[SOLVED] Ee442 lab 3- pulse shaping and matched filtering

Fall Semester
Telecommunications Circuits Laboratory EPFL

3 Pulse Shaping and Matched Filtering
Up to now we have only considered the transmission of complex data symbols over an AWGN channel. Of course, in reality there is no such thing as a complex symbol coming out of the antenna. Instead, the symbols have to be transformed into a time-continuous waveform. This transformation is called modulation.
3.1 Modulation
3.1.1 Transmission at Radio Frequency (RF)
The principle of digital communication systems is that only a finite number M of different symbols can be transmitted. Each symbol ai is mapped onto a pulse gi(t),i = 1,…,M. Since only finitely many pulses are possible, the receiver can recover the transmitted data if the signal is not too heavily distorted by the channel.
In this lab course, we only consider linear modulation schemes, where the pulses gi(t) can be expressed as scaled versions of the same base pulse: gi(t) = aig(t). Hence, the magnitude of the data symbol determines the amplitude of the transmitted pulse, and the argument determines its phase. (An example of a nonlinear modulation scheme is frequency shift keying, where the information lies in the frequency of the transmitted signal.)
Let T denote the duration of one data symbol, i.e., a new pulse is transmitted each T seconds. Note that the pulse g(t) can still be longer than T ; the pulses are allowed to overlap. The signal s(t) can then be written as the superposition of the pulses, scaled by the data symbols and delayed by multiples of T :
. (3.1)
The signal s(t) is called a complex baseband signal, because it is complex-valued and its spectrum is centered at f = 0. In order to transmit the signal over the air, it must be transformed into a real-valued signal and its spectrum must be shifted to a carrier frequency , where B is the bandwidth of the signal. This operation can be written as follows:
sRF(t) = <{s(t)e2πjfct}
(3.2) = <{s(t)}cos(2πfct) − ={s(t)}sin(2πfct).

Figure 3.1: Block diagram of a digital transmission system

Figure 3.2: Equivalent baseband system
3.1.2 Equivalent Baseband Model
We have seen that a description of the received samples is quite inconvenient if the digital/analog conversion and the upconversion to radio frequency is taken into account. Fortunately, it is possible to describe the whole system as a time-discrete model in the complex baseband (as long as one is not interested in the effects of the RF components, of course).
A block diagram of this simplified model is shown in Figure 3.2. First, the symbol stream is converted to sample rate by inserting L − 1 zeros between the symbols. (This implies that L is an integer, which can generally be assumed. In the lab, we use an oversampling factor of L = 4.) The sequence a[k] is then processed by the pulse shaping filter g[k]. The time-discrete transmitted signal s[k] is then transmitted over a time-discrete baseband channel and filtered by the receiver filter gMF[k]. Now the sample stream can be converted back to symbol rate by simply discarding three out of four samples. If the channel is an AWGN channel, which we assume up to now, the received symbols can then be written as z[n] = a[n]+w0[n], where w0[n] is the filtered noise.
3.2 Pulse Shaping Filters
3.2.1 Ideal Lowpass
We have not yet discussed the question how the pulse g(t) (or its time discrete version g[k] = g(kTs)) should look like. In order to use the available spectrum as efficiently as possible, we would like to use a pulse whose spectrum is as narrow as possible. In order to transmit at a symbol rate R = 1/T, the bandwidth B of the signal must be at least as large as R. The optimum filter would therefore be an ideal lowpass with the spectrum G(f) = rect(f/R), whose impulse response is known to be g(t) = si(πt/T), with si(x) = sin(x)/x. With an ideal lowpass, the transmitted signal can be written as
. (3.3)
It is important to note that g(t) becomes zero at multiples of the symbol duration T , except of t = 0:
g(nT) = 0, ∀n 6= 0. (3.4)
This is called the Nyquist criterion. If we sample this signal at the correct time instances t = nT, only the n-th pulse contributes to this sample, and we have recovered the transmitted symbols (except of noise).
3.2.2 Raised Cosine Filter
Such a filter is the raised cosine (RC). Its spectrum is given as
. (3.5)
The parameter α is the rolloff factor and determines the filter bandwidth. The two-sided bandwidth of the RC is B = R(1 + α). Thus, the bandwidth of the RC is by a fraction of α larger than that of the ideal lowpass. Figure 3.3 shows the transfer function and the impulse response

