Or xr (t ) x(nT ) n wcT sin wc t nT wc t nT The reconstructıon according to equation (4) with wc (4) ws is illustrated below: 2 Ideal band-limited interpolation in which the impulse train is replaced by a superposition of sinc functions. Interpolation using the impulse response of an ideal lowpass filter is referred to as “bandlimited” interpolation. If the crude interpolation provıded by zero-order hold is insufficient we can use a variety of smoother interpolation strategies ( higher order holds etc). Sampling of Signals in Time and Frequency Domains Sampling is the basic prerequisite for digital signal processing of continuous-time signals. The original signal is converted into a sequence of numbers by periodic sampling. Sampling may also be performed on the spectrum of a signal. For example if the signal is an aperiodic signal with finite energy, its spectrum is continuous and its computation is possible at a set of discrete frequencies. Since spectrum is observed at discrete frequencies such a process is called frequency domain sampling. Time Domain Sampling To process a continuous time signal by digital signal processing techniques it is necessary to convert the signal into a sequence of numbers. This can be achieved by sampling the analog signal x(t) periodically every T seconds to produce a discrete-time signal x[n] given by : xn xnT n These samples must then be quantized to a discrete set of amplitude levels. Let us now define some notation xt ........................................... is the ana log signal for CT xn xnT .................................. is the discrete time signal F .................................................. frequency domain symbol for ana log signal f ................................................... frequency domain symbol for discrete signal If the analog signal x(t) is an aperiodic signal with finite energy, its spectrum is given by X F xt e j 2Ft dt or similarly given spectrum X(F), signal is: xt X F e j 2Ft dF (1) The spectrum of a discrete –time signal x[n] obtained by sampling x(t) is given by: X w xne jwn or X f n xne j 2fn n The sequence x[n] can be recovered from its spectrum by the inverse FT : 1 xn 2 X we 1/ 2 jwn dw or X f e j 2fn (2) df 1 / 2 To find the relationship between the spectra of the discrete-time signal and the analog signal we note that `periodic sampling` imposes a relationship between independent variables t and n. t nT n Fs (3) Similarly in the frequency domain a relationship between F and f exists. Let us take equation (3) and substitude into (1): xnT X F e j 2F nT dF X F e j 2F n Fs dF (4) If we compare (2) and (4) we see that 1/ 2 xn X f e j 2fn df 1 / 2 X F e j 2F n Fs dF (5) (because x[n] = x(nT) ) . f here F Fs F is the frequency of the analog signal f is the frequency of the discrete signal Fs is the sampling frequency df dF Fs Using this new found factor we can rewrite equation (5) as shown below: xn F j 2 n 1 X e Fs Fs Fs Fs / 2 F Fs / 2 dF X F e j 2Fn / Fs dF (6) We note that the integral on the right hand side of (6) can be divided into an infinite number of intervals each of width Fs. Therefore we can represent the integral over the infinite range as a sum of integrals: xn X F e j 2Fn / Fs dF 1 k Fs 2 X F e j 2Fn / Fs dF (7) k 1 k Fs 2 We observe that X(F) in the frequency interval (k-1/2)Fs to (k+1/2)Fs is identical to X(F-kFs) in the interval - Fs/2 to Fs/2. 1 k Fs 2 j 2Fn / Fs dF X F e k 1 k Fs 2 Fs / 2 j 2Fn / Fs dF X F kFs e k Fs / 2 Comparing (8) to (7) and using (6) we note that j 2Fn / Fs dF (8) X F kFs e Fs / 2 k Fs / 2 F X Fs X F kFs k Fs X f Fs X f k F 9 s k This is the desired relationship between the spectrum of the discrete time signal and the spectrum of the analog signal. We note that the right hand side of (9) consists of a periodic repetition of the scaled spectrum Fs X F with period Fs . Example: Suppose that the spectrum of a band limited analog signal is as shown below in Fig .1(a). X(F) xa(t) 1 F t -B 0 B Fig. 1(a) Now if the sampling frequency Fs is selectedgreater than 2B the spectrum X F / Fs of the discrete time signal will be as shown in Fig. 1(b). X[f] x[n]=x(nT) 1 f n -B T Fig. 1(b) 0 B Thus if the sampling frequency Fs is selected such that Fs >= 2B then F X X f Fs X F Fs for F FS 2 In this case there is no aliasing . Spectrum of the discrete-time signal is identical to spectrum of the analog signal (within the scale factor Fs). If the sampling frequency Fs is less than 2B then we have the following spectra for the transform of x[n] : X[f] x[n]=x(nT) 1 n Fs/2 Fs T Fig. 1(c) Spectrum of the discrete time signal contains aliased frequency components of the analog signal spectrum X(F). This aliasing prevents the full recovery of the original signal x(t) from the samples. Given the discrete-time signal x[n] with spectrum X[f] as in Fig. 1(b) with no aliasing, it is possible to reconstruct the original analog signal from the samples x[n]. We know that 1 F X X F Fs Fs 0 Fs 2 F F s 2 F and also that xt Fs / 2 X F e Fs / 2 For our case if we take Fs = 2B j 2Ft dF x(t ) 1 F j 2Ft X e dF F Fs Fs / 2 s Fs / 2 1 Fs 1 Fs Fs / 2 j 2nF / Fs j 2Ft x[n]e e Fs / 2 n Fs / 2 n Fs / 2 x[n] Fs / 2 xnT n e n j 2F t Fs dF dF Fs / 2 xnT n e n j 2F t Fs T t nT T t nT sin (10) The seconstruction formula involves a sinc function appropriately shifted by nT where n = 0,±1, ±2, ±3,....... and weighted by the corresponding samples x(nT). Equation (10) is called interpolation formula for reconstructing x(t) from its samples. g (t ) T t T t sin Function g(t) is called the interpolation function.