Exercise 3.6Z: Two Imaginary Poles

Two imaginary poles and one zero

In this exercise, we consider a causal signal $x(t)$ with the Laplace transform

$$X_{\rm L}(p) = \frac { p} { p^2 + 4 \pi^2}= \frac { p} { (p-{\rm j} \cdot 2\pi)(p+{\rm j} \cdot 2\pi)} \hspace{0.05cm}$$

corresponding to the graph (one red zero and two green poles).

In contrast, the signal $y(t)$ has the Laplace spectral function

$$Y_{\rm L}(p) = \frac { 1} { p^2 + 4 \pi^2} \hspace{0.05cm}.$$

Thus, the red zero does not belong to $Y_{\rm L}(p)$.

Finally, the signal $z(t)$ with the Laplace tansform

$$Z_{\rm L}(p) = \frac { p} { (p-{\rm j} \cdot \beta)(p+{\rm j} \cdot \beta)} \hspace{0.05cm}$$

is considered, in particular the limiting case for $\beta → 0$.

Please note:

The exercise belongs to the chapter Inverse Laplace Transform.
The frequency variable $p$ is normalized such that time $t$ is in microseconds after applying the residue theorem.
A result $t = 1$ is thus to be interpreted as $t = T$ with $T = 1 \ \rm µ s$ .
The residue theorem is as follows using the example of the function $X_{\rm L}(p)$ with two simple poles at $ \pm {\rm j} \cdot \beta$:

$$x(t) = X_{\rm L}(p) \cdot (p - {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p\hspace{0.05cm} \cdot \hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it \beta}}+X_{\rm L}(p) \cdot (p + {\rm j} \cdot \beta) \cdot {\rm e}^{\hspace{0.03cm}p \hspace{0.05cm} \cdot\hspace{0.05cm}t} \Bigg |_{\hspace{0.1cm} p\hspace{0.05cm}=\hspace{0.05cm}{-\rm j \hspace{0.05cm} \cdot\hspace{0.05cm} \it \beta}} \hspace{0.05cm}.$$

Questions

Solution

(1) The suggested solutions 1, 3 and 4 are correct:

The following is obtained for signal $x(t)$ for positive times by applying the residue theorem:

$$x_1(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}1}} \hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t}\}= \frac {p} { p+{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}= \frac{1}{2} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\hspace{0.05cm} ,$$

$$ x_2(t)\hspace{0.25cm} = \hspace{0.2cm} {\rm Res} \bigg |_{p \hspace{0.05cm}= \hspace{0.05cm}p_{{\rm x}2}} \hspace{0.7cm}\{X_{\rm L}(p)\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm} t}\}= \frac {p} { p-{\rm j} \cdot 2\pi}\cdot {\rm e}^{\hspace{0.05cm}p \hspace{0.05cm}\cdot \hspace{0.05cm}t} \bigg |_{p \hspace{0.05cm}= -{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2\pi}= \frac{1}{2} \cdot {\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t} \hspace{0.05cm} .$$

$$\Rightarrow \hspace{0.3cm} x(t) = x_1(t) + x_2(t) = {1}/{2} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}+{\rm e}^{-{\rm j} \hspace{0.05cm} \cdot \hspace{0.05cm}2\pi t}\right ] = \cos(2\pi t) \hspace{0.05cm} .$$

(2) The suggested solutions 2 and 4 are correct:

In principle, this subtask could be solved in the same way as subtask (1).
However, the integration theorem can also be used.
This says among other things that multiplication by $1/p$ in the spectral domain corresponds to integration in the time domain:

$$Y_{\rm L}(p) = {1}/{p} \cdot X_{\rm L}(p) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t \ge 0:\quad y(t) = \int_{-\infty}^t \cos(2\pi \tau)\,\,{\rm d}\tau = {1}/({2\pi}) \cdot \sin(2\pi t) \hspace{0.05cm} .$$

Please note: The causal cosine signal $x(t)$ and the causal sine signal $y(t)$ are shown on the information page of Exercise 3.6 als $c_{\rm K}(t)$ and $s_{\rm K}(t)$, respectively.

(3) The suggested solutions 1 and 3 are correct:

A comparison with the computation of $x(t)$ shows that $z(t) = \cos (\beta \cdot t)$ holds for $t \ge 0$ and $z(t) = 0$ for $t < 0$ .
The limit process for $\beta → 0$ thus results in the step function $\gamma(t)$.
The same result is obtained by consideration in the spectral domain:

$$Z_{\rm L}(p) = \lim_{\beta \hspace{0.05cm} \rightarrow \hspace{0.05cm} 0}\hspace{0.1cm}\frac{p}{p^2 + \beta^2} = {1}/{p} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z(t) = \gamma(t) \hspace{0.05cm} .$$

	$x(t)$ is a causal cosine signal.
	$x(t)$ is a causal sinusoidal signal.
	The amplitude of $x(t)$ is $1$.
	The period of $x(t)$ is $T = 1 \ \rm µ s$.

	$y(t)$ is a causal cosine signal.
	$y(t)$ is a causal sinusoidal signal.
	The amplitude of $y(t)$ is $1$.
	The period of $y(t)$ is $T = 1 \ \rm µ s$.

	For $ \beta > 0$, $z(t)$ is cosine-shaped.
	For $ \beta > 0$, $z(t)$ is sinusoidal.
	The limiting case $\beta → 0$ results in the step function $\gamma(t)$.