Download A First Course in Digital Communications by Ha H. Nguyen, Ed Shwedyk PDF

By Ha H. Nguyen, Ed Shwedyk

Conversation know-how has turn into pervasive within the smooth global, and ever extra complicated. concentrating on the main easy principles, this conscientiously paced, logically based textbook is filled with insights and illustrative examples, making this an incredible advent to trendy electronic conversation. Examples with step by step strategies aid with the assimilation of theoretical rules, and MATLAB workouts enhance self assurance in utilising mathematical suggestions to real-world difficulties. correct from the beginning the authors use the sign house method of supply scholars an intuitive believe for the modulation/demodulation approach. After a assessment of signs and random strategies, they describe center themes and strategies akin to resource coding, baseband transmission, modulation, and synchronization. The booklet closes with assurance of complex themes reminiscent of trellis-coding, CMDA, and space-time codes to stimulate extra learn. this can be a fantastic textbook for an individual who desires to know about smooth electronic communique.

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22(a) shows a step function, which is not a finite energy signal. 14, it does not go to 0 as t → ∞. But like the previous two examples it has finite, nonzero average power since 1 lim T→∞ T T 2 − T2 1 s (t)dt = lim T→∞ T T 2 2 V 2 dt = 0 V2 (watts). 68) ∞ The direct approach to determine S(f ) = −∞ Vu(t)e−j2πft dt = 0 Ve−j2πft dt = V(1 − ej2π f ∞ )/2jπf fails because ej2π f ∞ is undefined. 22(c): s(t) = V V + sgn(t) ⇒ F{s(t)} = F 2 2 V 2 +F V sgn(t) . 69) We already know that F {V/2} = (V/2)δ(f ).

65a) ∞ −∞ Ve−j2πft dt = ej2π f ∞ − e−j2πf ∞ j2π f sin(j2π f ∞) . 65b) Unfortunately the above approach fails since sin(∞) is undefined. However, recall that S(f ) is a spectral density and given that the signal is a DC one it seems reasonable for all of it to be concentrated at f = 0. Therefore postulate that S(f ) = Vδ(f ) and check what time ∞ function corresponds to this S(f ). Evaluating −∞ Vδ(f )ej2π ft df we get V and therefore s(t) = V ←→ S(f ) = Vδ(f ), |S(f )| = Vδ(f ), S(f ) = 0. 66b) ∞ Note that the above implies that −∞ Ve−j2πft dt = Vδ(f ), or that the integral ∞ −j2πxy dy = δ(x).

6 Fourier series of a product of two signals To complete the discussion of periodic signals and their Fourier series representation we consider one last property. Namely, what is the Fourier series of a signal that is the product of two periodic signals with the same period. t 32 Deterministic signal characterization and analysis To this end, let s(t) = s1 (t)s2 (t), where signals s1 (t) and s2 (t) are periodic with the same fundamental period T or fr = 1/T. Substituting in the Fourier series representations of s1 (t), s2 (t), one has ∞ ∞ [s (t)] j2π kfr t s(t) = Dk 1 [s2 (t)] j2π lfr t e k=−∞ Dl e .

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