2.2. DTFT properties and tables#

2.2.1. Table of DTFT pairs#

\(\boldsymbol{x[n]}\)

\(\boldsymbol{X(e^{j{\hat{\omega}}})}\)

Condition

\(\displaystyle a^n u[n]\)

\(\displaystyle \frac{1}{1 - a e^{-j \hat{\omega}}}\)

\(|a| < 1\)

\(\displaystyle (n+1) a^n u[n]\)

\(\displaystyle \frac{1}{(1-a e^{-j \hat{\omega}})^2}\)

\(|a| < 1\)

\(\displaystyle \frac{(n+r-1)!}{n! (r-1)!} a^n u[n]\)

\(\displaystyle \frac{1}{(1-a e^{-j \hat{\omega}})^r}\)

\(|a| < 1\)

\(\displaystyle \delta[n]\)

\(\displaystyle 1\)

\(\displaystyle \delta[n - n_0]\)

\(\displaystyle e^{-j \hat{\omega} n_0}\)

\(\displaystyle u[n]-u[n-N]\)

\(\displaystyle \frac{\sin(N\hat{\omega}/2)}{\sin(\hat{\omega} / 2)} e^{-j(\frac{N-1}{2})\hat\omega}\)

\(\displaystyle \frac{\sin(\hat\omega_0 n)}{\pi n}\)

\( X(e^{j{\hat{\omega}}}) = \begin{cases} 1, & 0 \leq |\hat{\omega}| \leq \hat\omega_0 \\ 0. & \hat\omega_0 < |\hat{\omega}| \leq \pi \end{cases}\)

\(\displaystyle 1\)

\(\displaystyle 2 \pi \delta(e^{j\hat{\omega}})\)

\(\displaystyle u[n]\)

\(\displaystyle \frac{1}{1-e^{-j \hat{\omega}}} + \pi \delta(e^{j\hat{\omega}})\)

\(\displaystyle e^{j \hat{\omega}_0 n}\)

\(\displaystyle 2 \pi \delta(e^{\hat{\omega} - \hat{\omega}_0})\)

\(\displaystyle \cos(\hat{\omega}_0 n)\)

\(\displaystyle \pi \left[ \delta(e^{j(\hat{\omega}-\hat{\omega}_0)}) + \delta(e^{j(\hat{\omega}+\hat{\omega}_0)}) \right]\)

\(\displaystyle \sin(\hat{\omega}_0 n)\)

\(\displaystyle \frac{\pi}{j} \left[ \delta(e^{j(\hat{\omega}-\hat{\omega}_0)}) - \delta(e^{j(\hat{\omega}+\hat{\omega}_0)}) \right]\)

\(\displaystyle \sum_{k=-\infty}^{\infty} \delta[n - kN]\)

\(\displaystyle \frac{2 \pi}{N} \sum_{k=0}^{N-1} \delta\left(e^{j\left(\hat{\omega} - \frac{2 \pi k}{N} \right)} \right)\)

2.2.2. Table of DTFT properties#

For each property listed below, assume \(x[n] \stackrel{\text{DTFT}}{\longleftrightarrow} X(e^{j\hat\omega})\) and \(y[n] \stackrel{\text{DTFT}}{\longleftrightarrow} Y(e^{j\hat\omega})\):

Property

Time domain

Frequency domain

Linearity

\(\alpha x[n] + \beta y[n]\)

\(\alpha X(e^{j{\hat{\omega}}}) + \beta Y(e^{j{\hat{\omega}}})\)

Time Shifting

\(x[n-n_0]\)

\(X(e^{j{\hat{\omega}}}) e^{-j \hat{\omega} n_0}\)

Frequency Shifting

\(x[n] e^{j \hat{\omega}_0 n}\)

\(X(e^{j(\hat{\omega}-\hat{\omega}_0)})\)

Conjugation

\(x^*[n]\)

\(X^*(e^{-j\hat{\omega}})\)

Time Reversal

\(x[-n]\)

\(X(e^{-j\hat{\omega}})\)

Convolution

\(x[n]*y[n]\)

\(X(e^{j\hat{\omega}}) Y(e^{j\hat{\omega}})\)

Multiplication

\(x[n] y[n]\)

\(\displaystyle \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\theta}) Y(e^{j(\hat{\omega}-\theta)}) d\theta\)

Time Differencing

\(x[n]-x[n-1]\)

\((1-e^{-j\hat{\omega}})X(e^{j\hat{\omega}})\)

Accumulation

\(\displaystyle \sum_{k=-\infty}^{n} x[k]\)

\(\displaystyle \frac{X(e^{j\hat{\omega}})}{1 - e^{-j\hat{\omega}}} + \pi X(e^{j0}) \delta(e^{j\hat{\omega}})\)

Frequency Differentiation

\(n x[n]\)

\(\displaystyle j\frac{d X(e^{j\hat{\omega}})}{d\hat{\omega}}\)

Parseval Theorem

\(\displaystyle \sum_{n=-\infty}^{\infty} x[n] y^*[n]\)

\(\displaystyle \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\hat\omega}) Y^*(e^{j\hat\omega}) d\hat\omega\)

2.2.3. Use-of-tables example#

Consider the window signal \(w_{\lambda}[n] = \lambda^{|n|}\), where \(\lambda \in (0,1)\). Clearly, \(w_{\lambda}[n] \in \ell^1\) and hence its DTFT \(W_{\lambda}(e^{j\hat\omega})\) exists.

To use the tables above to find \(W_{\lambda}(e^{j\hat\omega})\), let \(\tilde{w}[n] = \lambda^n u[n]\) and notice that

\[\begin{align*} w_{\lambda}[n] &= \begin{cases} \lambda^n, & n \geq 0 \\ \lambda^{-n}, & n < 0 \end{cases} \\ &= \tilde{w}[n] + \lambda \tilde{w}[-n-1] . \end{align*}\]

Looking up the DTFT-pair table, \(\tilde{w}[n] \stackrel{\text{DTFT}}{\longleftrightarrow} \tilde{W}_{\lambda}(e^{j\hat\omega}) = \frac{1}{1 - \lambda e^{-j \hat{\omega}}}\).

Then using the linearity, time shifting, and time reversal properties, we get

\[\begin{align*} W_{\lambda}(e^{j\hat\omega}) & = \tilde{W}_{\lambda}(e^{j\hat\omega}) + \lambda e^{j\hat\omega} \tilde{W}_{\lambda}(e^{-j\hat\omega}) \\ & = \frac{1}{1 - \lambda e^{-j \hat{\omega}}} + \frac{\lambda e^{j\hat\omega}}{1 - \lambda e^{j \hat{\omega}}} \\ & = \frac{1 - \lambda^2}{1 - 2\lambda \cos \hat{\omega} + \lambda^2}. \end{align*}\]