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- Fourier Series
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- Laplace Transform
- Laplace Transforms
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- Difference between Z Transform and Laplace Transform
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- Relation between Laplace Transform and Fourier Transform
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- Z Transform
- Z-Transforms (ZT)
- Common Z-Transform Pairs
- Z-Transform of Unit Impulse, Unit Step, and Unit Ramp Functions
- Z-Transform of Sine and Cosine Signals
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- Z-Transforms Properties
- Properties of ROC of the Z-Transform
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- Conjugation and Accumulation Properties of Z-Transform
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- Initial Value Theorem of Z-Transform
- Final Value Theorem of Z Transform
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- Long Division Method to Find Inverse Z Transform
- Partial Fraction Expansion Method for Inverse Z-Transform
- What is Inverse Z Transform?
- Inverse Z-Transform by Convolution Method
- Transform Analysis of LTI Systems using Z-Transform
- Convolution Property of Z Transform
- Correlation Property of Z Transform
- Multiplication by Exponential Sequence Property of Z Transform
- Multiplication Property of Z Transform
- Residue Method to Calculate Inverse Z Transform
- System Realization
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- Discrete Fourier Transform
- Discrete-Time Fourier Transform
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- Inverse Discrete-Time Fourier Transform
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- Parseval’s Power Theorem
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- What is Mean Square Error?
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- Region of Convergence
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- Basic Elements to Construct the Block Diagram of Continuous Time Systems
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- Rayleigh’s Energy Theorem
Z-Transform of Sine and Cosine Signals
Z-Transform
The Z-transform (ZT) is a mathematical tool which is used to convert the difference equations in time domain into the algebraic equations in z-domain.
Mathematically, if x(n) is a discrete-time signal or sequence, then its bilateral or two-sided Z-transform is defined as â
$$\mathrm{Z[x(n)]\:=\:X(z)\:=\:\sum_{n=-\infty}^{\infty}\: x(n) z^{-n}}$$
Where, z is a complex variable.
Also, the unilateral or one-sided z-transform is defined as â
$$\mathrm{Z[x(n)]\:=\:X(z)\:=\:\sum_{n=0}^{\infty}\: x(n) z^{-n}}$$
Z-Transform of Causal Sine Sequence
The causal sine sequence is defined as,
$$\mathrm{x(n)\:=\:sin\:\omega n\:u(n)\:=\:\begin{cases}sin \:\omega n \:\:\text{for }\: n\:\geq\:0 \\\\ 0\:\:\text{for } n\:\lt\:0\end{cases}}$$
Therefore, the Z-transform of the sine sequence is obtained as follows â
$$\mathrm{Z[x(n)]\:=\:X(z)\:=\:Z[sin\omega n\:u(n)]}$$
$$\mathrm{\Rightarrow\:X(z)\:=\:\sum_{n=0}^{\infty}\:sin(\omega n)\:z^{-n}}$$
$$\mathrm{\because\:sin\:\omega n\:=\:\frac{e^{j \omega n} \:-\: e^{-j \omega n}}{2j}}$$
$$\mathrm{\therefore\:X(z)\:=\:\sum_{n=0}^{\infty} \left( \frac{e^{j \omega n} \:-\: e^{-j \omega n}}{2j} \right) z^{-n}}$$
$$\mathrm{\Rightarrow\: X(z) \:=\: \frac{1}{2j} \sum_{n=0}^{\infty} \:\left( e^{j \omega n} \:-\: e^{-j \omega n} \right) z^{-n}}$$
