New answers tagged homework
1
This exercise is meant to help the student appreciate the fact that if the response $y_1(t)$ of an LTI system to an input $x_1(t)$ is known, then the response to an input
$$x_2(t)=\sum_{k=1}^Ka_kx_1(t-t_k)\tag{1}$$
is given by
$$y_2(t)=\sum_{k=1}^Ka_ky_1(t-t_k)\tag{2}$$
which is a direct consequence of linearity and time-invariance.
Consequently, if an input ...
1
It must be added to the problem that $R(\omega)$ is a real-valued, possibly bipolar function. In that case, its inverse discrete-time Fourier transform must be even:
$$r[n]=r[-n]\tag{1}$$
From the given relation between $H(e^{j\omega})$ and $R(\omega)$ it is clear that
$$h[n]=r[n-25]\tag{2}$$
must hold. I'm sure that you'll manage to combine $(1)$ and $(2)$ ...
0
To answer your question in the context you have asked, you can look at it as follows:
You input signal is made up of two discrete-time sinusoids of equal length(5mins) $x_1[n]$ and $x_2[n]$. Supposing sampling rate such that 5mins correspond to $N$ samples and it is well above $400Hz$. So, your input signal will be of length $2N$.
Now, N-Point DFT of $x_1[n]$...
0
What you exactly you are going to see will depend a lot on the details and not only the overlap. What's your FFT size, what's your hop size, what's your window function, how do the sinusoids line up with the FFT frequency grid, how do changes in the signal line up with the individual frames and how the changes in frequency are actually implemented (i.e. what'...
0
So the sensory signal you are going to collect has the maximum component of 14.5 kHz. Let me guide you through my understanding and you can do the calculation yourself, testing your understanding.
As others have also implied, you need a sampling rate twice the bandwidth (you can check Nyquist Sampling Theorem to get it better).
And according to your question,...
1
It is not clear in which part of the problem statement you experience issues.
However, a few hints may help you get the gist of such problems
The sampling frequency, it is intuitive that in order for a system to monitor the state of another system it must poll it at higher rates than the highest rate in which the monitored signal is able to change its state....
0
Don't make this more complicated than it really is. $x(t)$ is non-zero in the interval $t\in[0,1]$, and $x(t+\tau)$ is non-zero in the interval $t\in[-\tau,1-\tau]$. The integrand is non-zero only if the two functions overlap. There is no overlap for $1-\tau<0$ and $-\tau>1$, i.e., for $|\tau|>1$. So for $|\tau|>1$ the autocorrelation is zero.
...
1
I would really recommend that you compute the DTFT of the sequences $x_1[n]=a^nu[n]$ and $x_2[n]=-a^nu[-n-1]$. This is very straightforward if you use the formula for the geometric series:
$$\sum_{n=0}^{\infty}q^n=\frac{1}{1-q},\qquad |q|<1\tag{1}$$
It will be important to consider the magnitude of $a$. Note that it's not for nothing that the table in ...
4
Purely by inspection of the block diagram the system is causal, because the output is the sum of the current input sample and stuff that's delayed -- there's no $z$ blocks in there to predict the future, just $z^{-1}$ block to react to the past.
Also by your method of finding the transfer function, the system is causal -- with a $3^{rd}$ order numerator and ...
6
Note that in this case you can see that the system is causal only from the given implementation. It's important to understand that you can't see it from the difference equation (if no initial conditions are given), and in general you can't see it from the transfer function either (if no region of convergence is given). The only case for which the expression ...
2
The system is causal, provided that the recursion is forward; i.e., it's recursed for increasing $k$.
Seeing that you are confused about causality tests, let me elaborate on it.
Let's put the definition of causality from Oppenheim's Signals & Systems book :
A system is causal if the output at any time depends only on values
of the input at the present ...
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