Z-transform of cos ωt

The z-transform of the sampled cosine is

$$X(z) = \frac{1 - z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} = \frac{z(z - \cos\omega T)}{z^2 - 2z\cos\omega T + 1}$$

The derivation follows.

Step 1: Sample the signal

With sampling period $T$, $x(kT) = \cos\omega kT$ for $k \ge 0$. Write it with Euler's formula:

$$\cos\omega kT = \frac{1}{2}\left(e^{j\omega kT} + e^{-j\omega kT}\right)$$

Step 2: Use the exponential pair

$$\mathcal{Z}\left[e^{-akT}\right] = \sum_{k=0}^{\infty}e^{-akT}z^{-k} = \frac{1}{1 - e^{-aT}z^{-1}}, \quad |z| > |e^{-aT}|$$

With $a = \mp j\omega$:

$$X(z) = \frac{1}{2}\left[\frac{1}{1 - e^{j\omega T}z^{-1}} + \frac{1}{1 - e^{-j\omega T}z^{-1}}\right]$$

Step 3: Combine the fractions

$$\begin{aligned} X(z) &= \frac{1}{2}\cdot\frac{(1 - e^{-j\omega T}z^{-1}) + (1 - e^{j\omega T}z^{-1})}{1 - (e^{j\omega T} + e^{-j\omega T})z^{-1} + z^{-2}} \\ &= \frac{1}{2}\cdot\frac{2 - 2z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} \\ &= \frac{1 - z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} \end{aligned}$$

Here $e^{j\omega T} + e^{-j\omega T} = 2\cos\omega T$ was used.

Multiplying numerator and denominator by $z^2$:

$$X(z) = \frac{z(z - \cos\omega T)}{z^2 - 2z\cos\omega T + 1}, \quad |z| > 1$$

Remarks

• The poles are at $z = e^{\pm j\omega T}$, on the unit circle. This is expected for an undamped oscillation.
• Check: $x(0) = \lim_{z\to\infty}X(z) = 1 = \cos 0$.
• In the same way, $\mathcal{Z}[\sin\omega kT] = \dfrac{z\sin\omega T}{z^2 - 2z\cos\omega T + 1}$.

Answer: $X(z) = \dfrac{1 - z^{-1}\cos\omega T}{1 - 2z^{-1}\cos\omega T + z^{-2}} = \dfrac{z(z-\cos\omega T)}{z^2 - 2z\cos\omega T + 1}$.