$$\mathrm{= \:\frac{1}{2j} \left[ \sum_{n=0}^{\infty} \left( e^{j \omega} z^{-1} \right)^n \:-\: \sum_{n=0}^{\infty} \left( e^{-j \omega} z^{-1} \right)^n \right]}$$
$$\mathrm{\Rightarrow\: X(z) \:=\: \frac{1}{2j} \left[ \frac{1}{(1 \:-\: e^{j \omega} z^{-1})} \:-\: \frac{1}{(1 \:-\: e^{-j \omega} z^{-1})} \right]}$$
$$\mathrm{\Rightarrow\: X(z) \:=\: \frac{1}{2j} \left[ \frac{z}{\left( z \:-\: e^{j \omega} \right)} \:- \:\frac{z}{\left( z \:-\: e^{-j \omega} \right)} \right] \:=\: \frac{1}{2j} \left[ \frac{z\left( z\:-\: e^{-j \omega} \right)\:-\:z \left( z \:-\: e^{j \omega} \right)}{\left( z\:-\: e^{-j \omega} \right)\left( z \:-\: e^{j \omega} \right)} \right]}$$
$$\mathrm{\Rightarrow\: X(z) \:=\: \frac{1}{2j} \left[ \frac{z\left( e^{j \omega} \:-\: e^{-j \omega} \right)}{z^2 \:-\: z \left( e^{j \omega} \:+\: e^{-j \omega} \right) \:+\: 1} \right]}$$
$$\mathrm{\because\: \left(\frac{e^{j \omega} \:-\: e^{-j \omega}}{2}\right) \:=\: \sin(\omega) \quad \text{and} \quad \left(\frac{e^{j \omega} \:+\:e^{-j \omega}}{2}\right) \:=\: \cos(\omega)}$$
$$\mathrm{\therefore\: X(z) \:=\: \frac{z \:\sin(\omega)}{z^2 \:-\: 2z\: \cos(\omega) \:+\: 1}}$$
This series converges for |z| > 1. Therefore, the ROC of the Z-transform of the causal sine sequence is |z| > 1. Hence, the Z-transform of a sine function along with its ROC is represented as,
$$\mathrm{\sin(\omega n) u(n)\:\overset{ZT}\longleftrightarrow\: \left(\frac{z \sin(\omega)}{z^2 \:-\: 2z\: \cos(\omega) \:+\: 1}\right); \quad \text{ROC } \:\rightarrow\: |z|\: \gt\: 1}$$
Z-Transform of Causal Cosine Sequence
The causal cosine sequence is defined as â
$$\mathrm{x(n) \:=\: \cos \omega\: nu(n) \:=\: \begin{cases}\cos \omega\:n\:\: \text{for } \:n \:\geq\: 0 \\\\ 0\:\: \text{for }\: n\:\lt\:0 \end{cases}}$$
Thus, the Z-transform of the cosine sequence is obtained as follows â
$$\mathrm{Z[x(n)]\:=\:X(z)\:=\:Z[cosin\:\omega\:n\:u(n)]}$$
$$\mathrm{\Rightarrow\:X(z)\:=\:\sum_{n=0}^{\infty}\:cos\:(\omega n)\:z^{-n}}$$
$$\mathrm{\because\:cos\:\omega n\:=\:\frac{e^{j\omega n}\:+\:e^{-j\omega\:n}}{2}}$$
$$\mathrm{\therefore\:X(z)\:=\:\sum_{n=0}^{\infty}\:\left( \frac{e^{j\omega n}\:+\:e^{-j\omega n}}{2}\right)z^{-n}}$$
$$\mathrm{\Rightarrow\:X(z)\:=\:\frac{1}{2}\:\sum_{n=0}^{\infty}\:(e^{j\omega n}\:+\:e^{-j\omega n})z^{-n}\:=\:\frac{1}{2}\left[\sum_{n=0}^{\infty}\:(e^{j\omega}z^{-1})^n \:+\:\sum_{n=0}^{\infty}\:(e^{-j\omega}z^{-1})^n\right]}$$
$$\mathrm{\Rightarrow\:X(z)\:=\:\frac{1}{2}\left[\frac{1}{(1\:-\:e^{j\omega}\:Z^{-1})}\:+\:\frac{1}{(1\:-\:e^{-j\omega}\:z^{-1})}\right]}$$
$$\mathrm{\Rightarrow\:X(z)\:=\:\frac{1}{2}\left[\frac{z}{(z\:-\:e^{j\omega})}\:+\:\frac{z}{(z\:-\:e^{-j\omega})}\right]\:=\:\frac{1}{2}\left[\frac{z(z\:-\:e^{-j\omega})\:+\:z(z\:-\:e^{j\omega})}{(z\:-\:e^{j\omega})(z\:-\:e^{-j\omega})}\right]}$$
$$\mathrm{\Rightarrow\:X(z)\:=\:\frac{1}{2}\left\{\frac{z[2z\:-\:(e^{j\omega\:-\:e^{-j\omega}})]}{z^2\:-\:z(e^{j\omega}\:+\:e^{-j\omega})\:+\:1}\right\}}$$
$$\mathrm{\therefore\:X(z)\:=\:\frac{z(z\:-\:cos\omega)}{z^2\:-\:2z\:cos\omega\:+\:1}}$$
This series also converges for |z| > 1. Therefore, the ROC of the Z-transform of the causal cosine sequence is |z| > 1. Hence, the Z-transform of a cosine function along with its ROC is represented as,
$$\mathrm{cos\omega n\:u(n)\:\overset{ZT}\longleftrightarrow\:\left[\frac{z(z\:-\:cos\omega)}{z^2\:-\:2z\:cos\omega\:+\:1}\right];\:\:\text{ROC }\:\rightarrow\:|z|\:\gt\:1}$$