Consider the cosine wave. Plot the spectrum of the discrete-time signal g (t) derived by sampling g(t) at the times t n = n=f s, where.

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1 Problem Consider the cosine wave g(t) =A cos(f 0 t) Plot the spectrum of the discrete-time signal g (t) derived by sampling g(t) at the times t n = n=f s, where n =0 1 and (i) f s = f 0 (ii) f s =f 0 (iii) f s =3f 0 g(t) =A cos(f 0 t) Hence, G(f) = A [(f ; f 0)+(f + f 0 )] G (f) = f s = Af s G(f ; mf s ) [(f ; f 0 ; mf s )+(f + f 0 ; mf s )] (i) f s = f 0 (ii) f s =f 0 G (f) = Af 0 G (f) =Af 0 [(f ; f 0 ; mf 0 )+(f + f 0 ; mf 0 )] [(f ; f 0 ; mf 0 )+(f + f 0 ; mf 0 )] 1

2 G(f) A= ;f 0 0 f 0 f G (f) Af 0 ;3f 0 ;f 0 ;f 0 0 f 0 f 0 3f 0 f Figure 1: f s = f 0 G (f) Af 0 ;7f 0 ;5f 0 ;3f 0 f 0 3f 0 5f 0 f 0 0 7f 0 f Figure : f s =f 0

3 (iii) f s =3f 0 G (f) = 3Af 0 [(f ; f 0 ; 3mf 0 )+(f + f 0 ; 3mf 0 )] G (f) 3Af 0 = f ;f 0 ;f 0 0 f 0 f 0 4f 0 5f 0 7f ;7f ;5f ;4f 0 Figure 3: f s =3f 0 Problem 4.1. The signal is sampled at the rate 50 samples per second. g(t) = 10 cos(0t) cos(00t) (a) Determine the spectrum of the resulting sampled signal. (b) Specify the cuto frequency of the ideal reconstruction lter so as to recover g(t). (c) What is the Nyquist rate for g(t). The signal g(t) is Hence, g(t) = 10 cos(0t)cos(00t) = 5[cos(0t) + cos(180t)] 3

4 G(f) =:5[(f ; 110) + (f +110)+(f ; 90) + (f +90)] Correspondingly the spectrum of the sampled version of g(t) with a sampling period T s =1=50s is given by G (f) = f s = 65 G(f ; mf s ) (f;110;50m)+(f +110;50m)+(f;90;50m)+(f +90;50m)] (b) the spectrum G(f) and G (f) are illustrated in Fig. (4). From this gure we deduce that in order to reconstruct the original signal g(t) fromg (t), we needtousealow-pass lter with a cuto frequency greater than 110Hz but less than 140Hz. (c) The highest frequency component ofg(t) is 110Hz. Hence, the Nyquist rate of g(t) is 0Hz. G(f) f, Hz G (f) Ideal reconstruction lter response Figure 4: Problem:

5 Problem A signal g(t) consists of two frequency components f 1 =3:9 khz and f =4:1 khz in such a relationship that they just cancel each other out when the signal g(t) is sampled at the instants t =0 T T, where T = 15s. The signal g(t) is dened by g(t) =cos f 1 t + + A cos(f t + ) nd the values of amplitude A and phase of the second frequency component. The signal at the sampling instants is g(nt ) = cos(f 1 nt + )+A cos(f nt + ) = 0 n =0 1 At n =0 cos( )+Acos() =0 Hence, A cos() =0 (1) At n = 1, with f 1 =3:9 khz, f =4:1 khz, and T =15s, wehave cos(0:975 + )+Acos(1:05 + ) =0 () From (1), we deduce that A must be nonzero. Hence, = =. Accordingly, () simplies as ; sin(0:975) A sin(1:05) =0 (3) But sin(1:05) =; sin(0:05) and sin(0:975) = + sin(0:05). To satisfy (3), we must therefore have A =1 and the ambiguous sign (of ) must be negative. That is = ;= Problem

6 Let E denote the energy of a strictly band-limited signal g(t). Show that E may be expressed in terms of the sample values of g(t), taken at the Nyquist rate, as follows: E = 1 W n=;1 where W is the highest frequency component of g(t). n g W If g(t) is band-limited to ;W f W,wemay express it as: The energy of g(t) is therefore g(t)= n=;1 g n W sinc(wt; n) E = = = Z 1 ;1 Z 1 ;1 n=;1 g(t)g (t)dt n g W n g W n=;1 sinc(wt; n) g k W Z 1 ;1 g k sinc(wt; k) W sinc(wt; n)sinc(wt; k) (1) But Z 1 ;1 sinc(wt; n)sinc(wt; k) = 8 < : 1 W 0 k = n k 6= n Hence, we may satisfy (1) as E = 1 W n=;1 n g W Problem Consider a continuous-time signal g(t) of nite energy, with a continuous spectrum G(f). Assume that G(f) is sampled uniformly at the discrete frequencies f = kf s, thereby obtaining the sequence of frequency samples G(kF s ), where k is an integer in the entire range ;1 <k<1, and F s is the frequency sampling interval. Show that if g(t) is duration-limited, so that it is zero outside the interval ;T < t < T, then the signal is completely dened by specifying G(f) at frequencies spaced 1=T hertz apart. 6

