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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: $$\mathcal{F}\big\{u(t)\big\}=U(\omega)=\pi\delta(... 6 Can one argue that discrete time-series coming from stocks or commodities (prices) are derived from a continuous-time process? I would say "no". The price of a stock is determined by a trade and each trade is a discrete event in time. I would argue that between trades the price is undefined since there is no experiment you could do to determine ... 5 The final value of the step response is the DC gain of the closed-loop transfer function, which is generally different from the open-loop DC gain. Assuming unity gain feedback, the feed-forward transfer function G(s) equals the open-loop transfer function, and the closed-loop transfer function is given by$$C(s)=\frac{G(s)}{1+G(s)}\tag{1}$$The final value ... 5 The answer is clearly 'no', except for trivial cases. Note that$$G(f)=|G(f)|e^{j\phi(f)}\tag{1}$$where \phi(f) is the phase of G(f). Rewriting (1) as$$|G(f)|=G(f)e^{-j\phi(f)}\tag{2}$$and taking the inverse Fourier transform gives$$\tilde{g}(t)=(g\star a)(t)\tag{3}$$where \tilde{g}(t) is the inverse Fourier transform of |G(f)|, and a(t) is ... 5 Concerning your first question, both, the Laplace and the Fourier transform, are frequency domain representations of a function or signal. In the Fourier transform we deal with a real-valued frequency variable \omega, whereas in the Laplace transform we have a generally complex-valued independent variable (usually s), the imaginary part of which equals ... 4 Open loop gain at DC is -3dB or .707 and 0 degrees. We don’t know the forward gain but assuming it is the open loop gain, the closed loop gain would be .707/(1+.707)= .4148, matching the first plot. (With 60 degrees of phase margin however I would have expected a response closer to the third plot, but as explained in the comments this is due to my ... 4 Some general hints for this type of problem If the function is defined piecewise than chances are your calculation and solution will also have to be done piecewise in sections. Determine first where the result is zero, i.e. where x(\tau) and h(t-\tau) don't overlap. This determines the start and end of the sections that you need to consider. Hint: if ... 4 Perfect recovery is one thing, niceness is another. Sampling above x2 Nyquist is sufficient for perfect recovery, after which we can FFT-upsample to make it look nice - which is more efficient than directly sampling at a higher rate (though it has its advantages). 4 First, the Nyquist rate is an absolute lower bound to what will possibly work if you're sampling a strictly bandlimited signal. The closer you get to the Nyquist rate when you're sampling, the more expensive your system gets in various ways. In particular, the closer you sample to the Nyquist rate, the sharper the cutoff you need in your reconstruction ... 4 From the transfer function alone it is generally impossible to say whether a system is causal or not. Only in combination with a given region of convergence (ROC), or, equivalently, with an assumption about stability, can we know for sure if a given system is causal or not. The given transfer function has a pole at s=1. There are two possible time domain ... 3 First, you are not trying to find the linearity of a signal, but the linearity of a system defined by the transform above between input and output signals. If you suspect your system is not linear (the different expressions depending on the sign of x(t) is a big red flag), then you only need a counterexample rather than a general proof of linearity. In ... 3 I would like to add to the previous answers that the difference between a second-order highpass and a second-order lowpass filter is generally NOT an allpass filter. The resulting transfer function$$H(s)=\frac{s^2-\omega_0^2}{s^2+\frac{\omega_0}{Q}s+\omega_0^2}\tag{1}$$has two zeros at s_0=\pm\omega_0 and two poles which are either real-valued (for Q\le\... 3 Just adding to the answer. The sign for recombining L/R filter alternates, so it's '-' for second order '+' for 4th order and so forth. The algebra is isn't all that pleasant so I will only go through the 2nd order. A second order L/R lowpass is simply the cascade of two first order Butterworth lowpass. The first order butterworths are$$L = \frac{1}{1+jx}, ...

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One first important matter is to understand how much an absolute value may change a signal, and his amplitude spectrum. Below are two examples: $s_1$ with alternating $-1/1$ (blue), and $|s_1|$ (red) which is constant flat; and a smooth sine $s_2$, with its bumpy magnitude counterpart. In plain text: an oscillating time signal can become either very calm ...

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$\omega$'s value does nothing for the convergence of the integral. $a$'s does. $a$ is not complex, it's real only. If $a$ is 0, the integral doesn't converge.

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Bottom line: it makes the math easy. In addition, there are certainly instances where we care that a signal has existed for some time in the past. But how does it even make sense to write a negative value for time, Think of $t=0$ as "now", $t < 0$ as "before now", and $t > 0$ as "later". Does that make sense? By ...

