# 5.6: Discrete -Time Fourier Transform (DTFT)

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- 1625

Learning Objectives

- Discussion of Discrete-time Fourier Transforms.
- Topics include comparison with analog transforms and discussion of Parseval's theorem.

The Fourier transform of the discrete-time signal **s(n)** is defined to be

\[S(e^{i2\pi f})=\sum_{n=-\infty }^{\infty }s(n)e^{-(i2\pi fn)}\]

Frequency here has no units. As should be expected, this definition is linear, with the transform of a sum of signals equaling the sum of their transforms. Real-valued signals have conjugate-symmetric spectra:

\[S(e^{-(i2\pi f)})=\overline{S(e^{j2\pi f}})\]

Exercise \(\PageIndex{1}\)

A special property of the discrete-time Fourier transform is that it is periodic with period one:

\[S(e^{i2\pi (f+1)})=S(e^{i2\pi f})\]

Derive this property from the definition of the DTFT.

**Solution**

\[S(e^{i2\pi (f+1)})=\sum_{n=-\infty }^{\infty }s(n)e^{-(i2\pi (f+1)n)}\]

\[S(e^{i2\pi (f+1)})=\sum_{n=-\infty }^{\infty }e^{-(i2\pi n)}s(n)e^{-(i2\pi fn)}\]

\[S(e^{i2\pi (f+1)})=\sum_{n=-\infty }^{\infty }s(n)e^{-(i2\pi fn)}\]

\[S(e^{i2\pi (f+1)})=S(e^{i2\pi f})\]

Because of this periodicity, we need only plot the spectrum over one period to understand completely the spectrum's structure; typically, we plot the spectrum over the frequency range

\[\left [ -\frac{1}{2},\frac{1}{2} \right ]\]

When the signal is real-valued, we can further simplify our plotting chores by showing the spectrum only over

\[\left [ 0,\frac{1}{2} \right ]\]

the spectrum at negative frequencies can be derived from positive-frequency spectral values.

When we obtain the discrete-time signal via sampling an analog signal, the Nyquist frequency corresponds to the discrete-time frequency **1/2**. To show this, note that a sinusoid having a frequency equal to the Nyquist frequency ** 1/2T_{s} **has a sampled waveform that equals

\[\cos \left ( 2\pi \times \frac{1}{2T_{s}}nT_{s} \right )=\cos (\pi n)=(-1)^{n}\]

The exponential in the DTFT at frequency 1/2 equals

\[e^{-\frac{i2\pi n}{2}}=e^{-(i\pi n)}=(-1)^{n}\]

meaning that discrete-time frequency equals analog frequency multiplied by the sampling interval

\[f_{D}=f_{A}T_{s}\]

**f _{D} and f_{A}** represent discrete-time and analog frequency variables, respectively. The aliasing figure provides another way of deriving this result. As the duration of each pulse in the periodic sampling signal

**p**narrows, the amplitudes of the signal's spectral repetitions, which are governed by the Fourier series coefficients of

_{Ts}(t)**p**

_{Ts}

**(t)**, become increasingly equal. Examination of the periodic pulse signal reveals that as

**Δ**decreases, the value of

**c**, the largest Fourier coefficient, decreases to zero:

_{0}\[\left | c_{0} \right |=\frac{A\Delta }{T_{s}}\]

Thus, to maintain a mathematically viable Sampling Theorem, the amplitude **A** must increase as ** 1/Δ**, becoming infinitely large as the pulse duration decreases. Practical systems use a small value of

**Δ,**say

**0.1.**

**T**and use amplifiers to rescale the signal. Thus, the sampled signal's spectrum becomes periodic with period

_{s }**1/T**Thus, the Nyquist frequency

_{s. }**1/2T**

_{s}**corresponds to the frequency**

**½**.

Example \(\PageIndex{1}\):

Let's compute the discrete-time Fourier transform of the exponentially decaying sequence

\[s(n)=a^{n}u(n)\]

where **u(n)** is the unit-step sequence. Simply plugging the signal's expression into the Fourier transform formula:

\[S(e^{i2\pi f})=\sum_{n=-\infty }^{\infty }a^{n}u(n)e^{-(i2\pi fn)}\]

\[S(e^{i2\pi f})=\sum_{n=-\infty }^{\infty }\left ( ae^{-(i2\pi fn)}\right )^{n}\]

This sum is a special case of the **geometric series**.

\[\sum_{n=0}^{\infty }=\forall \alpha ,\left | \alpha \right |< 1:\left ( \frac{1}{1-\alpha } \right )\]

Thus, as long as** ****|a|<1**, we have our Fourier transform.

\[S(e^{i2\pi f})=\frac{1}{1-ae^{-(i2\pi f)}}\]

Using Euler's relation, we can express the magnitude and phase of this spectrum.

\[\left | S(e^{i2\pi f})\right |=\frac{1}{\sqrt{\left ( 1-a\cos (2\pi f) \right )^{2}+a^{2}\sin ^{2}(2\pi f)}}\]

\[\angle \left ( S(e^{i2\pi f})\right )=-\tan^{-1}\left ( \frac{a\sin 2\pi f}{1-a\cos (2\pi f)} \right )\]

No matter what value of **a** we choose, the above formulae clearly demonstrate the periodic nature of the spectra of discrete-time signals. Fig. 5.6.1 below shows indeed that the spectrum is a periodic function. We need only consider the spectrum between **-****½**** and ****½** to unambiguously define it. When **a>0**, we have a lowpass spectrum—the spectrum diminishes as frequency increases from ** 0 to ½ **—with increasing

**a**leading to a greater low frequency content; for

**a<0**, we have a highpass spectrum as shown in Fig. 5.6.2 below.

Example \(\PageIndex{1}\):

Analogous to the analog pulse signal, let's find the spectrum of the length- **N** pulse sequence.

\[s(n)=\begin{cases} 1 & \text{ if } 0\leq n\leq N-1 \\ 0 & \text{ if } otherwise \end{cases}\]

The Fourier transform of this sequence has the form of a truncated geometric series.

\[S(e^{i2\pi f})=\sum_{n=0}^{N-1}e^{-(i2\pi fn)}\]

For the so-called finite geometric series, we know that

\[\sum_{n=n_{0}}^{N+n_{0}-1}\alpha ^{n}=\alpha ^{n_{0}}\frac{1-\alpha ^{N}}{1-\alpha }\]

for **all** values of α.

