Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap09

Continuous and Discrete Time Signals and Systems (Mandal & Asif) solutions - chap09

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Chapter 9: Sampling and Quantization Problem 9.1 (a) () 1 1 200 200 ( ) 5sinc 200 sinc 40 t xt t ππ × ⎛⎞ == ⎜⎟ ⎝⎠ . Therefore 1 400 1/40 200 1 ( )= rect 02 0 0 . 40 X ω π = > The maximum frequency is given by 100 Hz. Based on the Nyquist theorem, the maximum sampling period is 1 5 200 s Ts m s . (b) ( ) 2 max 100 max 50 max 100 ( ) 5sinc 200 8sin 100 freq Hz freq Hz freq Hz x ttt = =+ ±²³²´ ±²²³ ²²´ ±²²²²³²²²²´ . Therefore, maximum sampling period is given by 1 5 200 s m s . Fig. S9.1: A ( f ) denotes the spectrum of sinc(200 t ) and B ( f ) denotes the spectrum of sin(100 π t ). X 3 ( f ) is the sum of two shifted replicas of A ( f ), and is non-zero within the band [ 150,150] Hz. (c) Since x 3 ( t ) is a product of sinc(200 t ) and sin(100 π t ), the spectrum of x 3 ( t ) can be obtained by convolving the spectrums of sinc(200 t ) and sin(100 π t ). From the theory of CT convolution, it can easily be seen that (see Fig. S9.1), the maximum frequency present in x 3 ( t ) is 150 Hz. Therefore, the maximum sampling period is given by
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2 Chapter 9 1 3.33 21 5 0 s Ts m s == × . (d) Since x 4 ( t ) is a convolution product of sinc(200 t ) and sin(100 π t ), the spectrum of x 4 ( t ) can be obtained by multiplying the spectrums of sinc(200 t ) and sin(100 π t ). As the spectrum of sin(100 π t ) includes two impulses at 50 Hz and 50 Hz, the spectrum of x 4 ( t ) will only include two impulses at 50 Hz and 50 Hz. As the maximum frequency present in x 4 ( t ) is 50 Hz, the maximum sampling period is given by 1 10 25 0 s m s × . Problem 9.2 (a) From Table 5.2, we know that ( ) ( ) 2 sinc rect WW t W ω ππ . Applying the time-shifting property of the CTFT (see Table 5.4), we obtain ( ) ( ) (2 ) 2 2 sinc rect Wt j W W e −− . The uncertainty principle is satisfied as x 1 ( t ) is a bandlimited signal but NOT a time limited signal since a sinc function has infinite length. (b) From Table 5.2, we know that ( ) 2 () s inc t xt = . The uncertainty principle is satisfied as x 2 ( t ) is a bandlimited signal but NOT a time limited signal since a sinc function has infinite length. (c) From Table 5.2, we know that ( ) ( ) 2 sinc rect t W . Applying the frequency-shifting property of the CTFT (see Table 5.4), we obtain ( ) ( ) 0 0 2 sinc rect jt t W e ωω and ( ) ( ) 0 0 2 sinc rect t W e + . In other words, ( ) ( ) ( ) 00 2 3 0 ( ) sinc sinc sinc cos( ) t t t x te e t π =+ = , which is not a time-limited signal. The uncertainty principle is satisfied as x 3 ( t ) is a bandlimited signal but NOT a time limited signal since a sinc function has infinite length. (d) Express 0 0 1.5 40 0 12 ( ) ( ) ( 2 ) = rect 0e l s e w h e r e Xu u ωω ω << =− = . From Table 5.2, we know that ( ) ( ) 0 (0.5 ) ) sinc rect t . Applying the frequency-shifting property of the CTFT (see Table 5.4), we obtain ( ) ( ) 0 0 0 (0.5 ) (0.5 ) 1.5 1.5 sinc rect t e .
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Solutions 3 In other words, ( ) ( ) 00 0 0 (0.5 ) (0.5 ) 1.5 1.5 4 22 ( ) sinc sinc tt j tj t xt e e ωω ω ππ π == , which is not a time-limited signal. The uncertainty principle is satisfied as x 4 ( t ) is a bandlimited signal but NOT a time limited signal since a sinc function has infinite length Problem 9.3 (a) Express [ ] ( ) 10 0 2 ( ) cos( ) ( ) ( ) =cos( )rect t T t ut T ut T t =+ .
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This note was uploaded on 01/19/2012 for the course EE 421 taught by Professor Zhicheng during the Spring '11 term at Ohio State.

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Continuous and Discrete Time Signals and Systems (Mandal &amp; Asif) solutions - chap09

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