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Why does repeating periodically a signal means removing some frequency intervals from the single pulse spectrum? Because your signal is periodic. A periodic signal only contains frequencies that are integer multiples of the signal's repetition frequency. In other words, if the signal is periodic all of it's constituent parts must be periodic as well. In ...

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Usually, these kinds of questions are somehow intentionally ambiguous, just for you to take well-justified assumptions. Ideally, both solutions could give the same results. However, signals are usually not band-limited. What anti-alias filter is being used before the acquisition? The transition band of this filter, no matter how high its order is, will never ...

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From what you've said, you have two sample sets: $$x_n, n = 0 \ldots N-1$$ and $$y_m, m = 0 \ldots M-1$$ where $M \ne N$ and you want to compare the distributions of the underlying processes. Do you know anything about the expected distribution of the measurements? If they're Gaussian, then you can just calculate the sample mean and sample standard deviation ...

3

Yes and no. If you recall from Calculus, if $g(t) = \frac{d}{dt} f(t)$, then $\int g(t)\ dt = f(t) + C$. So, since the derivative of a sinusoid is a sinusoid of the same frequency, if $g(t)$ sinusoidal, its integral must be a sinusoid of the same frequency, plus a constant. Whether that "plus a constant" bit is going to make you consider the ...

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The official answer to your question is this : random processes (time-series) can be classified as being either continuous-time or discrete-time. Some of those discrete-time processes happen from sampling of continuous-time processes, but some of the discrete-time processes occur naturally, such as daily stock-market values, or Sun-Spot numbers. This ...

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For the world in the whole, with our current technical abilities we can't say for sure whether continuous time processes exist in the real world or not. The problem is degree of difference between time measurements. If your degree of difference is $\Delta t$ then you can't say something happened at the moment $t_0 + \Delta t/2$ or $t_0 + \Delta t$. From your ...

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Both results are correct. Hence, the equality $$\frac{1}{T}\sum_{k=-\infty}^{\infty}\frac{1}{1+ j \left(\omega - k\frac{2\pi }{T}\right)}=\sum_{n=0}^{\infty}e^{-nT(1+j\omega)}\tag{1}$$ holds true. One way to prove this is to realize that the term on the left-hand side of Eq. $(1)$ is periodic with period $2\pi/T$. Consequently, we can express this term by ...

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It is easy to see that using the trigonometric identity in Equation $(1)$ below $$\cos(\theta_1 \pm \theta_2) = \cos(\theta_1)\cos(\theta_2) \mp \sin(\theta_1)\sin(\theta_2) \tag{1}$$ We have \begin{align} f(t)\cos\big(\pi(t - 1)\big) &= f(t)\overbrace{\cos(\pi t - \pi)}^{\text{use Equation}\ (1)}\\ &\equiv -f(t)\cos(\pi t)\tag{2} \end{align} Which ...

2

D'oh. This turned out to be pretty obvious, and I should have gotten it from reading more closely. I would have deleted this out of shame, but instead will share the correction in hopes it saves someone else the same headache. The controlling equations for the LR2 high and low pass transfer functions are correct, but one of the signals needs to be ...

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The total power of an amplitude modulated signal is \begin{align}\overline{s^2_{AM}(t)}&=\overline{\big(A+m(t)\big)^2\cos^2(2\pi f_ct)}\\&=\frac12\overline{\big(A+m(t)\big)^2}+\frac12\overline{\big(A+m(t)\big)^2\cos(4\pi f_ct)}\tag{1}\end{align} The second term on the right-hand side of $(1)$ is zero if $m(t)$ is a lowpass signal, and if $f_c$ is ...

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You are right that the argument in the proof is not correct, or at least misleading. The fact that $e^{-st}$ becomes zero as $t\to\infty$ is true for any $s$ with $\textrm{Re}\{s\}>0$. However, the integral $$\int_0^{\infty}f(\tau)d\tau$$ might not exist, so we can't just compute the limit by claiming that the first term becomes zero. We have to compute ...

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The variance of the white noise will go down by $N$ through averaging (or standard deviation will go down by $1/\sqrt{N}$, where the standard deviation as a magnitude quantity will be visually consistent with the "spread" of noise on the graphic the OP is looking at. Thus to make it visually reduce by a factor of 10, the OP would need to average ...

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tl;dr: you can't. That's basics! longer story: You ask about symbols. When the symbol rate is much larger than the carrier frequency, you've basically lost, and cannot transport that many symbols per second; you don't have the bandwidth; you get at most twice the carrier frequency in bandwidth, namely from 0 Hz to twice the carrier frequency. However, you'...

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As said by @Juancho, the expression let us suspect that the system could be non-linear. So we could look for a counter-example, yet it is not evident how to find a good one. Anyway, lt us try to better understand the system. Since there is something around sign change. The insight is there to have a first signal that has a sign change, and a second that ...

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