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24

In signal processing, two problems are common: What is the output of this filter when its input is $x(t)$? The answer is given by $x(t)\ast h(t)$, where $h(t)$ is a signal called the "impulse response" of the filter, and $\ast$ is the convolution operation. Given a noisy signal $y(t)$, is the signal $x(t)$ somehow present in $y(t)$? In other words, is $y(t)$...


20

If the ratio between your sampling frequency and the frequency of your signal is irrational, you will not have a periodic discrete signal. Assuming you have a 1-kHz sine wave and you sample at 3000*sqrt(2) Hz. You will have approximately 4.2 samples per period. However you will not be able to sample the sine wave exactly at the same place. Hence your ...


19

Sampling at a higher frequency will give you more effective number of bits (ENOB), up to the limits of the spurious free dynamic range of the Analog to Digital Converter (ADC) you are using (as well as other factors such as the analog input bandwidth of the ADC). However there are some important aspects to understand when doing this that I will detail ...


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 ...


15

Your work is OK except for the problem that the Fourier transform of $\cos(2\pi f_0 t)$ does not exist in the usual sense of a function of $f$, and we have to extend the notion to include what are called distributions, or impulses, or Dirac deltas, or (as we engineers are wont to do, much to the disgust of mathematicians) delta functions. Read about the ...


14

This is called reconstruction.


13

You are correct. Your last equation is simply a fancy way of saying that $X(f)$ is real valued. In general: if it's real in one domain, it's conjugate symmetric in the other.


12

I've explained it once on StackOverflow. Your signal can be represented as a vector, and convolution is multiplication with a tridiagonal matrix. For example: Your vector/signal is: V1 V2 ... Vn Your filter (convolving element) is: [b1 b2 b3]; So the matrix is nxn: (Let it be called A): [b2 b3 0 0 0 0.... 0] [b1 b2 b3 0 0 0.... 0] [0 b1 b2 b3 ...


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_{\...


12

The two terms convolution and cross-correlation are implemented in a very similar way in DSP. Which one you use depends on the application. If you are performing a linear, time-invariant filtering operation, you convolve the signal with the system's impulse response. If you are "measuring the similarity" between two signals, then you cross-correlate them. ...


11

Suppose that the sampling is done at a rate of $1000$ Hz, one sample every millisecond. Suppose also that the signal being sampled is at $3200$ Hz, the first sample is at the peak of the sinusoid. The next sample, will be taken one millisecond later during which time the sinusoid will have gone through $3.2$ periods, and so the next sample will have the ...


10

Oppenheim and Willsky's Signals and Systems or Lathi's Linear Systems and Signals are intended for Sophomores who have only a single semester of differential equations under their belts, so it is a bit unfair to criticize them for leaving out the functional analysis and the conformal mapping. At the sophomore level my favorite book is Siebert's Circuits, ...


10

Depends on what you are controlling. For DC-motors it is the inertia of the device that acts as a low-pass filter of the PWM modulated signal resulting in a continuous motion. For most LEDs it is the human eyes that do the apparent low-pass filtering. If the PWM-frequency is not very high you can actually see this by moving your head from left to right ...


10

Let $h(t)$ denote the impulse response of an LTI system. Then, for any input $x(t)$, the output is $$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)\,\mathrm d\tau.$$ In particular, the response to the input $x(t) = \exp(j2\pi ft)$ is $$\begin{align} y(t) &= \int_{-\infty}^\infty h(\tau)\exp(j2\pi f(t-\tau))\,\mathrm d\tau\\ &= \exp(j2\pi ft)\int_{-\...


10

Well, first of all the Sound Level Pressure decreases by $6 \; \mathtt{dB}$ when doubling the distance - this plays a big role. We do also have sound attenuation coming from our medium - air. Let's take a closer look onto sound absorption coefficient for different frequencies: Knowing that human speech is mostly concentrated at the range of $300\;\mathtt{Hz}...


9

As Mohammad stated already the terms Continuous Wavelet Transforms (CWT) and Discrete Wavelet Transforms (DWT) are a little bit misleading. They relate approximately as (Continuous) Fourier Transform (the math. integral transform) to DFT (Discrete Fourier Transform). In order to understand the details it is good to see the historical context. The wavelet ...


8

A somewhat visual complement to the other answers You are talking about systems that are linear and time invariant. Exponential functions have one peculiar property (and can be actually defined by it): doing a time translation results in the same function multiplied by a constant. So $$ e^{t-t_0}=e^{-t_0}e^t$$ The red exponential could as well be the ...


8

For simplicity, let's talk about one-dimensional convolutions only. You need to understand that the result of a convolution is a function and the alleged convolution integral is actually an infinitude of integrals, one for each time instant. By this I mean that to compute the output $y(t)$ of a system with impulse response $h(t)$ to an input signal at ...


