jithin
• Member for 4 years, 11 months
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• Bangalore, Karnataka, India

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You have a set of message set $m_i$, $0 \le i \le N-1$. (For example, QPSK will be $N=4$). For the transmitted message $m_i$, the corresponding symbol vector is $\textbf{x}_i$, and the received symbol ...

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Circular convolution is just linear convolution aliased by DFT length $n$. The length of linear convolution of $a$ and $b$ will be $2n-1$. So take $FFTs$ of $a$ and $b$ , padding each of them to ...

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it seems like the bigger problem is that subsampling the signal would result in aliasing frequencies No, subsampling in frequency domain corresponds to aliasing in time domain. So the idea here is to ...

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Traditionally, OFDM became popular in WiFi and LTE because the channel model consisted of multi-path. That is, the radio signal transmitted in 1-6GHz frequencies bounced from various obstacles (walls, ...

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Convolving with Matched Filter is same as cross-correlation. Suppose say your known signal is $x[n]$ for $0 \le n \le N-1$. The matched filter is $$h[n] = x^*[N-n]$$ where $*$ denotes conjugate (...

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But I want to see why either the bin width is 1/T where T is the duration the samples are taken over, or why the DTF is Fs Hz periodic. (I think bin width as Fs/N is a red herring) The fourier ...

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Well, I think both are unrelated unless you have some constraints. The Nyquist channel capacity says you can transmit $2B\log_2(M)$ bits per second for given channel bandwidth $B$. It does not say ...

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$c$ need not be real number. For the LHS you would get $$|\langle s_1,s_2\rangle| = |\langle s_1,cs_1\rangle |=|c|\,|\langle s_1,s_1\rangle |= |c|\|s_1\|^2$$ For RHS, $$\|s_1\| \times \|c s_1\| =\... View answer 3 answers vote 187 views Accepted answer 2 votes For the classical baseband channel model for multipath, h(t)=\sum_ka_k\delta(t-\tau_k), the frequency response H(f)=\sum_k a_ke^{-j2\pi f\tau_k}. If K=1, single tap the phase would have been ... View answer 1 answers votes 164 views 2 votes Cross correlation is defined as R_{xy}(\tau) = E\{x(t)y^*(t+\tau)\}. You missed to take the conjugate of x_2. So x_1x_2^* = (c+id)(a-ib) = ac+bd +i(ad-bc). Cross PSD is the fourier transform of ... View answer 2 answers votes 361 views 2 votes halving the F_s (keeping T and N constant) would remove the right half of the bins (halving the maximum frequency) No. Number of bins remain same because you have not changed N. What will ... View answer 2 answers votes 71 views 2 votes See section 2.4.3 of this reference https://web.stanford.edu/~dntse/Chapters_PDF/Fundamentals_Wireless_Communication_chapter2.pdf If the doppler spectrum has to be gaussian, the auto-correlation of ... View answer 1 answers vote 136 views Accepted answer 2 votes Your understanding is correct and you can apply Hann windowing on discrete signals. The choice of your window depends on how easily you can visually resolve two sinusoids that are closely spaced in ... View answer 1 answers votes 468 views Accepted answer 2 votes In the line Inverse_GShifted=real(ifft(GShifted)); I have changed to Inverse_GShifted=real(fftshift(ifft(GShifted))); This is because the ifft output is from 0 \le n \le N-1, but your actual ... View answer 2 answers vote 94 views Accepted answer 2 votes The only reason I can think of may be to preserve the frequencies over which original signal had spectrum (before squaring). When you square a signal x[n], whose spectrum X(e^{j\omega}) extends ... View answer 1 answers votes 152 views Accepted answer 2 votes As mentioned in the comment, I modified the code given here and was able to adapt the LMS filter with error tapering to zero. The only assumption I made is that (since I am not an audio expert and do ... View answer 1 answers votes 170 views Accepted answer 1 votes For a SISO channel, assuming you have estimated channel coefficients for each sub-carrier k, the MMSE equalizer is$$ \hat{h}_k=\frac{h_k^*}{|h_k|^2 + \frac{\sigma_x^2}{N_0}} $$So you can see ... View answer 2 answers votes 162 views 1 votes I am not sure if you can use P_{err}=\text{erfc}(G/\sigma) because noise is not gaussian distributed. Here is my take on it. Assuming uniform probability of transmission for all 32 symbols x_i,... View answer 2 answers votes 285 views 1 votes From my experience EVM is defined as$$ EVM = \sqrt{\frac{1}{NP_{avg}}\sum_{n=0}^{n=N-1}(|x_n-x^*_n|^2)}\\ EVM_{\%} = EVM \times 100 where x_n is the equalized symbol, and x^*_n is the ... View answer 3 answers vote 148 views 1 votes Update: I feel there is no need to bring concept of Bandpass sampling (as mentioned in comments below) because this is specifically regarding a simple misunderstanding that OP has. The text referenced ... View answer 1 answers vote 70 views 1 votes I can answer for adaptive filter containing feedback path alone as I have implemented it recently (both Blind DFE and with Reference Symbols), but it should work for DFE containing feed-forward path ... View answer 3 answers votes 127 views 1 votes \begin{align} 1-e^{-j4\omega} &amp;=1-(\cos(4\omega)-j\sin(4\omega))\\ &amp;=1-\cos(4\omega)+j\sin(4\omega)\\ &amp;=1-(1-2\sin^2(2\omega))+j2\sin(2\omega)\cos(2\omega)\\ &amp;=2\sin(2\omega)(\sin(2\... View answer 2 answers vote 374 views 1 votes S_{xy}=S_{xx} is true but this does not mean F_x*F_y = F_x*F_x when y is random. The actual definition isS_{xy}(f)=\lim_{T\rightarrow \infty}\textbf{E}\{F^*_xF_y\}$$for random signals. ... View answer 1 answers votes 52 views Accepted answer 1 votes You need to zero pad the FFT so that it reveals more of the frequency response (increasing the resolution of the plot). I corrected your code (while taking fft, I took 1024 point FFT for windows as ... View answer 2 answers votes 28 views 1 votes One cycle duration of 50Hz is 0.02seconds. So this means you need to take 12 samples in a duration of 0.02s. In one second, there will be 600 samples (sample rate of 600 samples per second). View answer 2 answers votes 45 views 1 votes Taking the average values of these measurements should work. The estimator is$$ \hat{X}=\frac{1}{L}\sum_0^{L-1}K=\frac{1}{L}\sum_0^{L-1}(X+X*N) $$.To show that it is the unbiased estimator,$$ E(\...

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In contrast to your previous question(Why multi-path channel has linear phase within the coherence bandwidth?), the coherence BW in mmWave may not be relevant considering delay spread alone. If you ...

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For a finite energy signal (zero power), the time average is zero. For a periodic signal (finite power but infinite energy), MattL has already mentioned that frequency components are present at $f =k/... View answer 1 answers votes 451 views Accepted answer 1 votes Since this is not an LTI system, we cannot have$H(\omega) = K\Theta_1(\omega)$, where$K$is a scalar complex number. But as OP mentioned in the comment he has knowledge of$\Theta_1(\omega)\$, we can ...