4.2 Java1483-spectrum analysis using java, frequency resolution  (Page 11/19)

What can we learn from this?

Since we have concluded that the middle pulse in Figure 4 is sufficiently long to allow us to resolve the two peaks, let's see what we can learn from theparameters that describe that pulse.

Pulse length and frequency separation

To begin with, the length of the pulse is 100 samples.

What about the frequency separation of the two sinusoids? Recall that the frequency of one sinusoid is (0.0625 - 2.0/len) times the sampling frequency,where len is the length of the time series containing the pulse. The frequency of the other sinusoid is (0.0625 + 2.0/len) times the sampling frequency.

Thus, total separation between the two frequencies is 4/len, or 4/400. Dividing through by 4 we see that the separation between the two frequencies is1/100.

Eureka, we have found it

For the third pulse, the frequency separation is the reciprocal of the length of the pulse . Also, the length of the third pulse is barely sufficient to allow for separation and identification of the two peaks in the spectrum.

Thus, the two spectral peaks are separable in the absence of noise if the frequencyseparation is the reciprocal of the pulse length. (That is too good to be a coincidence. I must have planned that way.)

Thus, we have reached another conclusion. Under ideal conditions, the two peaks in the spectrum can be resolved when the separation between the frequencies of the two sinusoids is equal tothe reciprocal of the pulse length.

The general answer

There is no single answer to the question "how long must the operating bursts of this device be in order for you to resolve the peaks andidentify the enemy submarine under ideal conditions?"

The answer depends on the frequency separation. The general answer is that the length of the bursts must be at least as long as the reciprocal of thefrequency separation for the two sinusoids. If the separation is large, the pulse length may be short. If the separation is small, the pulse length must belong.

The program named Dsp032

As I indicated earlier, the plots shown in Figure 5 were the result of running the program named Dsp032 and displaying the data with the program named Graph03 .

The only thing that is new in this program is the code that generates the five pulses and saves them in their respective data arrays. Even that code isnot really new, because it is identical to the code shown in Listing 6 . Therefore, I won't discuss this program further in this module.

One more experiment

As you can surmise from the conclusions reached above, in order to be able to resolve the two peaks in the spectrum, you can either keep the pulse length thesame and increase the frequency separation, or you can keep the frequency separation the same and increase the pulse length.

Let's examine an example where we keep the pulse lengths the same as before and adjust the frequency separation between the two sinusoids to make it barelypossible to resolve the peaks for each of the five pulses.

We will need to increase the frequency separation for the first two pulses, and we can decrease the frequency separation for the fourth and fifth pulses. Wewill leave the frequency separation the same as before for the third pulse since it already seems to have the optimum relationship between pulse length andfrequency separation.