# how to find the frequency of this nonperiodic composite signal?

I'm new to electrical engineering, sorry if my question sound dumb. Below is a nonperiodic composite signal plot: As you can see this is a nonperiodic composite signal in 1s and I know that if the composite signal is nonperiodic, the decomposition gives a combination of sine waves with continuous frequencies.I don't know how to use fourier analysis to decomposite this composite signal into simple sine waves.

Can I just make a quick conclusion to say because the signal completes 4 cycles in one second, so the frequency is 4Hz?

PS: This questions originates from my textbook's description of signal received by an old-fashioned analog black-and-white TV. It says

An example of a nonperiodic composite signal is the signal received by an old-fashioned analog black-and-white TV.A TV screen is made up of pixels with each pixel being either white or black. The screen is scanned 30 times per second.If we assume a resolution of 525 × 700 (525 vertical lines and 700 horizontal lines), we have 367,500 pixels per screen. If we scan the screen 30 times per second, this is 367,500 × 30 = 11,025,000 pixels per second. The worst-case scenario is alternating black and white pixels. In this case, we need to represent one color by the minimum amplitude and the other color by the maximum amplitude. We can send 2 pixels per cycle. Therefore, we need 11,025,000 / 2 = 5,512,500 cycles per second, or Hz. The bandwidth needed is 5.5124 MHz.

So I'm going to simplify this and connect my textbook's description to my original question. Let's say I have a tiny TV that only has 8 pixels. So I am sending the signal in the picture, representing 8 pixels, the high amplitude represents "white" and low amplitude means "black", so the signal in the pictures represents the pixels(from pixel 1 to pixel 8) in the TV should be "wbwwbwww", then according to the textbook, We can send 2 pixels per cycle, Therefore, we need 8 / 2 = 4 cycles per second(4 Hz).

• Looks like a frequency + amplitude modulated signal, with 8Hz and 4Hz states, May 9, 2020 at 8:26
• You haven't got a repeating pattern in the sample you've shown. You'd need to see that complete pattern repeat before you could figure that out. May 9, 2020 at 8:39
• If the signal is nonperiodic, then what is your definition of frequency? Any number would work, so there must be something that is more useful to you than something else.
– pipe
May 9, 2020 at 8:40
• @Transistor I have added more context, could you have a check? May 9, 2020 at 9:02
• @pipe I have added more context, could you have a check? May 9, 2020 at 9:02

I think the question includes a contradiction, the frequency serves to explain the periodicity of a signal in the frequency domain. Then in case of non-periodic signals we use the term bandwith that is the term that indeed that book used, at the end of the paragraph it says, quoted "The bandwidth needed is 5.5124 MHz". The signal shown is simple so most likely a combination of a few sine waveforms, the highest frequency I see in there is 8Hz, it repeats evey square and there are 8 squares in one second. I think the textbook's intention when speaking about analog TV is a bit different to what you try to do.

• Thanks for your answer. Actually I'm confused with the term "cycle", isn't that only periodic signal has cycle, how come nonperiodic composite signal also has cycle? And why we need 11,025,000 / 2 = 5,512,500 cycles per second, that's what confused me the most, that's why I come up with my original question to try use 8 pixels TV May 9, 2020 at 10:43
• I recommend you to forget that example, from what I read it just confuses you. The frequancy domain is not easy to understand at the beginning. The word cycle is used to indicate that something, some event, a signal, whatever has a periodicity hence you are right when saying only periodic signals have a cycle. The analog TV signal, for instance is not periodic, is not cyclic, what happens with analog TV signal is that all the information necessary to watch standard TV can be sent within 5.5MHz. May 9, 2020 at 18:50

You have to use the Fourier transform. But Fourier transform is for a continuous time-domain signal. But when you will measure a signal using ADC you will get a discrete time-domain signal. Then you will need to do DFT (i.e. Discrete Fourier transform).

There are several ways to find the DFT sequence. The algorithms are commonly known as FFT (Fast Fourier transform). DIT FFT and DIF FFT are the simplest. This playlist has three youtube videos that may help you

This is an output of DIF FFT which I did two days ago. It's a 16 point DIF FFT. I mixed two signals and decomposed it into frequency domain using DIF FFT.

In the context of your TV line question, you are sending 8 pixel values per second.

You've chosen to bring the signal back to zero between each pixel, as so-called RZ (Return to Zero) code. The frequency content is always 'worst case' at 8 full cycles per second. This means you need a bandwidth of DC to 8 Hz to transmit the signal, if you want to preserve the waveform you've shown.

If instead you had used a NRZ code (non return to zero), then the worst case, as is described in your question, would be alternative BWBW pixels, for a frequency of only 4 Hz. This would need a bandwidth of only DC to 4 Hz to transmit.

This is why signals tend to be NRZ rather than RZ.

There's a major flaw in the question. Analog TV doesn't know about pixels. It knows about lines (PAL used 625, for example) but doesn't know about 'vertical lines' or columns. The signal is a continuous analog sweep. Figure 1. A composite video signal. Notice that the Active Video portion is always positive and that the high level stays on for as long as the image is "white". There is no periodic signal within the active video unless the image has a periodic pattern. Image source: National Instruments.

Your book then goes on to discuss black and white as colours! They're not. They're the same 'colour' (white) with full brightness or zero brightness and always ≥ 0 (and not going negative as shown in your graph).

I think there will be a better textbook out there. This one is gobbledygook.

• Many books about video engineering explain the concept of horizontal resolution like this Sure there are no pixels in the analog signal but if the analog signal is digitized or is output from digital frame buffer, then you have a resolution in pixels which relates to how much analog bandwidth the analog signal needs. If you have some bandwidth for video, a sine wave with that frequency can be represented as repeating two alternating digital codes, or pixels, and that tells you how many digital samples you can have per visible line or per total line, i.e the sample clock or pixel clock. Nov 4 at 12:25