12

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 ...


6

Answer : No, any causal LTI system with frequency response $H(f)$ cannot produce the output $y(t)$ in advance. And, the answer lies in the causality of input signal $x(t)$ being applied to $h(t)$. Any causal input $x(t)$ which has an identifiable beginning cannot truly be Narrow-Band or Band-Limited. It will have non-zero frequency content at all frequencies....


6

The DC value is simply the mean. Since the signal is periodic you only need to take the mean of one period. This can be simply done with $$h(t)=\left\{\begin{matrix} 1/T, 0 < t < T\\ 0, {\rm elsewhere} \end{matrix}\right. $$ Clearly (1) and (2) are different. (1) will not converge, so that doesn't work (2) is the best way for a function that's ...


6

Your problem comes about because you're pondering the convolution of a signal by itself: $$g(x) = f(x) * f(x)$$ So, yes, if you had a block diagram where you collect f(x) from the æther and then somehow magically convolve it by itself (which can't be done physically because it would require time-reversing $f(x)$, which requires looking into the future), then ...


5

A constant impulse response won't work because if the input signal has a non-zero DC component, the output will blow up. Note that the input signal has frequency components at DC and at integer multiples of $1/T$, the latter being its fundamental frequency. So you simply need a filter that retains the DC component and filters out all integer multiples of $1/...


5

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


5

This has to do with the way the Laplace transform and the $\mathcal{Z}$-transform are defined: $$\mathcal{L}\big\{x(t)\big\}=\int_{-\infty}^{\infty}x(t)e^{-st}dt\tag{1}$$ $$\mathcal{Z}\big\{x[n]\big\}=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\tag{2}$$ Of course, the question remains why they are defined that way. The answer lies in the fact that in continuous time ...


4

You are right that a distributed system could be "something like a transmission line". Note that the system $$y(t)=x(t-T)\tag{1}$$ is a simple model of a transmission line, where just a frequency-independent delay $T$ is taken into account, and the attenuation is neglected. Note that lumped electrical systems, described by resistors, capacitors and ...


4

Ah ha! There's a lot of obfuscation in the problem as stated, plus a bit of over-emphasis on the final value theorem. $$H(z) = \frac{z^{-1} \phi / (1 - \phi z^{-1})}{1 - z^{-1} \psi / (1 - \psi z^{-1})} \tag 0$$ OK, so far so good. It arises from the following feedback system $$Y(z)(1-z^{-1}) = z^{-1}F_{\psi}\cdot Y(z) + z^{-1}F_{\phi}\cdot X(z) \tag{1}$$...


4

Convolution of an input signal with a fixed impulse response is a linear operation. However, if the input-output relation of a system is $$y(t)=(x*x)(t)\tag{1}$$ then the system is non-linear, which is straightforward to show. Similarly, any convolution with a kernel that depends on the input signal is a non-linear operation. On the other hand, a system with ...


3

Here is a actual example with negative group delay that will provide further insight: Below is a plot of the output and input of a pulse through a realizable filter that has negative group delay: It seems like a complete violation of causality, but it is just a clever DSP magic trick. Let's explore further: The filter above that did this had the following ...


3

I would do the following, first design a filter that notch's out just the DC. A first attempt would say to place a zero at z = 1, or at the unit circle on the real axis where real part = 1, the transfer function for this is given as $$ H_{1}(z) = 1 - z^{-1} \tag{1}$$, the frequency response of this transfer function is shown below in the first figure. ...


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 ...


3

I think your understanding is correct. The steady state response of an LTI system is the part of the response that is caused by a steady excitation at the input, like a sinusoidal signal, or any other periodic signal. The transient response is caused by changes at the input, like switching on a signal, or changing the parameters of a periodic input signal, ...


3

You have to look at the autocorrelation function of $y(t)$: $$R_y(\tau)=E\{y(t)y(t+\tau)\}\tag{1}$$ with $$y(t)=(h_A\star a)(t) - (h_B\star b)(t)\tag{2}$$ where $\star$ denotes convolution. If you write out $(2)$ with integrals and plug it into $(1)$ then, with the given assumptions on $a(t)$ and $b(t)$, you'll see that the mixed terms with the negative sign ...


