DFT Example

The above

Eqs. (3-2) and (3-3) will become more meaningful by way of an example, so let’s go through a simple one step by step. Let’s say we want to sample and perform an 8-point DFT on a continuous input signal containing components at 1 kHz and 2 kHz, expressed as

(3-10)

To make our example input signal xin(t) a little more interesting, we have the 2 kHz term shifted in phase by 135° (3π/4 radians) relative to the 1 kHz sinewave. With a sample rate of fs, we sample the input every 1/fs = ts

seconds. Because N = 8, we need 8 input sample values on which to perform the DFT. So the 8-element sequence x(n) is equal to xin(t) sampled at the nts instants in time so that

(3-11)

If we choose to sample xin(t) at a rate of fs = 8000 samples/second from

Eq. (3-5), our DFT results will indicate what signal amplitude exists in x(n) at the analysis frequencies of mfs/N,

or 0 kHz, 1 kHz, 2 kHz, . . ., 7 kHz. With fs = 8000 samples/second, our eight x(n) samples are

(3-11′)

These x(n) sample values are the dots plotted on the solid continuous xin(t) curve in

Figure 3-2(a). (Note that the sum of the sinusoidal terms in Eq. (3-10), shown as the dashed curves in Figure 3- 2(a), is equal to xin(t).)

Figure 3-2 DFT Example 1: (a) the input signal; (b) the input signal and the m = 1 sinusoids; (c) the input

Now we’re ready to apply

Eq. (3-3) to determine the DFT of our x(n) input. We’ll start with m = 1 because the m = 0 case leads to a special result that we’ll discuss shortly. So, for m = 1, or the 1 kHz (mfs/N = 1·8000/8) DFT frequency term, Eq. (3-3) for this example becomes

(3-12)

Next we multiply x(n) by successive points on the cosine and sine curves of the first analysis frequency that have a single cycle over our eight input samples. In our example, for m = 1, we’ll sum the products of the x(n) sequence with a 1 kHz cosine wave and a 1 kHz sinewave evaluated at the angular values of 2πn/8. Those analysis sinusoids are shown as the dashed curves in

Figure 3-2(b). Notice how the cosine and sinewaves have m = 1 complete cycles in our sample interval.

Substituting our x(n) sample values into Eq. (3-12) and listing the cosine terms in the left column and the sine terms in the right column, we have

So we now see that the input x(n) contains a signal component at a frequency of 1 kHz. Using Eqs. (3-7), (3-8), and (3-9) for our X(1) result, Xmag(1) = 4, XPS(1) = 16, and X(1)’s phase angle relative to a 1 kHz cosine is Xø(1) = −90°.

For the m = 2 frequency term, we correlate x(n) with a 2 kHz cosine wave and a 2 kHz sinewave. These waves are the dashed curves in Figure 3-2(c). Notice here that the cosine and sinewaves have m = 2 complete cycles in our sample interval in Figure 3-2(c). Substituting our x(n) sample values in Eq. (3-3) for m = 2 gives

Here our input x(n) contains a signal at a frequency of 2 kHz whose relative amplitude is 2, and whose phase angle relative to a 2 kHz cosine is 45°. For the m = 3 frequency term, we correlate x(n) with a 3 kHz cosine wave and a 3 kHz sinewave. These waves are the dashed curves in

Figure 3-2(d). Again, see how the cosine and sinewaves have m = 3 complete cycles in our sample interval in

Our DFT indicates that x(n) contained no signal at a frequency of 3 kHz. Let’s continue our DFT for the m = 4 frequency term using the sinusoids in Figure 3-3(a).

Figure 3-3 DFT Example 1: (a) the input signal and the m = 4 sinusoids; (b) the input and the m = 5 sinusoids;

So

Our DFT for the m = 5 frequency term using the sinusoids in

Figure 3-3(b) yields

For the m = 6 frequency term using the sinusoids in Figure 3-3(c), Eq. (3-3) is

Figure 3-3(d), Eq. (3-3) is

If we plot the X(m) output magnitudes as a function of frequency, we produce the magnitude spectrum of the x (n) input sequence, shown in Figure 3-4(a). The phase angles of the X(m) output terms are depicted in Figure 3- 4(b).

Figure 3-4 DFT results from Example 1: (a) magnitude of X(m); (b) phase of X(m); (c) real part of X(m); (d)

imaginary part of X(m).

Hang in there; we’re almost finished with our example. We’ve saved the calculation of the m = 0 frequency term to the end because it has a special significance. When m = 0, we correlate x(n) with cos(0) − jsin(0) so that

Eq. (3-3) becomes

(3-13)

Because cos(0) = 1, and sin(0) = 0,

(3-13′)

Eq. (3-13′) is the sum of the x(n) samples. This sum is, of course, proportional to the average of x(n). (Specifically, X(0) is equal to N times x(n)’s average value.) This makes sense because the X(0) frequency term is the non-time-varying (DC) component of x(n). If X(0) were nonzero, this would tell us that the x(n) sequence is riding on a DC bias and has some nonzero average value. For our specific example input from Eq. (3-10), the sum, however, is zero. The input sequence has no DC component, so we know that X(0) will be zero. But let’s not be lazy—we’ll calculate X(0) anyway just to be sure. Evaluating Eq. (3-3) or Eq. (3-13′) for m = 0, we see that

So our x(n) had no DC component, and, thus, its average value is zero. Notice that Figure 3-4 indicates that xin (t), from Eq. (3-10), has signal components at 1 kHz (m = 1) and 2 kHz (m = 2). Moreover, the 1 kHz tone has a magnitude twice that of the 2 kHz tone. The DFT results depicted in Figure 3-4 tell us exactly the spectral content of the signal defined by Eqs. (3-10) and (3-11).

While looking at Figure 3-4(b), we might notice that the phase of X(1) is −90 degrees and ask, “This −90 degrees phase is relative to what?” The answer is: The DFT phase at the frequency mfs/N is relative to a cosine

wave at that same frequency of mfs/N Hz where m = 1, 2, 3, ..., N−1. For example, the phase of X(1) is −90

degrees, so the input sinusoid whose frequency is 1 · fs/N = 1000 Hz was a cosine wave having an initial phase

shift of −90 degrees. From the trigonometric identity cos(α−90°) = sin(α), we see that the 1000 Hz input tone was a sinewave having an initial phase of zero. This agrees with our Eq. (3-11). The phase of X(2) is 45 degrees so the 2000 Hz input tone was a cosine wave having an initial phase of 45 degrees, which is equivalent to a sinewave having an initial phase of 135 degrees (3π/4 radians from Eq. (3-11)).

When the DFT input signals are real-valued, the DFT phase at 0 Hz (m = 0, DC) is always zero because X(0) is always real-only as shown by Eq. (3-13′).

The perceptive reader should be asking two questions at this point. First, what do those nonzero magnitude values at m = 6 and m = 7 in Figure 3-4(a) mean? Also, why do the magnitudes seem four times larger than we would expect? We’ll answer those good questions shortly. The above 8-point DFT example, although admittedly simple, illustrates two very important characteristics of the DFT that we should never forget. First, any individual X(m) output value is nothing more than the sum of the term-by-term products, a correlation, of an input signal sample sequence with a cosine and a sinewave whose frequencies are m complete cycles in the total sample interval of N samples. This is true no matter what the fs sample rate is and no matter how large N is

in an N-point DFT. The second important characteristic of the DFT of real input samples is the symmetry of the DFT output terms.