At What Speed Will The Spot Appear To Rotate To The Human Observer?
ECET345 Signals and Systems Homework #4
Name of Student ______________________________
1. A shiny metal disk with a dark spot on it, as shown in figure below, is rotating clockwise at 100 revolutions/second in a dark room. A human observer uses a strobe that flashes 99 times/second to observe the spot on the metal disk (a strobe is a flashing light whose rate of flashing can be varied). The spot appears to the human observer as if it is rotating slowly.
Hint: This is equivalent to sampling a 100 Hz wave with a 99 Hz sampling rate. Because this represents a violation of the Nyquist theorem, an aliasing condition will result. The resulting aliased frequency observed at system output is given by (k * fs - f), where fs is the sampling frequency, f is the frequency of sampled signal, and k is an integer (positive or negative) such that (k * fs - f) produces a positive output less than or equal to fs/2. …show more content…
(a) At what speed (in revolutions per second) will the dark spot appear to rotate to the human observer?
Fs/2 = 50
K(fs)-fi=50
K(100)-90= 50 where K = 1.4
1.4 *2pie = 8.79
(b) Will the spot appear to rotate clockwise or anticlockwise? Briefly justify your answer.
Clock wise due to positive number between 100-90
(c) Suppose the observer changes the strobe frequency to 101 flashes per second. What differences in rotation will the observer note?
Will move anti clockwise due to 100-101 equalling negative number at 1.51
hz
(d) How will the spot appear to the observer if strobe frequency is exactly 100 flashes/second?
unmoving
2. (a) A system samples a sinusoid of frequency 480 Hz at a rate of 100 Hz and writes the sampled signal to its output without further modification. Determine the frequency that the sampling system will generate in its output.
Hint: keep in mind that no discrete system can generate an output greater than fs/2 where fs is the sampling frequency. Also, the frequencies produced by sampling a sinusoid are given by
However, the only frequency that will come out of the system will be the one that is between 0 and fs/2. So choose k, positive or negative, such that it yields a positive output frequency less than fs/2. Note that only one value of k will yield a positive frequency in the range 0 to fs/2.
K= 5fs-fin = 5(100)-480 = 20 Hz
(b) Find the output frequency when the input frequency is 425 Hz. Assume fs is unchanged at 100 Hz. Show the value of k chosen.
K= 4(fs)-425 = 4(100)-425= 25
3. The spectrum of an analog signal is shown below, containing . Such a signal is sampled by an ideal impulse sampler at a 100 Hz rate. List the first 10 positive frequencies that will be produced by the replication. (Hint: Follow the method outlined in the lecture for spectrum replication of sampled signals.)
75, 90, 110, 125,175,190,210,275,290,310 4. The spectrum of an analog signal is shown below. It is sampled, with an ideal impulse sampler, at a rate of 200 Hz.
(a) Draw the spectrum of the sampled signal and show the baseband and first two positive frequency replications with their end frequency values marked. (Hint: Follow the method outlined in the lecture.)
Points will be at 225,200 and 75
(b) State below the frequency range of the signal, in baseband, that will become corrupted and thus unrecoverable because the signal was undersampled. (Hint: The area of corruption is the area in which baseband and first positive frequency image overlaps.)
75 hz is when the undersampling will occur
5. Determine the Z transform of the signal,, shown below using the basic definition of Z transform . All values not shown can be assumed to be zero.
6. (a) A simulation diagram is shown below. Determine the difference equation associated with the diagram.
X.5+.8x(z)^-1+x(z)^-2+.25x(z)^-3=Y(n)+.75(y)^-1+.6Y(z)^-2
(b) A DE is shown below. Draw the simulation diagram corresponding to it.
Name of Student ______________________________
1. A shiny metal disk with a dark spot on it, as shown in figure below, is rotating clockwise at 100 revolutions/second in a dark room. A human observer uses a strobe that flashes 99 times/second to observe the spot on the metal disk (a strobe is a flashing light whose rate of flashing can be varied). The spot appears to the human observer as if it is rotating slowly.
Hint: This is equivalent to sampling a 100 Hz wave with a 99 Hz sampling rate. Because this represents a violation of the Nyquist theorem, an aliasing condition will result. The resulting aliased frequency observed at system output is given by (k * fs - f), where fs is the sampling frequency, f is the frequency of sampled signal, and k is an integer (positive or negative) such that (k * fs - f) produces a positive output less than or equal to fs/2. …show more content…
(a) At what speed (in revolutions per second) will the dark spot appear to rotate to the human observer?
Fs/2 = 50
K(fs)-fi=50
K(100)-90= 50 where K = 1.4
1.4 *2pie = 8.79
(b) Will the spot appear to rotate clockwise or anticlockwise? Briefly justify your answer.
Clock wise due to positive number between 100-90
(c) Suppose the observer changes the strobe frequency to 101 flashes per second. What differences in rotation will the observer note?
Will move anti clockwise due to 100-101 equalling negative number at 1.51
hz
(d) How will the spot appear to the observer if strobe frequency is exactly 100 flashes/second?
unmoving
2. (a) A system samples a sinusoid of frequency 480 Hz at a rate of 100 Hz and writes the sampled signal to its output without further modification. Determine the frequency that the sampling system will generate in its output.
Hint: keep in mind that no discrete system can generate an output greater than fs/2 where fs is the sampling frequency. Also, the frequencies produced by sampling a sinusoid are given by
However, the only frequency that will come out of the system will be the one that is between 0 and fs/2. So choose k, positive or negative, such that it yields a positive output frequency less than fs/2. Note that only one value of k will yield a positive frequency in the range 0 to fs/2.
K= 5fs-fin = 5(100)-480 = 20 Hz
(b) Find the output frequency when the input frequency is 425 Hz. Assume fs is unchanged at 100 Hz. Show the value of k chosen.
K= 4(fs)-425 = 4(100)-425= 25
3. The spectrum of an analog signal is shown below, containing . Such a signal is sampled by an ideal impulse sampler at a 100 Hz rate. List the first 10 positive frequencies that will be produced by the replication. (Hint: Follow the method outlined in the lecture for spectrum replication of sampled signals.)
75, 90, 110, 125,175,190,210,275,290,310 4. The spectrum of an analog signal is shown below. It is sampled, with an ideal impulse sampler, at a rate of 200 Hz.
(a) Draw the spectrum of the sampled signal and show the baseband and first two positive frequency replications with their end frequency values marked. (Hint: Follow the method outlined in the lecture.)
Points will be at 225,200 and 75
(b) State below the frequency range of the signal, in baseband, that will become corrupted and thus unrecoverable because the signal was undersampled. (Hint: The area of corruption is the area in which baseband and first positive frequency image overlaps.)
75 hz is when the undersampling will occur
5. Determine the Z transform of the signal,, shown below using the basic definition of Z transform . All values not shown can be assumed to be zero.
6. (a) A simulation diagram is shown below. Determine the difference equation associated with the diagram.
X.5+.8x(z)^-1+x(z)^-2+.25x(z)^-3=Y(n)+.75(y)^-1+.6Y(z)^-2
(b) A DE is shown below. Draw the simulation diagram corresponding to it.