Signal Discrete Fourier Transform

In summary, the conversation discusses a homework problem involving the discrete Fourier transform (DFT) of a sampled signal. The signal is defined and sampled at a given frequency and for a given time. The conversation explains how to determine the number of samples and how to plot the modulus of the DFT. The significance of using a rectangular window is also discussed, and various methods for applying a window to the signal are mentioned. The original poster later updates that they passed the exam.
  • #1
feelix
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Hi guys, I'm a bit embarassed that my first post here is in this section, but I'm taking a Elec. Eng. course abroad, which is out of my confort zone (i'm majoring in automotive eng.) and I'm trying to solve a few model problems. This one in particular deals with the DFT. Anyway:

Homework Statement



Consider the signal: y= 2*sin(1.5*2*pi*t)

the signal is sampled with a sampling frequency fc=8 Hz, for a time of 1 second and using a rectangular window. plot the modulus of the Discrete Fourier Transform.

-----------

I'm completely lost... I think I have to determine N, using the fc and sampling time, so something like this:

fc=8 Hz => 8=1/T and therefore T(sampling time) = 1/8 s

Since the total sampling time is 1s, for 1s we have 8 samples and so N=8 ?

Is this correct, and if so, where do I go from here?Thanks!
 
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  • #2
I'm not sure what the significance is of using the rectangular window. Hopefully, someone else will chime in and explain that.

In any case, you want to calculate what you'll get for the 8 samples. You'll have a sequence of 8 numbers {x0, x1, ..., x7}. You want to calculate the DFT of this sequence, which will result in a sequence of 8 complex numbers {y0, y1, ..., y7}. You want to plot the modulus of these numbers.
 
  • #3
you have a temporal signal y defined by:

f0 = 1.5
y(t) = 2*sin(2*pi*f0*t)

where f0 is the fundamental frequency in Hz. That is, the sin wave oscillates at f0 cycles per second.

You are told to sample at Fs = 8 (Hz) for 1 (s). Notice that Fs > 2*f0 so you *are* satisfying Nyquist.

As you stated, Ts, the sampling interval is:
Ts = 1/Fs = 1/8 (s)

so yes, the total sampling time is 1 (s) and the number of samples is N = 8

So you can do this a couple of ways:

n = {0, 1, … , N - 1} = {0, 1, …, 7}
Form the discrete signal:
y'[n] = y(n*Ts) = 2*sin(2*pi*f0*(n*Ts))

You now take the DFT of the discrete signal y' such that:
Y = DFT{ y' }

Note that Y is a complex signal, so you'll need to take the magnitude.
Are you supposed to work this out by hand? If you are using MATLAB, you would plot this as:

figure;
f = linspace(0, Fs - Fs/N, N)
plot(f, abs(fft(y', N))It's an odd way to phrase that you are using a rectangular window, seeing how it says to sample the signal, which you are doing. However, the process of obtaining y'[n] is equivalent to a rectangular window, since you are truncating the signal. Note that your original signal y(t) has an infinite time support (ie it is valid for all values of t)

If you are familiar with the rect function, then you are first forming a "new" time signal:

Tp = N*Ts
y'(t) = rect(t/Tp)*y(t)
and sampling this, and then performing the DFT.

You could also "chop" your original signal with a window (eg Hann, Taylor, ...). You could also create the y[n] signal and then apply the window in a discrete form.
 
  • #4
Cheers for the help guys, exam was today and I passed it!

RE: the window. Some example problems were for the Hann window, others for the rectangular.
 
  • #5


Hello,

First of all, don't be embarrassed about asking for help in a subject that is outside of your comfort zone. It takes courage to step out of your comfort zone and try new things. So, kudos to you for taking on this challenge!

Now, let's get to the problem at hand. The Discrete Fourier Transform (DFT) is a mathematical tool used to convert a signal from the time domain to the frequency domain. In simpler terms, it helps us understand the different frequencies present in a signal.

In this problem, you are given a signal y = 2*sin(1.5*2*pi*t), which is a sine wave with a frequency of 1.5 Hz. The signal is sampled at a frequency of 8 Hz, which means that the signal is measured 8 times in one second. The sampling time is 1/8 seconds.

To plot the modulus of the DFT, you first need to determine the number of samples (N) that you will use to calculate the DFT. You are correct in your calculation that for a sampling time of 1 second, with a sampling frequency of 8 Hz, you will have 8 samples. So, N = 8.

Next, you need to calculate the discrete frequencies at which the DFT will be calculated. These frequencies are given by the formula:

f_k = k*fc/N

where fc is the sampling frequency and N is the number of samples. So, in this case, the discrete frequencies will be:

f_0 = 0*8/8 = 0 Hz
f_1 = 1*8/8 = 1 Hz
f_2 = 2*8/8 = 2 Hz
f_3 = 3*8/8 = 3 Hz
f_4 = 4*8/8 = 4 Hz
f_5 = 5*8/8 = 5 Hz
f_6 = 6*8/8 = 6 Hz
f_7 = 7*8/8 = 7 Hz

Now, you can calculate the DFT using the formula:

X_k = sum(y[n]*e^(-i*2*pi*f_k*n))

where n is the sample number and X_k is the DFT value at frequency f_k.

Once you have calculated the DFT values at these discrete frequencies, you can plot the modulus of the
 

What is a Signal Discrete Fourier Transform?

A Signal Discrete Fourier Transform (DFT) is a mathematical tool used to analyze and represent signals, such as sound or electrical signals, in the frequency domain. It converts a signal from its original time domain into its component frequencies, allowing for further analysis and processing.

How does a Signal DFT work?

A Signal DFT works by breaking down a signal into its individual frequency components using complex mathematical formulas. It converts the signal from a continuous function in the time domain to a discrete function in the frequency domain, with each frequency component represented by a specific amplitude and phase.

What is the difference between a Signal DFT and a Continuous Fourier Transform?

The main difference between a Signal DFT and a Continuous Fourier Transform is that a Signal DFT operates on discrete data points, while a Continuous Fourier Transform operates on continuous data. This means that a Signal DFT is more suitable for analyzing digital signals, while a Continuous Fourier Transform is better for analyzing analog signals.

What are the applications of Signal DFT?

Signal DFT has many applications in various fields, including audio and video processing, communication systems, medical imaging, and digital signal processing. It is also used in the design and analysis of filters, equalizers, and other signal processing algorithms.

What are the limitations of Signal DFT?

One of the main limitations of Signal DFT is that it assumes the signal is periodic, which may not be true for all signals. It also has limited time and frequency resolution, meaning it cannot accurately represent signals with rapidly changing frequencies. Additionally, DFT requires a large amount of computation for longer signals, which can be a limitation for real-time signal processing applications.

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