3

Complex exponentials are eigenfunctions of LTI systems because they are eigenfunctions of the convolution operator: $$\begin{align}e^{j\omega_0t}\star h(t)&=\int_{-\infty}^{\infty}h(\tau)e^{j\omega_0(t-\tau)}d\tau\\&=e^{j\omega_0t}\int_{-\infty}^{\infty}h(\tau)e^{-j\omega_0\tau}d\tau\\&=e^{j\omega_0t}H(j\omega_0)\end{align}\tag{1}$$ where $h(t)$ ...


3

Your statement only holds true for a real sequence $h[n]$. The frequency response of $h[n]$ equals to $$H(e^{j\omega}) = \sum_{n=-\infty}^{\infty}h[n] e^{-j\omega n}$$ and if $h(n)$ is real, the conjugate symmetry condition holds $$H(e^{-j\omega}) = \sum_{n=-\infty}^{\infty}h[n] e^{j\omega n}=\big(H(e^{j\omega})\big)^*$$ So we can derive that $$\big|H(e^{j\...


2

Basically, you can solve this kind of EDO using two different techniques (which also have c different meaning). Classical Method - you will need the full EDO, containing input and outputs with initial conditions applied over the whole solution y(t), where y(t) = yhomogeneous(t) + yforced(t). The first term, you solve the homogeneous EDO considering NO ...


2

The region of Convergence of the system is $ \vert z \vert >2$. Sinusoids are eigen functions of LTI systems over infinite time, since the ROC does not include the unit circle hence system is not stable and hence an infintely running periodic signal will not converge. More info after comments The BIBO property of the system, if the input is unbounded ...


2

You are right that the system $h[n] = 2^n.u[n]$ is unstable in absolute summability sense. And absolute summability of impulse response of an LTI system is the required condition for BIBO stability. Meaning Bounded Input bounded output stability. Here the input is bounded and it is only corresponding to 4 frequencies in digital frequency domain $\omega = -\...


2

The integration in your question is equivalent to convolution with the unit step function: $$y(t)=\int_{-\infty}^tx(\tau)d\tau=(x\star u)(t)\tag{1}$$ This means that in the Fourier domain we have $$Y(j\omega)=X(j\omega)U(j\omega)\tag{2}$$ With $$U(j\omega)=\frac{1}{j\omega}+\pi\delta(\omega)\tag{3}$$ Eq. $(2)$ becomes $$Y(j\omega)=\frac{X(j\omega)}{j\...


2

I would not teach integrals this way. Issues might arise form the definition of terms. In standard calculus, a (note: "a", not "the) primitive function (also called antiderivative or indefinite integral) of a continuous function $f$ is a kind of converse of the concept of derivation. SO: If $F$ is "a" differentiable function whose derivative is equal ...


2

These problems are most easily solved by using the $\mathcal{Z}$-transform. But it is also possible to do it in the sample domain. You have two equations for expressing $w[n]$ and $w[n-1]$ in terms of $x[n]$ and $y[n]$: $$x[n]=w[n]-aw[n-1]\tag{1}$$ $$y[n]=w[n]+bw[n-1]\tag{2}$$ From $(1)$ and $(2)$, $w[n]$ is obtained as $$w[n]=\frac{1}{a+b}\big(bx[n]+ay[n]...


2

In complement to Matt L.'s as-usually-excellent-answer, some additional bits on the intuition, a simplification of the problem (to ease resolution) and the construction of a counter-example. They could be useful to understand and solve similar time-invariant/shift-invariant questions. First, on the intuition: the system contains a dilation on the time ...


2

You have a system with the following input-output relation: $$y(t)=\int_{-t}^{\infty}x(-3\tau)d\tau\tag{1}$$ In order to check whether the system is time-invariant or not, we need to compare the shifted output with the output resulting from a shifted input. The shifted output is $$y(t-T)=\int_{-(t-T)}^{\infty}x(-3\tau)d\tau\tag{2}$$ Shifting the input ...


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

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 ...


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

Sequences that are "time-invariant" must be constants (not varying with time at all). What the OP is actually asking about is a (discrete-time) system whose response $y$ to the input $x$ is defined to be $$y[n] = \begin{cases}\sum_{k=n_0}^n x[k], & n \geq n_0,\\??? & n < n_0\end{cases}$$ where the $???$ is there because according to some ...


2

Clearly, for negative values of $t$, the system needs to know the future in order to determine its output. Hence, the system can't be causal. Since the system is also time-varying (show it!), its response to an impulse doesn't say much about its general behavior, unlike it would be the case for a linear time-invariant (LTI) system. So the given system's ...


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