# Tag Info

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Your derivation doesn't give the correct result because the limit $\lim_{t\to\infty}e^{-j\omega t}$ doesn't exist. You seem to have confused that limit with $\lim_{t\to\infty}e^{-\omega t}$, which does exist and which equals zero for $\omega>0$. There are many ways to derive the Fourier transform of the unit step function $u(t)$ (see e.g., this answer), ...

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At first, the meaning of BIBO stability should be considered which really helps to understand what it actually means even though dealing with words rather than numbers seems abstract in terms of engineering. BIBO stability or Bounded Input-Bounded Output stability states that for every bounded input $x(t)$ i.e. $|x(t)| < \infty$, output of the system of ...

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Your derivation is correct. The impulse response is not absolutely integrable, hence the system is not BIBO stable. From the corresponding transfer function $H(s)=1/s$, you can see that there is a single pole at the origin. Systems with single poles on the imaginary axis, like the integrator in your example, are also called marginally stable. These systems ...

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BIBO stability means that if existe some bounded intput that will produce an unbounded output. Take for instance the step input, the output will be the convolution of the step with another step $$y(t) = \int u(\tau) u(t - \tau) d\tau = t$$ The output will grow indefinitely as the time passes, even though the input never exceeds $1$ In terms of poles, a ...

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I'm going to take a shot at this because I think your confusion is about how we use the word "bandwidth" and not necessarily a theoretical issue. When we say a signal has bandwidth, it is just as you might think: looking at it's Fourier transform will give us an idea of how much frequency content is in the signal. When we say something like "...

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The full question probes as far as "what is science?", so I'll try simplifying. Fourier Transform is a tool. A mathematical construct. The goal is to accurately describe reality. Suppose a swinging pendulum. Suppose we know it swings 3 times per sec because we designed a motor to drive it such. How do we describe this swinging mathematically? We ...

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Yes: a sufficient criterion for a valid wavelet is being zero mean in time domain, i.e. $\hat \psi (0) = 0$. For $\psi(0) = 0$ we require that $\hat \psi$ sums to zero; one example is higher order Generalized Morse Wavelets - from Olhede & Walden, (Though above is only approximately zero-sum). Such wavelets can be admissible and analytic, enabling CWT ...

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Yes, perfectly possible. That means that the mean of the function $\hat{\psi}(k)$ is zero.

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How is that possible to know? Two methods: Analyze the physics behind the process that generates the signal Looking at a lot of actual data Example for #1: if you have an AM radio transmitter, you know that the frequency range it can produce are the carrier frequency plus/minus the highest modulation frequency. The contraption is not physically capable of ...

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Some signals are generated by processes that are physically capable of generating frequencies only in a specific frequency range. For example, the human vocal tract can only generate signals between roughly 50 and 10,000 Hz. This does not mean that every human can achieve that range, or that there are people who can generate frequencies slightly above or ...

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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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Defining time domain signals for all time facilitates the understanding of basic principles and makes solving more challenging problems easier (just as do imaginary numbers, but maybe for the same reasons you have an objection to those as well?). A classic example is the relationship of even and odd functions in the Fourier Transform and how this can be used ...

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Only the first criterion is correct when referring to input-output stability. The second criterion is just a way to compute the poles, as mentioned in a comment. All realizable continuous-time systems using lumped elements have poles, which doesn't contradict the stability requirement. Poles in the right half-plane (RHP) make a causal system unstable. Note ...

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The simple answer is that $$e^{j\omega_0t}\neq e^{j(\omega_0+2\pi)t}\tag{1},\qquad t\notin\mathbb{Z}$$ Since $t$ is a real variable, the inequality is true for uncountably many values of $t$. Equality is only achieved for countably many integer values of $t$. From $(1)$ it follows that for real $t$, the function $e^{j\omega_0t}$ is not $2\pi$-periodic in $\... 0 But$t$can also have fractional values. If$t=n=$integer then$\omega_0$and$\omega_0+2\pi$are aliases for the same frequency. 0 I had good luck in the past with modelling thermals as an electric circuit (see for example http://www.ingaero.uniroma1.it/attachments/2176_Cap_3%20Thermal-electrical%20analogy.pdf). Heat (in Joules) is current , temperature difference is voltage, thermal resistance is a resistor and thermal mass is a capacitor. In your case the contribution from the ... 0 Open loop temperature correction techniques are very challenging as they have to account for both static errors and dynamic temporal effects. The dynamic temporal effects refer to the thermal memory in your device under test and its interaction with the environment such that there will be response time involved from any change in the environment or its own ... 1 From the title of the graph, it is seen that it is a 1Hz square wave and hence so your DFFT has a high magnitude in the initial bin (bin holding your 1Hz frequency component). Also, to be sure, perform FFT on another signal generated at a different frequency to check if that's the case. In analog signals, it is not always mandatory that the first harmonic ... 2 Simply put, a square wave can be decomposed into a weighted summation of series of sine waves, which is exactly what the Fourier transform does. Here are two animations that describe the synthesis of a square wave by accumulating sine waves. The amplitude of FFT results are the weights of each sine waves. As seen in Fig.1, the first order harmonic has the ... 0 The noise is actually never upconverted, it was already there in the bandpass channel. The textbook is simply modeling the noise as a quadrature signal. To recap: The transmitted passband signal is usually assumed to be noise-free. The passband signal is real, and it occupies frequencies from$-W_2$to$-W_1$and from$W_1$to$W_2$. The receiver adds noise ... 7 The integral doesn't converge in the conventional sense, so you can't solve it with standard methods. Assuming that you know (or can look up) the Fourier transform of the unit step function$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(... 3$\omega$'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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That's because Matlab preforms FFT, the formula that you referred to is the FT of your signal, the difference between the two can be derived by convolving the DTFT (which has the same form of the FT) with the sinc function, which in your case is close to a delta function. For that reason you get a slight different frequancy domain from the theory you would ...

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