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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 ...


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)));


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 ...


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. ...


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 $x(2t-t_0)$...


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-(-...


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}$...


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}, ...


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