# Tag Info

11

Absolutely! Conjugates are mentioned in textbooks because conjugation has no effect on real signals, but it does on complex ones. This way, formulations are more general and apply to both real and complex valued signals. Complex numbers don't exist themselves, they are a mathematical construct. Having said that, their mathematical properties can be ...

5

Software-defined radio (SDR) models real band-pass signals as complex baseband signals. All signals and filters operate on complex numbers.

3

The impulse response characterizes an LTI system completely, in the sense that the response to any input can be computed from the input and from the impulse response by convolution. That's a consequence of the system being linear and time-invariant. If we define stability as bounded-input bounded-output (BIBO) stability then stability can be determined from ...

2

$\Omega$ and $\omega$ are frequencies (in radians). Usually, $\omega$ is used, but if you need to deal with continuous-time as well as discrete-time systems, it's common to use one for discrete time and one for continuous time in order to distinguish the two. There is no real standard as to which one is which. I've seen $\Omega$ used for continuous time as ...

2

You should explore how are these complex time-domain symbols transmitted over channel (atmosphere or wire) using modulated waveforms. Also, a good starting point would be to figure out that complex numbers are nothing but 2 orthogonal/perpendicular dimensions. When we say $x = 3 + 3i$, we are basically saying we have a pair of numbers which lie in ...

2

Let's take a slightly more general input signal $$x(t)=\sin(\omega_0t)u\left(t-\frac{\theta}{\omega_0}\right),\qquad\theta\in [-\pi,\pi)\tag{1}$$ It's straightforward to show that the response of an ideal integrator to the input $(1)$ is given by $$y(t,\theta)=\frac{\cos(\theta)-\cos(\omega_0t)}{\omega_0}\tag{2}$$ The response $y(t,\theta)$ always consists ...

2

All the other responses are excellent, especially Envidia's, so not to take away from those but I want to add this very intuitive view that bottom lines it quickly: Consider the spectrums below that start with a real signal (positive and negative frequencies are complex conjugate symmetric). This is what we could measure with a single scope probe (one stream ...

1

Matlab can only deal with transfer functions that are rational ratios of polynomials in $s$ (or $z$, if you're in that domain). So you must approximate. You can either use the responses that @Matt L. has shown you, and convolve by the impulse response at each step, or you can write the partial differential equation (not an ODE) and solve it at each time ...

1

The inverse Laplace transform of $G(s)=e^{-\sqrt{s}}$ can be computed analytically: $$g(t)=\mathcal{L}^{-1}\left\{e^{-\sqrt{s}}\right\}=\frac{1}{2t\sqrt{\pi t}}e^{-\frac{1}{4t}},\qquad t>0\tag{1}$$ [I've cross-checked this result by looking it up in one of my old math books.] So now you have an analytical expression for the impulse response $g(t)$. The ...

1

Your solution looks correct. In any case, the given solution $y(t)=5e^{-3t}-4e^{-4t}$, $t>0$, must be wrong because it doesn't satisfy the initial condition on the derivative ($y'(0^+)=0$): $$y'(t)=-15e^{-3t}+16e^{-4t},\qquad t>0\tag{1}$$ So we have $y'(0^+)=1$. Your solution satisfies both initial conditions.

1

If your delayed input is $\hat{x}(t)=x(t-\alpha)$, then $\hat{x}(t-\tau)=x(t-\tau-\alpha)$, and not $x(t-\tau+\alpha)$. From here everything will work out as expected.

1

Although this question is over 2 years old now, I think it's interesting to consider the solution, assuming that the original interpretation was incorrect, and that $u[n]$ represents the system input, not the unit step sequence. If so, then the first thing to recognize is that the system is just an accumulator. For a time-domain demonstration, simply make ...

1

It looks like a Homework problem so I will only provide small hints: Check from the block diagram implementation of the system whether the current output sample $y[n]$ depends on only current input sample $x[n]$ and past input & output samples $x[n-n_0]$ and $y[n-m_0]$. If so, the system is causal, else it will be non-causal. You can use the ...

1

But if impulse response of a system is basically an input output relation and impulse response can be different for the same system depending upon our choice of output, so how does it completely characterize the system? Isn't it contradictory? Because any input signal can be form using impulses. If you know how the system may response to one impulse then you ...

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