Exercise \(\PageIndex{1}\)

Derive this formula for the finite geometric series sum. The "trick" is to consider the difference between the series' sum and the sum of the series multiplied by α.

**Solution**

\[\alpha \sum_{n=n_{0}}^{N+n_{0}-1}\alpha ^{n}-\sum_{n=n_{0}}^{N+n_{0}-1}\alpha ^{n}=\alpha ^{N+n_{0}}-\alpha ^{n_{0}}\]

which, after manipulation, yields the geometric sum formula.

Applying this result yields to Fig. 5.6.3,

\[S(e^{i2\pi f})=\frac{1-e^{-(i2\pi fN)}}{1-e^{-(i2\pi f)}}\]

\[S(e^{i2\pi f})=e^{-(i\pi f(N-1))}\frac{\sin (\pi fN)}{\sin (\pi f)}\]

The ratio of sine functions has the generic form of

\[\frac{\sin (Nx)}{\sin (x)}\]

which is known as the **discrete-time sinc function** ** dsinc (x)**. Thus, our transform can be concisely expressed as

\[S(e^{i2\pi f})=e^{-(i\pi f(N-1))}dsinc(\pi f)\]

The discrete-time pulse's spectrum contains many ripples, the number of which increase with **N**, the pulse's duration.

The inverse discrete-time Fourier transform is easily derived from the following relationship:

\[\int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-(i2\pi fm)} e^{i2\pi fn}df=\begin{cases} 1 & \text{ if } m=n \\ 0 & \text{ if } m\neq n \end{cases}\]

\[\int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-(i2\pi fm)} e^{i2\pi fn}df=\delta (m-n)\]

Therefore, we find that

\[\int_{-\frac{1}{2}}^{\frac{1}{2}} S(e^{i2\pi f})e^{i2\pi fn}df=\int_{-\frac{1}{2}}^{\frac{1}{2}} \sum_{_{m}m}s(m)e^{-(i2\pi fm)}e^{i2\pi fn}df\]

\[\int_{-\frac{1}{2}}^{\frac{1}{2}} S(e^{i2\pi f})e^{i2\pi fn}df=\sum_{_{m}m}s(m)\int_{-\frac{1}{2}}^{\frac{1}{2}} e^{(-(i2\pi f))(m-n)}df\]

\[\int_{-\frac{1}{2}}^{\frac{1}{2}} S(e^{i2\pi f})e^{i2\pi fn}df=s(n)\]

The Fourier transform pairs in discrete-time are

\[S(e^{i2\pi f})=\sum_{n=-\infty }^{\infty }s(n)e^{-(i2\pi fn)}\]

\[s(n)=\int_{-\frac{1}{2}}^{\frac{1}{2}} S(e^{i2\pi f})e^{i2\pi fn}df\]

The properties of the discrete-time Fourier transform mirror those of the analog Fourier transform. The DTFT properties table below shows similarities and differences. One important common property is Parseval's Theorem.

\[\sum_{n=-\infty }^{\infty }\left ( \left | s(n) \right | \right )^{2}=\int_{-\frac{1}{2}}^{\frac{1}{2}}\left | S(e^{i2\pi f}) \right |^{2}df\]

To show this important property, we simply substitute the Fourier transform expression into the frequency-domain expression for power.

\[\int_{-\frac{1}{2}}^{\frac{1}{2}}\left | S(e^{i2\pi f}) \right |^{2}df=\int_{-\frac{1}{2}}^{\frac{1}{2}}\sum_{nn}s(n)e^{-(i2\pi fn)}\sum_{mm}\overline{s(n)}e^{i2\pi fm}df\]

\[\int_{-\frac{1}{2}}^{\frac{1}{2}}\left | S(e^{i2\pi f}) \right |^{2}df=\sum_{n,m,n,m}s(n)\overline{s(n)}\int_{-\frac{1}{2}}^{\frac{1}{2}}e^{i2\pi f(m-n)}df\]

Using the orthogonality relation, the integral equals **δ(m-n)**, where **δ(n) **is the unit sample. Thus, the double sum collapses into a single sum because nonzero values occur only when ** n=m,** giving Parseval's Theorem as a result. We term

\[\sum_{nn}s^{2}(n)\]

the energy in the discrete-time signal **s(n)** in spite of the fact that discrete-time signals don't consume (or produce for that matter) energy. This terminology is a carry-over from the analog world.

Exercise \(\PageIndex{1}\)

Suppose we obtained our discrete-time signal from values of the product **s(t)****p**_{Ts}**(t)**, where the duration of the component pulses in **p**_{Ts}**(t)** is **Δ**. How is the discrete-time signal energy related to the total energy contained in **s(t)**? Assume the signal is bandlimited and that the sampling rate was chosen appropriate to the Sampling Theorem's conditions.

**Solution**

If the sampling frequency exceeds the Nyquist frequency, the spectrum of the samples equals the analog spectrum, but over the normalized analog frequency **fT**. Thus, the energy in the sampled signal equals the original signal's energy multiplied by **T**.

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