39

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 $y(t)$...


28

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 ...


21

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

If you imagine a Dirac delta impulse as the limit of a very narrow very high rectangular impulse with unit area centered at $t=0$, then it's clear that its derivative must be a positive impulse at $0^-$ (because that's where the original impulse goes from zero to a very large value), and a negative impulse at $0^+$ (where the impulse goes from a very large ...


17

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 ...


15

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_{-\...


15

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. ...


14

First of all the dirac delta is NOT a function, it's a distribution. See for example http://web.mit.edu/8.323/spring08/notes/ft1ln04-08-2up.pdf Treating it as a conventional function can lead to misunderstandings. Example: "informally" the dirac delta is often defined as "infinity at x=0 and zero everywhere else". Now let's look at a ...


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.


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}...


10

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-...


9

No doubt there are many links and expositions on wikipedia and elsewhere about convolution, its relation to linear systems, and its link to frequency analysis, however I do not think that is what you are after. Forget convolution for a moment. Just think of nothing but averages: If I give you a series of say, $10$ numbers, and ask you to take the average, ...


9

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 ...


9

Maybe a picture is worth a thousand words? Here's how a Gaussian pulse of variable width and its derivatives look like: As others have said, Dirac is a distribution, hence the Gaussian pulse, and its width gets narrower and narrower. The derivative of $$\mathrm{e}^{-x^2}=-2x\mathrm{e}^{-x^2}$$ Which says that the derivative is the same as the function, ...


8

The argument does not work in continuous time. In discrete time the argument is that $$e^{j\omega n}=e^{j(\omega+2\pi)n},\qquad n\in\mathbb{Z}\tag{1}$$ This is true because by definition $n$ is an integer. In continuous time we generally have $$e^{j\omega t}\neq e^{j(\omega+2\pi)t},\qquad t\in\mathbb{R}\tag{2}$$ because $t$ is a real variable. Eq. $(2)$ only ...


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, ...


8

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 ...


7

We can figure out what's going on if we first understand a simple identity and then just compute the Fourier transform of the periodic function. A useful identity First let's prove that $$D(\omega - \omega') \equiv \int_{-\infty}^\infty dt \, e^{i (\omega-\omega') t} = 2\pi \, \delta(\omega - \omega')\,. $$ We just use a test function $\tilde{g}(\omega)$: \...


7

@MathBgu I have read all above given answers, all are very informative one thing I want to add for your better understanding, by considering the formula of convolution as follows $$f(x)*g(x)=\int\limits_{-\infty}^{\infty}f(\tau)g(x-\tau)\,d\tau$$ and for the cross correlation $$(f\star g)(t)\stackrel{\text{def}}{=}\int\limits_{-\infty}^{\infty}f^*(\tau)g(...


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

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(...


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 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: $$R_x(\tau)=\lim_{T\to+\infty}\frac{1}{T}\int_{-T/2}^{+T/2}x(t)x^*(t-\tau)dt$$...


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

Duane Wise and i wrote a paper back in the 90s that we presented to an AES convention that spelled out how to model time-domain polynomial interpolation (of which linear interpolation is an example) in the frequency domain. i think you can get a copy here: Performance of Low-Order Polynomial Interpolators in the Presence of Oversampled Input


7

The integral doesn't converge in the conventional sense, so you can't solve it with standard methods. Assuming that you know (or can look up) the Fourier transform of the unit step function $u(t)$, it is straightforward to compute the Fourier transform of $\cos(\omega_0t)\,u(t)$ using the modulation property: $$\mathcal{F}\big\{u(t)\big\}=U(\omega)=\pi\delta(...


6

More generally, we have: $$y(t) = x(at + b)$$ You may want to combine both properties listed by Dilip Sarwate into one equation. Using the definition of the fourier transform, we can insert our input into the fourier integral: $$Y(jw) = \int^∞_{-∞}x(at+b)e^{jwt}dt$$ Using a change of variables such that T = at + b, we get: $$Y(jw) = \int^∞_{-∞}\frac{1}{...


6

Basically the dimensional number refers to the number of independent variables (input). In one dimensional signal $f(t)$, the amplitude $y=f(t)$ is the dependent variable (output), and there is only one independent variable $t$. In two dimensional image $f(x,y)$, the two independent variables are $x$ and $y$, and the dependent variable $f$ is also called ...


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

This sounds like a confusion in terminology. See http://en.wikipedia.org/wiki/Digitizing Digitization basically involves two steps: Discretization: Sampling the signal at discrete times Quantization: Turning the samples from (in theory) infinite resolution to finite resolution Look at all possible combinations you can have 4 different types of signal: ...


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