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**Given I know the period of a signal how do I find the frequency**

**Heres what I know**

Period = 10 years so using T = 1/f where f = frequency would that give me 1/10??

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- Thread starter andrey21
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Period = 10 years so using T = 1/f where f = frequency would that give me 1/10??

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Please any feedback would be great.

- #3

berkeman

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Given I know the period of a signal how do I find the frequency

Heres what I know

Period = 10 years so using T = 1/f where f = frequency would that give me 1/10??

The standard unit of frequency is Hz (Hertz), which is 1/seconds. Not 1/years.

Can you convert the period of 10 years into Hz? (It will be a very small number)

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Thank you for moving my post berkeman, I think I may have made an error with the question. I was asked to determine the frequency that corresponds to the period observed in a data set. Given the period is 10 years, im basically trying to find the cutoff frequency. If that makes any sense.

- #5

berkeman

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Doesn't make sense to me yet. What do you mean by cutoff frequency? I'm familiar with that term in filter design. What is the dataset?

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Yes well I have been asked to design a low pass filter, given that i know period of 10 years and sampling interval of 1 month.

- #7

berkeman

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What order digital filter do you plan to design? Which polynomial will you choose and why?

Other than the very low frequencies involved, it sounds like a straightforward digital filter design.

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Well ive been told either 4 or 5th order. What do you mean by polynomial?

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- #9

berkeman

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Well ive been told either 4 or 5th order. What do you mean by polynomial?

The filter functions are polynomial functions, and there are several different forms of the polynomials to choose from. You choose which one (like Cheby, Butterworth, etc.) based on what you are trying to optimize in your filter's response.

More here:

http://en.wikipedia.org/wiki/Filter_(signal_processing [Broken])

.

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Oh I understand I was planning on using a Butterworth filter.

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berkeman

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Well, what do you think you should choose for a cutoff frequency? What is the tradeoff with a Butterworth filter in choosing the cutoff closer and closer to the frequency region of interest?

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Im not to sure but I have to get it in terms of degrees, for example 0.2PI = 36 degrees etc...

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berkeman

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Im not to sure but I have to get it in terms of degrees, for example 0.2PI = 36 degrees etc...

Sorry, get what in terms of degrees? Degrees would relate to some phase. But you are asking about filtering in the time domain according to a frequency characteristic (the filter)...

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berkeman

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Can you post a copy of the figure?

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Figure? U mean the period? If so it is 11 years.

- #18

berkeman

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Using the figure(period) as a filter cutoff period together with known sampling interval (1 month) design a low pass filter to remove higher frequencies.

Figure? U mean the period? If so it is 11 years.

No, you said "using the figure", so I thought there was a figure that graphed the data as s function of time. It must be a semantics thing -- it looks like the word figure was used as a synonym of period?

So they want you to use the

Then just use your filter design software to design the coefficients for the Butterworth polynomial. Are you using an FIR or IIR digital filter?

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So I have to convert 1/11 years into Hz per second?

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berkeman

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So I have to convert 1/11 years into Hz per second?

1/(11 years) = 1/(how many seconds?) = ?? * 10^-?? Hz

- #21

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Basically I have:

A period of 11 years

A Nyquist frequency of 1/2 month (as sampling interval is one month, using 1/2(delta t)

I am asked to convert the cutoff frequency on the 0 to PI scale???

Does that sound correct to you?

Thanks in advance.

- #22

berkeman

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Basically I have:

A period of 11 years

A Nyquist frequency of 1/2 month (as sampling interval is one month, using 1/2(delta t)

I am asked to convert the cutoff frequency on the 0 to PI scale???

Does that sound correct to you?

Thanks in advance.

No, not sounding right. First, "frequency" is not plotted on any "0 to PI scale" that I'm familiar with. Phase angle can be plotted on such a scale, but I don't know what you mean by plotting frequency on a phase scale.

And when you make a lowpass filter to help avoid aliasing (or whatever), you want to pass the frequencies of interest, and block frequencies at the sampling frequency and above. The Nyquist frequency is twice the highest signal content frequency, so in this case it would be twice 1/T, or twice 1/11 years. They just happen to be oversampling in this problem, sampling at 12 * 11 the signal period.

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- #24

berkeman

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1/(11 years) = 1/(how many seconds?) = ?? * 10^-?? Hz

Correct. Usually frequency is expressed in Hz.

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