# DTFT angular frequency

Tags:
1. Sep 13, 2017

### MikeSv

Hello everyone.
Iam trying to understand the discrete time fourier transform for a signal processing course but Iam quite confused about the angular frequency.

If I have a difference equation given, what values should I choose for my angular frequency if I do
not know anything about the sample frequency?
Should they go from - pi to pi or from 0 to 2pi?

And what does it mean if The frequency is given in 'units of pi'?

Can I convert this into Hz?

Mike

2. Sep 13, 2017

### Baluncore

Angular frequency in radians per second, ω = 2 ⋅ π ⋅ Frequency ( in Hz. )
Time can be -π to +π or from 0 to 2π. Either is possible, depending on definition.
You should calibrate any DFT for phase and amplitude by generating a known input fundamental sine or cosine function and seeing what phase and amplitude it returns.

If you have n samples at a rate of r samples per second. Maximum frequency will be r/2 Hz.
The output will be frequency from 0 to fmax, but aliasing will wrap higher frequencies around through zero.
There will be n/2 discrete frequencies generated.

3. Sep 14, 2017

### MikeSv

So the maximum frequency Iam able to see in my DTFT is 1/2 of my sampling frequency?

That means I have to multiply my angular frequency by 1/2 the sampling frequency to get the frequency values in Hz, right? (In case the angular frequency is normalized)

But what if I have a sequence given without knowing anything about the sample rate? Can I get some useful information from my DTFT plot by just looking at the angular frequencies without knowing anything about my "frequency range"?

Thanks again,

Mike

4. Sep 14, 2017

### Baluncore

The DTFT computation uses the FFT algorithm. You provide n data points and it returns n/2 cosine terms and n/2 sine terms. That makes n/2 complex phasors. For example;
Sample 8 points in time at a rate of 8 samples per second, the DTFT will give 4 frequency bins. The acquisition time cycle wraps around at one second, so the frequency bins will each be 1/1sec = 1Hz wide.
The 8 DTFT outputs will make 4 complex numbers, or phasors, for frequencies of; 0, 1, 2, and 3. There is no frequency 4 as it is alias 0. The Cos(0) will be the DC offset, the Sin(0) should cancel to be zero.

Sampling data is also a form of harmonic mixing. If you digitise a 999kHz signal at 1MHz you will get a 1kHz waveform. When higher frequencies are present in the data, they will be mapped, or aliased, down into the fundamental spectrum. According to Shannon, you must sample at twice the highest frequency present.
https://en.wikipedia.org/wiki/Nyquist–Shannon_sampling_theorem

Yes, with a trap. The first element will be at frequency zero. The last frequency element will be the channel below Fsample/2 = Freq( (n/2) – 1). Remember the 0 to n–1, means you need to know n to scale frequency precisely. Discrete transforms have that digital counting problem.
Depending on how it is normalised you will need to multiply by n/2 and divide by the full scale value. You can only be sure if you calibrate the transform with a precise cosine wave and check that the “energy” ends up in the correct frequency bin, with the correct phase and amplitude.

If you know the input was a single cycle of a repeating signal then you can study the harmonic content of the waveform. Phase will be meaningless, so you must study the amplitude of the odd and even harmonic phasors to identify the signal.

5. Sep 14, 2017

### jasonRF

I agree with Baluncore on how to translate frequencies.

However, this conversation is a little confusing when it comes to samples, especially in frequency space. Are we talking about the Discrete Fourier Transform (DFT), or the Discrete Time Fourier Transform (DTFT)? Baluncore is clearly talking about the DFT, while I thought MikeSv was asking about the DTFT.

Jason