Generalized transforms

Laplace Transform

It turns out the CTFT is just a specific case of a more general transform called the Laplace transform.

Definition 32

X(s)=x(t)estdt for sCX(s) = \int_{-\infty}^{\infty}{x(t)e^{-st}dt} \text { for } s\in\mathbb{C}
Notice that the Fourier Transform is merely
. Like the Fourier Transform, the Laplace Transform does not always converge. The values of
for which the transform does converge is known as the region of convergence (ROC). If the ROC contains the imaginary axis, then the signal has a well-defined Fourier Transform.
Unlike the Fourier Transform, there is no easy way to take the Inverse Laplace Transform. However, for many LTI systems/signals, we can use partial fraction expansion and then use known Laplace Transform pairs in order to compute the inverse.
Two useful features of laplace transforms are their poles and zeros. Suppose
X(s)=N(s)D(s).X(s) = \frac{N(s)}{D(s)}.
The poles of the system are
{sD(s)=0}\{s | D(s)=0\}
and the zeros are
{sN(s)=0}\{s | N(s)=0\}

Properties of the Laplace Transform

The properties of the Laplace transform are largely the same as the properties of the Fourier Transform. One must just be careful about the Region of Convergence. For all these properties, assume that
x(t)X(s)x(t) \leftrightarrow X(s)
y(t)Y(s)y(t) \leftrightarrow Y(s)
with original region of convergence
. Linearity:
ax(t)+by(t)aX(s)+bY(s)ax(t) + by(t) \leftrightarrow aX(s) + bY(s)
Time Shift:
x(tt0)est0X(s)x(t-t_0) \leftrightarrow e^{-s t_0}X(s)
Time/Frequency Scaling:
x(at)1aX(sa), ROC: saRx(at) \leftrightarrow \frac{1}{|a|}X(\frac{s}{a})\text{, ROC: }s\in aR
x(t)X(s)x^*(t) \leftrightarrow X^*(s^*)
ddtx(t)sX(s),ddsX(s)tx(t)\frac{d}{dt}x(t) \leftrightarrow s X(s), \frac{d}{ds}X(s) \leftrightarrow -t x(t)
(xy)(t)X(s)Y(s)(x*y)(t) \leftrightarrow X(s)Y(s)
Frequency Shift:
es0tx(t)X(ss0), ROC: sR+s0e^{s_0 t}x(t) \leftrightarrow X(s - s_0)\text{, ROC: }s\in R+s_0

The Unilateral Laplace Transform

In the normal laplace transform, we integrate over all the real numbers. For the Unilateral Laplace Transform we only integrate over the right half of the real numbers.
The because the unilateral laplace transform is basically the same as the original, the properties of it are largely the same as well. However, Convolution and Differentiation change.
(x1x2)(t)X1(s)X2(s)(x_1*x_2)(t) \leftrightarrow \mathcal{X}_1(s)\mathcal{X}_2(s)
are right sided (i.e 0 for
This should be clear because the unilateral laplace transform only considers the right half of the reals, so unless the signal itself is right-sided, the convolution is going to blend together the left side of the signal with the right side.
dndtnsnX(s)k=1nsnkdk1dtk1x(0)\frac{d^n}{dt^n}\leftrightarrow s^n\mathcal{X}(s)-\sum_{k=1}^{n}{s^{n-k}\frac{d^{k-1}}{dt^{k-1}}x(0^-)}
Unlike the regular Laplace transform, differentiating a signal introduces initial conditions for the the Unilateral Laplace Transform. This actually makes it useful for solving differential equations because given the differential equation, we can take the Unilateral Laplace Transform of both sides, solve for our variable in the Laplace domain, and then take the inverse Laplace Transform to convert back to the time domain.

The Z Transform

Just like the Laplace Transform generalizes the CTFT, the Z-Transform generalizes the DTFT.
X(z)=n=x[n]znX(z) = \sum_{n=-\infty}^{\infty}{x[n]z^{-n}}
z=ejωz = e^{j\omega}
, we recover the DTFT
X(ejω)=n=x[n]ejωnX(e^{j\omega})=\sum_{n=-\infty}^{\infty}{x[n]e^{-j\omega n}}
We know that the DTFT will be well defined as long as the region of convergence of the Z-transform contains the unit circle (because that represents all points
. Unlike the ROC of the laplace transform, which is a rectangular section of the plane, the ROC of the Z-transform is a circular section. This means the only possible shapes for the ROC of a Z-transform are a ring, the interior of a circle, the exterior of a circle, or the empty set. For right-sided signals, the ROC is the exterior of the circle defined by the outermost pole. Like the Laplace transform, the best way to take the inverse Z transform is to use partial fraction decomposition.

Properties of the Z-Transform

ax[n]+by[n]aX(z)+bY(z)ax[n] + by[n] \leftrightarrow aX(z) + bY(z)
Time Shift:
x[nn0]zn0X(z)x[n-n_0] \leftrightarrow z^{-n_0}X(z)
z0nx[n]X(zz0)z_0^nx[n] \leftrightarrow X\left(\frac{z}{z_0}\right)
Time Reversal:
x[n]X(z1)x[-n] \leftrightarrow X(z^{-1})
nx[n]zdX(z)dz-nx[n] \leftrightarrow z\frac{dX(z)}{dz}
(xy)[n]X(z)Y(z)(x*y)[n] \leftrightarrow X(z)Y(z)

Theorem 8 (Initial Value Theorem)

, then
x[0]=limzX(z).x[0] = \lim_{z\rightarrow\infty}{X(z)}.

Unilateral Z-Transform

Like the Laplace Transform, the Z-Transform has a unilateral variant.
X(z)=n=0x[n]zn\mathcal{X}(z) = \sum_{n=0}^{\infty}{x[n]z^{-n}}
Like before, convolution with the Unilateral transform is only valid for right sided signals. The time-delay property also changes for the unilateral z-transform.
x[nn0]zn0X(s)+k=1n01zn0+kx[k]x[n-n_0] \leftrightarrow z^{-n_0}\mathcal{X}(s)+\sum_{k=1}^{n_0-1}{z^{-n_0+k}x[-k]}
Notice how this property can now help us solve difference equations because we can simply take the unilateral z-transform of both sides, solve for our output, and then convert back to the time domain.