Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Time Shifting and Frequency Shifting Properties of Discrete-Time Fourier Transform
Discrete-Time Fourier Transform
The Fourier transform of a discrete-time sequence is known as the discrete-time Fourier transform (DTFT).
Mathematically, the discrete-time Fourier transform (DTFT) of a discrete-time sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is defined as −
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$
Time Shifting Property of Discrete-Time Fourier Transform
Statement - The time-shifting property of discrete-time Fourier transform states that if a signal $\mathit{x}\mathrm{\left(\mathit{n}\right)}$ is shifted by k in time domain, then its DTFT is multiplied by $\mathit{e^{-j\omega k }}$. Therefore, if
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}$$
Then
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n-k}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{e^{-j\omega k }}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}$$
Where, k is an integer.
Proof
From the definition of discrete-time Fourier transform, we have,
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega }\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$
$$\mathrm{\therefore\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n-k}\right)}\right]}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n-k}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$
Substituting $\mathrm{\left(\mathit{n-k}\right)}\:\mathrm{=}\:\mathit{m}$ then $\mathit{n}\:\mathrm{=}\:\mathrm{\left(\mathit{m\mathrm{+}k}\right)}$ in the above summation, we get,
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n-k}\right)}\right]}\:\mathrm{=}\:\sum_{\mathit{m=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{m}\right)}\mathit{e^{-j\omega \mathrm{\left( \mathit{m}\mathrm{+}\mathit{k}\right)}}}\:\mathrm{=}\:\sum_{\mathit{m=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{m}\right)}\mathit{e^{-j\omega m}}\mathit{e^{-j\omega k}}}$$
$$\mathrm{\Rightarrow \mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n-k}\right)}\right]}\:\mathrm{=}\:\mathit{e^{-j\omega k}}\sum_{\mathit{m=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{m}\right)}\mathit{e^{-j\omega m}}}$$
$$\mathrm{\therefore \mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n-k}\right)}\right]}\:\mathrm{=}\:\mathit{e^{-j\omega k}}\mathit{X}\mathrm{\left(\mathit{\omega }\right)};\:\:\mathrm{\left ( Time\:delay \right )}}$$
Similarly
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n\mathrm{+}k}\right)}\right]}\:\mathrm{=}\:\mathit{e^{j\omega k}}\mathit{X}\mathrm{\left(\mathit{\omega }\right)};\:\:\mathrm{\left ( Time\:advance \right )}}$$
Frequency Shifting Property of Discrete-Time Fourier Transform
Statement - The frequency shifting property of DTFT states that the multiplication of a discrete-time sequence by $\mathit{e^{j\omega_{\mathrm{0}} n}}$ in time domain corresponds to the shift by $\omega _{\mathrm{0}}$ in the frequency domain. Therefore, if
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{\omega }\right)}}$$
Then
$$\mathrm{\mathit{e^{j\omega_{\mathrm{0}} n}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{\omega-\omega _{\mathrm{0}} }\right)}}$$
Proof
From the definition of DTFT, we have,
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{\omega}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$
$$\mathrm{\therefore \mathit{F}\mathrm{\left[\mathit{e^{j\omega_{\mathrm{0}} n}}\mathit{x}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{e^{j\omega_{\mathrm{0}} n}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-\mathit{j\omega n}}}}$$
$$\mathrm{\Rightarrow \mathit{F}\mathrm{\left[\mathit{e^{j\omega_{\mathrm{0}} n}}\mathit{x}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{e^{-j\mathrm{\left( \omega -\omega _{\mathrm{0}} \right )}\mathit{n}}}}$$
$$\mathrm{\therefore \mathit{F}\mathrm{\left[\mathit{e^{j\omega_{\mathrm{0}} n}}\mathit{x}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\mathit{X{\mathrm{\left( \omega -\omega _{\mathrm{0}} \right )}}}}$$
The time shifting property and frequency shifting property of DTFT are dual of each other.
