be 100 samples taken (100 x 0.02 = 2).
(d) When starting sampling at time zero for a cosine function, the values of each sample can be calculated using sin (2pi x t / T), where t is equal to each successive sample and T is equal to the period length, which in this case would be 0.02s. For example: Sample value at time 0: sin (2pi x 0/0.02) = sin (0) = 0; Sample value at time 1: sin (2pi x 1/0.02 ) = sin (100π) ≈ -1; Sample value at time 2: sin (2pi x 2/0.02 ) =sin200π≈-1 etc..
(e) If sampling a sine wave that oscillates at Nyquist frequency, then all sample values should have alternating positive and negative values, beginning with positive values for the first two samples and following an alternating pattern throughout all subsequent samples according to equation given above for part d). For example : Sample value 1 : Sin(2π*1/0.02)=sin(100π)= approx -1 ; Sample value 2 : Sin(2π*2/0.02)=sin200PI= approx +1 ;Sample Value 3:Sin 300PI=approx -1;etc...
(f) If sampling a cosine whose angular velocity ω't – 45°starting from time zero , then the corresponding value of each sample can again be calculated using equation mentioned in part d , but here shifted by 45 degrees i .e every odd number term should begin with negative sign while rest remain same as given in part d).For example:-Sample Value 1:-cos45°=- √21 ;Sample Value 2:-cos145°=- √21 etc...
(g) To ensure that there are 4 or more samples per period of the cosine being sampled ,the minimumrequired sample rate must satisfy T=period length which means it must cover 4 periods within total interval .In this case , since Nqyist Frequency is 100HZ hence requiredsample rate shoudlbe 400HZ i .e 400Samples Per Second.(h )If we consider sampling a Cosine whose frequency is 75 HZ than according to slide 15 & 21 other frequencies that will have same setofvalues ateachsample timewillbe 150HZ , 225hz ,300hz& 375 hz respectively .