7 Since g(t) = 0 outside the interval ;T <t<t,wemay express the Fourier transform of g(t) as G(f) = Z T ;T g(t) exp(;jft)dt (1) Expanding g(t) asafourier series, with period T, wehave jkt g(t)= c k exp T ;T <t<t where c k = 1 Z T g(t) exp T ;T Comparing (1) and (), we deduce that ; jkt T dt () c k = 1 k T G T = F s G(kF s ) where F s = 1 T Therefore, g(t)=f s G(kF s )exp(jkf s t) ;T <t<t (3) Substituting (3) in (1), we get We may thus state that G(f) = F s Z T = F s = ;T G G(kF s ) k T G(kF s )exp(jkf s t)exp(;jft)dt Z T ;T exp[;jt(f ; kf s )]dt sinc(tf ; k) (4) 1. The signal g(t), and therefore its spectrum G(f) is uniquely determined in terms of samples of G(f) taken at the rate F s =1=T.. Given the frequency samples fg(k=t )g, k =0 1, the original spectrum G(f) can be reconstructed without distortion. Problem

8 The spectrum of a band-pass signal occupies a band of width 0.5 khz, centered around 10KHz. Find the Nyquist rate for quadrature sampling the in-phase and quadrature components of the signal. g(t)=g I (t)cos( 10 3 t) ; g Q (t)sin( 10 3 t) where g I (t) and g Q (t) arelow-pass signals with a bandwidth: W = 1 0:5 =0:5 khz The Nyquist rate for g I (t) andg Q (t) is therefore W =0:5 khz Problem The signals g 1 (t) = 10 cos(100t) and g (t) = 10 cos(50t) are both sampled at times t n = n=f s,wheren =0 1, and f s = 75 samples per second. Show that the two sequences of samples thus obtained are identical. We note that 1. The Nyquist rate of g 1 (t) is 100 Hz hence, with a sampling rate of 75 Hz, the signal g 1 (t) is under-sampled by 5 Hz below the Nyquist rate.. The Nyquist rate of g (t) is 50 Hz hence, with a sampling rate of 75 Hz, the signal g (t) isover-sampled by 5 Hz above the Nyquist rate. 3. Although g 1 (t) and g (t) represent sinusoidal waves of dierent frequencies, by under-sampling g 1 (t) and over-sampling g (t) appropriately, their sampled versions are identical. Problem

9 G(f) G(0) f (Hz) Figure 5: Problem 4.4. Figure 5 shows the spectrum of a low-pass signal g(t). The signal is sampled at the rate of 1.5 Hz, and then applied to a low-pass reconstruction lter with cuto frequency 1 Hz. Plot the spectrum of the resulting signal. G (f) f (Hz) Reconstructed spectrum f(hz) Figure 6: Problem 4.4. Problem 4.5. This problem is aimed at investigating the fact that the practical electronic circuits will not produce a sampling function that consists of exactly rectangular pulses. Let h(t) denote some arbitrary pulse shape so 9

10 that the sampling function c(t) may be expressed as c(t) = n=;1 h(t ; nt s ) where T s is the sampling period. The sampled version of an incoming analog signal g(t) is dened by s(t) =c(t)g(t) (a) Show that the Fourier transform of s(t) isgiven by S(f) =f s n=;1 G(f ; nf s )H(nf s ) where G(f) =F [g(t)] H(f)=F[h(t)], and f s =1=T s. (b) What is the eect of using the arbitrary pulse shape h(t)? (a) The Fourier transform of the sampling function is C(f) = = H(f) = H(f) T s = 1 T s The Fourier transform of the sampled signal is H(f)exp(;jmfT s ) H exp(;jmft s ) f ; kts k T s f ; kts S(f) = C(f) G(f) = 1 T s 8 < : = 1 T s H k f 9 = ; kts T s k G f ; kts T s H G(f) (b) Using an arbitrary pulse shape h(t) means that the sampled spectrum is no longer periodic. Instead, each replica of G(f), centered at f = k=t s,ismultiplied by a frequency dependent constant H(k=T s ). However, 10

11 when the signal is reconstructed by alow-pass lter, all replicas are removed, leaving, T s ;1 H(0)G(f). Thus, except for a scaling factor, an arbitrary sampling function will not aect the reconstructed signal. Problem Twenty-four voice signals are sampled uniformly and the time-division multiplexed. The sampling operation uses at-top samples with 1 microsecond duration. The multiplexing operation includes the provision for synchronization by adding an extra pulse of sucient amplitude and also 1 microsecond duration. The highest frequency component of each voice signal is 3.4 khz. (a) Assuming a sampling rate of 8 khz, calculate the spacing between successive pulses of the multiplexed signal. (b) Repeat your calculation assuming the use of Nyquist rate sampling. (a) The sampling period is T s =1=8000 = 15s. There are 4 channels and 1 sync pulse. Hence the time allotted to each channel is T c = T s 5 =5s The pulse duration is 1 s, and so the time between pulses is 4s. (b) Assuming the use of sampling at the Nyquist rate (6.8 khz), the sampling period is Correspondingly, T s = 1 6:8 10 ;3 =0:147 10;3 s =147s T c = =6:68s Time between pulses = 5:68s 11