Convolution and Z-transform via convolution integral

Convolution

Convolution combines two signals to give the response of a linear time-invariant system. If $g$ is the impulse response and $x$ is the input, the output is the convolution $y = g * x$.

• Continuous time: $y(t) = \displaystyle\int_0^t g(t-\tau)\,x(\tau)\,d\tau$
• Discrete time (convolution sum): $y(k) = \displaystyle\sum_{h=0}^{k} g(k-h)\,x(h)$

The key property is that convolution in time becomes multiplication in the transform domain: $Y(s) = G(s)X(s)$ and $Y(z) = G(z)X(z)$. This is why pulse transfer functions are useful.

Convolution integral for the z-transform

The impulse-sampled signal is $x^*(t) = x(t)\sum_k\delta(t-kT)$, which is a product in time. Its Laplace transform is therefore a complex convolution of $X(s)$ with the transform of the impulse train, $\dfrac{1}{1 - e^{-Ts}}$:

$$X^*(s) = \frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty}X(p)\,\frac{1}{1 - e^{-T(s-p)}}\,dp$$

The contour is closed in the left half of the p-plane, which encloses the poles of $X(p)$. This is valid when $X(s)$ has at least two more poles than zeros. Setting $e^{Ts} = z$ gives:

$$X(z) = \sum_{\text{poles of } X(s)}\text{Res}\left[X(s)\frac{z}{z-e^{sT}}\right]$$

Apply to $X(s) = \dfrac{1}{s^2(s+1)}$

Poles: a double pole at $s = 0$ and a simple pole at $s = -1$.

Simple pole $s = -1$:

$$R_1 = \left[\frac{1}{s^2}\cdot\frac{z}{z - e^{sT}}\right]_{s=-1} = \frac{z}{z - e^{-T}}$$

Double pole $s = 0$:

$$\begin{aligned} R_2 &= \frac{d}{ds}\left[\frac{1}{s+1}\cdot\frac{z}{z - e^{sT}}\right]_{s=0} \\ &= \left[-\frac{1}{(s+1)^2}\cdot\frac{z}{z-e^{sT}} + \frac{1}{s+1}\cdot\frac{zTe^{sT}}{(z-e^{sT})^2}\right]_{s=0} \\ &= -\frac{z}{z-1} + \frac{Tz}{(z-1)^2} \end{aligned}$$

Sum of residues

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

Over the common denominator:

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

Check

$x(t) = t - 1 + e^{-t}$ (from partial fractions $\frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1}$). Its z-transform is $\frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}$, which is the same.

For $T = 1$ s: $X(z) = \dfrac{z(0.3679z + 0.2642)}{(z-1)^2(z-0.3679)}$.

Answer: $X(z) = \dfrac{z[(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})]}{(z-1)^2(z-e^{-T})}$.