An earlier post we discussed hard decision decoding for a Hamming (7,4) code and simulated the the bit error rate. In this post, let us focus on the soft decision decoding for the Hamming (7,4) code, and quantify the bounds in the performance gain.
We have discussed about probable transmit pulse shaping filter and have observed that raised cosine filtering filtering allows a simpler implementation, albeit at the cost of increased bandwidth. Let us know understand the eye diagram, which is a useful graphical tool to quantify the degradation of the signal due to filtering. Eye diagram An eye…
Given that we have discussed Binary to Gray code conversion, let us discuss the Gray to BInary conversion. Conversion from Gray code to natural Binary Let be the equivalent Gray code for an bit binary number with representing the index of the bit. 1. For , i.e, the most significant bit (MSB) of the Gray…
Post describes about the need for I-Q modulation by comparing the spectral efficiency of passband PAM and passband QAM.
My understanding of the CORDIC (Co-ordinate Rotation by DIgital Computer) thanks to the nice article in [DSPGURU-CORDIC].
In this post, the objective is to figure out the minimum separation between two sinusoidals having frequencies , of duration each to be orthogonal. Let the phase difference between the sinusoidals is where can take any value from to (Refer Example 4.3 [DIG-COMM-SKLAR]). For the two sinuosidals to be orthogonal,
Following a brief discussion with my friend Mr. Rethnakaran Pulikkoonattu on phase noise profiles, he pointed me to his write up on Oscillator Phase Noise and Sampling Clock Jitter . In this post, we will discuss the math behind integrating the phase noise power spectral density (in dBc/Hz) to find the root mean square jitter value.
Mr. Lealem Tamirat, in a comment on BER for BPSK in Rayleigh channel, wondered about the performance of an OFDM modulated system in a frequency selective Rayeligh fading channel. My response was that, Though the total channel is a frequency selective channel, the channel experienced by each subcarrier in an OFDM system is a flat…
In the past, I have wondered about discussing personal thoughts in this blog. The answer in my mind was NO and I ve been focusing only on technical topics till date. However, there is a change of mind, thanks to my friend Manoj. Thanks to him, I happened to see Prof. Randy Pausch’s brief 10…
Thanks to the nice article from Xilinx TechXclusives [XLNX-TECH], let us try to understand the probable digital implementation of resistor-capacitor based low pass filter. Consider a simple RC filter shown in the figure below. Assuming that there is no load across the capacitor, the capacitor charges and discharges through the resistor path. Figure: RC low…
IEEE 802.11ac Very High Throughput (for <6GHz band) is an upcoming standard which is development by IEEE standardization committee. The mandate of Task Group AC is supposed to enhance the High Throughput rates achieved by 802.11n. As described in the document VHT below 6GHz PAR plus 5C’s (802.11-08/0807r4) the group has the following objectives :
In previous posts, we had discussed equalization of a 2×2 MIMO channel with Zero Forcing (ZF) equalization and later, Zero Forcing equalization with successive interference cancellation (ZF-SIC). In this post, we will explore a variant of ZF-SIC called Zero Forcing Successive Interference Cancellation with optimal ordering. We will assume that the channel is a flat…
Question 39 on communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper. Q39. The signal as shown is applied both to a phase modulator (with as the phase constant) and a frequency modulator (with as the frequency constant) having the same carrier frequency. The ratio for the same maximum phase deviation is,…
Question 38 on Communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper. Q38. A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X=0)=9/10, then the probability of error for an optimum receiver will be (A) 7/80 (B) 63/80 (C)…