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Discrete-Time Fourier Transform
Discrete-Time Fourier Transform
A discrete-time signal can be represented in the frequency domain using discrete-time Fourier transform. Therefore, the Fourier transform of a discretetime sequence is called the discrete-time Fourier transform (DTFT).
Mathematically, if $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is a discrete-time sequence, then its discrete-time Fourier transform is defined as −
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$
The discrete-time Fourier transform X(ω) of a discrete-time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ represents the frequency content of the sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$. Therefore, by taking the Fourier transform of the discrete-time sequence, the sequence is decomposed into its frequency components. For this reason, the DTFT X(ω) is also called the signal spectrum.
Condition for Existence of Discrete-Time Fourier Transform
The Fourier transform of a discrete-time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ exists if and only if the sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is absolutely summable, i.e.,
$$\mathrm{\sum_{\mathit{n=-\infty}}^{\infty}\left|\mathit{x}\mathrm{\left(\mathit{n}\right)} \right|<\infty }$$
The discrete-time Fourier transform (DTFT) of the exponentially growing sequences do not exist, because they are not absolutely summable.
Also, the DTFT method of analysing the systems can be applied only to the asymptotically stable systems and it cannot be applied for the unstable systems, i.e., the DTFT can only be used to analyse the systems whose transfer function has poles inside the unit circle.
Numerical Example (1)
Find the discrete-time Fourier transform of the sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}$ .
Solution
The given discrete-time sequence is,
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\begin{cases} 1 & \text{ for } n\geq 0 \ 0 & \text{ for } n< 0 \end{cases}}$$
Now, from the definition of DTFT, we have,
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$
$$\mathrm{\therefore \mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{u}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}\:\mathrm{=}\:\:\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathrm{\left ( 1 \right )}\mathit{e^{-j\omega n}}}$$
$$\mathrm{\Rightarrow\mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\frac{1}{1-\mathit{e^{-j\omega }}}}$$
Numerical Example (2)
Find the DTFT of the sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left(\mathit{n-k}\right)}$.
Solution
The given discrete-time sequence is,
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left(\mathit{n-k}\right)}\:\mathrm{=}\:\begin{cases} 1 & \text{ for } n\geq k \ 0 & \text{ for } n< k \end{cases}}$$
Now, from the definition of DTFT, we have,
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$
$$\mathrm{\therefore \mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n-k }\right)}\right]}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{u}\mathrm{\left(\mathit{n-k}\right)}\mathit{e^{-\mathit{j\omega n}}}\:\mathrm{=}\:\:\sum_{\mathit{n=k}}^{\infty}\mathrm{\left ( 1 \right )}\mathit{e^{-j\omega n}}}$$
$$\mathrm{\Rightarrow \mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n-k }\right)}\right]}\:\mathrm{=}\:\mathit{e^{-j\omega k}\mathrm{+}\:\mathit{e^{-j\omega\mathrm{\left ( \mathit{k}+1 \right)}}\:\mathrm{+}\:\mathit{e^{-j\omega\mathrm{\left ( \mathit{k}+2 \right )} }}\:\mathrm{+}\:...}}}$$
$$\mathrm{\Rightarrow \mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n-k }\right)}\right]}\:\mathrm{=}\:\mathit{e^{-j\omega k}\mathrm{\left ( 1\:\mathrm{+}\:\mathit{e^{-j\omega}}\:\mathrm{+}\:\mathit{e^{-j\mathrm{2}\omega}}\:\mathrm{+}\:\mathit{e^{-j\mathrm{3\omega }}}\:\mathrm{+}\:... \right )}}}$$
$$\mathrm{\therefore \mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n -k}\right)}\right]}\:\mathrm{=}\:\frac{\mathit{e^{-j\omega k}}}{1-\mathit{e^{-j\omega }}}}$$
Numerical Example (3)
Find the DTFT of the sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{\delta}\mathrm{\left(\mathit{n-k}\right)}$.
Solution
The given discrete-time sequence is,
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{\delta}\mathrm{\left(\mathit{n-k}\right)}\:\mathrm{=}\:\begin{cases} 1 & \text{ for } n= k \ 0 & \text{ for } n
eq k \end{cases}}$$
Hence, from the definition of the discrete-time Fourier transform, we have,
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$
$$\mathrm{\therefore \mathit{F}\mathrm{\left[\mathit{\delta}\mathrm{\left(\mathit{n-k }\right)}\right]}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{\delta}\mathrm{\left(\mathit{n-k}\right)}\mathit{e^{-\mathit{j\omega n}}}\:\mathrm{=}\:\mathrm{\left [ \mathit{e^{-j\omega n}} \right]}_{\mathit{n=k}}}$$
$$\mathrm{\Rightarrow \mathit{F}\mathrm{\left[\mathit{\delta }\mathrm{\left(\mathit{n-k }\right)}\right]}\:\mathrm{=}\:\mathit{e^{-j\omega k}}}$$
Numerical Example (4)
Find the discrete-time Fourier transform $\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathrm{\left\{1,3,-2,5 \right\}}$.
Solution
The given discrete-time sequence is,
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathrm{\left\{1,3,-2,5,2 \right\}}}$$
The DTFT of a sequence is defined as −
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$
$$\mathrm{\Rightarrow \mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\mathit{x}\:\mathrm{\left(\mathrm{0}\right)}\:\mathrm{+}\:\mathit{x}\mathrm{\left(\mathrm{1}\right)}\mathit{e^{-j\omega }}\:\mathrm{+}\:\mathit{x}\:\mathrm{\left(\mathrm{2}\right)}\mathit{e^{-j\mathrm{2}\omega}}\:\mathrm{+}\:\mathit{x}\:\mathrm{\left(\mathrm{3}\right)}\mathit{e^{-j\mathrm{3}\omega }}\:\mathrm{+}\:\mathit{x}\:\mathrm{\left(\mathrm{4}\right)}\mathit{e^{-j\mathrm{4}\omega }}}$$
$$\mathrm{\therefore \mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:1\:\mathrm{+}\:3\mathit{e^{-j\omega }}-2\mathit{e^{-j\mathrm{2}\omega}}\:\mathrm{+}\:5\mathit{e^{-j\mathrm{3}\omega }}\:\mathrm{+}\:2\mathit{e^{-j\mathrm{4}\omega}}}$$
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