Communication Systems 3 Tutorial 1

Communication Systems 3 1. a) Continuous time is used in analog and is a constant moving time that forms a smooth line when plotted Discrete time is when you sample a continuous source at a discrete sampling rate and get a bunch of small dots of time that you can logically link together

c) The Nyquist sampling theorem states that to cleanly sample a wave discretely our sampling frequency has to be double the highest frequency, which upon further thought makes sense as if we do not sample at least twice as fast as the highest frequency we would only get one discrete point on information per wave for the highest frequency, which would be unusable

Q2a) $DynamicRange(dB)=20{\log }_{10}[\frac{V_{max}}{V_{res}}]$ therefore

$$V_{res}=\frac{5V}{10^{36/20}}$$

b) $V_{rms}=\frac{V_{amp}}{\sqrt{2}},Vamp=1/2\ast Vp2p$ $V_{rms}=\frac{5.0-0.2}{2\sqrt{2}}=1.697V$

$ENOB=\frac{20{\log }_{10}\frac{1.697}{0.1}-1.76}{6.02}=\frac{24.594-1.76}{6.02}=3.79$

V_{res} = \frac{\Delta V}{2^{M}} = \frac{4.8}{2^4} = 0.3V $$ but our dynamic range states we need at least 6 bits therefore

V_{res} = \frac{\Delta V}{2^{M}} = \frac{4.8}{2^6} = ??V

$$Q3a)QAM,becauseourSNRisveryhighsowecancleanlytrasmitmanybitspersecond‘‘‘mathematica‘‘‘$$