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1

There is no real shortcut for computing the impulse response. Partial fractions is the standard way to do it. However, in the case of a second-order transfer function as in your example, the result can be found in tables of $\mathcal{Z}$-transform pairs. E.g., if you use the last two correspondence of this table, you can pretty quickly write down the result. ...


0

Yes it's a little confusing at first. And it only deepens if you also take an LTI system perspective into account, which imposes its own constraints into the corresponding difference equation structure. Consider the following difference equation structure for a causal system: $$\sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{M} b_k x[n-k]$$ with auxiliary ...


4

It's true that the algebraic expression for the $\mathcal{Z}$-transform is generally not sufficient for computing the corresponding time-domain sequence. The additional information we need is the region of convergence (ROC). The ROC together with the expression for the $\mathcal{Z}$-transform uniquely determines the corresponding time-domain sequence. For ...


1

This is all about consequently substituting variables. Shifting by $t_0$ means replacing the variable $t$ by $t-t_0$. If $y(t)=x(2t)$, then $y(t-t_0)=x(2(t-t_0))=x(2t-2t_0)$. If you have a shifted input $x_2(t)=x(t-t_0)$, then the output is $y(t)=x_2(2t)$, i.e., the variable $t$ is multiplied by $2$. Replacing $t$ by $2t$ in $x(t-t_0)$ results in $x(2t-t_0)$...


3

I believe that h(-t) means a "time-reversed" version of h(t). Your command: 'y = conv(r,-h);' computes the convolution of 'r' and negative 'h', and you don't want that. I think you want: y = conv(r,conj(fliplr(h)));


0

Assume that the input to a simple up-sampler (say by $L$) is $x[n]$, hence $$y[n] = \begin{cases} { x[n/L] ,\qquad n=kL \\ 0 , \qquad \qquad \ \text{ else } }\end{cases}$$ If you compute the auto-correlation sequence, defined as $$\phi_{xx}(i,j) = \mathbb{E}(x[i]x[i+j])$$ for an iid white Gaussian process, we have that \begin{align} \phi_{xx}(i,j) &= ...


1

A more general expression states that for $ M \geq N$: $$ \sum_{n= N}^{n = M} c = (M-N+1) \cdot c $$ where the derivation simply relies on fact that the epxression has (M-N+1) terms : $$ \sum_{n= N}^{n = M} c = \{ c + c + ... + c\} = (M-N+1) \cdot c $$ And when applied for your particular case (with $N = -M$) it becomes: $$ \sum_{n= -M}^{n = M} c = (M-(-...


0

For instance, from $-3$ to $3$, you have $-3,\,-2\,-1,\,0\,1,\,2,\,3$, hence $2\times 3+1$ terms. More generally, the sum from $-M$ to $M$ is composed of $2M+1$ terms: indices with $m$ strictly negative (a total of $M$), those which $m$ strictly positive (a total of $M$), plus one at zero ($1$). If all terms are the same constant $c$, the total is $(2M+...


0

You want to take a signal with sample rate 30 Hz and upsample the signal to 3 kHz rate and then pass it through a low pass filter to perform the sinc interpolation. In MATLAB, you have a couple options: You could use the built-in interp function. You will need to pass it your signal and the upsample factor (3,000 / 30 = 100). The function will figure out a ...


1

You may know that one important property of linear time-invariant (LTI) systems is that the complex exponential $e^{j\omega_0}$ is an eigenfunction, and the corresponding eigenvalue is given by the system's frequency response evaluated at $\omega_0$. So the response to an input signal $$x[n]=e^{jn\omega_0}$$ is given by $$y[n]=H(e^{j\omega_0})e^{jn\omega_0}$...


0

This feels a lot like there's quite a bit of homework you should do about this, but let's give you a start: Write down your signal explicitly as $$s(t) = p(t) \cdot e^{\lambda t}\text,$$ with $p$ being your periodic oscillation. You sample that, and divide it into vectors. Each vector has $N=128$ samples. Let's write it this: \begin{align} s_k[n] &= ...


1

You are right and the manual is wrong. Given $S_1, S_2, S_3$ and the respective input-output signals as below : $$\\x[n] \rightarrow \boxed{S_1} \rightarrow v[n] \rightarrow \boxed{S_2} \rightarrow w[n] \rightarrow \boxed{S_3}\rightarrow y[n] \\$$ $\\$ $ x[n] \xrightarrow{S1} v[n] = \begin{cases} x[n/2] & \text{if n is even;}\\ 0 & \text{if n is ...


1

You put zeros in between to show delay between the channel taps. I don't know what $M$ is in your code but suppose $M=10$ and say your sample rate is 10 MHz. Then you can interpret mc = [1 zeros(1, M) 0.28 zeros(1, 2.3*M) 0.11]; as you get the first multipath component with zero delay (gain = 1), then 10 zeros later (which is equal to 1 microsecond) you get ...


1

The idea here is that any Linear Operator is Homogeneous though not every Homogeneous Operator is Linear. The classic proof indeed is to build the zero term as a sum (Difference) of 2 other elements in the domain. Then use Linearity to show it must be zero: $$ T \left( 0, 0 \right) = T \left( {m}_{1} - {m}_{1}, {n}_{1} - {n}_{1} \right) = T \left( {m}_{1}, ...


2

Let's assume our data is in finite dimension. So $ x \left[ m, n \right] \in \mathbb{R}^{M \times N} $. So it can be written as a matrix $ X \in \mathbb{R}^{M \times N} $. Using the SVD Decomposition the matrix can be written as: $$ X = \sum_{i = 1}^{n} {\sigma}_{i} {u}_{i} {v}_{i}^{T} $$ Seprable 2D Matrix is defined as a rank 1 Matrix which can be ...


0

You just have to checkof the rank of the matrix. When you look a discrete signals, it is customary to express their product the following way. If vectors are considered "column-wise", then you typically use the transpose operation: $$ x[n_1,n_2] = f[n_1]^T g[n_2]$$ This implies that the 2D signal is of rank one or less (rank being the maximal number of ...


0

The best answer I could give is just a hint: think about what is the inverse discrete time Fourier transform of $X(e^{j\omega})$. Constants in the frequency domain are what in the time domain? Answering that will lead you to the answer of this question.


1

Note that if you upsample a sequence $x[n]$ by a factor $M$ you have $$y[Mn]=x[n]\tag{1}$$ The DFT of $x[n]$ is $$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N}nk}\tag{2}$$ The DFT length of $y[n]$ is $MN$, so the DFT of $y[n]$ is $$Y[k]=\sum_{n=0}^{MN-1}y[n]e^{-j\frac{2\pi}{MN}nk}\tag{3}$$ Since only every $M^{th}$ sample of $y[n]$ is non-zero, $(3)$ ...


0

Of course, if you have a formula describing the system's response to an arbitrary input $x[n]$, you will obtain the response to an impulse if you choose $x[n]=\delta[n]$. However, I suppose that the idea of the exercise is to obtain a closed-form expression for the impulse response. You should recognize that the given input-output relation is just a ...


1

First of all, as you said the sampling rate is probably $12$ Hz, rather than $12$ kHz, and perhaps they want to demonstrate an aliasing example. Given a bandlimited continuous-time periodic signal $$x(t) = \cos(16\pi t + \phi)$$ the samples taken at the rate $F_s = 12$ Hz will be denoted as $x[n]$ and will be obtained by via $x[n] = x(t_n)$ with $t_n = n ...


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