16

One thing that really helped me understand poles and zeros is to visualize them as amplitude surfaces. Several of these plots can be found in A Filter Primer. Some notes: It's probably easier to learn the analog S plane first, and after you understand it, then learn how the digital Z plane works. A zero is a point at which the gain of the transfer ...


13

The Fourier and the Laplace transform obviously have many things in common. However, there are cases where only one of them can be used, or where it's more convenient to use one or the other. First of all, even though in the definitions you simply replace $s$ by $j\omega$ or vice versa to go from one transform to the other, this cannot generally be done ...


12

I think there are actually 3 questions in your question: Q1: Can I derive the frequency response given the poles of a (linear time-invariant) system? Yes, you can, up to a constant. If $s_{\infty,i}$, $i=1,\ldots,N,$ are the poles of the transfer function, you can write the transfer function as $$H(s)=\frac{k}{(s-s_{\infty,1})(s-s_{\infty,2})\ldots (s-s_{\...


8

Why is the fourier transform a special case of the laplace transform? The Laplace transform produces a 2D surface of complex values, while the Fourier transform produces a 1D line of complex values. The Fourier transform is what you get when you slice the Laplace transform along the jω axis. For instance, a simple lowpass filter $H(s)=\frac{1}{s+1}$ has a ...


8

The system $$y[n]=y[n-1]+x[n]\tag{1}$$ is an ideal accumulator, i.e., it computes the cumulative sum of the input samples: $$y[n]=\sum_{k=-\infty}^nx[k]\tag{2}$$ It is in a way analogous to a continuous-time integrator, but this doesn't mean that you will necessarily obtain an ideal integrator by transforming the discrete-time system to a continuous-time ...


7

Assume you have a $N\times M$ sized image. If you know take what is classically used, a square filter kernel, of let's say size $L\times L$, you'd need to convolve that with the picture – which gives you $N\times M$ pixels, each needing $L^2$ multiply-accumulates. So you end up with $A_{2D}=L^2MN$ operations. Now, if you can decompose that filter into an $...


6

As pointed out by Peter K., it is important to distinguish between Laplace and Fourier transforms. The first few transform pairs in your question are Fourier transform pairs, whereas the last pair is a correspondence of the unilateral Laplace transform: $$F(s)=\int_{0}^{\infty}f(t)e^{-st}dt$$ In the last transform pair in your question the $\mathcal{F}$ ...


6

Both notations are common and correct. As pointed out by Yuri Nenakhov, the advantage of the argument $j\omega$ is that it coincides with the complex (Laplace transform) variable $s$ when its real-part is zero. Note that in the complex $s$-plane the frequency axis is the imaginary axis. So $j\omega$ has nothing to do with complex frequency (which makes no ...


6

The Laplace Transform is more representative of real systems that have a starting point, which is why the integral starts at 0, and also why the unit step function is generally talked about alongside the Laplace Transform. With the Laplace Transform, we can examine the transient and steady-state behavior of a system. Using $e^{st}$ instead of $e^{iwt}$ ...


6

Laplace transforms can capture the transient behaviors of systems. Fourier transforms only capture the steady state behavior. Of course, Laplace transforms also require you to think in complex frequency spaces, which can be a bit awkward, and operate using algebraic formula rather than simply numbers. If you want to see the power distribution of a signal ...


5

We usually talk of $j\omega$ when we're also interested in the Laplace transform of a signal / system, but want to just talk about the frequency response. The physical meaning of the imaginary part is that it refers to purely sinusoidal signals and are constant "amplitude". The real part refers to signals for which the "amplitude" decays or grows ...


5

In engineering practice, the complex inversion integral is hardly ever used. As an engineer, you will almost exclusively need to invert rational functions, and this can be done by partial fraction expansion and elementary inversions. So first I'll show you how to obtain the inverse Laplace transform by partial fraction expansion, then I'll explain the ...


5

The short answer is yes, if you have the Laplace or Z-transform of a function you do not need the Fourier transform. This is because the CFT is a special case of the Laplace transform and the DTFT is a special case of the Z transform. The Fourier transform is used to find the complex sinusoids that compose a function, whereas the Laplace transform finds ...


5

You cannot make conclusions about the stability of a system by only considering its transfer function evaluated on the imaginary axis $s=j\omega$. Replacing $s$ by $j\omega$ in the transfer function only makes sense for a stable system, otherwise you get a function of $\omega$ that does not describe the system, but another (stable) system. Let me explain ...


5

The problem is that you took the derivative of the function $$\hat{x}_u(t)=2e^{-3t}-e^{-4t}\tag{1}$$ whereas using the Laplace transform you implicitly assumed that $x_u(t)$ equals zero for $t<0$: $$x_u(t)=\hat{x}_u(t)u(t)=(2e^{-3t}-e^{-4t})u(t)\tag{2}$$ where $u(t)$ is the unit step function. If you take the derivative of $(2)$ then you'll get the ...