8

For one-dimensional Fourier transforms, $$\begin{align*} x(t) &\leftrightarrow X(f)\\ x(t-a) &\leftrightarrow X(f)\exp(-j2\pi fa)\\ x(bt) &\leftrightarrow \frac{1}{|b|}X\left(\frac{f}{b}\right) \end{align*}$$ all of which are proved by change of variables in the definition of the Fourier transform $$X(f) = \int_{-\infty}^{\infty} x(t)\exp(-j2\...


8

Yes, if eqs. (2) and (3) hold for any "type of signal" (which they do), then (5) must hold. Inserting (4) into (2) we get $$ \mathscr{F}\big\{x^*(t)\big\} = X(-f) $$ and using (3) $$ X(-f) = X^*(-f) $$ If we substitute $f = - g$ we get $$ X(g) = X^*(g) $$ which, as Hilmar has already observed, means that $X(f)$ is real-valued. This is to be expected as, ...


7

A necessary (but not sufficient) conditions for $f_2$ to be a temporally scaled version of $f_1$ is that a spectral representation with a logarithmic frequency scale (such as the constant-Q transform) of $f_1$ is a translation of a log-frequency spectral representation of $f_2$. Practically, given two signals, you can perform the test and evaluate $a$ by ...


7

Yes, it's possible to analyse sound the way ears do. For example, you could compute the DFT of a signal continuously using several Goertzel filters. $$ y_k[n] = e^{j2\pi k/N} y_k[n-1] + x[n] $$ where $k= 0,1,\ldots, N-1$, so that $y_0$ is the DC or zero frequency term. Of course, this is an unstable filter, so some resetting or forgetting factor is ...


7

From your solution: I followed the following algorithm: $$ y(t) =x(2t) $$ $$ y_1(t) = x_1(2t)$$ Let $$x_2(t) = x_1(t-t_0) ~~~\text{and}~~~ y_2(t) = x_2(2t) $$ On this following step (time sampling of the shifted argument) you make the usual mistake: $$\implies y_2(t) = x_1(2(t - t_0))$$ which should instead be : $$\implies y_2(t) = x_2(2t) = x_1(t - ...


7

The answers by @Deve and @Hilmar are technically perfect. I would like to provide some additional insights, with a few questions. First, do you know of a signal satisfying this reversed-time/conjugate identity: $$x(−t)=x^*(t)\,?$$ A first obvious idea is to choose among real and symmetric signals. A natural one in the Fourier framework is the cosine. ...


7

If you multiply a continuous-time finite energy signal $f(t)$ with an impulse train you get $$\tilde{f}(t)=\sum_{n=-\infty}^{\infty}f(nT)\delta(t-nT)\tag{1}$$ where $T$ is the sampling interval and $\delta(t)$ is the Dirac delta impulse. Note that the "signal" $\tilde{f}(t)$ is a mathematical fiction, it cannot exist in practice, and it cannot even be ...


7

There are 4 versions of Fourier transforms that are all close cousins. It's all due to the basic property that "sampling in one domain corresponds to periodicity in the other domain". If a time signal is periodic, the frequency domain signal is discrete. If a frequency domain signal is periodic, the time domain signal is discrete. So you have four ...


6

Frequency is more likely decreasing (period is getting longer). This is a sweep sine signal or as some people used to call it - chirp tone. No point in rewriting the literature, so here are some links with very good explanation for both linear and exponential frequency change. For linear sweep you get: $$s(t)=\sin\left(\omega_1t+\frac{\omega_2-\omega_1}{T}\...


6

Let $y_1(t)$ be the response to the signal $x_1(t)$: $$y_1(t)=x_1(-t)\tag{1}$$ Now let $x_2(t)$ be a shifted version of $x_1(t)$: $$x_2(t)=x_1(t-T)\tag{2}$$ The response to $x_2(t)$ is $$y_2(t)=x_2(-t)=x_1(-t-T)\tag{3}$$ If the system were time-invariant, its response to $x_2(t)$ should be a shifted version of its response to $x_1(t)$: $$y_2(t)=y_1(t-...


6

A signal is indeed a function. Given a signal $f(x)$, according to whether continuous or discrete for both the variable $x$ and the function $f(x)$, there are four types of combinations: (1) $\mathbf{continuous}$ $x$ and $\mathbf{continuous}$ $f(x)$ This is the most common $\mathbf{analog}$ signal. (2) $\mathbf{continuous}$ $x$ and $\mathbf{discrete}$ $f(...


6

The definition of the autocorrelation function $R_x(\tau)$ depends on the nature of your $x$. If $x$ is a deterministic signal with finite energy then: $$R_x(\tau)=\int_{-\infty}^{+\infty}x(t)x^*(t-\tau)dt$$ If $x$ is a deterministic signal with finite average power$^{(1)}$ then: $$R_x(\tau)=\lim_{T\to+\infty}\frac{1}{T}\int_{-T/2}^{+T/2}x(t)x^*(t-\tau)dt$$...


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