You are designing a digital speed-monitoring system for the cruise control of a new automobile. A tachometer sensor produces a square wave signal with a 50% duty cycle. Each pulse corresponds to one full rotation of the rear right tire, The tires are 24 inches in diameter. The vehicle’s absolute top speed is 100 mph.

The sensor signal is low-pass filtered with a cutoff frequency between the tenth and eleventh harmonic of the signal. The minimum sampling frequency (samples per second) required to avoid aliasing when the vehicle is at its top speed is most nearly.

A.   23.4
B.   234
C.   467
D.   1,000

According to the Nyquist theorem, in order to avoid aliasing, the sampling frequency must be at least twice the highest frequency of the signal to be sampled.

We will first need to calculate the frequency of our signal.

We are told that each pulse corresponds to one full rotation of the rear right tire. We are also told that the tires are 24 inches in diameter and the vehicles absolute top speed is 100 miles per hour. 

From this information, we want to figure out how quickly the tire completes one full rotation at a speed of 100 miles per hour. To do so, we need to first calculate the circumference of the tire. The circumference of a circle is as follows:

$$\begin{gathered} C = \mathrm{\pi d} \\ \mathrm{C} = \mathrm{Circumference} \\ \mathrm{d} = \mathrm{Diameter} \end{gathered}$$

Our wheel's circumference is therefore:

$$\begin{gathered} C = \mathrm{\pi d} \\ \mathrm{C} = \left(\mathrm{\pi }\right)\left(24\mathrm{inches}\right) \\ \mathrm{C} = 75.398\mathrm{inches} \end{gathered}$$

Therefore, during one full rotation of the rear tire, the car will travel a distance of 75.398inches.

Since we know that 1 mile is equal to 63,360 inches, our car's top speed is 63,360*100 = 6,336,000 inches per hour. If we divide this number by 3600, since there are 3600 seconds in an hour, we know the car is moving at 1,760 inches per second.

Therefore, in order to complete one full rotation of the rear tire, and therefore one full pulse:

$$\begin{gathered} v = \frac{d}{t} \\ 1760=\frac{75.398}{t} \\ t = \frac{75.398}{1760} \\ t = .04283977s \end{gathered}$$

Therefore the period of our signal is .043 seconds. To calculate frequency:

$$\begin{gathered} f = \frac{1}{T} \\ f = \frac{1}{.04283977s} \\ f = 23.343Hz \end{gathered}$$

Since the cutoff frequency is described to be somewhere between the 10th and 11th harmonic, the highest frequency of the signal is the 10th, shown below:

$$\begin{gathered} f_{10th-harmonic}=10f \\ {f}_{10th-harmonic}=10(23.256Hz) \\ f = 233.43Hz \end{gathered}$$

Now let's multiply this by two to get our Nyquist frequency, which will be our answer:

$$\begin{gathered} f_{Nyquist}=2f \\ {f}_{Nyquist}=2(233.43Hz) \\ {f}_{Nyquist}=466.86Hz \end{gathered}$$

Therefore, our minimum sample frequency to avoid aliasing is therefore 467Hz, which corresponds to answer C.