12

You need to define what you mean by "invertible". Do you mean invertible by a causal and stable system? If yes, then any system that is not minimum-phase is not invertible (because the inverse system can't be causal and stable). Example of a system that cannot be inverted by a causal and stable system: a simple delay $y(t)=x(t-T)$, $T>0$, could only be ...


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 ...


7

A necessary condition for invertibility is that any output has only one possible input (or injectivity, as proposed in comments). Since we are looking at counterexamples, we can look at when this condition is not satisfied. The null system, that turns every signal into a zero flat line, is not invertible, but a bit trivial. A system that computes a ...


6

No. It's only LTI (Linear and Time-Invariant) systems that can be modeled with convolution through a unique single impulse response. For example the systems $$ y(t) = g(t) x(t) $$ or $$ y[n] = \sum_{k=0}^{k < n} x[n-k] $$ are both linear but not time-invariant and their output $y[n]$ cannot be computed with the convolution operation ( $\star$ denoting ...


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 ...


5

Although what @Fat32 wrote is correct, I think the potential instability of IIR filters is not the main reason for the instability of an adaptive IIR filter. After all, we can calculate the poles in each iteration and put a hard constraint to avoid poles out of the unit circle. Even within the case of the FIR filters - which are unconditionally stable- we ...


5

The IIR filter doesn't have to be unstable, but it has the potential of being so; unlike the FIR case which doesn't have even the potential. One reason for the (potential) unstability of an IIR (adaptive) filter is the numerical issues due to coefficient quantization. When the poles are closer to unit circle this will be critical. This is especially ...


5

the Fourier Series for a single periodic function, having period $P$, $$ x(t) = x(t+P) \qquad \forall \ -\infty < t < +\infty $$ is $$ x(t) = \sum\limits_{k=-\infty}^{\infty} c_k \, e^{j(2\pi k/P)t} $$ each harmonic $c_k$ is labelled with its harmonic number, $k$, in the above. the actual frequency of each harmonic, in Hz or units of 1/time, is $\...


5

You cannot sum $X_1[k]$ and $X_2[k]$ (the discrete spectra of the two signals) element-wise because for them $k$ represents different angular frequencies $\frac{k\,2\pi}{P_1}$ and $\frac{k\,2\pi}{P_2}$ in radians. Time domain addition will only translate to frequency domain addition if your frequency variable is equally proportional to angular frequencies ...


5

For continuous-time systems, a pole at location $s_0=\sigma_0+j\omega_0$ will create a time-domain contribution of the form $$e^{s_0t}=e^{\sigma_0t}e^{j\omega_0t}\tag{1}$$ which is a damped oscillation if the pole is in the left half-plane (i.e., $\sigma_0<0$), and if the pole is not on the real axis (i.e., $\omega_0\neq 0)$. For $\omega_0=0$ there is ...


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.


4

No; the system given by $$ y[n] = x[n] + n $$ is time-varying, due to the added term $n$. Your mistake is in the line : $$\ y_2[n] = x_2[n] + n = x[n-k]+(n-k) $$ which should be instead $$\ y_2[n] = x_2[n] + n = x[n-k]+n$$ and therefore implies that $$y_2[n] \neq y[n-k] $$.


4

I would take approach based on Blind Deconvolution. Since we're dealing with ill posed problem some assumptions should be made. The intuitive approach would be using the information as a prior for the signal. Another idea is to add LPF assumption of the Filter by setting the sum of its coefficients to be 1 and non negative. Yet since we have Discrete ...


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 ...


3

Any LTI system can be completely characterized (among other things) by it's transfer function or it's impulse response. If your filter represents an LTI system, that you can calculate it's output by either convolving the input with the impulse response or multiplying the transfer function with the spectrum of the input signal. In theory these things are ...


3

(Adapted from this answer on dsp.SE) The reason that the impulse response (also called the unit pulse response for discrete-time systems) determines the output for arbitrary input $x$ to an LTI system is that The output of a linear time-invariant system in response to input $x$ is the sum of scaled and time-delayed versions of the impulse response. ...


3

Yes your computation is correct. Indeed, you could have seen this even easier, had you considered that the output computation sum was $$ y[n] = T\{x[n]\} = x[n-1] + x[n] + x[n+1] + x[n+2] .$$ And this is clearly an LTI (linear time-invariant), FIR (finite impulse response) system with associated impulse response $$h[n] = \delta[n-1] + \delta[n] + \delta[...


3

Whether LTI or not all systems are invertible if unique (distinct) inputs produce unique (distinct) outputs Causality and stability are later concerns for making sense of the obtained inverse system. For example the inverse to the delay system $$y[n] = x[n-d] $$ is $$y[n] = x[n+d] $$ Which is clearly noncausal for $d > 0$, and is not ...


3

In addition to all the answers that are correct in a mathematical sense, in a practical sense, a system whose frequency response goes below some finite but small-enough value will not be usefully invertable, even if a simple mathematical analysis would suggest that it is. In frequency-domain terms, the frequency response of a system's inverse will have gain ...


3

For a parallel system, the individual subsystems just add, so you simply have $H_{||} = H_A + H_B$ . That goes for both the transfer function and the impulse response, so you can simply do $h_b = h_{||} - h_a$


3

Your proof is correct, and the result is that the system is time varying because the response to $x(t-t_0)$ does not in general equal a delayed response to the input $x(t)$. Of course, for the special case $t_0=2\pi k$ the delayed response to $x(t)$ equals the response to $x(t-t_0)$, but for time-invariance this equality must hold for any $t_0$.


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, ...


2

As the book's title is control theory; a controller is the general broad name given to any piece of hardware (or sowftware) that aims to control some processes, in your aerodynamics field probably the flight process; by menas of receiving signals from sensors and sending commands to propellers - actuators, and in between is the control hardware or software......


2

For an LTI system with impulse response $h(t)$ and transfer function $H(s)=\mathcal{L}\{h(t)\}$, an input signal $x(t)=e^{st}$ (note: no multiplication with the unit step function $u(t)$), results in an output $$\begin{align}y(t)&=(x\star h)(t)\\&=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau\\&=\int_{-\infty}^{\infty}h(\tau)e^{s(t-\tau)}d\tau\\&...


2

For the biquad section that is cascaded, the quantization issues regarding the pole locations are well understood. For a biquad transfer function: $$\begin{align} H(z) &= \frac{b_0+b_1z^{-1}+b_2z^{-2}}{1+a_1z^{-1}+a_2z^{-2}} \\ \\ &= \frac{b_0z^2+b_1z+b_2}{z^2+a_1z+a_2} \\ \\ &= b_0\frac{z^2+\frac{b_1}{b_0}z+\frac{b_2}{b_0}}{z^2+a_1z+a_2} \...


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