5

Let $H(s)$ be a transfer function of the form $$H(s) = \frac{1}{s-p}$$ where $p$, which is a pole of $H(s)$, can be written as a complex number $a+jb$. Taking the inverse Laplace transform of $H(s)$ gives the corresponding impulse response $h(t)$ (that is, the output of your system when given $\delta(t)$ as input). Noting $\mathcal{L}^{-1}$ the inverse ...


5

Let $s = \sigma + j\omega$, the inverse Laplace transform of $f(t+a)$ is given by $$f(t+a) = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s)e^{s(t+a)} \mathrm{d}s = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s)e^{sa}e^{st} \mathrm{d}s.$$ Hence the bilateral Laplace transform of $f(t+a)$ is $F(s)e^{sa}$ where $F(s)$ is the Laplace ...


5

The widespread use of the unilateral Laplace transform reflects the fact that in practice we often deal with causal systems and signals that have a defined starting time (usually chosen as $t_0=0$). The Fourier transform is mainly used for analyzing ideal signals and systems, such as ideal filters (e.g., low pass, high pass, etc.) and ideal signals such as ...


5

First of all, it's important to understand that there is no single best way to transform a continuous-time system to a discrete-time system. The method you're using is called backward Euler method, and it is defined by the mapping $$s\leftarrow\frac{1-z^{-1}}{T}\tag{1}$$ Note that in $(1)$ you scale by $1/T$, where $T$ is the sampling interval (i.e., $1/T$ ...


4

$X(j \omega)$ (frequency response) is a Fourier transform of system's impulse response. It's actually a function of frequency ($\omega$) but usually is written as $X(j \omega)$ because replacing $j \omega$ in the formula with $s$ will give you system's Laplace transform $X(s)$ without any additional conversions. (This works in the opposite direction as well: ...


4

If you have an understanding of Fourier transforms then you probably already have a conceptual model of transforming signals into the frequency domain. The Laplace transform provides an alternative frequency domain representation of the signal - usually referred to as the "S domain" to differentiate it from other frequency domain transforms (such as the Z ...


4

Matt is correct that the sign is convention. I think that there is a reason for it beyond that though. If we look at complex frequencies in the complex plane, they look like a constant vectors that rotate in one direction or another. Positive frequencies rotate counter-clockwise, negative frequencies rotate clockwise, and "0 Hz" frequencies don't rotate ...


4

You are right that the (bilateral) Laplace transform can be interpreted as the Fourier transform of $e^{-\sigma t}f(t)$. However, I think that the significance of the Laplace transform only becomes clear when $s=\sigma+j\omega$ is viewed as a complex variable because then we can study the analytic properties of the system function. E.g., electrical networks ...


4

Your simple integrator is called a "Rectangular Rule" integrator. There are more complicated (and more accurate) integrators called "Trapazoidal Rule", "Simpson's Rule", and "Tick's Rule" integrators.


4

Both transforms are equivalent tools, but the Laplace transform is used for continuous-time signals, whereas the $\mathcal{Z}$-transform is used for discrete-time signals (i.e, sequences). You can see that they are equivalent by using the continuous-time representation of a discrete-time signal, and then applying the Laplace transform to that signal. The ...


4

The answer to your last question is definitely 'no'. The point hotpaw2 makes in his answer is very relevant: the FFT is an efficient implementation of the DFT, and there are no equivalently efficient implementations for the numerical computation of the $\mathcal{Z}$-transform or the Laplace transform. But that's not the only reason. There are important ...


4

It's important to realize that generally the differential equation (DE) alone doesn't tell us anything about causality. You claim that the system given in your question is causal. However, an anti-causal system with impulse response $h(t)=-u(-t)$ (where $u(t)$ is the step function) also satisfies the very same DE! So the given DE actually describes two ...


4

You're comparing the transforms of two different functions. You consider the Fourier transform of the function $x_1(t)=\cos(\omega_0 t)$, but you took the Laplace transform of the function $x_2(t)=\cos(\omega_0t)u(t)$, where $u(t)$ is the unit step function: $$X_1(j\omega)=\int_{-\infty}^{\infty}x_1(t)e^{-j\omega t}dt\\ X_2(s)=\int_{0}^{\infty}x_2(t)e^{-st}...


4

This is because the impulse response of an integrator is $h(t)=u(t)$. The output which is the convolution with the impulse respoponse is $$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$$ and with $h(t)=u(t)$ it becomes $$\begin{align} y(t)&=\int_{-\infty}^{\infty}x(\tau)u(t-\tau)d\tau\\ &=\int_{-\infty}^{t}x(\tau)d\tau\tag{1}\end{align}$$ where $(...


4

Because, the region of convergence in the Laplace transform $$ X(s) = \int_{-\infty}^{\infty} x(t) e^{-st} dt $$ is related to the weighting provided by the real part of the complex $s = \sigma + j \omega$; as this will yield the weight $|e^{-st}| = e^{-\sigma t}$ applied on the input signal $x(t)$, and is a function of $\sigma$ alone and is a ...


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