Figure 3.3: Transfer function (a) and impulse response (b) of the raised cosine filter
of the raised cosine filter for α ∈ {0.2,0.5,0.8}. We see that the bandwidth increases with increasing rolloff factor, but the impulse response converges faster to zero. For α = 0, the raised cosine becomes an ideal lowpass. Note that the raised cosine fulfills the Nyquist criterion, i.e. becomes zero at multiples of the symbol duration.
3.2.3 Matched Filtering and Root Raised Cosine Filter
In the receiver, the signal is filtered by another filter. It can be shown that in case of an AWGN channel, the SNR after the receive filter is maximized if this filter is the complex conjugated and reverted (in time direction) transmit filter:
gMF(t) = g∗(−t). (3.6)
This filter is matched to the transmit filter and is therefore called the matched filter (MF). The total effective filter of the transmission stem is then the combination of transmit and receive filter g(t) ∗ gMF(t), where ∗ denotes convolution. It is important to note that this effective filter (and not the individual filters) must fulfill the Nyquist criterion. We can achieve this if both filters have a transfer function that is equal to the square root of that of the raised cosine filter. Such a filter is therefore called a root raised cosine (RRC). The combination of both RRC filters then becomes a raised cosine and thus fulfills the Nyquist criterion. Furthermore, since the filters are real-valued and symmetric, the RRC is its own matched filter. The impulse response of the
RRC filter is
(3.7)
Note that si( .
3.2.4 Summary
When we use the RRC both as pulse shaping filter and matched filter, the effective filter fulfills the Nyquist criterion (3.4) and makes ISI-free reception possible. Also, the RRC converges faster to zero than an ideal lowpass and therefore suffers less from the drawbacks of the ideal lowpass discussed above.
The output of the matched filter can be written as
(3.8)
or in time-discrete samples as
, (3.9)
Unfortunately, (3.9) only holds if the receiver is perfectly synchronized to the incoming signal, which is never the case in a real system. In the next chapters, we will therefore examine some synchronization algorithms.
3.3 Your Tasks
A3T1 For the first task we provide an already shaped signal. The system model is slightly different as we now use a oversampling factor of L = 4. The pulses are RRC shaped with a roll-off factor of α = 0.22. Add an AWGN to the system. Afterwards write a matched filter (MF) to demodulate the pulses. Demap the symbols and decode the image. Keep the number of receive filter taps as a variable. The frame synchronization that you implemented in the previous exercise can not be used with oversampled signals. For this reason, a modified frame synchronization function is provided. We also provide a function rrc that can be used to generate the root raised cosine filter.
A3T2 Now implement a complete baseband transmission system to measure the BER:
(a) Generate a random bitstream.
(b) Convert to QPSK symbols.
(c) Upsample by a factor L = 4.
(d) Shape RRC pulses with a roll-off factor of α = 0.22 and 41 filter taps.
(e) Add AWGN.
(f) Apply matched filter (MF) with a variable number of filter taps.
(g) Revert upsampling (i.e., downsample by a factor L = 4).
(h) Demap the symbol stream.
(i) Calculate the BER.
Note: Convolutions introduce heads and tails. Take special care to have a correct symbol alignment for BER calculations.
A3T3 A longer FIR receive filter (i.e., a filter having more taps) will lead to a lower bit error rate, but will also require more multiplications, which increases the energy consumption of the (battery driven) device. Your task is to determine the minimum length of the MF that exceeds a specified BER performance. The provided script compute the BER obtained at an SNR of 8dB for different filter length. Choose the minimum MF length so that the BER is better than 7 · 10−3 , i.e. the implementation loss is less than 0.2 dB (c.f. BER of assignment 1).