4/1/10 1 MS in Telecommunications TCOM 500: Modern Telecommunications Dr. Bernd-Peter Paris George Mason University Spring 2009 MS in Telecommunications Outline • Detailed treatment of modern digital radios. • A few preliminaries • Sampling Theorem • In-phase and quadrature channel • Homodyne and heterodyne transmitters and receivers. • Linear modulation and commonly used signal sets. • Signal space • Anatomy of a modern digital radio. Paris 2 TCOM 500: Modern Telecommunications4/1/10 2 MS in Telecommunications Reminder • The spectrum of a signal can be translated to a new location. • This is accomplished by multiplying (mixing) the message signal by a sinusoidal carrier. • Demodulation is the process of recovering the original message signal. • It is also accomplished by mixing the received signal by a sinusoidal carrier, followed by low-pass filtering. Paris 3 TCOM 500: Modern Telecommunications m(t) cos(2πfct) s(t) s(t) cos(2πfct) m(t) Lowpass Filter MS in Telecommunications THE SAMPLING THEOREM Paris 4 TCOM 500: Modern Telecommunications4/1/10 3 MS in Telecommunications Sampling and Discrete-Time Signals • Digital processing systems, cannot process continuous-time signals. • Instead, digital systems require the continuous-time signal to be converted into a discrete-time signal. • The conversion process is called sampling. • To sample a continuous-time signal, we evaluate it at a discrete set of times tn = nTs, where • n is a integer, • Ts is called the sampling period (time between samples), • fs = 1/Ts is the sampling rate (samples per second). Paris 5 TCOM 500: Modern Telecommunications MS in Telecommunications Sampling and Discrete-Time Signals • Sampling a sinusoidal signal results in a sequence of samples x(nTs) = A · cos(2πfnTs + φ). • Note that the independent variable is now n, not t. • We focus on sinusoidal signals, because general signals can be expressed as a sum of sinusoids. • To emphasize that this is a discrete-time signal, we write x[n] = A · cos(2πfnTs + φ). • Sampling is a straightforward operation. • However, we will see that the sampling rate fs must be chosen with care! Paris 6 TCOM 500: Modern Telecommunications4/1/10 4 MS in Telecommunications Experiment • Compute sampled versions of three different sinusoids: 1. x(t)=cos(2πt+π/4) 2. x(t) = cos(2π9t − π/4) 3. x(t) = cos(2π11t + π/4) • The sampling rate for all three signals is fs = 10. • Compute the first three samples (n=0, 1, 2) for each sinusoid. Paris 7 TCOM 500: Modern Telecommunications MS in Telecommunications Experiment 0 2 4 6 8 10−1−0.8−0.6−0.4−0.200.20.40.60.81Time (s) f=1f=9f=11Paris 8 TCOM 500: Modern Telecommunications • All three sinusoids yield exactly identical samples!? • Which frequency do you actually observe?4/1/10 5 MS in Telecommunications What happened? • The samples for all three signals are identical: how is that possible? • Suspicion: The problem is related to our choice of sampling rate. • To test this suspicion, repeat the experiment with a different, higher sampling rate – fs=100. • We also reduce the duration to keep the number of samples constant - that keeps the plots reasonable. Paris 9 TCOM 500: Modern Telecommunications MS in Telecommunications Sampling with High Sampling Rate 0 0.2 0.4 0.6 0.8 1−1−0.8−0.6−0.4−0.200.20.40.60.81Time (s) f=1f=9f=11fs=10Paris 10 TCOM 500: Modern Telecommunications • Now all three sinusoids are clearly distinguishable. • Samples from first experiment are marked with black circles.4/1/10 6 MS in Telecommunications Oversampling and Undersampling • The critical quantity in the previous examples is the ratio of signal frequency f and sampling rate fs. • When f/fs is less than ½, the signal is oversampled. • This is what we want! • The samples accurately reflect the continuous-time signal. • When the signal frequency f is greater than ½ fs, then undersampling occurs. • The sampled signal does not accurately reflect the original signal. • The sampled signal appears to have the wrong frequency. Paris 11 TCOM 500: Modern Telecommunications MS in Telecommunications Aliasing and Folding • Undersampled signals appear to have come from sinusoids of different, lower frequency. • It can be shown that the “apparent frequency” of a sampled sinusoid depends on the fractional part of the ratio f/fs. • The fractional part is the part after the decimal point – denote this frational part as fp. • If fp < 1/2, then apparent frequency is fp×fs. • This is called aliasing. • If fp > 1/2, then apparent frequency is (1-fp)×fs. • This is called folding. Paris 12 TCOM 500: Modern Telecommunications4/1/10 7 MS in Telecommunications Example • In our earlier example, we considered sinusoids of frequencies: 1. f=1 2. f=9 3. f=11 • The sampling rate was fs=10. • The apparent frequencies for the three cases are: Paris 13 TCOM 500: Modern Telecommunications f f/fs fp Apparent Frequency Category 1 0.1 0.1 1 oversampling 9 0.9 0.9 1 Undersampling, folding 11 1.1 0.1 1 Undersampling, aliasing MS in Telecommunications Apparent Phase • In addition to creating an apparent frequency, the phase may also be misrepresented in the sampled signal. • Assume the original sinusoid has frequency f and phase ϕ. • Then the apparent phase of the sampled signal depends on the ratio f/fs. Paris 14 TCOM 500: Modern Telecommunications Category Apparent Phase Oversampling ϕ Undersampling, folding -ϕ Undersampling, aliasing ϕ4/1/10 8 MS in Telecommunications Back to Example • The above discussion, explains why these three sinusoids produce identical samples: 1. x(t)=cos(2πt+π/4) Oversampling: apparent frequency and phase are correct. 2. x(t) = cos(2π9t − π/4) Undersampling, folding: apparent frequency is 1 and apparent phase is +π/4 (phase is reversed). 3. x(t) = cos(2π11t + π/4) Undersampling, aliasing: apparent frequency is 1 and apparent phase is +π/4. Paris 15 TCOM 500: Modern Telecommunications MS in Telecommunications Summary f/fsfa/fs0 0.5 1.0 1.5 200.51Over-sampling Folding Aliasing Foldingfafd,0fd,1fd,2fd,3Paris 16 TCOM 500: Modern Telecommunications4/1/10 9 MS in Telecommunications