NRZ vs. Manchester pulse shaping filter (Modulator)

Question

I'm trying to visualize the differences on the output waveform of a modulator when using NRZ or a Manchester pulse shaping filter. To keep it simple, let's think of a binary PAM signal with an input bit sequence 101011.

My thoughts are that with a Half Sine or a Raised Cosine pulse shaping filter, I will see the transitions between the 1's and 0's smooth out, but I'm struggling to think about the differences between the NRZ and the Manchester.

Answer 1

Manchester coding effectively creates a bit sequence that is twice as long as the given bit sequence and then applies NRZ pulses to the resulting bit stream. There are two types of Manchester coding. In one form, each $0$ in the original sequence is replaced by $01$ and each $1$ by $10$ while the other form complements these bit patterns. Using the first form, $$101011 \to 10\,01\,10\,01\,10\,10$$ while the second form gives the complementary bit sequence $01\,10\,01\,10\,0101$. If each data bit is of duration $T$, the NRZ waveform with Half Sine pulses has sines of period $2T$ so that each bit pulse has duration half a period. Thus, that leading $1$ gets modulated to $\sin(\pi t/T)$ lasting from $t = 0$ till $t = T$. With Manchester coding, each channel bit (of duration $T/2$) gets a Half Sine of a sine of period $T$. Put another way,

• with plain NRZ, that first $1$ in your bit sequence will have pulse $\sin(\pi t/T)$ lasting from $t = 0$ to $t = T$. This is a positive going pulse rising from $0$ at $t=0$, peaking at $1$ at $t=T/2$, and falling to $0$ at $t = T$
• with Manchester coding, that data $1$ will become channel $10$ giving a positive pulse $\sin(2\pi t/T)$ from $0$ to $T/2$. The next channel bit $0$ has the delayed half-sine pulse $\sin(2\pi (t-T/2)/T$ from $t=T/2$ to $t = T$, but with a negative amplitude (the channel bit during this interval is $0$ instead of a $1$) which works out to be $$\begin{align*} -\sin(2\pi(t-T/2)/T) &= -\sin(2\pi t/T - \pi)\\ &= -\sin(2\pi t/T)\cos(\pi) + \cos(2\pi t/T)\sin(\pi)\\ &= \sin(2\pi t/T) \end{align*}$$ lasting from $t = T/2$ to $t = T$. In short, during the entire data bit interval $[0,T]$ the signal can be expressed as $\sin(2\pi t/T)$ and is one cycle of the sine wave of period T. If the data bit were a $0$ instead of a $1$, the channel bits would be $01$ and we would get $-\sin(2\pi t/T)$ from $t=0$ to $t=T$. In short, Manchester coding with half-sine pulses effectively creates a phase-modulated signal from the data sequence with a carrier frequency of $1/T$ Hz and each data bit interval contains exactly one period of the sinusoid. If you look at the figure for Thomas's Manchester coding in Wikipedia, what we have is the $\pm$ rectangular pulses comprising each data bit interval, which look like a hard-limited sinusoid of period $T$, being replaced by the sinusoid.

The OP says, "My thoughts are that with a Half Sine .... pulse shaping filter, I will see the transitions between the 1's and 0's smooth out,..." but this does not happen, there will be abrupt reversals of carrier phase at those data bit boundaries where there is a transition from $0$ to $1$ or from $1$ to $0$, pretty much as happens in plain PSK.