Numerical Example (1)
Using the time shifting property of DTFT, find the DTFT of sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left( \mathit{n}-\mathrm{1}\right )}\:\mathrm{+}\:\mathit{u}\mathrm{\left( \mathit{n}\mathrm{+}\mathrm{2}\right )}$.
Solution
The given discrete-time sequence is,
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{u}\mathrm{\left( \mathit{n}-\mathrm{1}\right )}\:\mathrm{+}\:\mathit{u}\mathrm{\left( \mathit{n}\mathrm{+}\mathrm{2}\right )}}$$
Since the DTFT of a unit step sequence defined as −
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\frac{1}{1-\mathit{e^{-j\omega }}}}$$
Hence, using the time shifting property of DTFT $\mathrm{\left [ i.e,\mathit{x}\mathrm{\left(\mathit{n-k}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{e^{-j\omega k }}\mathit{X}\mathrm{\left(\mathit{\omega }\right)} \right ]}$ ,we get,
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n -\mathrm{1}}\right)}\right]}\:\mathrm{=}\:\mathit{e^{-j\omega }}\mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\frac{\mathit{e^{-j\omega }}}{1-\mathit{e^{-j\omega }}}}$$
And
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n \mathrm{+}\mathrm{2}}\right)}\right]}\:\mathrm{=}\:\mathit{e^{j\mathrm{2}\omega }}\mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\frac{\mathit{e^{j\mathrm{2}\omega }}}{1-\mathit{e^{-j\omega }}}}$$
$$\mathrm{\therefore\mathit{F} \mathrm{\left [ \mathit{u}\mathrm{\left(\mathit{n -\mathrm{1}}\right)}\:\mathrm{+}\:\mathit{u}\mathrm{\left(\mathit{n \mathrm{+}\mathrm{2}}\right)}\right ]}\:\mathrm{=}\:\frac{\mathit{e^{-j\omega }}}{1-\mathit{e^{-j\omega }}}\:\mathrm{+}\:\frac{\mathit{e^{j\mathrm{2}\omega }}}{1-\mathit{e^{-j\omega }}}}$$
Numerical Example (2)
Using the frequency shifting property of DTFT, find the DTFT of the sequence $\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{e^{j\mathrm{2\omega }}}\mathit{u}\mathrm{\left(\mathit{n}\right)}$.
Solution
The given discrete-time sequence is,
$$\mathrm{\mathit{x}\mathrm{\left(\mathit{n}\right)}\:\mathrm{=}\:\mathit{e^{j\mathrm{2\omega }}}\mathit{u}\mathrm{\left(\mathit{n}\right)}}$$
Since the DTFT of a unit step sequence is given by,
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{u}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\frac{1}{1-\mathit{e^{-j\omega }}}}$$
Now, using the frequency shifting property of DTFT $\mathrm{\left [ i.e,\mathit{e^{j\omega_{\mathrm{0}} n}}\mathit{x}\mathrm{\left(\mathit{n}\right)}\overset{\mathit{FT}}{\leftrightarrow}\mathit{X}\mathrm{\left(\mathit{\omega-\omega _{\mathrm{0}} }\right)} \right ]}$,we get,
$$\mathrm{\mathit{F}\mathrm{\left[\mathit{e^{j\mathrm{2\omega }}}\mathit{u}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\mathrm{\left [ \frac{1}{1-\mathit{e^{-j\omega }}} \right ]}_{\omega =\mathrm{\left ( \omega -\mathrm{2} \right )}}}$$
$$\mathrm{\therefore \mathit{F}\mathrm{\left[\mathit{e^{j\mathrm{2\omega }}}\mathit{u}\mathrm{\left(\mathit{n }\right)}\right]}\:\mathrm{=}\:\frac{1}{1-\mathit{e^{-j\mathrm{\left ( \omega -\mathrm{2} \right )} }}}}$$
8K+ Views
